Eliminating the variables

Windy used to have a great tagline, "opening the envelope". I use it all the time. It simply states that certain systems (in stunt in particular) allow you to extend the level of performance and perform in a greater array of challenging condions. Improvements such as stronger control systems, exotic materials, greater power to weight ratios, higher quality components (glow plugs and fuel for example), hard point handles, aft CG's, thick airfoils, etc all help to "open the envelope". I recommend all very highly.

I would like to start my own tagline, "eliminating the variables". It seems to me there are very few fixed knowns in a lot of what we do. Mostly we have knowledge passed from one generation to the next and a lot of hopeful copying. The varaiables have never really been fixed in space very well, so that we can work around them.

I called Frank Williams the other night to give him an update on my engine testing. We, of course, were talking about the "finding the back side of the curve" that I spoke of in an earlier post. I told Frank that I have no doubt that what he is proposing is the proper way to set up a stunt engine. I also said that it is apparent to me that all successful stunt engines already use the back side of the curve to some extent. There simply is no other explanation for the behavior.

The thing I did not completely agree with is that the power curve is fixed in space and the engine is tracking nicely up and down this curve all the way through the pattern. This simply is not happening, NO MATEER WHAT KIND OF TUNED PIPE SYSTEM YOU ARE USING!

For example, Frank sent me the instructions for the "prop dyno" system that he uses to test engines. Right there in the directions he made it clear that the fuel head feeding the engine must be kept constant or the OUTCOME OF THE CURVE WILL BE SUSPECT. Right there, Frank has eliminated a variable to indentify a trend.

Len made some excellent points in dispute of the "back side" theory. One point that really hit home was the fact that stunt engines pick up RPM when they are pointed up. In a nutshell, the fuel head has changed in an instant, and there was resultant change in RPM. In the case of a typical 2 stroke stunt engine, the result would be a change from a 4 cycle to a 2 cycle or 4-2 break.

Well......right there the curve just moved, didn't it?

Specifically, the curve will move up and to the right without a significant drop in compression.

Brett on the other hand (paraphasing) said that all changes in the RPM of the motor are load related, and are simply a function of drag and load on the prop.

What happens when you set 2 cycle engine at max RPM on a suction tank, then point the nose up (combat guys?). It sags doesn't it? In a flat two stroke the engine has no "second gear". What happens on a bladder? Guess...

What happens to a 4 cycle when you set it to max RPM and then point the nose up? It sags.

Well......right there the curve just moved, didn't it?

Maybe a lot, maybe a little, depending on the prop, RPM, and fuel, but it will sag. Sometimes it will continue to sag until it literally starts to fall of the planet.

Specifically, the curve just moved to the left in the overlean condition. In the worst case the curve will continue to move the left until you allow the engine to richen again. Since the curve is moving left the output begins to drop as the engine struggles to maintain RPM.

Until the variables are eliminated, the curve is secondary function...

>What happens when you set 2 cycle engine at max RPM on a
>suction tank, then point the nose up (combat guys?). It
>sags doesn't it? In a flat two stroke the engine has no
>"second gear". What happens on a bladder? Guess...
>
Depends a lot on the engine and prop. A well broken in ABC engine with the correct size/pitch of prop will just keep running when you tip the nose up. Too big a prop/too much pitch and it will slow down, that's for sure.

On a bladder pretty much the same thing. If the prop is correct it will slow down some in maneuvers but not go over lean. Too big a prop and esp. too much pitch makes the motor slow down a lot in maneuvers. Then it takes a lot longer to get back up to speed. That was a favorite setup back in the early 70's- G-21 35 running an 8/8 prop for maximum level flight speed.

>Depends a lot on the engine and prop. A well broken in ABC
>engine with the correct size/pitch of prop will just keep
>running when you tip the nose up. Too big a prop/too much
>pitch and it will slow down, that's for sure.

If we are indeed talking about simply tipping a statically loaded engine, the load is constant. The mixture is the only thing changing. The effect is still there with the lower load but reduced.

You could also add enough nitro to the larger prop combination until the effect diminished, but it will still be there.

I'm with you, Brad. Load AND tank location make a big difference.

In the case you sited, hold the airplane level, lean it out to max, then tip it up (while still holding it) and it will go overlean but the load does not change. Do the same in level flight with it slightly rich, then tip it up and it leans out to greater power. Again the load remains constant. This is the way I set a Fox 35 on the ground. Set it so it is 4-cycling while level, but breaks to a 2-cycle while you hold it at about a 45 degree angle pointed up. This portion of the equation is strictly a factor of fuel draw only, and has nothing to do with load. (You may be able to do the same thing by grabbing the spinner while level and adding more load, but that is factor number 2).

Leonard you are 100% correct. You hold it and it will break when pointed up and the load has not changed. Then it does the very same thing in flight. And this is where the load is supposed to change right? If it were so then it would act differently in the air right?

I agree with you all the way...

What you describe as moving the curve left or right in response to fuel pressure changes, to me seems quite reasonable and logical, and perhaps just another way of expressing it.

Loading the engine moves the RPM along the curve and moving the curve via mixture changes has similar effects.

[photo not recovered: 3e69bb3870f12ad5.jpg]

"Therapy helps, but screaming obscenities is cheaper."

LAST EDITED ON Nov-20-03 AT 12:53 PM (CST)
 
>Loading the engine moves the RPM along the curve and moving
>the curve via mixture changes has similar effects.

Yes.

The point being that depending on the curve alone will call for a solution that keeps the mixture constant. Siphon feed systems will never do that.

I wish I knew enough about this "stuff" as you guys, then I could comment.
But..... I just know that when I am running a 4-2 engine and it comes on when I point the plane up and shuts off (back to 4cycle) when It starts down, its working.
As for my pipe ships, I look for the same *type* senario, but the .51 and .61 just seem to run the same. Just cranking out the right amount of power to do what it needs to do as long as I got a good prop on them.
Haven't tried the 4S, yet. Anyone want to donate a 56 with some props for me to experiment with?? LOL!!
Bill <><

Godzilla,

Maybe I misunderstood you, but I think you are, at least in current context, looking at "the curve" in the wrong way. As the motor in question increases or decreases in rpm while running, the spot where it is on the curve is what we need to describe, not the curve shifting about, as it is a constant.

Yes, I have come to believe, and understand why, a torque curve can be caused to move left or right with our compression-ignition engines run on different blends of fuel. But that is not what you said.

Am I missing something?

>Yes, I have come to believe, and understand why, a torque
>curve can be caused to move left or right with our
>compression-ignition engines run on different blends of
>fuel. But that is not what you said.
>
>Am I missing something?

Yes.

(BTW this is my theory but I will stand behind the logic, and even prove it when I get the chance).

The curve definitely moves with mixture and combustion temperature, both typically causing the curve to move to the left. Too rich or too lean both cause the engine to lose power and RPM (to different extents on 2 and 4 cycle engines).

The curve is definitely shifting with changes in mixture and heat, or rephrased, the engine has jumped from one curve to another. The nitro can be used to move to a move favorable curve (or simply compensate for mixture changes) but will never stop the engine from shifting the curve.

For example:

Take and engine, any engine. Set the carb to the "perfect" setting. Dyno the engine. Look at the curve. Now set the same engine rich and then lean and repeat the test. Look at the curves. They will be different!!!! The rich and lean setting curves will be left and below the curve at the stoichiometric setting.

In stunt engines the curve is moving due to a constantly changing mixture. No matter how slight, the curve is moving. The only way to stay on one curve is to regulate the fuel intake and get rid of the siphon feed fuel intake.

Godzilla,

I am sorry, but you have gone well off the reservation when it comes to your dyno-room nomenclature.

But I did get the props! Thanks.

>Godzilla,
>
>I am sorry, but you have gone well off the reservation when
>it comes to your dyno-room nomenclature.
>

What does that mean?

Is he not correct in his talk about what will happen is the engine is dynoed in a rich and lean and optimum setting?

Here in the world of siphon feed our fuel delivery is ever changing so the engines ability to produce power changes also. They go hand in hand. If you really want to figure things out you will have to come to grips with what is happening with fuel delivery first. I mean that is the foremost changing effect in all of the stunt runs of all time. The fuel delivery is always changing. I mean think about it for a minute. The fuel delivery is so sensitive that the run will change if the tank is to low or high by as much a 1/32 of an inch in level flight. Well in the maneuver this effect is even more so. In level flight the delivery is pretty consistant. You can think you have the tank close. Then you do a maneuver and the different runs show up. That tiny little differnce in tank height will have an effect on the engines performance right? I know you agree with me on that. Eveyone goes through the on initial flights. If this tiny little amount will have a noticable effect on the engine run and how much more power it will put out one way as compared to the other it proves out exactly what he is saying. Who cares about dynos and curves and stoichmetric-whatever it really means nothing. Fuel draw is king in our world. And dealing with it propoerly is the name of the game. Those who control it with the most consistancy usually are the winners.(That can mean the break run or the fat run or the lean run but the most consistancy with it will yeild the best patterns) As the attitude changes the fuel draw changes and this changes the amount of power sent to the prop. In the 4-2-4 it gains power as the motor leans out. It is coming from a rich setting.(or could be called the idle setting) In the slight break pipe setups it is less noticable and can be controlled very nicely. In the 4s it can lead to overheating and power losses by the end of the sqr 8 like never before. When the engine is lean it just simply cant deliver the power to the prop. It just cant. It doesnt have an extra gear so to speak.

I think talk of dyno and curves and all that is fun for sure but is it real? We cant measure anything in the air. We have no idea what is really going on up there. We have to go by our ear and our feel. And our feel and ears are telling exactly what is happening. And I have been running a great deal of tests that are showing this to be very true.

Eliminate the variables...eliminate what we cant really know for sure and deal with the facts. Fact is when it is lean it wont put it to the prop and the same goes for rich.(I am talking 4s here)

Look at Drag racers they jet and rejet the carb to get the optimum setting for max power. There is no use in running way rich it helps nothing and lean it will just be down on power. Say you are in a Prostock 68 Camaro with 500 incher pounding out 1310 hp and you have just left the line and are about to put in a 6.89 at 209 mph and at the 1000ft mark your motor loses a 1/3 of its fuel to the carb. What do think will happen? It falls straight on its ass. Same goes if you were to leave the line rich and smoking and then all of sudden you get the right setting at half track and it just jumps up and you are gone. The attitude of the engine means nothing it is the sudden loss in fuel or sudden gain in fuel to the amount of air entering the system that really makes things hay wire. In my case it was a just a bit rich and then way advance the timing...40 degrees. The 454 just loved that from 4500 to 7800 rpms below that and it smoked all the time. But start hammering it and was off to the races.

In a 2s we leave the line a little rich and use that as our idle speed and then we use the attitude of the plane to adjust the fuel mix and its off to the races. Our prop dictates the how much it will speed up. In a 4s there is nowhere to go if you set it way lean like we do on the ground. I mean you pretty much have to to run it for stunt.

To really eliminate this variable it will be a completely seperate electric pump. It will also have a return system to the tank with a smaller line much similar to your car's system. Using pumps that rely on the muffler pressure or crankcase pressure are still relying on rpm to establish pressure and that wont totaly seperate the effect. I guess a bladder would do the same thing with alot less trouble.

>Len made some excellent points in dispute of the "back side"
>theory. One point that really hit home was the fact that
>stunt engines pick up RPM when they are pointed up. In a
>nutshell, the fuel head has changed in an instant, and there
>was resultant change in RPM. In the case of a typical 2
>stroke stunt engine, the result would be a change from a 4
>cycle to a 2 cycle or 4-2 break.
>
>Well......right there the curve just moved, didn't it?

Not necessarily. All we know is that the engine RPM increased. It could have increased while still staying on the curve, i.e. it was making exactly the same amount of torque (and therefore power) the curve predicts it was making both before and after the RPM increase.

I don't think the curve is fixed in RPM-torque space, either. But I think what it'd take to move it would be a change in venturi size, an increase/decrease in nitro, or perhaps running very lean (and overheating). I don't think running rich would do it very much.

>>Well......right there the curve just moved, didn't it?
>
>Not necessarily. All we know is that the engine RPM
>increased. It could have increased while still staying on
>the curve, i.e. it was making exactly the same amount of
>torque (and therefore power) the curve predicts it was
>making both before and after the RPM increase.

Sorry, I think I am right (at least in the case of the typical stunt run). BTW I am wrong a lot, but I think I am right in this case.

I think Frank's dyno testing also proves what I am saying. Frank does not dyno the engine "running in a 2 stroke". The engine is set to a typical "stunt mixture" and dyno'd at that point. If set to peak RPM (flat 2 stroke) the curve will look much different.

If the typical 2 stroke stunt engine is set to transisition for "4 stroking" to "2 stroking" SIMPLY DUE TO THE FUEL HEAD CHANGING THEREBY CHANGING THE FUEL MIXTURE, the engine is changing from one curve to another.

If the same engine changes from "4 stroking" to "2 stroking" due to the load changing, you would be on the same curve.

>I don't think the curve is fixed in RPM-torque space,
>either. But I think what it'd take to move it would be a
>change in venturi size, an increase/decrease in nitro, or
>perhaps running very lean (and overheating). I don't think
>running rich would do it very much.

I agree. That is basically what I said. Running rich and lean changes the curve. Changing the fuel head changes the mixture, thereby changing the curve.

Changing the nitro is somewhat similar to changing venturis (or opening the throttle) but not exactly.

The 4 cycle in particular is very susceptible to overheating in the lean state due to its high exhaust temperatures. It also does not have the excess of unburned fuel and oil to cool it like a typical 2 stroke.

Doug and Brad (and all), what I really see here is an effective adjusting of the mixture when we point the nose up. Because the location of the tank affects the fuel draw, we start out rich so that it can lean out to where we want it in those more critical moments. I don't know that this is affecting the horsepower or torque curve, however. (Remember, I said "I don't know". This is not a definitive answer.)

When we run too rich, we will never reach the peak of the power curve, much less fall off. And when we go too lean, well we are not even allowing it to reach its peak. It is sort of like stalling a wing. (I liked Igor's explaination of that.)

On another note, but following the subject, I have often thought of using a pusher configuration and how this would really mess things up. Can you imagine running with the tank in front? You point the nose up (engine down) and just at the point when more thrust is demanded, the engine goes rich. You point the nose down (engine up) and all of a sudden it leans out and goes wild. Did you ever wonder why you don't see many pushers in a stunt ship? (It might be perfect for a canard except for this "minor" glitch.)

>Godzilla,
>
>I am sorry, but you have gone well off the reservation when
>it comes to your dyno-room nomenclature.

Thanks for setting me straight. I guess I can stop now.

>>Godzilla,
>>
>>I am sorry, but you have gone well off the reservation when
>>it comes to your dyno-room nomenclature.
>
>Thanks for setting me straight. I guess I can stop now.

You are most welcome. But don't stop now.

There is indeed some sort of miscommication here. Look, I don't speak jive, but I can speak dyno, and you are sounding to me as if you're speaking jive. Maybe the problem is on my side of the equation, but until we have some common ground here, I am having trouble following what you are trying to say.

Ya know, it is winter, time for building and developing Killer Bits for engine and models. You guys keep talking about fuel feed and such, yet all that known-to-work stuff I sent to you sits around getting all clogged with old fuel.

Friends--in particular one Mr. Moon--I once set to Brad a fuel regulator which had been used, in a contest, in Mr. Zilch, between a bladder tank and a Fox 35. Not wholly original work on my part, PW had years earlier done the same thing with an ST 60. In fact, Paul readily supplied the last bit of knowledge required to make mine actually work in a Stunt application.

I don't know if such a setup is The Answer, rather think not, in fact. But it might be cool, and I am hoping you guys will invest 1,380 hours of effort into further development of a system I will in the future be able to buy from UHP for, oh, $14.95 and which not only feeds vitamin N to the motor but also allows me to put entire load of fuel spot on the balance point.

Okay, make it $19.95, no problem.

Dan

P.S. If you need more hand-wound springs for the regulator, you are more than welcome to come sweep my shop floor at your convenience.

P.P.S. Those who don't get the joke should know these springs were made from a single strand of .018 7-strand cable. Clip, clip, clipping lengths of spring fabricated by tightly wrapping wire onto a mandrel would send these things sproinging off all over the place. Finding the exact one you had just cut to an exact length was horribly frustrating.

>Friends--in particular one Mr. Moon--I once set to Brad a
>fuel regulator which had been used, in a contest, in Mr.
>Zilch, between a bladder tank and a Fox 35. Not wholly
>original work on my part, PW had years earlier done the same
>thing with an ST 60.

Don't worry I remember... That little sucker may come in handy real soon.

Dan,

Since you know Dyno why dont you enlighten us? Tell us how fuel mixture will not change the output of the motor? Who cares about dyno and curves. The truth is the lean 4s makes less power than its level flight situation. Why do you think PW first 56 would run a 4.80 lap and real slow maneuver? It would go lean and lose power to a point where it would run the maneuver. I know this becuase I was running these 5 years ago and running into the same thing. I didnt prefer this type of run.

Please talk dyno. And dont forget when the nose is up the mixture changes. Please tell me how that doesnt effect the output.

I am all ears.

PS I wish talk of where the motor is on the curve would stop. No one knows where the motor is in the air. There is no way to monitor it. Just talk of power loss/gain due to mixture.

PSS I heard about the stuff you sent but never saw it or had any want to fly with a bladder. But it would be a good test. Maybe I can try it sometime.

>Dan,
>
>Since you know Dyno why dont you enlighten us? Tell us how
>fuel mixture will not change the output of the motor? Who
>cares about dyno and curves.

Snip, snip...

I don't believe I have questioned the specific output (dyno talk, hope it doesn't confuse you) of a given motor falling or rising due to changes in fuel mixture.

Besides, your last comment, as quoted above, really makes any further discussion pointless. And maybe that is as it should be.

After all, not many of us have toy-airplane motor dynos in their inventory of Cool Stuff. Frank does, but he is a bit too far away for me to drop in on when I have a question as to the latest tune-up on my 56 or 72. Ted's got one, but the last time he brought up the subject, we--the collective we--chased him back in the house with well-intentioned, and conflicting, advice. Haven't heard a word from him since on this subject.

So we're all kinda shooting in the dark here. Making inaccurate statements as to cause and effect all the more confusing.

>After all, not many of us have toy-airplane motor dynos in
>their inventory of Cool Stuff.

I do. You may borrow it if you'd like.

>>After all, not many of us have toy-airplane motor dynos in
>>their inventory of Cool Stuff.
>
>I do. You may borrow it if you'd like.
>
>Preston

This has got to be a head fake. You're using the shop of one H. Rush, he surely knows of this dyno of yours, and it is still available for the use of others?

My man! I love ya! And truly regret those things I said last time we were together and you were out trying to hold tight to a Nelson-powered AMA Combat plane. Actually, you appeared to know *exactly* what you were doing, my remarks to the contrary.

Anyway, you're on. And thanks.

>This has got to be a head fake. You're using the shop of one
>H. Rush, he surely knows of this dyno of yours, and it is
>still available for the use of others?

I guess we're overwhelmed with other projects right now,
so the dyno's available. I'll send you a note.

>My man! I love ya! And truly regret those things I said last
>time we were together and you were out trying to hold tight
>to a Nelson-powered AMA Combat plane. Actually, you appeared
>to know *exactly* what you were doing,

Of course I did - I was holding on for dear life, like usual.

>
>I don't believe I have questioned the specific output (dyno
>talk, hope it doesn't confuse you) of a given motor falling
>or rising due to changes in fuel mixture.

Well you told Brad he was off the reservation with his dyno talk and you know dyno. So talk dyno.

I am not confused just reading what you typed. I am also perfectly willing to listen to anyone talk about this stuff and put logic to it. It seems logical to me that the fuel draw or lack there of sometimes seems to play a much larger role in the run than anything else that is discussed.

>
>Besides, your last comment, as quoted above, really makes
>any further discussion pointless. And maybe that is as it
>should be.

I resent that. You have made no points as to why Brad is incorrect. Then you stand off and say he is off the reservation. I was simply trying to get you to explain your position but you seem to refuse to do so. You say you know about dynos and then refuse to talk about it. What gives?

>
>After all, not many of us have toy-airplane motor dynos in
>their inventory of Cool Stuff. Frank does, but he is a bit
>too far away for me to drop in on when I have a question as
>to the latest tune-up on my 56 or 72. Ted's got one, but the
>last time he brought up the subject, we--the collective
>we--chased him back in the house with well-intentioned, and
>conflicting, advice. Haven't heard a word from him since on
>this subject.

I know I dont have one either. And what would it prove if we did have one? Not much I think. You have to test it out in lean and rich situations with the carb on it. The motor runs differently with it on there. Oh well..

>
>So we're all kinda shooting in the dark here. Making
>inaccurate statements as to cause and effect all the more
>confusing.
>
>
>Dan

What a contridiction. If we are shooting in the dark then how can one say the other is inaccurate? Doesnt make much sense to me.

I know full well we are shooting in the dark. That is why I keep saying quit talking curves and dynos and stuff and talk about the real effect you feel in the pattern. However, I would like to hear about your dyno experience real life or whatever. I have none and find them very interesting.

Once on this board someone wrote that when the motor (a 2s) goes lean with the nose pointed up it is a load situation more than it is a fuel draw(mixture) issue. I used to think that for about as long as I have been flying and really trying to figure out motors. But I should have seen along time ago this is not entirely true. Just today I went out back and bolted up the FP 40 and got it running in a nice fast 4s. Then I rolled the stand to 90 degrees. This has no effect on the load. However, the motor goes lean in a fast 2s. The load is not changed but the motor is lean(read that change of mix due to poor fuel draw) So I reach in there and open the needle 1.5 turns while in this situation and it returns to it original rich run. Only now the motor is pointed straight up. Then I pointed it back to level. It went really rich and almost died. The load never changed since the motor is never allowed move forward and unload but the mix changed drastically as the motor had to pull fuel uphill. When I run this test on 4s engines they act the opposite and get very hot in the 90 degree postion. The draw uphill hurts the motors ability to turn the prop. This is where this moving curve comes from. Pretty simple huh? If the power was not effected by the draw/mix then the prop would keep on spinning at the same rpm and the motor would be fine. You would have to dyno the motor level then dyno it turned up but at the same needle setting to see if it really moves left and down. Do we care? I dont know. Will it help to figure out how to run it better? Maybe?

Brad, You get those props yet? I want to try this out and see what happens.

>Just today I went out
>back and bolted up the FP 40 and got it running in a nice
>fast 4s. Then I rolled the stand to 90 degrees. This has
>no effect on the load. However, the motor goes lean in a
>fast 2s. The load is not changed but the motor is lean(read
>that change of mix due to poor fuel draw)

Seems like a good explanation to me.

> So I reach in
>there and open the needle 1.5 turns while in this situation
>and it returns to it original rich run. Only now the motor
>is pointed straight up. Then I pointed it back to level.
>It went really rich and almost died.

Yep, this makes sense too.

> When I run this test on 4s engines they act the
>opposite and get very hot in the 90 degree postion.

OK, but isn't this a result of the engine leaning out,
just like your test with the FP40?

> The draw uphill hurts the motors ability to turn the prop.
> This is where this moving curve comes from. Pretty simple huh?

Well, no. I was following along fine, but your leap to
"the draw uphill hurts..." seems perhaps unwarranted.
To me, the uphill draw leaned the engine out, it overheated
and began to sag. But maybe these are just slight wording
differences.

Indeed, I think a lot of the difficulty in this
discussion could be solved if we were all in the same room
and could draw pictures. Without pictures, we must write
(and read) very carefully, with patience, and some amount
of generosity. (don't just assume the other guy is really
as idiotic as he seems to sound; instead, be generous and
assume you've probably misunderstood him and vice versa)

Now, what I think you're suggesting is that we put an engine
on a dyno, set the fuel mixture, plot the power as a function
of rpm, and label this curve as A, then reset the needle
(say a little leaner) and make a new plot labelled B.
Then compare the two curves and see what we see.

I think your hypothesis is
(begin Moonzilla hypothesis)
The horsepower peak shown
in curve B occur at a higher rpm for a 2-stroke engine
(that is, the general shape of the curve will appear to have
shifted to the right on the graph, assuming the rpm increases
as we move right). Furthermore, the behavior might be different
for a 4-stroke engine.
(end Moonzilla hypothesis)

Seem about right?

If so, this seems straightforward to test.
Probably take a couple of months, but straightforward.

If not, then I'm just lost, lost, lost. As usual.

LAST EDITED ON Nov-22-03 AT 11:34 PM (CST)
 
>>Just today I went out
>>back and bolted up the FP 40 and got it running in a nice
>>fast 4s. Then I rolled the stand to 90 degrees. This has
>>no effect on the load. However, the motor goes lean in a
>>fast 2s. The load is not changed but the motor is lean(read
>>that change of mix due to poor fuel draw)
>
>Seems like a good explanation to me.

Yep that is what I thought too. On the same page there.

>
>> So I reach in
>>there and open the needle 1.5 turns while in this situation
>>and it returns to it original rich run. Only now the motor
>>is pointed straight up. Then I pointed it back to level.
>>It went really rich and almost died.
>
>Yep, this makes sense too.

Yep we are thinking alike here also.

>
>> When I run this test on 4s engines they act the
>>opposite and get very hot in the 90 degree postion.
>
>OK, but isn't this a result of the engine leaning out,
>just like your test with the FP40?

Yes and the reason, I didnt make clear, is that we already have to start with the 4s at a very lean or near full peak rpm in level position. So if it has to draw fuel uphill it will hurt the power output to be any leaner, no more room for more rpm. In a 2s it is not at full lean mixture and can handle the lean out pretty easily. Does this clear up that staement a little?

>
>> The draw uphill hurts the motors ability to turn the prop.
>> This is where this moving curve comes from. Pretty simple huh?
>
>Well, no. I was following along fine, but your leap to
>"the draw uphill hurts..." seems perhaps unwarranted.
>To me, the uphill draw leaned the engine out, it overheated
>and began to sag. But maybe these are just slight wording
>differences.

Yes they are directly realted. One happens because of the other but they dont happen seperate. See note above. If it didnt have to draw fuel uphill it wouldnt heat and sag. Also you would think if you start out a little rich it would gain rpm during the lean out and run just fine. But as I found it doesnt seem to be so. When it leans due to uphill draw/and load it just revs until it heats up and sags. The 2s can be started out so much richer and still produce enough usable power that when it is leaned out it doesnt hurt the run. That is how I am seeing it. Do you think so too?

>
>Indeed, I think a lot of the difficulty in this
>discussion could be solved if we were all in the same room
>and could draw pictures. Without pictures, we must write
>(and read) very carefully, with patience, and some amount
>of generosity. (don't just assume the other guy is really
>as idiotic as he seems to sound; instead, be generous and
>assume you've probably misunderstood him and vice versa)

That is a very true statement and couldnt be more right on the money.

>
>Now, what I think you're suggesting is that we put an engine
>on a dyno, set the fuel mixture, plot the power as a
>function
>of rpm, and label this curve as A, then reset the needle
>(say a little leaner) and make a new plot labelled B.
>Then compare the two curves and see what we see.
>
>I think your hypothesis is
>(begin Moonzilla hypothesis)
>The horsepower peak shown
>in curve B occur at a higher rpm for a 2-stroke engine
>(that is, the general shape of the curve will appear to have
>shifted to the right on the graph, assuming the rpm
>increases
>as we move right). Furthermore, the behavior might be
>different
>for a 4-stroke engine.
>(end Moonzilla hypothesis)
>
>Seem about right?
>
>If so, this seems straightforward to test.
>Probably take a couple of months, but straightforward.
>
>If not, then I'm just lost, lost, lost. As usual.
>
>Preston

I think you followed along my post just as I intended. I dont think you are lost unless I am lost and then we are just plain old lost together...

>I think your hypothesis is
>(begin Moonzilla hypothesis)
>The horsepower peak shown
>in curve B occur at a higher rpm for a 2-stroke engine
>(that is, the general shape of the curve will appear to have
>shifted to the right on the graph, assuming the rpm
>increases
>as we move right). Furthermore, the behavior might be
>different
>for a 4-stroke engine.
>(end Moonzilla hypothesis)

Preston gets it.

Most specifically the 4 cycle curve will down and to the left in curve B. It will also continue to fall down and left with time (and temperature).

It's easy to test, but I can tell you from just watching tachs and doing a lot of needling on engines, the hypothesis is correct. We have also done preliminary tests.

I think it's simpler than this. On the 4 stroke, you're running too large a venturi for consistent fuel draw, and the power you're making depends on having that large venturi. The large venturi, however, results in leaning out when you point the nose up, or conversely, it floods horizontally when you set the needle correctly for the vertical orientation.

On the 2 stroke, you're running a venturi that is small enough to easily draw fuel in the vertical orientation without having to set the needle so that it sputters in the horizontal.

In other words, the ONLY difference between the two is the way the venturi was chosen. In other words, you've chosen to run the 4 stroke with a huge venturi because you "need" the extra power, while, on the 2 stroke, you don't "need" the extra power, and can live with the smaller venturi. With the 4 stroke, you're hurting for power, so you run a large venturi, on the 2 stroke, you've got so much extra power, you can afford to back off and run a smaller venturi.

Or to put it another way.. given the same output the way you've set the engines up now, the 2 stroke has scads more power than the 4 stroke does when you choose the correct venturi size!

>I think it's simpler than this. On the 4 stroke, you're
>running too large a venturi for consistent fuel draw, and
>the power you're making depends on having that large
>venturi. The large venturi, however, results in leaning out
>when you point the nose up, or conversely, it floods
>horizontally when you set the needle correctly for the
>vertical orientation.
>
>On the 2 stroke, you're running a venturi that is small
>enough to easily draw fuel in the vertical orientation
>without having to set the needle so that it sputters in the
>horizontal.
>
>In other words, the ONLY difference between the two is the
>way the venturi was chosen.

Well... no.

The reason for samller venturis on 2 strokes is to limit the RPM (typically) at least with the modern engines that have higher RPM's.

There simply is no reason to limit the 4 stroke with venturi, the RPM range is already prettty much ideal for stunt. Closing the venturi simply reduces the power.

We did all our preliminary tests on the 4 stroke with a very small venturi. The effect is still there. It really gets no worse with the larger venturi. The burndown is actually less with the larger venturi because the engine breaths better.

>The reason for samller venturis on 2 strokes is to limit the
>RPM (typically) at least with the modern engines that have
>higher RPM's.
>
>There simply is no reason to limit the 4 stroke with
>venturi, the RPM range is already prettty much ideal for
>stunt. Closing the venturi simply reduces the power.
>
>We did all our preliminary tests on the 4 stroke with a very
>small venturi. The effect is still there. It really gets
>no worse with the larger venturi. The burndown is actually
>less with the larger venturi because the engine breaths
>better.

Brad, I have no experience and very limited knowledge on the 4-stroke, so cannot comment on that. On the 2-stroke, however, it would not be correct to say that the reason for smaller venturis on 2-strokes is to limit the rpm. Certainly some people have been known to try to choke down the venturi and pile in the head shims on a Schnurle engine to "tame" it (Usually very unsuccessfully), but that is not the norm. I have a "formula" that I use in making venturis for various engines that is very reliable, unless the engine has been modified in some special way. Pat Johnston was the one who first published this in testing a number of engines. He sent his findings to me to see if I agreed, and they paralleled pretty much what I had also discovered. The size of the venturi used on the 2-stroke affects not only the available power, but the engine run itself, and this becomes the limiting factor.

Matt just recently got a new 65. Running it with the same pipe and same prop that he was using with the 61, he used a larger venturi with the larger engine (number 12 venturi with the 61, number 10 venturi with the 65. And both numbers may change in hotter weather). If the size of the venturi was chosen to limit the rpm, then the opposite would seem to be true.

Venturi size on the 2-stroke is simply a function of the cubic inches available and the fuel draw that results in order to obtain proper combustion for our applications. Size can be adjusted within certain limits, and we will do this on occasion because of power needed or conditions (high altitude, heat). However, if we went to a super large venturi on the 2-stroke, we would lose a lot of the characteristics of the engine run--including fuel draw. We might gain some in rpm. But it would not be a stunt friendly engine. A speed engine, for instance, would lean horribly if you were to point the nose up because of inadequate fuel draw. But you are correct that we could limit rpm by choking it down and adjusting the fuel mixture accordingly. It would be like going to half throttle (or whatever). But that is not what we are normally attempting to do.

I am curious, however, in asking you what size venturi you are using (or have used) on the 72 Saito, along with the size of the needle restricting it as well as the prop size. By comparison, do you know what the open choke area of the carb was with this same engine? (total throat area minus any protusions for the needle body).

Again, I would caution that carb size is not necessarily chosen to give a "stunt run", and some carbs on some engines are much larger than what we would use for venturis on our stunt engines. (As are some venturis. Just because a manufacturer provides a "CL" version of their engine does not mean that they intend it for "stunt" when using that venturi.)

>I am curious, however, in asking you what size venturi you
>are using (or have used) on the 72 Saito, along with the
>size of the needle restricting it as well as the prop size.

With the UHP manifold, the .265 with the OS Max nedle is the most popular.

I am currently running the .310 with the same.

I like this discussion very much. there are 2 things I would like to add. .

1. Fuel draw. The engine will react with a lag to changing mixtures, while it will react immediately to changing load. I think this an important difference in the discussion. When the path of the mixture is long, the lag will be considerable. If the path is small(e.g Fox 35) the lag is small.

2. Heath. Aeromodeller magazine has published the load consumption of several commercially available props via a grafic over a broad RPM range. This way it is possible to determine the powercurve of your engine without a dyno. You can do this with several needle settings. This way I have been experimenting a lot with different props with the engine in the testbench. I did notice however that the engines behaviour in the air was different than expected from my experiments.
The fact of changing the props allowed the engine to cool down and after restarting the engine, the engine was allowed to reach a stable RPM, hence a stable temperature.
I think heath is an extra important variable. Normally engine torquecurves are a set of measure points where each point represents a stable situation (Load, mixture, temperature of the engine). However if the engine is going from one load condition to another, the temperature change will be slower than the load change. So although the backside of the powercurve is rather steep on paper, the behaviour of the engine can be totally different because of the heath. E.g.when the engine is 2 stroking in a high load condition (climbing in the wingover) It might not slow down on the downwards path immediately. It might keep on 2 stroking for a few more secs because of the heath buildup in the engine. So really torque or power curves should have an extra dimension: Engine temperature.

Eliminating variables? Very difficult!

fwiw
Paul

P.s. I've been measuring ST 51 (with blocked boostport and hemo head) and ST46. I can try to publish the results here.

>I like this discussion very much. there are 2 things I would
>like to add. .
>
>1. Fuel draw. The engine will react with a lag to changing
>mixtures, while it will react immediately to changing load.

Yes, Yes. Yes!!!! I said the same thing. That is why there is still so much fiddling to be done, even with the piped engines. The main contributing factor is the FUEL DRAW and its effects, the secondary effect is load effects due to the power curve.

"Charging on the bottoms?" Not due to load at all, delayed burst from changing the mixture.

>2. Heat.
>The fact of changing the props allowed the engine to cool
>down and after restarting the engine, the engine was allowed
>to reach a stable RPM, hence a stable temperature.
So really torque or power curves should have an
>extra dimension: Engine temperature.

I like everything you said here. I agree completely. This is why I say lean mixtures are never really good for what we do. Lean leads to hot.

>Eliminating variables? Very difficult!

Some maybe...

LAST EDITED ON Nov-23-03 AT 11:10 PM (CST)
 
>"Charging on the bottoms?" Not due to load at all, delayed
>burst from changing the mixture.

How would it charge on the bottom from mixture change. The mix is richening up on the bottoms. It is returning to its initial setting. During a hard corner it is very loaded to pull the plane through as it loads it charges. Load and mix go together. One isnt without the other. You say it reacts to load very quickly and this would have to be a very fast reaction but then say it is a mixture issue. Hmmmm. Anyway how can you say for certain this is what it is without some sort of test to tell you so.

>I like everything you said here. I agree completely. This
>is why I say lean mixtures are never really good for what we
>do. Lean leads to hot.

I disagree with this as we talked about it earlier. Lean up in 2s that are controlled are very good for stunt. I have been able to get it to work very well. According to some I cant really think for myself so I just copy everyone else.

>So we're all kinda shooting in the dark here. Making
>inaccurate statements as to cause and effect all the more
>confusing.

That really chaps my butt.

What positive have you brought to this converstation?

Not one darn thing. So far all you have managed to do is make slighted remarks as to what is being said with abvsolutely no qualification.

Since you do not feel inclined to explain you remarks, maybe you should just keep your mouth shut...

BTW I am not shooting in the dark. Maybe you just don't understand what I am saying.

LAST EDITED ON Nov-22-03 AT 03:23 AM (CST)
 
The rich and lean
>setting curves will be left and below the curve at the
>stoichiometric setting.

Ron B.
F4Fguy

I may be nitpicking, but stoichiometric mixture will never give best power or torque.Best power will always be on the rich side of stoichiometric,and best fuel consumption always on the lean side.Torque will always peak at whatever speed the the most air is consumed per cycle.Power will always peak at the point where the most air is consumed per unit time.
Attached are some curves illustrating this.Yes they're old and for gasoline,but an otto cycle is an otto cycle.the main differences re ours are size(making ours less efficient),and the curious semi-compression ignition system which makes ours pretty much single speed devices.For a given advance(ours is essentially fixed)the effect of mixture strength is going to be the same.In respect to varying mixture strength there's also no material difference between two and four stroke engines,they're both operating on the same cycle.
As to fuel itself,there's obviously going to be an effect on power.Methanol has less heat per pound than gas,but best power mixture is more than double,and the range of combustible mixture is far wider. In addition the heat of vaporization improves mixture density giving more power at the expense of fuel economy.This cooling also has the effect of retarding the "spark" for a given compression ratio.This is also what we're really doing by putting in extra head shims.

These data are from: Combustion Engine Processes -Lester C.Lichty,Professor Emeritus,Yale Univ. !967 edition.

Your attachments do not seem to be HTML files.. (?)

[photo not recovered: 3e69bb3870f12ad5.jpg]

"You do ill if you praise, but worse if you censure, what you do not understand." -Leonardo da Vinci

LAST EDITED ON Nov-22-03 AT 10:34 AM (CST)
 
I like the idea of a miniature fuel pump being used to supply fuel to the venturii. I can see where it wouldn't be that hard to build such a unit, complete with smaller return line to the tank. Fuel delivery to the venturii would not be variable, or affected by the planes attitude. Seems to me that would negate the problem, especially with the 4 strokers, of the engine going off speed in the manuevers that's been noted in this forum.

The drawbacks, of course, are extra weight, and more complication.

Someone with a few machining skills should find it rather easy to build up the pump cavity. a small battery driven electric motor could be mounted to drive the pump, independent of RPM. The connection at the venturii could be as simple as a T connector, with a slight restriction on one side, the side where the return line connects, to having the NVA having the restricted return line within the body.

I wish I had the machining equipment, and the skills to try this out.

I just thought of another possible benifit. Making it possible to turn off the pump during flight, either by bellcrank movement, timer, or remote. This would allow the engine to be shut down. This might save an over run, an engine when it goes off maladjusted, or after a prop strike on take off. Of course under FAI rules, it wouldn't be allowed to have this type of control over the fuel delivery.

I'm wondering why everything is spinning around?

Pumps have existed for a long time. The Perry Pump works off engine vibration, and might be the easiest thing to use if you are using a 4 stroke.

LAST EDITED ON Nov-22-03 AT 12:10 PM (CST)
 
>Your attachments do not seem to be HTML files.. (?)
>
>[photo not recovered: 3e69bb3870f12ad5.jpg]
>
>"You do ill if you praise, but worse if you censure, what
>you do not understand." -Leonardo da Vinci

Ron B.
F4Fguy

Sorry about that!

Brad
I can give you a bladder to try a pressure system,
if I can get out to the field.
Andrew Raney

Hi Guys,

Sorry if this repeats some of the above but I'm a little short on time. Here's my take on what I see to be the question: hp/torque curves and mixtures, etc.

One of the things that the scientific mind set deems most important in dyno testing is to only take readings at peak rpm for a given combination of factors; load, compression ratio, ambient conditions, etc. The reason is basic. There is no question that the hp/torque curve produced once you start messing with the variables is going to change! Simple as that.

The classic dyno test is used to determine the maximum horsepower and torque of an engine. Ultimately, these are the important issues and alternative curves based on less the optimum combinations of variable are what the testers/engine builders are trying to eliminate.

The thing that makes our issues as stunt drivers so much different and more complex is that we (with two strokes) almost never run the engine at what the dyno would show produces the most horsepower...we're not running dragsters or crotch rockets, we're trying to get our stunt ship to do tricks as easily and accurately as possible.

The act of lowering or raising the tank (or turning the test stand on its tail as has been discussed) is really nothing more than richening and leaning the needle valve. Doing so will, in fact, "move the curve" because the revs under a given load (your test prop)will change based on whether the engine is running at optimum at the start of the test or running rich. As soon as the revs change for a given load the torque changes by definition.

Having said that, this really has no real world application in our stunters. Under our flight conditions the "fuel head" the engine feels has only a small fore and aft component. Due to the centrifugal/centripital forces that result from tethered flight, for the most part the engine will "feel" the fuel pick up is a constant distance behind it and the more influential effect will be the feeling the engine has that the pick up is "outboard" of the engine. Due to G forces the sideways displacement of the fuel pick up (pretty much constant) is a greater force than is the fore and aft. Thus mixture changes due to tank/engine relationship are significantly different under flight conditions than on the test stand or when you point the nose up prior to takeoff.

This is where the subject of load enters the picture. The fuel draw conditions in flight, as we've demonstrated are only modestly variable and, therefore, have only a small amount to do with the four/two/four break as a result. The much larger component is the fact that when we start to climb we put a greater load on the engine and, if the mixture and combustion cylinder pressure conditions are in the proper relationship to one another the resulting load will cause the engine to break into a two stroke. Simulate this on the test stand by putting a load on the spinner (your choice how to do this) and listen to the four stroking engine break into a two stroke even as the actual revs decrease.

This can also be demonstrated by running the engine on one prop and setting it at your usual RPM for launch. Now, take that prop off and progressively put on larger props of the same nominal brand, etc. Don't touch the needle and you'll discover that as the load gets greater the revs will go down but the engine will at some point break into a two stroke and, as the load is increased more, will over heat and quit from a lean two stroke condition. Thus, we demonstrate the effect of "load" as a 4/2/4 controller rather than simply fuel mixture.

This final scenario is what led Brett Buck to suggest dyno testing in such a manner as opposed to the "conventional" method testing only at peak. The fact is, we operate our engines under varying load conditions and generally at less than optimum mixture for launch. Running tests with varying prop sizes and a constant mixture setting would approximate the conditions the stunt engine sees in flight. The resulting torque/hp curves would be more reflective of our real world conditions as a result.

Four strokes are a bit different animal in this regard. We do, in fact, launch them at pretty much peak and often do experience the sag that results from loading them to the point that the mixture goes to the lean side of optimum and the engine sags (exactly as Doug and Brad have suggested happens if you turn the four stroke test stand on its tail) In my opinion, of course, in flight the sag and heat comes from the increased load rather than the tank/engine relationship.

This appears to be more of a problem to overcome than with the two strokes and is probably the reason we find our selves going to such huge displacements with them to fly the same airplanes that many fly competitively with VF .40s for instance. They need something with enough torque to effectively shrug off the load from vertical maneuvers, etc.

I also suspect that part of the issue with the four strokes is that they only have one intake stroke in every two revolutions and thus load/mixture effects are exacerbated.

By the way, I think the above addresses the issue of why the clunk tank is so magically superior to a conventional vented tank with the four stroke. As we discussed the centrifugal forces in flight are considerable. In addition to those forces, in hard maneuvering we get substantial vertical forces on the fuel as well. In a hard inside corner for instance, there might be 15 or 20 positive Gs acting on the airframe and everything attached to it including the fluid in the tank. Thus, the clunk goes to where the head pressure at the fuel inlet is the greatest. This pressure would, for instance, be the greatest at the bottom outside corner of a tank during a hard inside corner. Ta Da! Exactly where the clunk went. Although head pressure is not a huge factor, in increase will tend to richen the mixture. This is exactly what we need to keep the engine in a happier relationship vis a vis load/mixture during the corner.

Enough.

>So we're all kinda shooting in the dark here. Making
>inaccurate statements as to cause and effect all the more
>confusing.
That really chaps my butt.

What positive have you brought to this converstation?

Not one darn thing. So far all you have managed to do is make slighted remarks as to what is being said with abvsolutely no qualification.

Since you do not feel inclined to explain you remarks, maybe you should just keep your mouth shut...

BTW I am not shooting in the dark. Maybe you just don't understand what I am saying.

The City Smasher

Godzilla,

Sorry to hear of your butt getting chapped. Stuff happens.

I don't think my absence over the weekend should distress you in any way, nor should you make any assumptions based upon same. I've been making a second batch of sauerkraut and a bunch of pumpkin pies. The pies starting with actual Sugar Baby pumpkins from my garden. Even you don't rate higher than those two activities, I am sorry to relate.

Keeping my mouth shut is not an option, you know that, although once I had a line on use of Preston's dyno I figured on coming back to the table armed with more information.

It is quite possible that I don't understand what you are saying, never said otherwise. But your cavalier use of terminolgy to describe specifics has not helped in any way I can see.

And--short-term only--I had kinda tossed in the towel, hoping to see some input from, among others, F4FGuy. As you can see, he has done so; I hope the information has been fully digested.

I see Ted has checked in, although I have not read his post yet. As Ted is known to have a dyno, even though he has not mentioned it in some time, I am real interested in what he has to say.

Hugs and kisses,

I am reminded of a favorite dyno story.

It was a number of years ago, my family of four was still racing RC cars and I had a close--some said too close--relationship with the Campbell family at Delta Mfg. in Iowa.

Delta had been importing Picco 21s for a year or so, our little race team was using them exclusively. Chassis had caught up to the current power levels, we were looking for more grunt and over-rev ability/reliablility at the top end. The latter was solved--no details will be furnished at this time--by switching from Picco's bushed rods to a Delta-produced aluminum rod without any bushings whatsoever. Instantly we went from 10-hour rods to pieces which could safely be used for 20 hours of operation.

(Bushed rods are not always the hot tip, especially when in order to fit the bushing one needs to punch a bigger hole in body of rod, effectively weakening it.)

More power was needed, Delta's low-oil fuel netted us some when we got down to 7% total oil content, having tested at a mere 5% oil content without breaking stuff.

At the time Kevin Orton had gone to work for Delta. Kevin was a specialist in 1/12-scale electric racing. Smart kid. He was the first to develop what "everybody knew" to be impossible, a fully-automatic charger for sub-C nicads, a device which needed only the push of one button to fully peak a set of batteries. Yeah, everybody sells 'em now; but Kevin and Delta were first. Kevin eventually moved on to found Tekin Electronics, supplying all sorts of chargers, speed controls, etc., etc. to RC car guys.

But during this time Kevin was being converted to 1/8-scale racing and it bothered him to not know all the specifics concerning power output of the Picco motors. He first wanted a solid baseline and then to grind on the things, checking to see what worked and what didn't.

Needed a dyno. So he built one. Pretty simple device, a fairly small prop just for cooling purposes, belt-driven jackshaft assembly running in very high quality bearings, this driving a standard Pylon electric starter which was being used as a brake.

Ken Campbell, owner of Delta, told him it wouldn't work, any old 21 would spin the starter almost as if it wasn't even there. Kevin fired the motor. Ken was right.

"No problem," Kevin said. "I'll just back-drive that sucker with voltage; that will give us plenty of load!"

And it did. The poor old starter motor almost instantly turned red hot, Ken ripped the whole test rig off the bench, throwing it in a bucket of water.

Saying, "Hey, this isn't a bad idea after all. We'll measure the temperature of the water in that bucket. You run your dyno setup for an amount of time certain. Then we'll toss the starter in the water, measure the increase in temperature and we'll have the data we need!"

Ron B.
F4Fguy

YEAH,YEAH,that's the ticket! We'll use a calibrated bucket to get the mass,then we'll time the temp rise and get the horsepower!

This is getting close to my "caloremeter" science project I built in high school. The idea was to heat water in a container by burning something in a vessel in the water and then measure the heat rise of the water. The container was the can from a hand-crank ice cream freezer, the vessel was a film can, and the fuel I would test was a generous gob of match heads.(Do not try this at home!) An expensive thermometer sat in a hole drilled in the can to read the heat rise. But how to ignite it? Being a C/L type of kid, I used a Champion VG-2 wired through a hole in the top of the container to a dry cell battery. The result was, as the Moon boys often say, "Awesome!" No one was hurt, the expensive thermometer was broken, the table was wet, and the lab reeked of sulfur smoke. A genuine "stink bomb". My science teacher did record a heat rise, however! PS-I got the P/L just fine, Mr. Dan.

LAST EDITED ON Nov-24-03 AT 03:11 PM (CST)
 
>Hugs and kisses,
>
>Dan

Smoochy, smoochy!!!

It's really hard to "think out of the box" when it seems we don't agree on where the "box" is...

>>Hugs and kisses,
>>
>>Dan
>
>Smoochy, smoochy!!!
>
>It's really hard to "think out of the box" when it seems we
>don't agree on where the "box" is...

Agreed.

And keep your tongue in your own mouth next time.

Ted that was a great post. Very informative and easy to follow.

However I do want to ask you a question.

If it is truly load that is causing the break and not mix then would a bladder fed 2s with a regulator still break when the nose is pointed up?

I was leaning to the fact that the two mix and load were intertwined during out flight.

What do you think?

Doug,

I've been trying to say the same thing as Ted, in different ways, for quite a while... No halo, nor the urge to work hard enough to earn one, so my remarks may easily be dismissed. No sweat.

LOAD is what makes a tuned pipe engine shift from strangled 'above the boost' RPM to lower, higher boost RPM. LOAD comes from a rapid, drastic rise in INDUCED drag over the wing to make lift for the 20 g to 35 g corner, or the 10 g round. The induced drag rises as the SQUARE of the change of lift coefficient. Lift coefficient rises directly as the change in required g for a maneuver. Go to a 10 g round, induced drag increases 100 times. In a 35 g corner, induced drag increases 1,225 times! (The 35 g max value is from Wild Bill N's remarks over the years...) Slam that much drag onto the model and it MUST slow down, some.

The slowing IS the "load" Ted, I, and a few others are talking about. At easy, level cruising flight, the prop generates little thrust, so far as the model feels things. The "net" thrust applied to the model is only enough to match the total drag of lines, skin friction, and induced (related to lift) drags -- not much, iow.

The load on the fuel in level flight is 3 to 5 g 'centrifugal' outboard, and 1 g vertically down. Draw that rectangle. The diagonal from inboard upper corner to outboard bottom corner is the "resultant" force on the fuel delivery system, AND THE ANGLE ON WHICH IT ACTS.

Figure the "height" along THAT angle as the fuel head. In maneuvers, added g are perpendicular to the wing's span, but a "vector" solution is still simple. The length of the fuel plumbing, from pickup end inside the tank to the fuel jet in the spraybar, is largely irrelevant. Where it DOES matter is that it is just about constant in terms of line loss!

Flight is a DYNAMIC condition; static fuel head is another animal entirely. We CAN go through some middle-school/high-school arithmetic to get an idea of the dynamic conditions... Or we can just point the model's nose up and say, "Yup, it gets leaner..." Hasn't anyone else had a model chirp to 2-stroke on turning DOWN into an outside square entry???? Nose down means rich, right? So, WHY a chirp to lean on a tight turn downwards, eh?? (My answer is the induced drag change and consequent slowing, causing prop load to increase, briefly...)

Dyno testing is to put together a curve touching the maximum torque and power values of an engine across its range of usable RPM. To get the maxima, we have to VARY the loads AND check to make sure the ultimate of Torque and Power are adjusted-in by mixture tuning. Trying to define a "stunt conditions" percentage of max values, and all on one prop? LOL.

On the bench, I've observed that a Fox Stunt 35 has less than 500 RPM on a 'good' stunt prop between blubbering rich and sagging lean. (Thus, its 'break' is not a thundering slam of gobs of power.) Other engines have over 1,000 RPM in 'wet' to 'lean' 2-cycle -- on one prop...

Change props and we have different conditions. A Fox 35, reworked, will 4-cycle at 13,500 to 15,000 RPM on a small enough prop... It will also never 4-cycle on a large enough prop, even at 6,000 RPM.

The ultimate torque and horsepower curves are useful because we can select prop, fuel, setting to:

-- cruise above torque peak RPM, at an RPM stabilized by mixture setting,

-- "shift" down to lower RPM (nearer torque peak RPM) under maneuvering loads 'felt' at the prop (as the drag increase FORCES the model to slow down.),

-- use the increased torque called upon to reduce speed loss and to power the prop around to regain cruise/near cruise airspeed more quickly, and

-- return to higher RPM, unloaded, 4-stroking when loads fall as we regain level cruising conditions.

I've learned that many things that seem 'intuitively" obvious are not at all accurate -- consequently, I seek to eliminate things that do NOT seem involved. Some cannot be eliminated, and, on further study, make more sense than what seems a simpler answer. Just one example is the difference between the 'static' and 'dynamic' conditions. We can't even simulate the dynamics on the ground, unless we can move at the model's airspeed on a path with the same radius AND angles from the horizontal...

Just because a thing seems to work for the present model, we should still be skeptical about it being a universal truth... it may work backwards, cause everything to get worse, even, on the next, nearly identical model... Cutting through to some deeper basics can help us avoid disasters like that...

>Doug,
>
>I've been trying to say the same thing as Ted, in different
>ways, for quite a while... No halo, nor the urge to work
>hard enough to earn one, so my remarks may easily be
>dismissed. No sweat.
>
>LOAD is what makes a tuned pipe engine shift from strangled
>'above the boost' RPM to lower, higher boost RPM. LOAD comes
>from a rapid, drastic rise in INDUCED drag over the wing to
>make lift for the 20 g to 35 g corner, or the 10 g round.
>The induced drag rises as the SQUARE of the change of lift
>coefficient. Lift coefficient rises directly as the change
>in required g for a maneuver. Go to a 10 g round, induced
>drag increases 100 times. In a 35 g corner, induced drag
>increases 1,225 times! (The 35 g max value is from Wild Bill
>N's remarks over the years...) Slam that much drag onto the
>model and it MUST slow down, some. (snip)
>
>The load on the fuel in level flight is 3 to 5 g
>'centrifugal' outboard, and 1 g vertically down. Draw that
>rectangle. The diagonal from inboard upper corner to
>outboard bottom corner is the "resultant" force on the fuel
>delivery system, AND THE ANGLE ON WHICH IT ACTS.
>
>Figure the "height" along THAT angle as the fuel head. In
>maneuvers, added g are perpendicular to the wing's span, but
>a "vector" solution is still simple. The length of the fuel
>plumbing, from pickup end inside the tank to the fuel jet in
>the spraybar, is largely irrelevant. Where it DOES matter is
>that it is just about constant in terms of line loss!
>
>Flight is a DYNAMIC condition; static fuel head is another
>animal entirely. We CAN go through some
>middle-school/high-school arithmetic to get an idea of the
>dynamic conditions... Or we can just point the model's nose
>up and say, "Yup, it gets leaner..." Hasn't anyone else had
>a model chirp to 2-stroke on turning DOWN into an outside
>square entry???? Nose down means rich, right? So, WHY a
>chirp to lean on a tight turn downwards, eh?? (My answer is
>the induced drag change and consequent slowing, causing
>prop load to increase, briefly...) (snip again)

To both Lou and Ted: I am with Doug on this one. I read Ted's carefully thought out response with much agreement on some areas but still disagreement in others. It appears that both of you are almost completely dismissing the effect of mixture and trying to transfer the total effect on load alone. What I would like to see is someone build a pusher stunter with the tank in front and get the same effect you describe. (no, I am not going to do it, because I do not believe it will work.)

Ted mentions the fuel head forcing the fuel to the outside, so (correctly) it is not computed from the back end of the pick up tube, but somewhere up the side of the tank. Now we all know what happens on a standard vented tank. The plane leans out during the run--even if flying straight and level--because of the change in the pressure from the fuel head. On such a tank situation, as the fuel is consumed, there is less and less centrifrugal force induced pressure and a leaning condition results. This is made worse when we go into maneuvers and the fuel is thrown (somewhat) to the rear (as in a straight up climb). In the cited example there would be 1 force of gravity pointing down and triple that (perhaps) pointing out. Vector it, and that is your fuel force. As the fuel is consumed, the top of that head moves farther and farther away from the front of the tank. With the uniflow tank the internal forces resulting in fuel consumption are more closely smoothed out because of a changing partial vacuum in the tank, so...

Let's assume uniflow here. No matter what we do, there is some fuel draw needed from the outside of the tank to the engine spray bar. At most this is an inch, increased up to, perhaps, three times in force by centrifrugal force. This force remains fairly constant so does not affect the engine run. On the other hand we have three inches (approximately) from the spray bar to the front of the tank. So, assuming a full load, or at least a fuel head that reaches the front of the tank, we still have 3 inches of difference from flying level to when the nose is pointed up. That will still affect the engine run no more and no less than when we are holding the plane with a running engine in our hand and pointing the nose up. That change does NOT disappear. Or, if it does, I would like for someone to explain to me how.

Now, I am accepting a change in load as "a" factor in the engine run. I am not disputing that this can and does affect it. However, when we enter our maneuvers, a square corner, for instance, you place the change solely on a change in load. I believe that is wrong.

You mentioned what the plane does, but have you truly considered what the fuel is doing under those circumstances? When, from flying level, we point the nose up in a violent corner, the fuel wants to continue its path of level flight. (Newton, "a body in motion stays in motion") The force of that action actually pulls the fuel in a 90 degree vector away from the engine. The same happens when the plane is flying level inverted at the top of the square and suddenly points down. The fuel is suddenly and violently pointed in a 90 degree vector away from the engine's venturi. Certainly the plane points down when making that turn from inverted at the top of the inside square. But the corner is often a softer corner, and, under any circumstances, the load increase on the engine is not nearly as severe as when going from level upright to the first corner "up". In the first example, where we are flying level and then turning "up", we have the total impact of the weight of the plane adding to the load. In the inverted turn to "down" we have the total weight of the plane reducing the drag. Yet the engine will react very similarly, due, primarily to the momentary change in fuel load.

I think it needs to be pointed out that the engine is in no way "required" to bring the plane up to any desired speed. I don't see the load on the engine being increased in such a dramatic fashion as you depict here (Oh, the plane slowed down, I have to pull harder.) The engine is turning the same propeller in the same way as when we hold the plane level in our hands and then aim it up. Now the engine may "unload" when the plane gets up to speed while flying, so, in that sense, we are "loading" it when we turn the corner, or whatever. But if this is a "load" on the engine, then the "load" on the engine is greatest when we are holding the plane in our hands and the plane is not allowed to go anywhere. (The plane now has a 220 pound gorrilla on its back). But, even while we are holding it, when we return the plane to level, the motion of the plane (going nowhere) remains the same, the "load" on the engine and prop remains the same, but it returns to its 4-stroke mode.

Thus, you have not convinced me in any way that load is the primary force. Load may enter the picture, but I am more convinced after this discussion that it is still the leaning effect of fuel that causes the major change in engine run.

>To both Lou and Ted: I am with Doug on this one. I read
>Ted's carefully thought out response with much agreement on
>some areas but still disagreement in others. It appears
>that both of you are almost completely dismissing the effect
>of mixture and trying to transfer the total effect on load
>alone. What I would like to see is someone build a pusher
>stunter with the tank in front and get the same effect you
>describe. (no, I am not going to do it, because I do not
>believe it will work.)

Or simply put the same engine on a pressure feed system or pump and see if it still reacts the same.

Leonard is right. The 4 cycle especially, because of its fuel draw issues, reacts differently. We have yet to test the 2 cycle.

Any combat flyer out there should be able to attest to the difference between siphon and force feed.

No one will ever be able to convince me that our engines react to load alone. If that is so then why does 1/32" of tank shim make any difference? OR changing from unflow to suction, or from pressure to atmosphere, or from standard venturis to spigots? All of these things have to do with draw not load.

It's both working in concert.

Len,

(didn't want to get into this... much subtle and louder discussion yrs back on ModelNet...)

With all friendly thoughts, I agree we see it a bit differently. I enjoy sharing understandings.

To me, ignition depends on both mixture setting and load. What makes our "easy" running cruise setting what it is - IMVHO - is that the piston passes through optimum combustion chamber conditions too quickly under light load, given the damping effects of a rich mixture, to fire strongly. If it did fire "hard" we'd see either RPM rise, or sag when it got to over-lean.

The mixture setting, whether trad 4-2-4 or "wet 2" settles RPM at the point where the piston's descent opens the chamber volume significantly before a 'full burn' can occur. Similarly, but for different reasons, a pipe goes to RPM beyond max torque boost in 'cruise' or other unloaded conditions, then prop load from turning drags RPM down toward max boost. That happens very quickly -- a matter of a few shaft rotations, at 175 to 200 /second, doesn't take long. Combustion chambers only remain in optimum conditions less than an eighth of a rotation, possibly much less. We may be talking about a shift of the time in which strong burning can occurs as changing from 2 or 3 ten-thousandths of a second to 4 or 5 tenths. That is apparently (for me at least) enough to alter the burn mode in the way we are used to hearing. And, I don't omit mixture setting -- it is crucial to the process. A rich mixture has plenty of fuel and air to produce power, but the soggy f/a mix dampens the rate of burn, prolongs the time needed to complete it.

Without a pipe, combustion's response to load is a mite slower switching to 2, and much slower returning to 4. We don't notice this as much with a steady 'wet 2' run as with trad 4-2-4...

Considering a sudden large increase in Induced Drag (can we agree that occurs?) fuel in the supply line from pickup to jet DOES tend to keep going. But that is FORWARD in the metal and plastic line for the slug of fuel there. Why, then, doesn't the engine go rich on corner entry? Fuel pressure in the tank is held in balance by the uniflow venting, and if the mass of fuel outside the supply plumbing does slosh in any direction I don't expect it to pull the fuel out of the supply tubing (pickup to jet.)

The bench test, squeezing a spinner, say, shows a switch from 4 to 2 if the engine was in 4-cycle, even though a tach shows reduced -- braked -- RPM. These ideas led to my comments about prop load being much more a factor than the g forces. In maneuvers, the g forces act at right angles to the direction of fuel flow in the tubing (pickup to jet). The fuel is subject to greater g, so it in effect weighs more, but the copper and silicone tubing support that weight. The "rise" from the pickup's spanwise position to the jet, in the direction of the resultant g vector, is still an inch or less. To me, THAT is the "head" the venturii draws against...

OBTW, I mentioned to Bill Netzeband, at a NATS or VSC years back, that I had come to believe that stunt engines are at their highest RPM in level flight and/or in long downhill paths like the back of the hourglass. He'd figured that out much earlier...

Whatever works, works. More important: the loose nut on the back of the handle <G> has so much to do with performance that lumbering atrocities of aircraft have been flown quite well. Today, we are blessed to know so much more about how to make good flying stunters... Any and all methods that work reliably are good. ...Even if we find someone who does it the exact opposite from our approach. As long as we believe it works, it works. I may be totally wrong, and accept that so long as it works for me.

I've done some experiments running a stunt engine from a bladder. Since I wasn't using a pressure regulator is was a bit tricky getting the setting exactly right, but the engine(40LA) could be set to a nice solid four cycle. Normally it would hold that setting through a large portion of the flight(some changes due to the bladder changing pressure). On a few flights, when it was set just right, it had a nice 4-2 break in the maneuvers.

So my conclusion is that the change in load from maneuvering is enough, all by itself, to trigger the 4-2 break. That doesn't rule out the mixture changing too though when you are using a conventional tank setup.

>Ted that was a great post. Very informative and easy to
>follow.
>
>However I do want to ask you a question.
>
>If it is truly load that is causing the break and not mix
>then would a bladder fed 2s with a regulator still break
>when the nose is pointed up?
>
>I was leaning to the fact that the two mix and load were
>intertwined during out flight.
>
>What do you think?

HI DOUG,

I DON'T DISAGREE. PLEASE NOTE THAT IN MY POST I STATED THAT "...Under our flight conditions the "fuel head" the engine feels has only a small fore and aft component..." THERE IS, OF COURSE, AN EFFECT THAT IS DUE TO HEAD PRESSURE AND, AS 'ZILLA COMMENTS IN ANOTHER POST, THIS DOES HAVE AN EFFECT ON THE WAY THE ENGINE RUNS.

MY ONLY POINT IN THAT PARAGRAPH WAS THAT THE POINT AT WHICH THE ENGINE "THINKS" THE FUEL PICK-UP IS LOCATED CHANGES SIGNIFICANTLY WITH THE INFLIGHT FORCES. ONCE THAT POINT IS ESTABLISHED THE POSITION RELATIONSHIP BETWEEN THE ENGINE AND THAT "NEW" POINT WILL HAVE PREDICTABLE RESULTS AS THE ORIENTATION CHANGES BETWEEN THE TWO POINTS.

I DO FEEL THIS IS A MUCH SMALLER FACTOR IN OUR INFLIGHT CHANGES THAN IS LOAD. YOU CAN ANSWER YOUR OWN QUESTION ABOUT A BLADDER FED FOUR STROKE BY REPEATING THE TEST I MENTIONED. GET THE ENGINE RUNNING IN A SOLID FOUR STROKE. SHUT IT DOWN BY PINCHING THE LINE. INSTALL A SIGNIFICANTLY LARGER PROP (GREATER LOAD) AND RESTART THE ENGINE AT THE SAME MIXTURE SETTING. THE LIKELY RESULT, DEPENDING ON JUST HOW RICH THE ORIGINAL SETTING, WILL BE A LOWER RPM AT A TWO STROKE!

EVEN MORE TELLING, PUT A LARGE ENOUGH PROP ON IT AND YOU'LL NOT ONLY GET A SLOWER RPM AND A TWO STROKE, YOU'LL FIND YOU NEED TO RICHEN THE MIXTURE TO KEEP THE SETTING FROM BEING TOO LEAN FOR THE LOAD!

WE'VE ALL, I THINK, WITNESSED THE MOST COMMON DEMONSTRATION OF THIS EFFECT DURING THE TAKE OFF AND LEVEL FLIGHT SEGMENTS OF MANY A TWO STROKE PATTERN. ESPECIALLY ON UNPIPED 4/2/4 ENGINES, IT IS COMMON TO RELEASE THE AIRPLANE, HAVE THE ENGINE BREAK INTO A TWO STROKE AND STAY THERE FOR A "CERTAIN" LENGTH OF TIME BEFORE "SETTLING" INTO THE DESIRED FOUR STROKE. IN MY TIGER .46 LIFETIME I DEPENDED ON THAT PHENOMENOM TO LET ME KNOW HOW GOOD MY NEEDLE SETTING WAS FOR THE UPCOMING FLIGHT. TOO LONG IN THE TWO AND IT WAS GOING TO BE FAST AND LEAN; TOO SHORT AND I COULD COUNT ON A SOFT FLIGHT.

THIS IS A CLEAR DEMONSTRATION OF THE EFFECTS OF LOAD AS THE AIRPLANE ACCELERATES. AT REST STATIC THRUST IS GREAT AND LOAD IS HIGH. AS THE AIRPLANE ACCELERATES THRUST AND LOAD BOTH DECLINE AND EVENTUALLY THE LOAD DECREASES TO THE POINT THAT THE ENGINE CAN FOUR STROKE. LOAD IT AGAIN AS IN HIGH "G'S" OR A CLIMB, AND THE AIRSPEED DECREASES BACKING BACK INTO THE RANGE AT WHICH THE ENGINE TWO STROKED DURING THE TAKEOFF ACCELERATION. THUS, IT BREAKS TO A TWO STROKE AND, WHEN THE DRAG DECREASES THE AIRPLANE ACCELERATES, THE PROP UNLOADS AND THE ENGINE GOES BACK TO A NICE FOUR.

YOU CAN FIND FURTHER SUPPORT FOR MY ACCEPTANCE OF THE TANK AND GRAVITY EFFECTING RUNS BY REREADING MY LAST PARAGRAPH OF THE PREVIOUS POST WHEREIN I POSTULATE ABOUT THE CHANGE IN FUEL HEAD PRESSURE DURING MANEUVERING WITH AND WITHOUT A CLUNK TANK. ALSO WHY I FEEL THE ISSUE IS SO MUCH GREATER WITH A FOUR CYCLE ENGINE THAN WITH A TWO STROKER.

GOOD DISCUSSION.

BOY, THIS IS JUST GREAT STUFF FOR WORKING THE OLD THINK TANK!

AGAIN, RE THE EFFECTS OF APPARENT TANK LOCATION RELATIVE TO THE ENGINE. WHATEVER THE SOURCE, FUEL HEAD PRESSURE IS SIMPLY ANOTHER SOURCE OF MIXTURE CONTROL. NO DIFFERENT THAN RAISING AND LOWERING THE TANK OR CHANGING THE EFFICIENCY OF THE VENTURI TO ALTER SUCTION.

NOTE THAT IN THE FOUR STROKE/CLUNK TANK COMBINATION THE SYSTEM APPEARS ON THE SURFACE TO ACT COUNTER TO THE "NOSE HIGH OR LOW" SYNDROME WE USE AS EVIDENCE OF THE NEED TO PULL FUEL TO A GREATER OR LESSER HEIGHT RELATIVE TO THE ENGINE.

WHEN WE THROW THE "G'S" TO THE ACFT IN AN INSIDE MANEUVER, FOR INSTANCE, THE CLUNK IN THE TANK GOES IMMEDIATELY TO THE BOTTOM OF THE TANK...TECHNICALLY, "LOWER" IN RELATION TO THE ENGINE. YET, MAGICALLY IT DOESN'T GO "LEAN" AS A RESULT...VERY IMPORTANT WE'VE LEARNED FOR FOUR STROKE "PULL" IN MANEUVERS.

I WOULD POSTULATE THAT THE REASON THIS IS SO IS BECAUSE THE FUEL HEAD PRESSURE IS SIMULATANEOUSLY RAISED AS THE "G" FORCES ACT ON THE FUEL MASS RIGHT ALONG WITH THE CLUNK--THE COMBINATION OF FORCES OBVIATING THE INDEPENDENT INFLUENCE OF ONE VERSUS THE OTHER AND THE ENGINE STAYS IN THE NICE POWERFUL MIXTURE SETTING WE HAD IN LEVEL FLIGHT. OF COURSE, ANY CHANGES THAT WOULD THEN HAPPEN WOULD LIKELY BE LOAD RELATED.

OH BOY, MY HEAD IS SPINNING!

This is the sort of discussion that convinces one successful flight of a CL Stunt model must be impossible, let alone worrying about (relatively) minor variations in the power and running characteristics of our engines.

Ted's head is spinning, mine hurts...

I am, however, reminded of a tour through a 60-powered tether car, one Luke Roy expounding on the technicalities. I don't believe it was actually Luke's car, although he was certainly active in tether car racing at the time.

As he got to the tank setup--a standard-looking hard tank--the mounting of same was pointed out and he was just getting into the fine adjustments required in moving the tank left or right in the car.

Somebody walked up, an experienced modeler of some sort, and expressed dismay that the car had a "suction" tank where he was expecting a tank pressurized by puffs from the crankcase, a bladder system, at the least some pipe pressure.

An exact quote will not be forthcoming, but Luke eyed this guy and explained how, what with running near 200 mph, a short (35-foot?) chunk of what seemed to be monstrous piano wire being required to restrain the car with a reasonable margin of safety, there was plenty of pressure developed within the tank.

I don't believe the guy got it, but I didn't forget.

>I DON'T DISAGREE. PLEASE NOTE THAT IN MY POST I STATED THAT
>"...Under our flight conditions the "fuel head" the engine
>feels has only a small fore and aft component..." THERE
>IS, OF COURSE, AN EFFECT THAT IS DUE TO HEAD PRESSURE AND,
>AS 'ZILLA COMMENTS IN ANOTHER POST, THIS DOES HAVE AN EFFECT
>ON THE WAY THE ENGINE RUNS.
>
>MY ONLY POINT IN THAT PARAGRAPH WAS THAT THE POINT AT WHICH
>THE ENGINE "THINKS" THE FUEL PICK-UP IS LOCATED CHANGES
>SIGNIFICANTLY WITH THE INFLIGHT FORCES. ONCE THAT POINT IS
>ESTABLISHED THE POSITION RELATIONSHIP BETWEEN THE ENGINE AND
>THAT "NEW" POINT WILL HAVE PREDICTABLE RESULTS AS THE
>ORIENTATION CHANGES BETWEEN THE TWO POINTS.
>
>I DO FEEL THIS IS A MUCH SMALLER FACTOR IN OUR INFLIGHT
>CHANGES THAN IS LOAD. YOU CAN ANSWER YOUR OWN QUESTION
>ABOUT A BLADDER FED FOUR STROKE BY REPEATING THE TEST I
>MENTIONED. GET THE ENGINE RUNNING IN A SOLID FOUR STROKE.
>SHUT IT DOWN BY PINCHING THE LINE. INSTALL A SIGNIFICANTLY
>LARGER PROP (GREATER LOAD) AND RESTART THE ENGINE AT THE
>SAME MIXTURE SETTING. THE LIKELY RESULT, DEPENDING ON JUST
>HOW RICH THE ORIGINAL SETTING, WILL BE A LOWER RPM AT A TWO
>STROKE!

Sure. Higher loads always require a richer mixture.

There are two independent factors (or variables-the original intent of the post) affecting the way the engine (in the case of a 2 stroke stunt engine) 4-2 breaks (switching from rich to lean).

1. Load. Due only to the changing drag.
2. Fuel draw due from the engine. The natural draw of the engine is (typically) insufficient to keep the mixture supplied to the engine from constantly changing.

I would agree that the load is more significant (in a 2 stroke), but maybe not by more than a factor of two. So that would mean that 33% of the "switching" of the engine is due to changing fuel head.

ALSO

I think that the fuel draw could be the DEFINING factor in achieving the last little tiny bit of critical switching. The fuel draw can literally be the "straw that broke the camel's back" so to speak.

That is why I think it would be nice to eliminate suction draw (or at least minimize its effects). This would be the first step to isolating the two variables of load and fuel draw.

>The fuel draw can literally be the "straw that broke the
>camel's back" so to speak.
>
>That is why I think it would be nice to eliminate suction
>draw (or at least minimize its effects). This would be the
>first step to isolating the two variables of load and fuel
>draw.

'zilla,

kind of hard to disagree with that. I'm sure a significantly pressurized fuel cell would drastically moderate any mixture changes from the effects of gravity although the +/- 1.0 G will always be some definable percentage of the ultimate pressure at the fuel pick-up.

My personal opinion is that the problems associated with high pressure fuel delivery (bladders, etc) might well outweigh any benefits derived from same.

As I recall witnessing the flights of the B-17 which utilized both regulators and a high pressure bladder system, there was still a distinct change in sound when loaded or unloaded. Of course, the four little putt-putts were running close to peaked out from the get go so there would have been no 4/2/4 break per se, but they seemed to clearly "peak" during loaded conditions. Paul could be more definitive.

Ted

p.s. this thought occurs to me while discussing the subject. I'm sure someone has actually measured the pressure obtained from mufflers, pipes, crankcases, etc. I'd be interested in knowing just how much if any there is from the conventional exhaust sources.

>My personal opinion is that the problems associated with
>high pressure fuel delivery (bladders, etc) might well
>outweigh any benefits derived from same.

Don't start thinking that I am advocating that we all go to bladder tanks. Quite the contrary. DA Dirt sent me some bladder stuff out of the goodness of his heart, and I totally balked.

I would think that the effect is subtle enough in 2 strokes to warrant some less drastic measures.

In a nutshell, you have to ask yourself, "how I make the fuel flow easier to the engine?" There are some very simple ways to do this.

Secondly, you have to ask yourself,

1. What are the goals?
2. What are the variables?
3. Why am I doing the way I am doing it now?

Isolate the variables and change the parts of the equation that affect those variables.

Questions to you:

Which is easier to suck through a straw, water or a milkshake?
Whithout changing the milkshake, how could you make it easier to suck through a straw?

Ted, I have to get back in here again, too. That earlier part of your comments that I did not include from your quote we all agree to. Put a larger prop (load) on the engine, and it will affect the mixture that is needed. Certainly adding a larger prop, with no mixture change, will cause the engine to run leaner. But we are not changing props on the plane. And, with an adequate engine for the size of the plane and prop for the size of the engine, I see the load effect on the engine as being much less than imagined when we are going through our maneuvers.

We (personally--Matt and I) tend to load our engines pretty heavily. On our 35 FP set-up, for instance, when we were running the 12.5 x 5.5 Bolly prop, the engine would wind up pretty hard on the ground. It reminded me of a pipe engine, even though we are running a 5.5 pitch. Well, the engine would wind up on the ground, but as soon as the plane was released it jumped into a lean 4-cycle and never, ever, went back to that high pitch again. It would pick up in rpm when pointed up, and drop back when level or aimed down, but never would it break into a lean 2-cycle. Thus the heaviest load would appear to be at rest, and it never returned to it. So, load has an effect, but let's move on.

>WE'VE ALL, I THINK, WITNESSED THE MOST COMMON DEMONSTRATION
>OF THIS EFFECT DURING THE TAKE OFF AND LEVEL FLIGHT SEGMENTS
>OF MANY A TWO STROKE PATTERN. ESPECIALLY ON UNPIPED 4/2/4
>ENGINES, IT IS COMMON TO RELEASE THE AIRPLANE, HAVE THE
>ENGINE BREAK INTO A TWO STROKE AND STAY THERE FOR A
>"CERTAIN" LENGTH OF TIME BEFORE "SETTLING" INTO THE DESIRED
>FOUR STROKE. IN MY TIGER .46 LIFETIME I DEPENDED ON THAT
>PHENOMENOM TO LET ME KNOW HOW GOOD MY NEEDLE SETTING WAS FOR
>THE UPCOMING FLIGHT. TOO LONG IN THE TWO AND IT WAS GOING
>TO BE FAST AND LEAN; TOO SHORT AND I COULD COUNT ON A SOFT
>FLIGHT.

This is a function of a certain type of prop and engine combo. If I understand what Igor was trying to say (on another post and private e-mail) a 6 inch pitch prop will be stalled when the plane is at rest. Thus the engine would actually develop greater thrust as it starts to move forward before it begins to decrease in thrust. A 4 inch pitch prop, as on a pipe engine, will decrease in thrust from the very start. So, it may be merely the pitch of the prop, etc. that is affecting this.

>THIS IS A CLEAR DEMONSTRATION OF THE EFFECTS OF LOAD AS THE
>AIRPLANE ACCELERATES. AT REST STATIC THRUST IS GREAT AND
>LOAD IS HIGH. AS THE AIRPLANE ACCELERATES THRUST AND LOAD
>BOTH DECLINE AND EVENTUALLY THE LOAD DECREASES TO THE POINT
>THAT THE ENGINE CAN FOUR STROKE. LOAD IT AGAIN AS IN HIGH
>"G'S" OR A CLIMB, AND THE AIRSPEED DECREASES BACKING BACK
>INTO THE RANGE AT WHICH THE ENGINE TWO STROKED DURING THE
>TAKEOFF ACCELERATION. THUS, IT BREAKS TO A TWO STROKE AND,
>WHEN THE DRAG DECREASES THE AIRPLANE ACCELERATES, THE PROP
>UNLOADS AND THE ENGINE GOES BACK TO A NICE FOUR.

I don't understand your illustration here, Ted, except to say it is prop related. If the load is greatest at rest, then the engine should not be breaking into a 2-stroke at all while accelerating if you launched it at a 4-cycle. There has to be something more involved here.

But let's move back to the classic 4-2-4 engine in the above illustration, and take the Fox 35 of which I am am most familiar. When I set this engine, I set it for a 4-cycle run when the plane is sitting level, but breaking into a 2-cycle when I aim it up at about a 45 degree angle. So it is running at a 4-cycle when level when held in the hand. When the plane is launched, lifts off, and climbs into the air (at a shallow angle), it continues in that 4-cycle. Not a lot of change in run or sound. But as soon as you point the nose up, then it breaks into the 2-cycle.

The load on the engine at that point is less than if you were holding it in your hand. Remember, the engine is supposed to be developing its greatest thrust at rest (unless the blades are stalled). If you held the airplane out the window of a car and began to accelerate bringing the plane up to speed, the engine would have less load on it as it nears flying speed than if it were at rest. The same thing happens even when it is flying by itself.

When the airplane is in motion there is less drag on the engine than when it is at rest. It matters not that the airplane is a "drag" on its thrust. The engine doesn't know this. There is an airplane in motion behind it, and the prop is merely "digging" through the air. If it were load alone, then we should be able to duplicate the effect by holding the plane level out the window of our car and simply speed up or slow down the car. It doesn't happen. But aim the nose up while you do this, and it will break into the 2-cycle once again.

Although load is a factor, I still maintain that the greatest factor is fuel location. The engine leans out when the fuel drops below the engine. It happens on all of them.

>
>YOU CAN FIND FURTHER SUPPORT FOR MY ACCEPTANCE OF THE TANK
>AND GRAVITY EFFECTING RUNS BY REREADING MY LAST PARAGRAPH OF
>THE PREVIOUS POST WHEREIN I POSTULATE ABOUT THE CHANGE IN
>FUEL HEAD PRESSURE DURING MANEUVERING WITH AND WITHOUT A
>CLUNK TANK. ALSO WHY I FEEL THE ISSUE IS SO MUCH GREATER
>WITH A FOUR CYCLE ENGINE THAN WITH A TWO STROKER.

Once again, I think this disputes your own (correct) comments about fuel head. If fuel head is a factor (it is) then the location of the end of the pick up tube is immaterial, so long as it is somewhere under the fuel head. If you siphon water out of a water barrel, it makes no difference if the end of the pick up tube is barely into the water or submerged totally all the way to the bottom of the barrel. The water pressure will remain the same, so long as the height and drop of the tube remain the same. The "height" and "drop" of the tube in this case is determined by the routing of the feed tube to the venturi. Where you submerge the other end in the fuel is totally irrelevant, so long as it is, indeed, submerged.

Your turn.

>GOOD DISCUSSION.
>
>TED

You are right.

Len,

Dare I offer another thought??? <G>

Our uniflow tank venting 'reference' head is the opening of the wet end of the air vent tube inside the tank shell. Change in fuel head, whatever direction it points, is tied to the spanwise distance between that vent tube end and the fuel pickup for the venturii jet. A small distance.

Pressure inside a uniflow tank ought to be slightly less than atmospheric, as air drawn in to replace fuel drawn out has to push past fuel in the vent tube before it releases into the tank volume. That, however, can by affected by ram air on the forward facing dry end of the vent tube.

Most of the uniflow tanks I've made or heard ran with negligible apparent RPM (or laptime) change as the fuel load was consumed. The change comes at the end of the run, when bubbles can reach the pickup-to-jet plumbing. I get a slight richening, followed by the expected lean shutdown. Critical tank height tuning is more precise, and has to be, with uniflow venting. Moving the tank actually moves the reference head! When it is right, any load/attitude variations are at least symmetrical for inside and outside g...

OBTW, re: Induced Drag, it figures as a fraction of an ounce at one g in level cruise. At 10 g (round loops) Induced Drag goes well over one pound, and in square turns at 35 g, it goes 15 lb or more... to the rear, of course. We start these maneuvers abruptly. In rounds, we reach fairly stable conditions (between 9 and 11 maneuvering g) -- they last longer. These drag loads come on sorta like popping brake chutes. Load should affect prop RPM, as Ted mentioned for his 'ST46 life' experience. Many of my Fox35 models launch warbling 2/4 and settle to 4 in about one lap. In both cases, mixture and load settle out as the prop UNloads... The D(i) numbers fit stunters from Nobler to Modern sizes, weights and airspeeds. Actual values are probably worse: aerodynamic characteristics decay more the smaller we go. NACA/NASA 'charts' were designed around 1:1 people-carrying aircraft sizes.

Bless!

>Len,
>
>Dare I offer another thought???
>
>Our uniflow tank venting 'reference' head is the opening of
>the wet end of the air vent tube inside the tank shell.
>Change in fuel head, whatever direction it points, is tied
>to the spanwise distance between that vent tube end and the
>fuel pickup for the venturii jet. A small distance. (snip)

Isn't the reference head the spanwise width of the fuel in a normally aspirated tank? And then it goes to zero as the fuel is used up. Thus 2 inches wide would be multiplied by 3 times the force of gravity in a 3 g pull to make the equivilent of 6 inches to nothing. On a uniflow tank it is changed by how deep the uniflow tube is "buried" in the fuel, and, again, would have no reference to where the pick up tube is located. If the uniflow vent is at the outside edge near the pickup, then it produces a partial vacuum in the tank (air having to go through the fuel) which offsets the physical pressure of the fuel itself. This partial vacuum gradually increases as the fuel is consumed and ends up at atmospheric. The biggesst problem then is not the fuel head, but the distance between the pickup where it exits the fuel and the inlet to the engine--approxomately 3 inches or more, which is the amount you are dropping the fuel below the engine when you aim the engine up.

>
>OBTW, re: Induced Drag, it figures as a fraction of an ounce
>at one g in level cruise. At 10 g (round loops) Induced Drag
>goes well over one pound, and in square turns at 35 g, it
>goes 15 lb or more... to the rear, of course. We start these
>maneuvers abruptly. In rounds, we reach fairly stable
>conditions (between 9 and 11 maneuvering g) -- they last
>longer. These drag loads come on sorta like popping brake
>chutes. Load should affect prop RPM, as Ted mentioned for
>his 'ST46 life' experience. Many of my Fox35 models launch
>warbling 2/4 and settle to 4 in about one lap. In both
>cases, mixture and load settle out as the prop UNloads...
>The D(i) numbers fit stunters from Nobler to Modern sizes,
>weights and airspeeds. Actual values are probably worse:
>aerodynamic characteristics decay more the smaller we go.
>NACA/NASA 'charts' were designed around 1:1 people-carrying
>aircraft sizes.
>
>Bless!

Just remember, if you are putting 35 g on the plane in a turn, you are putting the same pressure on the fuel. So, if it is that great--even momentarily, our three inch fuel drop mentioned above becomes over 100 inches in effect. That, even if only momentary, can certainly cause a "go lean" condition in our engines in the corner turn. I am still thinking that the increased load on the prop (or decreased unloading) is less important than fuel draw, for it is never going to be as much as if the plane were simply standing still. The air flows through the prop either way.

Leonard Neumann

Len,

RE: uniflow tank head reference: Pet water bottles, for example, don't empty themselves after being filled. A thirsty critter displaces a check valve, and vent air enters an inverted, otherwise sealed, bottle.

When water weight balances the pressure drop which it causes in air trapped above the spout, a small diameter spout may not even need a check valve. Dry-feed hoppers can neck down above a bowl or pan. When Rover eats enough, more sifts down until the neck is plugged, and it doesn't overflow.

Large-critter water dishes? When the (product) level falls below the spout or outlet, more flows to the "dish." Granular feed works by simple gravity and blockage; liquids work by pressure balancing and blockage. Flow stops when pressure drop in the chamber above the neck supports liquid weight, and liquid level in the "pan" closes the spout or neck, blocking further "vent air" from entering the flask.

In our uniflow tanks, atmospheric pressure does not act on the fuel's physical surface. Vent air only gets in through the uniflow vent. I've run tests, some with static heads (to initial liquid surface) greater than 1 foot. Classic method: -- observing how far liquid streams out of an exit tube low in the container. With uniflow venting simulated, little 'throw' occurs. It can be varied a bit by changing the height of the air outlet above the exit tube, but is never as forceful as open vessel stream throw. Fuel seeks the same level inside the vent tube as in the rest of the tank. So, fuel volume drawn out by the engine reduces tank pressure enough to pull air past the fuel standing in the vent tube, else no air could enter to replace fuel volume removed.

Stream throw, or trickle, was about constant until the physical fuel surface went below the vent outlet. Then free venting began, first with greater throw, diminishing as expected for "open vessel" static head conditions. There were some experimental setup issues to resolve before tests repeated consistently.

So that's why IMHO the actual fuel surface inside the uniflow tank is not the "head" reference -- rather, the "height" between the vent tube air outlet under the fuel's surface and the outlet at the "bottom" of the tank, viewed in the direction of the resultant of forces on that region of the model.

I like to approach things like these in terms of vectors and resultants of forces. The pictures you and I see of the forces on models, wings, fuel and props, etc., ARE not identical. That matters less than getting reliable results from methods we trust. If so, we're doing well.

Sorry to be so wordy, but I have done some thinking on these things over the years... <g>

Hi Lou,

I completely agree with your posting. What has been puzzling me since a few years: I am using a uniflow setup: clunk tank with fixed uniflo tube. At first I had the uniflow tube against the tankwall at the righthand side (outside of the circle). THe classic position to have a stable regime until the last few seconds of the flight, I assumed. THe amazing thing is that the engine was gradually running richer towards the end of the flight with that setup. The cure for me was to stay away 2 cm from the tank wall with the uniflow tube.
Till now I have been unable to explain the phenomenon. I have been doing the same kind of experiments as you have been doing, but still haven't an answer. Muffler pressure even seems to make this effect more pronounced.
Any clues?

Regards!
Paul

LAST EDITED ON Nov-28-03 AT 07:36 PM (CST)
 
Lou, I don't have time at the moment to write up what I have been wanting to do (and promising on the web site) since the last Century. I would like to write a clear explainatin of the uniflow tank, since people keep asking the question. However, let me make a couple of comments to what you are here saying:

>Len,
>
>RE: uniflow tank head reference: Pet water bottles, for
>example, don't empty themselves after being filled. A
>thirsty critter displaces a check valve, and vent air enters
>an inverted, otherwise sealed, bottle.
>
>When water weight balances the pressure drop which it causes
>in air trapped above the spout, a small diameter spout may
>not even need a check valve. Dry-feed hoppers can neck down
>above a bowl or pan. When Rover eats enough, more sifts down
>until the neck is plugged, and it doesn't overflow.
>
>Large-critter water dishes? When the (product) level falls
>below the spout or outlet, more flows to the "dish."
>Granular feed works by simple gravity and blockage; liquids
>work by pressure balancing and blockage. Flow stops when
>pressure drop in the chamber above the neck supports liquid
>weight, and liquid level in the "pan" closes the spout or
>neck, blocking further "vent air" from entering the flask.

Up to this point I am following you perfectly. I understand the principle of the automatic water feeders, and this is what makes our uniflow tanks work as well. If the water bottle in our water feeder were, say, 50 feet high, then the water would drop out of the bottle when tipped upside down, even when the neck was submerged in water. However, it would drop down to 30 some feet with a perfect vacuum above it, the height of the column being supported by the weight of atmospheric pressure against the water down below. (Slightly less than 15 pounds per square inch, which would be the weight of the 30 plus foot column of water.) The only way for the water column to drop any further would be to either decrease the air pressure on the outside, or allow some air to enter the bottle. It is the latter that we do with our uniflow tanks, and which is the principle which is applied to the "chicken hopper" automatic waterers. Using the chicken feeder as an example, when the chicken drinks the water around the neck of the bottle, the water level drops to the point of allowing some air to enter. This increased pressure above the water in the bottle then allows some water to scape again until the levels is raised to the point where no more air can enter. When that point is reached, the process stops.

If, however, in the case of the automatic watering trough ("Chicken Hopper or whatever") we were to insert a tube into the bottle where we could allow more air to come in, then more water would flow out and the water trough surrounding the bottle would over flow. This is the principle of our uniflow tanks...in a way.

Basically the tank is sealed, and the fuel level in flight, due to centrifrugal force, is more akin to the tank standing on the wedge. If we were to simulate this on our test stand, we would place the tank at a slight, approximate 3:1 angle, with the wedge down. So a 2 inch wide tank has a 2 inch "high" fuel head at the very beginning.

The air inlet is placed at the very bottom, and the air inlet is vented to atmosphere or the muffler (it makes no difference, so slong as it remains constant), and any air entering the tank must pass through the pressure of the fuel. It is only 2 inches, so it is not much, but it is 2 inches. And when we run the engine in non-uniflow style (open vented) on the test stand with a 2 inch high tank we can tell a vast difference in the run from beginning to end. The uniflow vent, however, with the air having to pass through the (ever so slight) increased pressure of the 2 inch high fuel head, will be reduced by the exact same pressure as caused by that fuel itself. Thus, as the fuel decreases, and the pressure head with it, the air inside the tank increases and the "Fuel head pressure" to the engine remains relatively constant.

Now, for this to work, the air inlet must be at the very "bottom: of the tank, because that is where the greatest pressure of the fuel is. And the fuel pick up must also be at the "bottom" of the tank if we wish to use all of the fuel that is in the tank. If the fuel pick up uncovers, it will draw air. If the uniflow tube uncovers, it will vent total atmospheric pressure to the tank. And this is why you will see both very nearly at the rear of the tank, but located in the outside wedge (bottom of the tank while in flight).

Now, as long as the fuel is being shoved to the outside along the wedge, it wouldn't make any difference if the pickup or the uniflow tube were in the back, the middle, or the front. You could have one in the back and one in the front and it would still work the same. However, since we do our maneuvers and tend to slosh the fuel around, we generally put both tubes as closely as possible towards the back of the tank, with the pick up tube the farthest back so it can draw the very last drop of fuel, while the uniflow tube (slightly inboard and ahead) will uncover first, giving a slight difference in run just before running out when we are done with the pattern anyway.

And for those who are concerned about the fuel inlet drawing bubbles from the uniflow tube, Think again of the tank being located on the test stand with the wedge end hanging down. The affect of gravity causes the air bubbles to go up and away from the pick up tube. "Up", whle flying, remember, is towards the inside of the tank, and thus, even with the pickup and uniflow tubes being separated by merely 1/4 to 3/8 of an inch, this is far enough.

But, yes, the atmospheric pressure does act on the fuel's physical surface, and that is why I dispute the following:

>In our uniflow tanks, atmospheric pressure does not act on
>the fuel's physical surface. Vent air only gets in through
>the uniflow vent. I've run tests, some with static heads (to
>initial liquid surface) greater than 1 foot. Classic method:
>-- observing how far liquid streams out of an exit tube low
>in the container. With uniflow venting simulated, little
>'throw' occurs. It can be varied a bit by changing the
>height of the air outlet above the exit tube, but is never
>as forceful as open vessel stream throw. Fuel seeks the
>same level inside the vent tube as in the rest of the tank.
>So, fuel volume drawn out by the engine reduces tank
>pressure enough to pull air past the fuel standing in the
>vent tube, else no air could enter to replace fuel volume
>removed.
>
>Stream throw, or trickle, was about constant until the
>physical fuel surface went below the vent outlet. Then free
>venting began, first with greater throw, diminishing as
>expected for "open vessel" static head conditions. There
>were some experimental setup issues to resolve before tests
>repeated consistently.
>
>So that's why IMHO the actual fuel surface inside the
>uniflow tank is not the "head" reference -- rather, the
>"height" between the vent tube air outlet under the fuel's
>surface and the outlet at the "bottom" of the tank, viewed
>in the direction of the resultant of forces on that region
>of the model.
>
>I like to approach things like these in terms of vectors and
>resultants of forces. The pictures you and I see of the
>forces on models, wings, fuel and props, etc., ARE not
>identical. That matters less than getting reliable results
>from methods we trust. If so, we're doing well.
>
>Sorry to be so wordy, but I have done some thinking on these
>things over the years... <g>
>

So, Lou, let me say once again, it is not the distance between the uniflow tube inlet and the pick up tube that makes the difference, but the location of the uniflow tube in relation to the "top" of the fuel head. The depth of the fuel head to the uniflow tube is what determines whether or if the uniflow tank is going to work.

Len....I agree with your explanation, however the 30ft. plus coulumn of water will exert a force of approx. 14.7 psi which is atmospheric pressure at sea level (not 30 psi as stated) .

>Len....I agree with your explanation, however the 30ft. plus
>coulumn of water will exert a force of approx. 14.7 psi
>which is atmospheric pressure at sea level (not 30 psi as
>stated) .

Absolutely correct, my goof. I had meant 15 pounds, or slightly less than, but got the 30 foot column (30 plus, whatever) in my head and 30 got recorded erroneously a second time.

Thanks for the correction. I edited my response to correct it there as well.

Len,
Very briefly...

EXCELLENT description of the uniflow situation, although we differ slightly about the reference points.

Quick example: It is possible to create a 'static' head by a length of fuel line in the fuel out tube, by pointing the tube down. THEN, at the right length, weight of fuel in the line can overbalance the slight pressure drop in the tank. Different heights from end of this tubing to tank outlet tube can yield: no flow, an occasional drop, steady dripping, and even a slow but steady stream. As the VERTICAL height of this fuel line governs the result, you can do this with about a foot of medium silicone held at various angles between the tank fuel outlet height and straight down.

Glad to see you appreciate the vector picture! However, it tells me that the "head" in the line to the NVA is in line with the resultant force. The fuel line mostly sees gravity itself as a point, or zero length vector (no magnitude), in level flight (tube horizontal/gravity vertical). Hydraulic pressure on remaining fuel also increases in this line under g effects, so, even there the effect is reduced.

For Paul's richening: When (freely entering) air bubbles reach the uniflow interior tip, the slight pressure drop we'd set the needle by is gone. The pressure difference vanishes. We're vented at atmospheric pressure (plus any ram air on the vent outer tip). The needle was set richer for the "sealed" uniflow condition -- to draw past the pressure difference. That seems enough to overcome the free vented tank's tendency to go lean toward the end. Have you ever had an overflow cap fall off in flight? Same riching occurs...

I enjoy this, hope you do, too, and truly admire the excellence (overall) of SSW.

Joyous Holidays!

Great discussion.

Hi Lou,

One remark on the phenomenon of richening of the engine at the end of the run, seconds before quitting. I have been wondering what was causing this. I think I got the answer when testing an engine on the testbench. The tube connecting the tank to the engine was about 10 inches. The engine was running in a fast 4stroke. When the tank became empty, air was sucked into the feed tube. From that moment onwards, the engine started to get richer until the fuel line was completely empty and the engine leaned out and cut.
What happened?
From that last drop in the tank onwards, the length of fuel in the feed tube is diminishing until length zero. When the length of fuel is diminishing, basically the length of tube that is responsible for the flowresistance is getting smaller. You might see it as the fuelline actually getting shorter.
From Fluidomechanic laws we know that a shorter fuel line means less resistance to flow. Less resistance to flow results in a faster flow for a fixed pressure difference. So in other words: the dynamic pressure drop in the fuel feed line is responsible for the richening at the end of the flight.

I don't think the richening has anything to do with the uniflow tube becoming uncovered by the fuel; An explanation I've seen frequently seen published.

FWIW
Paul

Don't start thinking that I am advocating that we all go to bladder tanks. Quite the contrary. DA Dirt sent me some bladder stuff out of the goodness of his heart, and I totally balked.

The City Smasher

Snipped, and snipped some more...

Brad,

I would expect some contrary opinions as to sentence above, even though you are so very correct about the goodness of my heart.

I'm not advocating bladder tanks either, by the way. I first made the conversion with my second Smoothie and it was pretty much done on a lark. Then Don McClave, seeing all the plumbing where there used to a hard tank, said "That's a solution to no known problem."

It was only at that point in time that making a bladder/regulator setup work in a CL Stunt application, on a Fox 35 no less, became something truly worth doing.

But that aspect of the deal I know you understand exceedingly well...

And in another thread, height differences aside, I see where you obviously married up. So did I. So did we all.

Ted wrote:

>The act of lowering or raising the tank (or turning the test
>stand on its tail as has been discussed) is really nothing
>more than richening and leaning the needle valve. Doing so
>will, in fact, "move the curve" because the revs under a
>given load (your test prop)will change based on whether the
>engine is running at optimum at the start of the test or
>running rich. As soon as the revs change for a given load
>the torque changes by definition.

True, but how do we know that (until you get to extremely lean or rich conditions, anyhow) the torque doesn't follow the torque curve in torque-RPM space? I suppose this assumes that we're still in positive slope territory (i.e. below the torque peak).

>Having said that, this really has no real world application
>in our stunters. Under our flight conditions the "fuel
>head" the engine feels has only a small fore and aft
>component. Due to the centrifugal/centripital forces that
>result from tethered flight, for the most part the engine
>will "feel" the fuel pick up is a constant distance behind
>it and the more influential effect will be the feeling the
>engine has that the pick up is "outboard" of the engine.
>Due to G forces the sideways displacement of the fuel pick
>up (pretty much constant) is a greater force than is the
>fore and aft. Thus mixture changes due to tank/engine
>relationship are significantly different under flight
>conditions than on the test stand or when you point the nose
>up prior to takeoff.

I'm not convinced of this any longer, for this reason. Since tanks are mounted more or less in line with the engine's crankcase (except in profiles, where they are mounted slightly outboard of the engine), the difference in "height" in the outboard direction (i.e. considering outboard wingtip as "down") is small. Compare this with the difference in "height" in the tail's direction (counting "tail" as "down") and you get a fairly large difference between the fuel level and the needle valve (at least 3-4 inches). Consider that the difference between going up and going down, the acceleration acting on the fuel will change by 2g (it's +1g going up, -1g coming down, 0g in level flight). Changes in head can be calculated using the formula rho*acc*height, in this case, acc changes by 2G.

What does a stunt model pull in g attributable to its circular path (i.e. excluding side thrust and aerodynamic effects)? This can be calculated (I'll let someone else do it), and be plugged into the same equation for the fuel head we used earlier, taking into account that the difference in "height" between the fuel and the needle is much smaller in this direction. Hence, we can then compare the magnitude of this in relation to the changes in head when pointed up and down, to see how they compare to each other.

Note that the "height" of the fuel in the up-down direction (the plane's floor is now "down") is more or less the same as the outboard "height", and the plane turns much smaller radii in this direction when in a turn, so...

>This is where the subject of load enters the picture. The
>fuel draw conditions in flight, as we've demonstrated are
>only modestly variable and, therefore, have only a small
>amount to do with the four/two/four break as a result. The
>much larger component is the fact that when we start to
>climb we put a greater load on the engine and, if the
>mixture and combustion cylinder pressure conditions are in
>the proper relationship to one another the resulting load
>will cause the engine to break into a two stroke. Simulate
>this on the test stand by putting a load on the spinner
>(your choice how to do this) and listen to the four stroking
>engine break into a two stroke even as the actual revs
>decrease.

Yeah, I agree this happens, but I wonder if it is ALL that happens. I used to think so, now I'm not so sure.

>By the way, I think the above addresses the issue of why the
>clunk tank is so magically superior to a conventional vented
>tank with the four stroke. As we discussed the centrifugal
>forces in flight are considerable. In addition to those
>forces, in hard maneuvering we get substantial vertical
>forces on the fuel as well. In a hard inside corner for
>instance, there might be 15 or 20 positive Gs acting on the
>airframe and everything attached to it including the fluid
>in the tank. Thus, the clunk goes to where the head
>pressure at the fuel inlet is the greatest. This pressure
>would, for instance, be the greatest at the bottom outside
>corner of a tank during a hard inside corner. Ta Da!
>Exactly where the clunk went. Although head pressure is not
>a huge factor, in increase will tend to richen the mixture.
>This is exactly what we need to keep the engine in a happier
>relationship vis a vis load/mixture during the corner.

This I don't agree with. The position of the pickup shouldn't have any bearing on fuel head as seen by the needle valve, as long as it's submerged in the fuel. You can think of it this way. If you have a well that's 30 feet deep, with the water level 5 feet below ground, and your pump at ground level, you'd see a head of minus 5 feet when you tried to pump the water, regardless of whether the pump's pickup was 6 feet below ground level, 10 feet below ground level, or 30 feet below ground level. When I was in grad school, one of my colleagues was building something that used a laser, which was mounted six feet above a bucket containing the cooling water. He found that the pump he was using wasn't giving enough flow, so he was inspired to move the pump down towards the bucket. He found no improvement whatsoever - what he didn't realize was that the laser was still six feet over the bucket, and the water in the return hose was in equilibrium head-wise with the water coming up the feed hose, and it didn't matter where you put the pump. (If I recall correctly, this state of affairs is true until you get a water column of more than 32 feet, at which point not all the suction in the world will draw water up.. you'd need to pump from below).

I've got my own theory. About why uniflo tanks don't work as well on 4 strokes, given what Doug Moon's been saying about the marginal fuel suction with the large venturis. And that is simply that, in the uniflo configuration, the tank gives LESS of a fuel head than does a normally vented tank (i.e. in the "outboard wing is down" direction). A uniflo tank holds the outboard of the tank (where the uniflo vent debouches) at a constant 1 atmosphere. This point is BELOW the venturi (in the outboard direction) or in line with it (in the up/down direction).

A normally vented tank has 1 atmostphere at the fuel-air interface. The fuel-air interface changes position, but is, at least for most of the run, ABOVE the venturi (in both the outboard and up/down directions).

So, for most of the run, a normally vented tank provides MORE fuel head to the needle valve.

This is all, literally, "Rocket science" (i.e. simple mechanics), but really I don't think it's too difficult to work out, given that we have dimensions of the typical stunter, and most of the data we need (i.e. airspeed, flying circle radius, etc.). One thing we do need but don't have is fore-aft acc(i.e. dec)eleration in the turns due to the increase in induced drag. Time someone put accelerometers on board a Stunter.

>I've got my own theory. About why uniflo tanks don't work as
>well on 4 strokes, given what Doug Moon's been saying about
>the marginal fuel suction with the large venturis. And that
>is simply that, in the uniflo configuration, the tank gives
>LESS of a fuel head than does a normally vented tank (i.e.
>in the "outboard wing is down" direction). A uniflo tank
>holds the outboard of the tank (where the uniflo vent
>debouches) at a constant 1 atmosphere. This point is BELOW
>the venturi (in the outboard direction) or in line with it
>(in the up/down direction).

>So, for most of the run, a normally vented tank provides
>MORE fuel head to the needle valve.

Ding! Ding! Ding!

Isky wins again.

>I don't think the richening has anything to do with the
>uniflow tube becoming uncovered by the fuel; An explanation
>I've seen frequently seen published.

The uniflo vent being uncovered would make the engine run LEAN, not rich. The uniflo vent, when covered, is at 1 atm., and the head at the fuel pickup will be 1 atm + rho*g*h, where h is the distance between the uniflo vent and the pickup, small though it is. Once the vent is uncovered, you have 1 atm at the fuel-air interface, and the fuel head at the vent would now be 1 atm + rho*g*x, where x is the distance between the fuel-air surface and the pickup. Note that x is, by definition, smaller than h. Thus the plane SHOULD go lean.

I do like your explanation about why it richens - never actually thought about it before.

Hi. It should indeed go lean. Not jumpy but gradually.

>Hi. It should indeed go lean. Not jumpy but gradually.

You are correct in that it will go lean, but not gradually. The uniflow process maintains a slightly lowered atmospheric pressure inside the tank which is equal to the pressure of the fuel head. The atmospheric pressure increases, as the fuel head decreases, maintaining a (nearly) uniform fuel pressure from beginning to end of flight. However, the fuel does slosh around during maneuvers (wouldn't you if you were inside the tank?) and the size, or shape of the fuel head thus changes. So the atmospheric pressure gradient inside the tank is not always perfectly aligned with the pressure exerted by the fuel, itself (which also changes slightly). However, when the uniflow tube becomes uncovered, at that point the air pressure inside the tank immediately returns to full outside pressure and, even that slight difference, will cause the engine to run slightly rich. However, this is also the point where the pick up tube begins to uncover and to draw air, and the resultant mixture of fuel and air can sometimes cause the engine to go lean for as much as a lap or two before the engine quits.

My uniflow tanks ALWAYS went rich when the uniflow line was uncovered, ALWAYS. I had to run extra fuel so the clover wouldnt be in a rich situation. I run uniflow on an LA 25 set the needle. Then switch the vent tubing around and start the motor at the same needle and it is way richer, not leaner. Went to clunks and never looked back.

Hi Doug,

Interesting. One question: How far apart where the feedexit and the uniflowexit? I mean the distance towards the outside of the circle. Usually in wedge tanks this distance is 1 or 2 mm. In that case uncovering the uniflow exit also practically means uncovering the feedexit.

Paul

>Hi Doug,
>
>Interesting. One question: How far apart where the feedexit
>and the uniflowexit? I mean the distance towards the outside
>of the circle. Usually in wedge tanks this distance is 1 or
>2 mm. In that case uncovering the uniflow exit also
>practically means uncovering the feedexit.
>
>Paul

Paul I dont know. On the next plane I went to clunk tanks. That uniflow was in a 94 Buc 740. The next year I went to an OPS 40 in a Buc 746. I have tried uniflow many more times on many more planes but it doesnt seem to be as trouble free for me. Its all in what you know and what you are used to...

Hi Len,

Thanks for the reaction.

>>Hi. It should indeed go lean. Not jumpy but gradually.

What I meant was that in my setup, where the uniflow exit is about 2 cm above the feedline, the engine starts to lean out gradually after the uniflow exit is uncovered. From that point onwards the uniflow principle is lost and the tank will start behaving as a classic vented tank.

I see no reason for the engine to start running richer this way.

I think that the sloshing of the fuel is probably too fast to have effect on the mixture. On top: as the uniflow exit is uncovered by sloshing, the fuelhead on that spot in the tank is also very low. I feel that in average this should result in a stable pressure at the feedpipe.

fwiw
Paul

I realize this is 2 stroke experience, but some might carry over to strokers as well.

It would seem to me that the simple answer could be that your uniflow tube is in such a position exiting your fuse to be in dead air or maybe even offset to the airflow enough to be on the edge of siphoning. Thus you uncover the uniflow and it goes rich. I had a Jamison do this "go rich at the end" until I bent the uniflo slightly. Fortunately it was long enough to do. You could experiment with placement with some blue/pink fuel line...

Other more common problems I've had with uniflo when the uniflo uncovers, it really leans out like mad on the last laps. Also have had it force me to set the needle much leaner on the ground to get the run I wanted in the air. These both were solved with the little "restricter with a hole drilled in it" stuck in a piece of fule line shoved on the uniflo trick. Worked like a champ.

For my 4strokes, I've used standard venting as has been suggested by Brad, and never tried uniflo. They have all worked perfectly. The only trouble I've had is with the tiny clunk tank for my OS26FS in a Ringmaster that I hard plumbed the pickup with copper (the flex tube wouldn't flex enough it was so short) and that tank still gives me weird very short runs and lot's of fuel left in the tank that I have not taken the time to solve yet.

>I realize this is 2 stroke experience, but some might carry
>over to strokers as well.
>
>It would seem to me that the simple answer could be that
>your uniflow tube is in such a position exiting your fuse to
>be in dead air or maybe even offset to the airflow enough to
>be on the edge of siphoning. Thus you uncover the uniflow
>and it goes rich. I had a Jamison do this "go rich at the
>end" until I bent the uniflo slightly. Fortunately it was
>long enough to do. You could experiment with placement with
>some blue/pink fuel line...
>
>Other more common problems I've had with uniflo when the
>uniflo uncovers, it really leans out like mad on the last
>laps. Also have had it force me to set the needle much
>leaner on the ground to get the run I wanted in the air.
>These both were solved with the little "restricter with a
>hole drilled in it" stuck in a piece of fule line shoved on
>the uniflo trick. Worked like a champ.(snip)
>
>Eric Viglione

What you are suggesting here could be the answer to why some people experience different results with the same tank set up. It could be that the inlet tube location is causing the problem. I have never personally had any good results with the restrictor tube as you describe, however, and have even tried a small piece of brass tubing that I could squeeze. Even squeezing it didn't seem to make much difference. On the "go lean" situation at the very end, we have had experience with this for a last lap or two, but never over lean, nor for an extended period of time. What we have had problems with is the shut off, where it stops, starts again, stops, starts again, stops...oh, about four times it does this, and each time the judges get up out of their seats thinking they are about to judge the landing, when, it starts...

One argument I have heard for the richening effect is surface tension. As long as the uniflow vent is covered, the air that enters has to make a "bubble". This actually means that the fluid pressure, just outside the bubble, is slightly less than the airpressure in the bubble (otherwise the bubble doesn't get biiger). One the vent is uncovered, the vent opens up directly into the air volume, so now that fuel surface is at atmospheric. So the fuel now has a slightly higher pressure, giveing an intital richening. Depending on how much fuel is left in the tank, you would expect the the mixture to start leaning out as the fuel level continues to drop.

So what we see is the competition between two small effects. The surface tension effect might vary depending upon fuel mixture and possible whether there are anti-foaming agents in the fuel, which I think act to lessen surface tension--but I am a bit hazy on that point.

LAST EDITED ON Nov-27-03 AT 09:22 PM (CST)
 
>Hi Len,
>
>Thanks for the reaction.
>
>>>Hi. It should indeed go lean. Not jumpy but gradually.
>
>What I meant was that in my setup, where the uniflow exit is
>about 2 cm above the feedline, the engine starts to lean out
> gradually after the uniflow exit is uncovered. From that
>point onwards the uniflow principle is lost and the tank
>will start behaving as a classic vented tank.

Hmmmm... 2cm is quite a bit. That's something like 0.8 inches.

>I see no reason for the engine to start running richer this
>way.
>
>I think that the sloshing of the fuel is probably too fast
>to have effect on the mixture. On top: as the uniflow exit
>is uncovered by sloshing, the fuelhead on that spot in the
>tank is also very low. I feel that in average this should
>result in a stable pressure at the feedpipe.

OK, here's one reason why you would get richening at the end of the tank, especially if you're still flying maneuvers and are not just flying out the tank level. It's got to do with fuel sloshing.

Imagine a situation where the uniflo vent is still being covered by fuel. The pressure above the fuel would be:

1 atm - rho*g*x

where in this case, x is the distance between the uniflo vent and the fuel surface. This results in something less than 1 atmosphere, so that the pressure at the uniflo vent is 1 atm.

OK, supposing now you pull into a corner of a maneuver. Fuel sloshes downwards. The uniflo vent is uncovered momentarily. What happens? Air rushes in, and the pressure in the tank above the fuel is now 1 atm. When the plane returns to level flight, what happens now? The pressure at the uniflo vent is no longer 1 atm, but is above 1 atm, because the pressure in the tank's airspace is 1 atm. The pressure at the uniflo, which normally should be held to 1 atm, is now:

1 atm + rho*g*x

At the fuel pickup, the pressure would, under normal circumstances be:

1 atm + rho*g*y

where y is the distance between the uniflo and the pickup, but now it's

1 atm + rho*g*(x+y)

So the engine runs rich for a while, until the fuel reaches the uniflo vent's level, at which point it's running normally for a moment, and then it'll go lean.

As to why it'd go lean suddenly.. sloshing would perhaps be one reason, the other would be that the level of the fuel drops very quickly at this point, considering that only the wedge part of the tank has fuel in it, and dh/dv (i.e. the change in height per change in volume) gets larger and larger the further down the wedge you get...

>OK, here's one reason why you would get richening at the end
>of the tank, especially if you're still flying maneuvers and
>are not just flying out the tank level. It's got to do with
>fuel sloshing. (snip)
>
>OK, supposing now you pull into a corner of a maneuver. Fuel
>sloshes downwards. The uniflo vent is uncovered momentarily.
>What happens? Air rushes in, and the pressure in the tank
>above the fuel is now 1 atm. (snip)

Without all the equations, I can understand and accept this as an explaination for those who are experiencing such a situation. I guess we run with enough fuel at the the end that we don't experience such a situation, at least under normal conditions. But, yes, if the tank were constructed in such a way that it allowed the tank to vent to full atmospheric pressure before all, or nearly all of the fuel were consumed, then I could see this happening. It might happen especially with a clunk tank where the uniflow was set stationary, halfway in the middle, or in a metal tank where the vent tube was not all the way to the wedge or close to the rear. I guess we should ask these questions (along with how the vent is exposed to the outside air) just to get clarification.

Perhaps two things come into play here. One would be if the uniflow vent were in a place other than adjacent to and immediately ahead of the pickup.

I did not get enough time to read so many inputs earlier, so I do some notes now:

There is no nose up / nose down which makes more or less fuel pressure. Pressure is force on an area. Fuel is liquid; a screw cannot mount it, so the only way to change pressure in liquid is acceleration different to fuel and different to the frame or tank. If you have model in gravity field, the gravity is applied to the fuel the same way as to the frame, so there is NO static pressure in fuel, and thus also nose up, nose down does not make any difference. What you explore on ground is force/acceleration of your hand applied to the frame and not to the fuel in tank what makes pressure in fuel. There is acceleration by your hand applied to the frame equivalent to the gravity acceleration.

But in flight, there is not any “hand” making an acceleration or force. There are only aerodynamic and centrifugal forces it means:

1/ If model flights constant speed level flight (with normal pressure at NVA), and then it points its nose up, nothing happens. The pressure is still the same, because there is NO acceleration. But model loses its speed (thanx gravity) and thus prop pulls more - because of higher AoA on prop blades at lower speed. THAT is that aerodynamic acceleration which can cause the fuel pressure difference. But AGAIN – engine MUST PULL MORE. That is also answer why rich 2C engine accelerates (there is a potential to do it as they are rich and theay sre also sensitive to fuel pressure) and also answer why lean 2C or 4C engine does not stop (leaning makes it pulling less – less leaning, still working) – even they would stop on ground. All depends on power train if in flight nose up pressure difference is more or if it is less that 1G on ground. There is nothing forcing it to be exactly 1G … BESIDE our wantage to have constant speed uphill.

2/ The wing makes acceleration too and much much stronger. The wing in corners makes lot of force and thus tiny shimming makes so differences in engine run. If you fly 25G corners, 1mm thick shimming has the same effect like 5cm tank altitude difference on stand.

3/ There are also forces, which are NOT perfectly perpendicular to axes. For example lift and drag in corner. Model has some AoA in corner. If you count lift, fuselage AoA and drag in corner, the residual force or acceleration can very modify pressure at NVA. I did and calculation in another thread some months later, showing that drag with conjunction of some flaps very close to those we use, giving an angle of attack gives very good balance. It means that the engine in corner does not go rich and also does not go lean. If you miss something, your model will go out of balance and thus rich or lean in corners. (Lou I think you asked “WHY”).

Well … and … I found some “miracle” of richening on end of flight – did someone think of heating of fuel? It makes the fuel thin, so just try to separate tank from hot cooling air, it always helps.

LAST EDITED ON Nov-29-03 AT 10:22 PM (CST)
 
Igor,
I am afraid you are just plain wrong on point 1. Gravity as a force and (at least in the Chicago Area) points downward. Since it is a conservative force, you can associate a potential energy with it. When you point tht nose up, you have now put a few inches (=~10cm) of distance between the uniflow vent opening and the spraybar. You now have to pump the fuel up this distance, this is a greater distance than when the nose is horizontal and the distance is vertical distance is a few mm at most. This extra height translates into less pressure at the spraybar--and less fuel being sprayed into the venturi--the engine leans out.

You have to remember that you (or the engine) can't tell the difference between an actual kinematic acceleration and a constant gravitational field. Besides, you must hve raised the nose of your plane vertically at some point and noticed that the engine leans out. That's how i set my needle on my rc plane--point the nose vertical and adjust rpm on the rich side of of max rpm.

Now when the plane is in flight, and you pop a corner, two things happen, (after the fuel stops sloshing around):1) fuel pressure drops at the spraybar, so less fuel goes into the venturi 2) due to loss of airspeed, the prop load goes up. Both these two items affect the engine rpm and load, but I admit, I don't know which effect is dominant especially since both go in same direction.

added point: When the corner is popped, the airplane, which now has lost some of it's level flight velocity, is certainly not in a "free fall" situation (downward acceleration just equal to "g") . I am not sure whether it is accelerating upwards due to the leaner engine run (even more positive acceleration which adds to gravity), or is slightly decelerating (subtracting from gravity), or holding a constant velocity, but I am pretty positive that there is a net gravitational effect.

Exactly. Isostatic pressure at a given point in any fluid medium depends on several things (is it enclosed? are the walls pressing in, such as in a bladder?). For our purposes, the fuel can be regarded as a free-standing column. Pressure at any given point depends on the height to the free surface, the acceleration acting on the fuel (if it's at rest, on Earth's surface, this acceleration is 1 g, i.e. 10 ms-2) and the fuel's density, plus the pressure at the surface, i.e. 1 atmosphere. The equation is a very simple one:

P = rho*g*h + 1atm

where rho=density, g=acceleration, h=height of fuel column.

Since we're dealing with three dimensions, the calculation will have to involve three component vectors, in the three directions (which we can define). For instance, in a bucket of water at rest, all the pressure is contributed by acceleration in the z direction (i.e. straight down), and because there is no acceleration in the x and y directions, the x and y contributions go to zero. We can calculate all of these for fuel in a tank at most points in a flight. It's fairly easy for level flight and straight up and down flight (assuming near-constant speeds due to low pitch high RPM setups), and should be fairly easy during segments of loops, but it gets hairy in the corners, since we don't know really know what fore-aft acceleration is.

Perhaps it's time someone mounted accelerometers in a stunter?

>>>For our purposes, the fuel can be regarded as a free-standing column.<<<
OK, and forget atmospheric pressure ok? It will be simpler.

>>>Pressure at any given point depends on the height to the free surface<<<
OK

>>>the acceleration acting on the fuel (if it's at rest, on Earth's surface, this acceleration is 1 g, i.e. 10 ms-2) and the fuel's density<<<
OK

>>>P = rho*g*h<<<
WHY??? It is not enough. I know it is from high school book, but it is really not true.

Look:

The column has known height. OK. But we have some area. If you take it, you know also the volume of the liquid and thus its mass thanx the density. But even if we know the mass and acceleration, we cannot say anything about FORCE. If we would know the force, and we know also area, we can really calculate the pressure. The equation above does not contain the area, just because it is cancelled from both sides of equation.

But again again – WE DO NOT KNOW THE FORCE. I know you would like to say the force is mass x acceleration, but it is really not true every time. It is true only in ONE CASE: if the mass object has constant speed and that is not true in our case. So that equation is really not valid for our purpose. The speed is constant only in case that if the model is nose up, the prop makes acceleration exactly 1G and that is what no one can guarantee and what is in reality far from true. So the pressure difference really depends on engine pull and not on gravity or nose direction. That is why I wrote:
>>>There is nothing forcing it to be exactly 1G … BESIDE our wantage to have constant speed uphill.<<<

>>>Gravity as a force and (at least in the Chicago Area) points downward.<<<
Gravity is not force, it is acceleration. It applies to both tank and the fuel. If you apply it to the tank and also to the fuel, it does not bring any force between fuel and tank (that is what is necessary for pressure).

>>>Since it is a conservative force, you can associate a potential energy with it.<<<
You cannot until you do not know the mass. If you know the mass you know also kinetic energy. The trick is, that kinetic energy converts to the potential energy and back. But nor the conversion, nor absolute value of energy does not give force between tank and fuel and thus pressure.

>>>When you point tht nose up, you have now put a few inches (=~10cm) of distance between the uniflow vent opening and the spraybar.<<<
Distance does not make pressure, you really need force.

>>>You now have to pump the fuel up this distance, this is a greater distance than when the nose is horizontal and the distance is vertical distance is a few mm at most.<<<
Once again, pressure has nothing with the distnace you need FORCE and AREA. Nothing else. The pressure under the tall column of liguid os not because of distance, it is because of mass over the area and acceleration to the mass against the area under. If you apply the same acceleration to the column and to the body making the area (the tank) you do not have any force and thus also pressure.

If you like Sci-Fi books you will know a book written by J. Verne – the trip to moon in gun bullet. Once they are launched they do not know where is up and down anymore. Do you know it?

If the same accelearton applies to the bullet and also to the body inside, you do not have any force between them and thus you do not have any pressure in container with any liquid.

Poor Sir Isaac is rolling over in his grave right now!!!

I really disagree with your ideas regarding Newton's Laws and I think Old Isaac would agree with me.

Quick refresher Newton's Second Law F=ma right?

If I know the mass and the acceleration that the mass is undergoing then I surely can calculate the force that produced that acceleration.

Gravity is a force.

Look at the fuel system as a complete system from tank to needle valve assy. Also assume that we have a normal tank and engine configuration. Tilting the nose up causes a change in the static head pressure as measured from the needle valve to the top of the fuel in the tank. The pressure starts to fall as we rotate the nose up which reduce the height difference between the needle valve and the top of the fuel level. As soon as the fuel level in the tank is below the needle valve, the static head pressure at the needle is negative and if we don't change the needle valve setting, the fuel flow will be reduced and the engine leans out... period. No Voodo - - - no mysterious forces!

Sorry but I really think that we sometimes try to create "variables" and mystery forces that don't exist!

>>>I really disagree with your ideas regarding Newton's Laws and I think Old Isaac would agree with me.
Quick refresher Newton's Second Law F=ma right?
<<<

Yes it is, I see you learned it well, but you did not learn conditions for that equation. It is true only for objects without any other force like drag, friction, magnetic and ESPECIALLY GRAVITY field. If you wanted to speak about gravity acceleration it is little different:

F = g * m
There are also conditions: no magnetic field, no friction, no daemon sitting on top and especially no SPEED ACCELERATION.

So what you wrote does not have any sense without all others variables.

If you want really name someone known whose theory apply here better, than name Einstein. Take his famous lift from theory of relativity, you will understand better:

The real force in our example: means object moving in gravity field is:

F = g * m – a * m

Where “g” is gravity acceleration, “m” is mass of object and “a” is acceleration of its speed in direction of gravity field (+ means acceleration up).

Got it already? If not and if you like laws, you will certainly know, that every force has counterbalancing force. So if you have an object in gravity field, and you say (I do not say Newton, because he never told it separately – it is just wrong application) the force is “F=m * a” then where is that counter balancing force???

That is exactly like Iskandar wrote:

P = rho*g*h

Yes, it is true, but you need also know conditions. One of them is that the area under the column is supported somehow in gravity field, means you have force against. Otherwise you do not have any pressure.

Yes, I speak about that “a * m”, do you know what I mean already?

LAST EDITED ON Nov-30-03 AT 07:51 PM (CST)
 
Igor wrote:

>Look:
>
>The column has known height. OK. But we have some area. If
>you take it, you know also the volume of the liquid and thus
>its mass thanx the density. But even if we know the mass and
>acceleration, we cannot say anything about FORCE. If we
>would know the force, and we know also area, we can really
>calculate the pressure. The equation above does not contain
>the area, just because it is cancelled from both sides of
>equation.

OK, let's see if I remember how to do this (dimensional analysis)

P = rho*g*h

P (pressure) = F * Area = F * (L^2)
F (force) = M * Acc
A (acceleration) = L(T^-2)
F = ML(T^-2)

So on the left side of the equation:

P = M(T^-2)(L^-1)

On the right side of the equation:

rho (density) = M(L^-3)
g (acceleration) = L(T^-2)
h = L

rho*g*h = ML(T^-2)

So dimensionally, the left and right sides of the equations match and the equation is correct.

>But again again – WE DO NOT KNOW THE FORCE. I know you would
>like to say the force is mass x acceleration, but it is
>really not true every time. It is true only in ONE CASE: if
>the mass object has constant speed and that is not true in
>our case. So that equation is really not valid for our
>purpose. The speed is constant only in case that if the
>model is nose up, the prop makes acceleration exactly 1G and
>that is what no one can guarantee and what is in reality far
>from true. So the pressure difference really depends on
>engine pull and not on gravity or nose direction. That is
>why I wrote:
>>>>There is nothing forcing it to be exactly 1G … BESIDE our wantage to have constant speed uphill.<<<
>

The force is easy to calculate - just take the calculated pressure and multiply by the area you want to apply it to.. Simple hydraulics.. and this is, incidentally, why hydraulic jacks and car brakes work the way they do.

Yeah, it's true that the pressure isn't going to remain constant - it'll change according to the acceleration forces acting on the plane. But isn't this the entire crux of the current discussion?? Whether or not the engine can or does richen or lean out according to the varying acceleration taking place on the plane? We're not really interested in absolute pressure and accelerations here - more in comparisons of the states of affair at various points in aerobatic flight.

Dick Fowler wrote:

>Poor Sir Isaac is rolling over in his grave right now!!!
>
>I really disagree with your ideas regarding Newton's Laws
>and I think Old Isaac would agree with me.
>
>Quick refresher Newton's Second Law F=ma right?
>
>If I know the mass and the acceleration that the mass is
>undergoing then I surely can calculate the force that
>produced that acceleration.
>
>Gravity is a force.

Actually, Igor's right, I think, at least, in the way I was taught Mechanics (high school physics class). Gravity is an acceleration. It PRODUCES a force, which we call "weight" (measured in Newtons, or Poundals, or Slugs - pounds or kilograms are actually units of "mass"). In the case of gravity, you'd re-term Newton's second law as:

Weight = Mass * g

where g is the gravitational constant, and has the units of acceleration, i.e. L(T^-2), being 9.8 m(s^-2).

I think some of the confusion comes about because there are two different frames of reference used to teach Mechanics. They're both internally consistent, but in one, gravity is called a "force", in the other, an "acceleration". In one, a body at rest on the surface of the earth is considered under acceleration (because "weight" is exactly the same as though the body were being subject to an upward acceleration), in the other, it is at rest (the force produced by gravity is balanced by the force exerted on the object by the ground). I can't remember the details, but there was a huge flame war over this on rec.models.rockets years and years ago.

LAST EDITED ON Nov-30-03 AT 08:44 PM (CST)
 
>Dick Fowler wrote:
>
>>Poor Sir Isaac is rolling over in his grave right now!!!
>>
>>I really disagree with your ideas regarding Newton's Laws
>>and I think Old Isaac would agree with me.
>>
>>Quick refresher Newton's Second Law F=ma right?
>>
>>If I know the mass and the acceleration that the mass is
>>undergoing then I surely can calculate the force that
>>produced that acceleration.
>>
>>Gravity is a force.

Iskander then replied:
>
>Actually, Igor's right, I think, at least, in the way I was
>taught Mechanics (high school physics class). Gravity is an
>acceleration. It PRODUCES a force, which we call "weight"
>(measured in Newtons, or Poundals, or Slugs - pounds or
>kilograms are actually units of "mass"). In the case of
>gravity, you'd re-term Newton's second law as:
>
> Weight = Mass * g
>
>where g is the gravitational constant, and has the units of
>acceleration, i.e. L(T^-2), being 9.8 m(s^-2).
>
>I think some of the confusion comes about because there are
>two different frames of reference used to teach Mechanics.
>They're both internally consistent, but in one, gravity is
>called a "force", in the other, an "acceleration". In one, a
>body at rest on the surface of the earth is considered under
>acceleration (because "weight" is exactly the same as though
>the body were being subject to an upward acceleration), in
>the other, it is at rest (the force produced by gravity is
>balanced by the force exerted on the object by the ground).
>I can't remember the details, but there was a huge flame war
>over this on rec.models.rockets years and years ago.

Actually..... gravity is one of the fundamental weak forces. As a Physicist,( granted it's been 40 years since I've graduated) I think the proper perspective is to consider that the force of gravity is an atttraction of two bodies whose magnitude is a function of the masses of the two bodies and the distance between their center of masses. So think of it as a variable force that produces a constant acceleration on all masses acted on by the mass of the earth.

Just think..... as our little models fly up over our heads they weight less than they do in level flight. These are the sort of variables that are absolutely useless for any analysis we want to do

Final comment - I don't think we need to resort to using quantum mechanics to analyze our toys ..... unless the pipe guys really start going fast

Enough - enough my head is starting to hurt!

LAST EDITED ON Dec-01-03 AT 03:58 AM (CST)
 
>>>Enough - enough my head is starting to hurt!<<<

… I understand, I hope this will enlighten it enough:

Do you see bullet as recommended traveling to Moon by J. Verne? Guys inside are not forced to ground of bullet; even they are in gravity field

[photo not recovered: 3fcb0f294d223e50.jpg]

… until external force and then acceleration appears (or decceleration?)

[photo not recovered: 3fcb0f854d4d4eb7.jpg]

Following picture from my child book shows it well … if box is supported from its bottom side, you can sit on the box and you are forced to its side. But if the box is not supported anymore, you are not forced to the box – even you are in gravity field. That is exactly what is happening to our model, it is free fall until you apply a force – by wing or prop.

At no time is a Stunt plane in free fall, though. The point I'm trying to get at isn't that the acceleration varies in the length direction - it is that the orientation of the length direction with respect to gravity changes when you go from downward flight to level flight to upward flight, and since the lenth "height" from the tank to the needle valve is a significant amount (far more than the outboard "height" or floor "height"), it will cause a pretty big change in head. Even if the plane is not in flight, if you tilt the nose up, and you tilt the nose down, the pressure of fuel at the needle valve will change a huge amount.

The ONLY case where this will not be true is if the plane is in free fall going up AND coming down (in which case the acceleration in the length direction is zero in both cases, AND in level flight). And we know for sure that Stunt planes are never in free fall going up or down.

>>>At no time is a Stunt plane in free fall<<<
It is free fall plus aerodynamic forces nothing else. I see you agree that in the free fall you do not have any pressure changes is it true?

Then what else I say if I say that the engine thrust makes the pressure difference?

>>>The ONLY case where this will not be true is if the plane is in free fall going up AND coming down<<<
Not necessary, question is what you call free fall. It enough if it decelerates (mechanically). The question is how much - if it is 10 m/s/s then you are in free fall even you are going up (and you have canceling forces thrust = drag) – like it is on the first picture – the bullet is going up pretty well and it is still in free fall. It can land to the Moon and until landing (hard, but landing) it is all way at free fall.

I will say it another way: If your engine does not pull enough, then you decelerates (nose up) – say 1 m/s/s then the pressure difference is not the same like on ground in your hands, but 10% less. If you slow down by 2 m/s/s then it is 20% less. If you decelerate 10 m/s/s then you have no difference at all. And you are still in the same gravity field 10 m/s/s.

If your engine gets more power lean and model accelerates 1 m/s/s then you have 10% MORE fuel pressure difference than in your hands on ground. That is what I say.

LAST EDITED ON Dec-01-03 AT 05:48 AM (CST)
 
>>>>At no time is a Stunt plane in free fall<<<
>It is free fall plus aerodynamic forces nothing else. I see
>you agree that in the free fall you do not have any pressure
>changes is it true?

True, BUT it is, by definition, not in free fall.

>Then what else I say if I say that the engine thrust makes
>the pressure difference?

No, propeller thrust. Positive AND negative (it brakes on the way down).

>>>>The ONLY case where this will not be true is if the plane is in free fall going up AND coming down<<<

>Not necessary, question is what you call free fall. It
>enough if it decelerates (mechanically). The question is how
>much - if it is 10 m/s/s then you are in free fall even you
>are going up (and you have canceling forces thrust = drag) –
>like it is on the first picture – the bullet is going up
>pretty well and it is still in free fall. It can land to the
>Moon and until landing (hard, but landing) it is all way at
>free fall.

Free fall is acceleration towards the source of gravity at the same magnitude of gravitational acceleration. On Earth, that's towards the ground at 10 m(s^-2). In other words, the only force acting on the airplane is its weight (which is produced by gravitational attraction). The same reason why there is "zero gravity" in orbit (which isn't true, it just seems that way when you're in orbit).

>I will say it another way: If your engine does not pull
>enough, then you decelerates (nose up) – say 1 m/s/s then
>the pressure difference is not the same like on ground in
>your hands, but 10% less. If you slow down by 2 m/s/s then
>it is 20% less. If you decelerate 10 m/s/s then you have no
>difference at all. And you are still in the same gravity
>field 10 m/s/s.

>If your engine gets more power lean and model accelerates 1
>m/s/s then you have 10% MORE fuel pressure difference than
>in your hands on ground. That is what I say.

Sure. But we do know that the acceleration going up and going down is going to be less than g. And is, in fact, something fairly easy to approximate or measure in the vertical legs of a square loop. (Constant speed being a good first order approximation, given the engines and props people run these days.) Acceleration in the hard corners, though - that's something else.

>>>But we do know that the acceleration going up and going down is going to be less than g. <<<
Why LESS THAN 1G? If you have rocket vertically in gravity field in space without the aerodynamic drag. If engines are pulling by the force equal to weight of the rocket, then the acceleration you feel inside is exactly that 1 G. It engines pull more you feel more than 1G and if less then you feel less than one G. Thus also liquid pressure in the same column will be different at different pull.

What is happening if the same rocket is in space without gravity? EXACTLY THE SAME. The gravity has nothing to do with forces inside the body. It is really only question of external forces – pull of the engine in our case, and that pull in our flying model can be more or less than weight of the model minus drag. So the only pull of engine is determining the pressure nothing else (if do not count another aerodynamic forces). But again I really mean pull –> force on shaft, not the throttle, mixture, AoA or whatever.

You feel engine pull, not gravity, you cannot feel gravity, you can feel only force/acceleration acting against, thus also liquid pressure can not be altered by gravity, only by force against = engine thrust.

That is exactly like centrifugal acceleration – you do not feel it until you act against. That is whole story.

>>>Acceleration in the hard corners, though - that's something else.<<<
Yes, clear.

I have been watching this for a while and I think we are talking about two different things. I think it is like the argument "Is it centrifugal force?" or "Is it centripetal force?" Each side argues its point and we all know that "the force is with us."

Here we argue that it is either load or it is gravity affecting what we all know is happening to the engine. And, although at the start of this I was willing to concede that load had a portion to do with it, I get less and less convinced that load has much of anything to do with it.

Throwing all of the formulas aside--I really get nothing out of these. They are just points for argument--let us look at what is happening. What I (and Iskandar and others) have beem arguing is that when the nose is pointed up the engine must "pull" the fuel up a longer column against the "force" of gravity. Certainly the tank (being moved by the plane being moved by the prop being moved by the engine) is moving the fuel along at whatever speed the plane is flying. But all that does it put everything back as if it were all at rest.

When the nose of the airplane is pointed up, the engine still must pull the fuel up that same 3 or so (and sometimes 4 or 5 or 6) additional inches that it did not have to when the engine was level with the tank. That would have the same effect as if we somehow lowered the tank by three inches while the plane was flying level. That movement of the fuel level produces a leaning of the engine. If the engine is running too lean in level flight, then it will sag when the engine is pointed up. It loses power (and that is why running the engine on the back side of the curve doesn't work on non-piped engines. And it works on pipe engines only if the engine is allowed to run rich and fall off of the pipe when it over revs.)

Now, when we run the engine rich in level flight (which is the normal setting) then when the nose is pointed up--at rest or in flight--the engine will lean because the fuel level is lowered in relation to the engine. This leaning effect will increase power or rpm or whatever, depending on how the engine is set up, and that gives us our increase. If we have the engine set up properly (not too "zippy") then this increase in rpm should allow it to move the airplane along at its same speed which means that load has absolutely NO EFFECT on the engine in this situation.

The engine does not know load. It only knows speed at which the airplane is moving. When you move the airplane faster behind the engine you can "unload" the engine, but only to a certain degree. If you slow the airplane, I suppose we can say we "reload" the engine. But if the airplane is allowed to maintiain its same speed (due to the increase in power or rpm or whatever in the engine) then there is no change in the load on the engine. In a perfect world there would be no change in speed of the airplane in flight. In the real world there is some change, but we would like to keep it minimal. And that minimal amount that the airplane changes speed in flight thus has minimal effect on the engine. When we set these things up properly, it is not load, but change in tank location that causes it all to come about.

However, just as in the argument of "Is it centrifugal force?" or "Is it centripetal force?", we all know that the effect is there. The whole start of this discussion was to determine whether or not keeping the fuel flow more constant in the case of the 4-stroke engine which does not seem to pick up rpm when pointed up, but does lean out, would benefit the run. I happen to believe it would. In fact, I am more convinced than ever that it would.

So, Brad, go for it.

>>>it is like the argument "Is it centrifugal force?" or "Is it centripetal force?"<<<
Yes Len it is, until we speak about pressure in column standing on ground. Then it is really like speaking 2+2=4 or 2*2=4 even we all know it must be equivalent.

>>> Each side argues its point and we all know that "the force is with us." <<<
I do not see it that way; I think arguments are pretty correct. I think the difference is in application I would really like to know where we differ.
>>>Here we argue that it is either load or it is gravity affecting what we all know is happening to the engine. And, although at the start of this I was willing to concede that load had a portion to do with it, I get less and less convinced that load has much of anything to do with it. <<<
My point of view and understanding is, that it is ONLY the thrust (I think that is what you call load) what can make pressure difference and even more – if you engine quit at all, the pressure will be even higher, not smaller – it is because drag of model which will cause running the fuel to the NVA, even nose up.

>>>arguing is that when the nose is pointed up the engine must "pull"<<<
I would be careful with that “MUST” – where is the reason? If you fly at velocity “X” where your thrust is equal to the drag, and you point that nose up and you still fly that “X” speed (even with deceleration – but at least at begin) then you do NOT have any force which can lean the mixture. The only reason why your engine can pull more, is lower speed, lower speed means deceleration and deceleration is counter action to the gravity. If you go over and over you will come to the conclusion that the prop pull makes that pressure.

>>>When the nose of the airplane is pointed up, the engine still must pull the fuel up that same 3 or so (and sometimes 4 or 5 or 6) additional inches that it did not have to when the engine was level with the tank.<<<
Not needed, the mass of inertia is enough to force the fuel to the nose if you decelerate. If you do not decelerate, you re right, but in that case you must pull and we are back what I say, the engine pull is force making pressure difference.
>>>That would have the same effect as if we somehow lowered the tank by three inches while the plane was flying level.<<<
Definitely, but here we have wing lift making that force.
>>> when the nose is pointed up--at rest or in flight--the engine will lean because the fuel level is lowered in relation to the engine.<<<
Here I do not agree, the leaning is because of engine thrust. You must remember, the fuel has mass inertia, if you decelerate, you have enough pressure.
>>This leaning effect will increase power or rpm or whatever, depending on how the engine is set up, and that gives us our increase. If we have the engine set up properly (not too "zippy") then this increase in rpm should allow it to move the airplane along at its same speed which means that load has absolutely NO EFFECT on the engine in this situation. <<<
It is little different, if you point up, the model starts to decelerate, because the thrust is able to balance only drag, but we have also gravity. As you are slowing down, the prop starts to pull more (thanx AoA) and only that pull is reason for leaning. If not then you say that thrust=drag kills both and model is in free fall (no lift fron wing) and we know that pressure in free flight is 0. So because of this positive feedback, you can reach acceleration better than 1G. Richer run has more potential for acceleration. It is because if you once get to or over the best mixture, you cannot accelerate anymore because the feedback is negative (more pull, leaner mixture, more sag).
>>>The whole start of this discussion was to determine whether or not keeping the fuel flow more constant in the case of the 4-stroke engine which does not seem to pick up rpm when pointed up, but does lean out, would benefit the run.<<<
You see the answer in statement before?

>>>>The whole start of this discussion was to determine whether or not keeping the fuel flow more constant in the case of the 4-stroke engine which does not seem to pick up rpm when pointed up, but does lean out, would benefit the run.<<<
>You see the answer in statement before?

No.

The lean 4C engine deccelerates uphill more, it keep mixture richer, so it does not overlean.

>The lean 4C engine deccelerates uphill more, it keep mixture
>richer, so it does not overlean.

Igor, if, as you have stated here, we have deceleration causing the fuel to move forward and thus richening the mixture, would not the fuel be flowing foward inside the tank and away from the pickup tube entirely? What I am saying is that even before the plane has a chance to decelerate to any significant degree we have lowered the tank and caused a leaning out of the engine. If this leaning out of the engine increases power (rpm, whatever) on the engine such that there is enough added thrust, then there will be no deceleration.

It is possible to even make the airplane move faster up hill due to a greater than desired increase in power (lower the compression too much so that it picks up too much rpm when it breaks).

Remember, the engine does not see load in the sense of added weight or whatever. It only sees the affect of airspeed. If the airspeed can be kept constant (through an increase in rpm, and thus thrust, when the plane is rotated up) then there is in effect no increase in load caused by the airplane (still traveling at the same speed). Any increase in the load is merely the cause of the engine increasing is speed which is the result of something other than aircraft load. (In this case the lowering of the tank in relation to the engine.)

I am trying to reason this from the perspective of maintaining flying speed in all directions. If we are maintaing our speed, then the effects of raising the nose are the same as lowering the tank, and this, and this alone, is what is acting on the engine.

A good experiment to try would be to see just exactly how much affect slowing the airplane down would have on the engine run. If we could mount a running engine on the hood of our car (or top, or bed of our pick-up truck or...) and then put a tach on it and drive our car up to 50 mph and read the tach and then drop back to 40 mph and read the tach I am wondering just exactly how much difference in rpm we would really see. Will the engine pick up in rpm because of load? And if so, how much, really? Does this really explain what is happening when we point the nose of the airplane up and the engine picks up in rpm and there is no apparent loss in flying speed? Hmmm?

>>>Igor, if, as you have stated here, we have deceleration causing the fuel to move forward and thus richening the mixture, would not the fuel be flowing foward inside the tank and away from the pickup tube entirely?<<<

No, only in case that the deceleration is more than 10 m/s/s means more than gravity. So it can happen if your engine quit. Then the drag causes moving the fuel to the front.

Try to fillow my #108, I know it is lot of crazy letters, but I think it is not so difficult to understand, especially its conclusion.

>>>A good experiment to try would be to see just exactly how much affect slowing the airplane down would have on the engine run. If we could mount a running engine on the hood of our car (or top, or bed of our pick-up truck or...) and then put a tach on it and drive our car up to 50 mph and read the tach and then drop back to 40 mph and read the tach I am wondering just exactly how much difference in rpm we would really see. Will the engine pick up in rpm because of load? And if so, how much, really? Does this really explain what is happening when we point the nose of the airplane up and the engine picks up in rpm and there is no apparent loss in flying speed? Hmmm?<<<

It is nice, but it is not effect due to the leaning richening itself. If you presume you have constant speed, then all is true the same way as on stand on ground and the pressure is really hydrostatic caused by gravity. But that presumption “constant speed” is not valid, you can not use rules for hydrostatic pressure. The “constant speed” is what we are questioning.

>
>>>>A good experiment to try would be to see just exactly how much affect slowing the airplane down would have on the engine run. If we could mount a running engine on the hood of our car (or top, or bed of our pick-up truck or...) and then put a tach on it and drive our car up to 50 mph and read the tach and then drop back to 40 mph and read the tach I am wondering just exactly how much difference in rpm we would really see. Will the engine pick up in rpm because of load? And if so, how much, really? Does this really explain what is happening when we point the nose of the airplane up and the engine picks up in rpm and there is no apparent loss in flying speed? Hmmm?<<<
>
>It is nice, but it is not effect due to the leaning
>richening itself. If you presume you have constant speed,
>then all is true the same way as on stand on ground and the
>pressure is really hydrostatic caused by gravity. But that
>presumption “constant speed” is not valid, you can not use
>rules for hydrostatic pressure. The “constant speed” is what
>we are questioning.

I am merely suggesting that we drive at flyinjg speed of approximately 50 miles per hour or so and then hit the brakes to drop back to 40 or even 30. This would have the same effect as increasing the "load". I don't believe the results would be nearly as dramatic as when we point the nose up or drop the level of the tank. And ANY lowering of the tank, I don't care if we drop the speed to half, ANY lowering of the tank is going to lean the engine. If we lean the engine, that will increase the thrust if the engine is set up properly (as we do) and will compensate for increased gravitational effect on the plane (not load on the engine). The load on the engine will increase because it leans and runs faster, but it will not increase if the flying speed is not significantly reduced.

We are just arguing two different points, Igor. And all the formulas mean nothing if we don't agree on the points we are arguing.

>>>ANY lowering of the tank is going to lean the engine<<<

I recommended to express it formally, because this statement is not clear. I do not know what you understand by “lowering”. If you mean it vertically, I do not agree. You must mean “lowering” in net acceleration direction – and its orientation of course – if tank is higher than nva and acceleration points down then you have still less pressure at NVA.

I tried to separate gravity and thrust only, not any corner effects or lift or so. Thus I spoke only about straight vertical flight.

The only what I say is, that the pressure difference at NVA depends on fuel column height, its density and prop thrust, nothing else.

I completely agree with you and everybody else that the prop makes thrust. I really believe the prop pulls up, really … But it does not change what I wrote originally.

>>>>ANY lowering of the tank is going to lean the engine<<<
>
>I recommended to express it formally, because this statement
>is not clear. I do not know what you understand by
>“lowering”. If you mean it vertically, I do not agree. You
>must mean “lowering” in net acceleration direction – and its
>orientation of course – if tank is higher than nva and
>acceleration points down then you have still less pressure
>at NVA.

And here we agree. That would be "lowering" it by the intent of my statement. In inside loops or plain level flight, "lowering" the tank would be towards the wheels. In outside loops or inverted flight, "lowering" it would be away from the wheels.

>I tried to separate gravity and thrust only, not any corner
>effects or lift or so. Thus I spoke only about straight
>vertical flight.

Me, too. Keep it simple for now until we understand each other's terminology.

>The only what I say is, that the pressure difference at NVA
>depends on fuel column height, its density and prop thrust,
>nothing else.

When we talk about fuel column height, isn't this what I hand others have been saying? If bringing the tank below the engine as we do when we point the nose up is not the determining factor, or if this is to be ignored, why, then, do we not mount the tank on the center of gravity of the plane? It would certainly feed just as well when running level, and if it is some other factor and not the "lowering" of the tank that affects the engine when we point the nose up, why not have it farther away so that it stays on the cg?

>I completely agree with you and everybody else that the prop
>makes thrust. I really believe the prop pulls up, really

Hey, Igor, we are pretty much agreeing on this one. I have a couple of more questions which I shall e-mail to you because I think we are confusing ourselves with terminology.

>… But it does not change what I wrote originally.
>
>igor

>>>When we talk about fuel column height, isn't this what I hand others have been saying? If bringing the tank below the engine as we do when we point the nose up is not the determining factor, or if this is to be ignored, why, then, do we not mount the tank on the center of gravity of the plane? It would certainly feed just as well when running level, and if it is some other factor and not the "lowering" of the tank that affects the engine when we point the nose up, why not have it farther away so that it stays on the cg?<<<

OK, may be I did not wrote it clear enough. Yes, I fully agree that you HAVE level difference IN DIRECTION OF ACCELERATION, and agree that it IS determining factor. I say that it is not the ONLY determining factor. The other factor is that acceleration. There is still density, but it is not variable, so for us it is NOT determining factor.

I say that if we exclude wing lift, AoA, drag and another unimportant things the pressure difference to the normal pressure in level steady flights is:

Pdif = Height_difference * density * prop_thrust / mass_of_model

And my original message also noted that there is NO GRAVITY in the equation.

So if you have constant distance between NVA and tank uniflow vent, your model does not change mass, then the pressure difference and thus mixture is affected by the prop thrust ONLY. And as far as there is no gravity in equation, this is true in every direction.

I agree with you and all others with fact that gravity affects flying speed and thus prop blades AoA and thus its thrust. And thrust as we know makes that leaning, but there is NO automatic relation between pressure and gravity beside previous.

And if someone wants to tell me that prop makes thrust equal to model weight then I do not agree at all. It can be true on ground but not in flight. If I take “some” 2 blade prop 12x4 and do calculation, then I found that the thrust at speed 10% lower than level is only 4Newtons what make acceleration 4N/1.4Kg=2.8m/s/s what is only quarter of gravity so the leaning in hands on ground is 4 x more effective.

>(snip) I say that if we exclude wing lift, AoA, drag and another
>unimportant things the pressure difference to the normal
>pressure in level steady flights is:
>
>Pdif = Height_difference * density * prop_thrust /
>mass_of_model
>
>And my original message also noted that there is NO GRAVITY
>in the equation.
>
>So if you have constant distance between NVA and tank
>uniflow vent, your model does not change mass, then the
>pressure difference and thus mixture is affected by the prop
>thrust ONLY. And as far as there is no gravity in equation,
>this is true in every direction.
>
>I agree with you and all others with fact that gravity
>affects flying speed and thus prop blades AoA and thus its
>thrust. And thrust as we know makes that leaning, but there
>is NO automatic relation between pressure and gravity beside
>previous.

I guess this is where we are arguing different points and losing each other. You are arguing from equations, but equations are only valid if we include all of the discernable facts. One discernable fact as I see it is we cannot take gravity out of the equation. We are stuck in its grasp. Everything, including fluids, are affected by gravity. No matter what other figures we put into or take out of the equation, we still must leave gravity there. Even when we add other forces pushing fuel up or pulling it down, we still must leave gravity there.

We know that when the nose is pointed up in flight, the fuel still remains in the back of the tank (engine keeps running, observations from camera), so there is not sufficient deceleration (if any) to cause the fuel to move forward. Deceleration (if any) could reduce the effects of gravity, but not eliminate it unless the deceleration were so severe as to cause the fuel to move to the front of the tank.

However, it is not so much deceleration that is happening in a corner, but a change in direction. If the plane could do an immediate 90 degree turn, it would have to accelerate up, but there would be no deceleration in that direction as it never was moving in that direction in the first place.

I will grant that there will be a momentary loss of speed in the change of direction in the corner. These corners are not nearly as violent, however, as they appear to the naked eye. An 18 foot corner is a lot different from a 5 foot corner. Still, turning that corner would cause the fuel to flow away (in a different direction) from the engine. And that flowing away will, in itself, cause leaning.

And, granting that there is a slowing in the corner, when the plane begins the straight path up and the engine accelerates it back up to speed, during the acceleratin portion, acceleration is affecting the reduction in fuel flow (leaning) along with gravity. Once the plane is up to speed again, if we can get it back to the same speed as in level flight, it is gravity alone that is affecting the run. So gravity is there in all cases. And acceleration is aiding gravity in causing the leaning of the engine.

And, in either case, it is not just the increased load on the prop (caused by a slowing down of airspeed only, not by other factors), but the "pull" of the fuel away from the engine (caused by acceleration of the plane in a new direction or by gravity) that is causing the engine to lean. I am ever more convinced that these two, far more than any increased load on the prop, are the primary cause of the engine leaning. So, again, going back to Brad's original question on this thread that if we could keep the fuel pressure constant we could better control the leaning of the 4-cycle engine. I agree.

Just one last comment: If gravity is not the major force, or is so insignificant that we can leave it out of the equation, then why are we so concerned about putting the tank as close to the engine as possible? Why do we not locate the tank in the center of gravity of the plane so as not to disturb the balance point as it empties, if the location of the tank below the engine is not going to affect the run?

>And if someone wants to tell me that prop makes thrust equal
>to model weight then I do not agree at all. It can be true
>on ground but not in flight. If I take “some” 2 blade prop
>12x4 and do calculation, then I found that the thrust at
>speed 10% lower than level is only 4Newtons what make
>acceleration 4N/1.4Kg=2.8m/s/s what is only quarter of
>gravity so the leaning in hands on ground is 4 x more
>effective.

But you are assuming a greater slowing of the airplane and acceleration. If there is a greater slowing than 10%, then the affect is greater. If there is acceleration, then this must be factored in with gravity for it will have the same affect on the fuel. (2 gs acceleration = 2 gs gravity.

Igor, if we can hold the airplane straight up in our hands and let go and the prop can pull it into an immediate wing over (which, with a good competitive model, it can) then the prop has more thrust than the weight of the model.

I totally agree that the faster the plane moves the less effective thrust we have until at some point thrust equals weight and drag and the plane will cease to accelerate any further (motion continues, however). But if the engine is caused to speed up when the nose is pointed up (and it does, simply from the leaning out of the engine in the way that we have our planes set up--be it gravity alone, or gravity and acceleration) then it will produce an increase in thrust which can overcome some, if not all of the deceleration caused by the additional affect of gravity.

How much this is true will certainly depend on the weight (and drag) of the airframe, the size (power available) of the engine, size of the prop (bigger prop will have bigger reserve or require less rpm increase), among other factors (timing, compression, fuel used, venturi size, etc. all affect how much, if any, "boost" we will get out of the engine in these conditions). It is possible, even, to adjust these variables to such a degree that the engine will speed up enough to cause it to fly faster up hill than in level flight.

The ideal (never fuly achieved) is to maintain constant speed, so that we neither lose, nor gain speed in the various legs of the maneuvers. But it is possible to gain as well as lose speed, as is evidenced when some engines "go wild" in the maneuvers.

Good discussion. I would still like to see someone mount an engine on a test stand in their car or truck and then drive at flying speed and suddenly decelerate and accelerate to see what affect air speed alone has on the run of the engine. This is the effect that the engine sees in the maneuver. This is the only effect. It does not see or feel weight or drag or anything else. It only sees the slowing down or the speeding up of the air through the propeller.

Of course, one flaw in all this is that we could see what affect slowing down in the air has on the prop. But if we suddenly accelerate, then we will be throwing the fuel to the back and adding the affect of gravity to the run. This would cause a leaning to the engine, when, in flight, we actually experience an unloading of the prop as it gets up to speed (usually happening shortly after take off.)

>The ideal (never fuly achieved) is to maintain constant
>speed, so that we neither lose, nor gain speed in the
>various legs of the maneuvers. But it is possible to gain
>as well as lose speed, as is evidenced when some engines "go
>wild" in the maneuvers.

Wow this has been a great thread. Some of the posts I have not even made it all the way through.

But what I was wondering is a constant speed what we really want?

When I was purchasing the YS the guy behind the counter was a CL guy and said that for stunt the constant speed setup might actaully hurt us stunt guys. No push on the way up and no slowing effect on the way down. Brad says we make enough power now not to need a boost on the way up. Maybe so but I am thinking of a vert 8 in some pretty strong winds and that little boost is pretty nice and the slowing effect on the way down seems to help kill wind up. We will ever really know? I dont know. But it will be fun to try to get there...at least it has been so far...

>Wow this has been a great thread. Some of the posts I have
>not even made it all the way through.
>
>But what I was wondering is a constant speed what we really
>want?
>
>When I was purchasing the YS the guy behind the counter was
>a CL guy and said that for stunt the constant speed setup
>might actaully hurt us stunt guys. No push on the way up
>and no slowing effect on the way down. Brad says we make
>enough power now not to need a boost on the way up. Maybe
>so but I am thinking of a vert 8 in some pretty strong winds
>and that little boost is pretty nice and the slowing effect
>on the way down seems to help kill wind up. We will ever
>really know? I dont know. But it will be fun to try to get
>there...at least it has been so far...

This, again, is one of those "we say something and it is interpreted another way" thingies. I am agreeing with you, Doug, that constant speed on the engine is not, necessarily, what we want. I was referring to constant speed in the maneuvers (something I guess I didn't express very well.) It is something where you can count "one, two, three, four" as you do the corners of the square and get in a rhythm. If the plane is going at different speeds on different legs, that can really mess up your timing.

constant speed in the maneuvers is achievable either through an increase in rpm on the up legs (as you noted) or in producing so much power (available thrust) that the slow downs are minimal. I would want to keep the flying speed as constant as possible, and we can achieve it with a boost in our 2-cycle engines at the necessary times.

LAST EDITED ON Dec-02-03 AT 03:17 PM (CST)
 

>A good experiment to try would be to see just exactly how
>much affect slowing the airplane down would have on the
>engine run.

Just watch Bob Reeves video... The fuel never leaves the back of the tank.

>
>>A good experiment to try would be to see just exactly how
>>much affect slowing the airplane down would have on the
>>engine run.
>
>Just watch Bob Reeves video... The fuel never leaves the
>back of the tank.

I agree with that Brad. That is why I am also agreeing that lowering the tank will affect the mixture, hence increase the rpm (leaning it) on the engine.

My suggestion for the experiment would be to see what effect slowing the airplane down in flight would have on the "load" of the propeller, and how much of a part this really does play. I am firmly convinced that the biggest factor is the lowering of the tank.

>The lean 4C engine deccelerates uphill more, it keep mixture
>richer, so it does not overlean.

Gosh, if that were only true!

>>>Yeah, it's true that the pressure isn't going to remain constant - it'll change according to the acceleration forces acting on the plane. But isn't this the entire crux of the current discussion??<<<
YES!!!
>>> I think some of the confusion comes about because there are two different frames of reference used to teach Mechanics. They're both internally consistent, but in one, gravity is called a "force", in the other, an "acceleration". In one, a body at rest on the surface of the earth is considered under acceleration (because "weight" is exactly the same as though the body were being subject to an upward acceleration), in the other, it is at rest (the force produced by gravity is balanced by the force exerted on the object by the ground). I can't remember the details, but there was a huge flame war over this on rec.models.rockets years and years ago. <<<
That is exactly what is going on. While gravity applies on both tank and fuel and while there is no other external force, there is NO FORCE BETWEEN tank and fuel and thus no pressure. In our example, when the uniflow went in level of NVA in level flight, you can think there is “no” pressure. If you point up and if your prop pulls exactly as the drag is, then the model is in reality in free fall, even the model goes up, just because the mechanic deceleration is equal to the acceleration called gravity. In this case you really do not get any pressure difference even the nose still points up. The engine must pull more than drag is; only then we have some pressure difference.

While the gravity applies to all particles same way (Ok Dick I presume we have small bodies and we are far from closest black hole), it does not make any internal forces – that is necessary for pressure (something against something). Only mechanical acceleration can make internal forces: you press bottom of liquid column (area) against mass of liquid (area * height * density). Does not matter if you do it dynamically (you really accelerate motion) in space free of gravity or if you do it in gravity field statically (you apply counter acceleration to keep it on one place). It is because you are forcing to only SOME of particles against another – that is where the pressure comes from.

The Mechanics book says a liquid column makes static pressure in gravity field according its height, but it is big simplification, which cannot be applied to our model. I reality it is not the height what makes the pressure, it is weight on the area. But the area is the same on both sides (as Iskanda checked) so it is first simplification, which is applicable in our example too. The second is, that it is mechanical force of area to the liquid what makes the pressure, not the gravity. But that force to that mass is exactly equal to the gravity acceleration … well again it is STATIC … again SATIC pressure. This simplification is well usefull for static column on ground, but not useful on flying model.

It means if the engine is very sensitive to lean mixture and if the level setting is exactly that spot of optimal mixture, and if nose points up, the engine will sag and it will keep exactly that pull=drag decelerating mode and it will keep the “level” pressure at NVA itself – even nose up.

It would certainly quit on ground in your hands (they make acceleration up – against the gravity – makes pressure difference), but it will not quit in flight – even nose up.

>The Mechanics book says a liquid column makes static
>pressure in gravity field according its height, but it is
>big simplification, which cannot be applied to our model.

The only way that this is a simplification is that it's in one dimension, and we have to deal with three. And that's not really that hard to do, if you can calculate the accelerations involved in each direction.

>I
>reality it is not the height what makes the pressure, it is
>weight on the area. But the area is the same on both sides
>(as Iskanda checked) so it is first simplification, which is
>applicable in our example too. The second is, that it is
>mechanical force of area to the liquid what makes the
>pressure, not the gravity. But that force to that mass is
>exactly equal to the gravity acceleration … well again it is
>STATIC … again SATIC pressure. This simplification is well
>usefull for static column on ground, but not useful on
>flying model.

No, it IS the height (plus the density, and the acceleration). Given constant acceleration, density and height, the pressure at the bottom of the column is going to be the same, no matter what the area of the column. And it IS extremely useful here, even if the pressure changes constantly, if you can calculate or measure accelerations at any given time (and the height and density are held as constants). The pressure at any given moment in time is the result of the accelerations at that given moment in time.

>It means if the engine is very sensitive to lean mixture and
>if the level setting is exactly that spot of optimal
>mixture, and if nose points up, the engine will sag and it
>will keep exactly that pull=drag decelerating mode and it
>will keep the “level” pressure at NVA itself – even nose up.

>It would certainly quit on ground in your hands (they make
>acceleration up – against the gravity – makes pressure
>difference), but it will not quit in flight – even nose up.

Sure. But you'd have to accelerate downwards (i.e. decelerate upwards) at a significant proportion of g for this to happen. How slow does a modern stunter get at the top of a square loop? And even if it doesn't quit, it'd still run leaner than it did in level flight, right?

I'm not being a smart a** but I think a refresher might help here.

This is not a bad place to review the concepts.

http://www.glenbrook.k12.il.us/gbssci/phys/Class/newtlaws/u2l1a.html

I think you are getting confused by thinking about the accelerations when you should focus on the forces acting on the object. It's kind of cart before the horse (or for Bill, mule!) Thinking from the standpoint of the sum of the forces forces creating a resulting acceleration is the "classical method"

Draw a free-body diagram of the forces acting on the plane. Gets easier that way.

Most people struggle with circular motion because of the uniform acceleration component. Much less intuative than linear motion.

PS - There really isn't a " centrifugal force" as such.

Ok guys, I quit my arguments, because I see my theoretical arguments, examples and esoteric pictures can not convince you, so let’s speak formally:

We have a liquid column. The height is H, the density is R, the gravity acceleration is G and column is fixed on one place on ground.

The question is what is the pressure on bottom of column P.

I thing we will agree that it is

P = H * R * G

OK?

And now, my question is, what is the pressure in the same column if it is in M heavy vertically flying vehicle, if engine thrust is T drag of vehicle is D.

No do the math and we will see.

I think it will satisfy you both, Isky an Dick, may be I will agree, may be I will see error, and may be we will see where we speak about different things.

>>>
We have a liquid column. The height is H, the density is R, the gravity acceleration is G and column is fixed on one place on ground.
And now, my question is, what is the pressure in the same column if it is in M heavy vertically flying vehicle, if engine thrust is T drag of vehicle is D.
<<<

I see no one want to post the solution so I do it, I hope it will say what I mean:

1/ I would like to go the way my opponents:
If the wehicle will be static on one place, there will be static pressure coming from gravity:

P = H * R * G

2/ But model moves, the acceleration / decceleartion changes the pressure also, so let us analyze speed and acceleration:

The force acting to the model is:

F = Fthrust - Fgravity

Fthrust = T – D and Fgravity = M * G

So the force acting to the model is:

F = T – D - M * G

So thanx Newton’s laws the acceleration is:

a = F / M = (T – D - M * G) / M = (T – D) / M - G

quite clear I think, so thus the pressure difference thanx acceleration is:

P = H * R * a = H * R * ( (T – D) / M – G) =
= H * R * (T – D) / M – H * R * G

So now we know acceleration of the model movement and so we can apply it to the column also.

Now take both components together and the residual pressure is:

P = H * R * G + H * R * (T – D) / M – H * R * G =
= H * R * (T – D) / M

If you take closer look, the (T – D) / M is exactly the acceleration of the engine thrust minus drag on the model, so it clearly shows that the pressure depends ONLY on engine thrust and is independent on gravity:

P = H * R * pull / M

OK?

>A good experiment to try would be to see just exactly how
>much affect slowing the airplane down would have on the
>engine run.
Just watch Bob Reeves video... The fuel never leaves the back of the tank.

The City Smasher

And so it came about that Mr. Reeves was witness to the very high demand for his videos instantly dropping to nothing.

Snip

>The force acting to the model is:
>
>F = Fthrust - Fgravity

You are accounting for Gravity here! (Fgravity)

>
>Snip
>

>P= H * R * (T – D) / M – H * R * G

Agree here as well!

>
>So now we know acceleration of the model movement and so we
>can apply it to the column also.
>
>
>Now take both components together and the residual pressure
>is:

This is what I don't understand. Why are you adding Gravity back in here when it is already accounted for in the above equation?
>
>P = H * R * G + H * R * (T – D) / M – H * R * G =
>= H * R * (T – D) / M
>
This equation states that if T=D, there would be no pressure. If the plane were pointed up, and T=D, and there was no acceleration of the plane relative to earth, there would still be 1g on the plane, thus I believe there would still be a "P" there. It seems to me that there is an additional component of Gravity in your formule that cancels out this effect. If you are in space under no "gravity" then your equation makes sense to me.

snip

Help me out here please.

>
>P = H * R * G + H * R * (T – D) / M – H * R * G =
>= H * R * (T – D) / M
>
This equation states that if T=D, there would be no pressure. If the plane were pointed up, and T=D, and there was no acceleration of the plane relative to earth, there would still be 1g on the plane, thus I believe there would still be a "P" there. It seems to me that there is an additional component of Gravity in your formule that cancels out this effect. If you are in space under no "gravity" then your equation makes sense to me.
<<<

I wanted to start from point of view Isky and Dick who say the gravity is acting on fuel and may be plus some acceleration from movement. I see both components static pressure and mechanic acceleration of movement and its application to the mass of model and thus resulting acceleration you accept, so here I continue more detailed:
The acceleration is

a = (T – D) / M – G

It is actual thrust minus drag – means real force to the nose applied to the mass of model, minus gravity acceleration acting back. It is real movement acceleration acting to the liquid column thanx Newton law. So it must make some additional pressure contribution to the static pressure from gravity. That contribution is:

Pdiff = H * R * a

If you add this contribution to the original static pressure:

P = Pstat + Pdiff

You will get:

P = H * R * G + H * R * a =
= H * R * G + H * R * {(T – D) / M – G } =
= H * R * G + H * R * (T – D) / M – H * R * G =

= H * R * (T – D) / M

…. Hhhhhh … too many letters, I hope I did not do mistake

>>>If the plane were pointed up, and T=D, and there was no acceleration of the plane relative to earth<<<
Exactly, it is free fall.

>>>there would still be 1g on the plane<<<
Definitelly, and also to the fuel inside, therefore no force pushig it to the wall. Just imagine – no external mechanical force (both are canceling each other), only gravity – to the frame as well as fuel. Both are either decelerating the same way or falling down same way – no force between.

>>>I believe there would still be a "P" there.<<<
There is missing force, therefore – sorry no pressure. May be atmospheric, but this we do not count in equation – it is from boths side equal.

>>>It seems to me that there is an additional component of Gravity in your formule that cancels out this effect.<<<
Tell me where? Try to read message #124 it shows exactly this math on Lens example. May be will understand each other easier.

"AND A LIGHT COMES ON"

Igor...I think what you are trying to say is that we must also consider what is happening to the fuel when we change direction (an acceleration) in flight or accelerate the model from a rest to flying speed.

Simple example is:

Place a 10 gram mass in a rocket on a scale. At rest, the scale will read 10 grams (acceleration due to gravity is 9.8 m/sec/sec). If this mass is a liquid (make it water) and I install a suction pump, I can lift this liquid by suction to a vertical height of approx. 9.75 meters.

If I fire the rocket and it produces an accelerates that is three times that of gravity, then my scale reads 30 grams. But now I can only lift the liquid by suction to a height of 3.25 meters.

Because Static Fluid pressure = density x acceleration x depth

Think in terms of having changed the density of the fluid. Its apparent mass is now 3 times what it was at rest for the same volume… hence its density seems to be three times greater (Even though it is not compressible in a classic sense).

Our engines are suction devices….. so we pull less fuel and the engines goes lean.

Just a couple of comments.

Without running the numbers I think that this change has less effect on an engine’s fuel flow than does pitching the model up or down which changes the needle valve to fuel level height. Why you ask? The duration of the forces acting to produce these accelerations are short and by the time the engine sees the change the model is in a different attitude but remains nose up or nose down for longer periods.(Thinking of a square).

Someone in a post commented that at launch, their engine would go lean for a lap or two then settle into a steady “four stroke”. Could this be the effect of the acceleration as described above? Probably.

This makes a case for pressurized fuel systems. Someone mentioned this in a post a couple of weeks ago. If we can get the fuel system pressure up high enough then all these other forces are less influential.

Finally…. Igor, thanks for being so determined in trying to get us to see what you were trying say. I’m not sure I would have been as determined and would have just given up. I’m sure that language differences are big part of the problem but perseverance paid off.

Thanks for getting me to think outside of the box!

And I will add an "amen" to what Dick said about thanking Igor. Sometimes we talk from different view points and have a hard time understanding each other. Sometimes we even maintain these view points. But in a couple of private e-mails to and from Igor he has been helpful, and there is much common ground of agreement here, even though at times it does not look like that.

We just sometimes have a hard time understanding each other.

Thanks, Igor (and Dick, and all the rest.)

Sorry, it is somehow messed in upper messages, this is aresponce to #119 where Len did very nice example which can clearly say who and what means:

>>>Igor, if we can hold the airplane straight up in our hands and let go and the prop can pull it into an immediate wing over (which, with a good competitive model, it can) then the prop has more thrust than the weight of the model.<<<

Ok so let’s do it. It can happen that:

1/ The prop thrust is exactly the weight of the model.

In this case the pressure is exactly as you keep it in hands you say it is is because of gravity, OK I agree. I just mean it is hand pushing up - but this is only action and reaction which must be opposite and equal, but it does not change the reality.

2/ The prop thrust is 10% MORE than the weight of the model.

You have the same pressure like in first point because of gravity, but after the start you accelerates up. The acceleration is known. The excessive force is 0.1 of model weight so it is 10% of gravity acceleration and that bring +10% of pressure. That is several times mentioned Newton law I hope everyone agree. So the resulting pressure is 1.1 of that in you hands. Even the distance is the same and gravity also.

3/ The prop thrust is 10% LESS than the weight of the model.

You have the same pressure like in first point, but after the start your model falls down. The acceleration is also known. The acceleration down is also 10% of gravity acceleration and that bring -10% of pressure. So the resulting pressure is 0.9 of that in you hands.

4/ If the engine does not have prop at all … surprise

Model is in free fall and there is no pressure at all, because mechanic acceleration is equivalent to gravity - so there is nothing forcing the fuel against the wall.

Do you all agree? I hope so, it is your point of view.

I just note that if you take closer loot you will see that the pressure is linear to prop thrust – it tells lot … but continue.

Now guess what will happen if the gravity will be 10X more and thrust is still the same?

I will do it only for the first point, I believe everyone can continue:

1/ The prop thrust is exactly the weight of the model in normal gravity but actual gravity is 10x more. I note I mean the very first moment after releasing, so there no speed which can change prop thrust yet.

You have 10 x more pressure like in first point because of 10x more actual gravity, but after releasing you start fall down. The acceleration is known. The excessive force is 9 x of model weight in regular gravity (would be 10x in 10x more gravity, but it is less 1x gravity because of thrust equal to that normal gravity) so it is 9x normal gravity acceleration and that bring -9 x original pressure. So the resulting pressure is 10-9 = 1 x of that in you hands – means exactly the same pressure in 10x more gravity as was the pressure in regular gravity.

So conclusion is: pressure is linear to thrust and INVARIANT to the GRAVITY.

And now do not kill me please, I agree that the thrust in 10x more gravity would be higher, just because we would probably use better engine and prop, but it does not change fact, that the pressure depends on that thrust, not on that GRAVITY.

I think there was whole misunderstanding.

Yes, I agree with you, Igor, that much of our problem is in the terminology that we are using. You are not denying gravity. You are just using different terms. And you use engineering terms, and I use...er, well, something I can understand.

One more thing, since (in your e-mail) you asked my reaction to what you posted here, I will give it. I think we are agreeing on most things, just using different terminology. However, I used this as an illustration to you in an e-mail, and you agreed on it at the time. So I will elaborate on it now and you can tell me how this fits in.

What if we were in the elevator of a big skyscraper--150 stories tall--and the elevator is made to go fast so you can reach the top before night fall.

You are holding the airplane in your hand, and before the elevator even moves you start the engine. Hold it level and you get the normal 4-cycle. Raise the nose and it breaks into a 2-cycle. Everything is working just like "normal". Gravity.

Now the elevator starts to move. Going up. Going up fast. As the elevator begins to move, you raise the nose and the engine once again leans out. This is acceleration combined with gravity. We are still agreed. OK. But once acceleration is over and the elevator is going up at a constant speed you raise the nose again and...it still leans out. Gravity. Just gravity. Isn't that what is happening to the airplane when it is flying straight up once it has reached a constant speed?

If I were in this elevator going lickety split and there were no gravity, I could float all over the place. I could walk on the ceiling. But with gravity, I am still confined to walking on the floor. Once the airplane, that has its nose pointed up, has reached a constant velocity, no acceleration, no deceleration, all that remains is gravity. And it is still working just like before. Point the nose up and the engine leans out. Isn't that what is happening? And it doesn't make any difference whether the elevator is going up at 20 miles per hour, 40 miles per hour, or 60 miles per hour. It still has the same effect.

I know I skipped the part about needing more thrust than before in order to maintain the same speed while going up, and that is still important. But if the engine leans out we can gain more thrust. And, with the proper amount of thrust, we can maintain speed. So, making those assumptions, aren't we still talking gravity?

As always, thanks for your input.

LAST EDITED ON Dec-02-03 AT 07:15 PM (CST)
 
Leonard.... great example!

Your point is correct. The gravity component is always present in our system and always influences the fuel system regardless of other forces acting on the system.

If that evelvator was dropping in free fall and you tilt the nose up.... the engine goes lean. Can't escape that old gravity.

I have to think of something else to talk about.

Uh, no... If the elevator were in free fall, it's the same as though you were in orbit, or even somewhere in deep space - i.e. "no gravity". There will no longer be an up or a down. You can tilt the nose in any direction you want, and the engine's not going to go lean. At least, not until the free-floating pockets of air in the tank wander along and find the fuel pickup...

As for the elevator going up.. the faster it accelerates upwards, the leaner the engine's going to get when you point the nose up. If it isn't accelerating at all (i.e. going up at constant velocity), it'll be just like it would be on stable ground, when you tilt the nose up.

Yes Len you are right, what you say is exactly my point 2/ when lift starts to move – it pulls for example 10% more than weight of lift + your body + model to start moving up. If it once moves, it is my point 1/ when the engine of lift pulls exactly the weight of lift + you + model.

When it starts to move, the fuel pressure is affected 10% more than in constant speed or if standing on one place – means then the gravity is. But always, the pressure is linear to engine of lift pull. Does not matter what the gravity is, or if you are falling down or accelerating up.

I do not say there is NO gravity I say that if the engine pulls somehow, the fuel pressure is some, but if you change gravity (do not ask me how) and you do not change engine pull; the pressure is still the same. …. Even I agree that the engine pull will probably change, but still – the pressure is linear to PULL.

One example: imagine a rocket. Its engine pulls exactly the weight of the rocket on ground and let us assume, that it does not change mass of fuel with burning (it is not true, but let us expect it). So if you launch it, it will hang on one place.

You will say that the pressure in tank is still the same, because of gravity. OK I will not try to argue against. It is really exactly like on ground.

But now - imagine you give it very small short impulse up, so the rocket will slowly move up by constant speed. The engine pull is still constant, the speed is constant so the pressure will be still the same. OK.

But now – as rocket slowly but safely goes out of Earth, the gravity is descending. But the engine still pulls the same way. Do you still think that the pressure in tank is lower because the gravity is lower? I say not, because the engine pull is still the same.

The smaller gravity just allows better mechanical acceleration, but inside the rocket you feel still the same force as you would expect from gravity – even far away from earth even out of solar system, and INDEPENDENTLY from real actual gravity.

I think the original point of this thread was to say that gravity or acceleration affects the engine run. When the airplane is pointed up, then the tank drops below the engine and the fuel must be "lifted" by that amount (3 inches, 6 inches, whatever) and the engine will lean out. Brad was seeking an answer to that problem with his 4-stroke and I agree that this is undoubtedly the major problem that he is experiencing with his 4-stroke engine.

If the airplane is accelerating while in the climb, then acceleration adds to the problem, but the fuel still must be "lifted" that same 3 inches, 6 inches, whatever, and it is now lifting against both gravity and the additional affects of acceleration. This could happen. But gravity is still there.

If the airplane is decelerating (not braking, just decelerating because the added "weight" of gravity now demands more thrust than the engine is presently handing out so that the airplane is slowing down) then both the airplane and the fuel will be decelerating at the same rate and the fuel will still need to be lifted the same 3 inches, 6 inches, whatever. This could happen. But gravity is still there.

If, somehow, we could cause the airplane to brake on the way up--throw out some drag flaps or something--then the airplane would slow while the fuel did not slow to the same amunt. This does not happen while the airplane is moving straight up. But if it did, then, and only then, would the effects of gravity be reduced (not necessarily eliminated). See my related comments concerning flying into the wind below.

The above illustration of braking does happen (in a sense) when the airplane turns a corner. Then there is an increase in drag which acts as a braking force. However, it is a braking force only in the direction that the plane was previously traveling, not in the new direction. So, as the airplane rotates from level flight to a climb, for instance, there is a "braking force" in the previous direction of level flight and the fuel will want to continue moving in that direction. But there will be no affect on the force of gravity in the new direction as the level of fuel is lowered below the engine (3 inches, 6 inches, whatever). This could happen, but, again, the effects of gravity are still there.

>Yes Len you are right, what you say is exactly my point 2/
>when lift starts to move – it pulls for example 10% more
>than weight of lift + your body + model to start moving up.
>If it once moves, it is my point 1/ when the engine of lift
>pulls exactly the weight of lift + you + model.
>
>When it starts to move, the fuel pressure is affected 10%
>more than in constant speed or if standing on one place –
>means then the gravity is. But always, the pressure is
>linear to engine of lift pull. Does not matter what the
>gravity is, or if you are falling down or accelerating up.

Here, I think, is where we may be missing the point (I am not understanging your point and I have a feeling you are not understanding mine. So please bear with me while I simplify.)

The engine knows nothing of the airplane behind or beneath it (as in the case of a climb). It only knows the speed of the air through which it moves. And this will change with the amount of drag that the airplane is producing, the effects of gravity, and the amount of thrust that the engine is capable of producing. If the engine's speed increases, then it can carry more weight, or more drag and maintain the same (or similar) flying speed. If the engine's speed does not increase (or increase sufficiently) then with added weight or added drag the airplane will slow down and the engine will sense a slower airspeed. But the affects of "load" on the engine are felt only because of the speed of the air through which it is passing and it knows nothing of the airplane that may be causing this airspeed to decrease. The same thing happens when we fly into or out of the wind in our tethered circle. The engine knows only the speed of the air passing through it, and nothing of the cause of that speed increase or decrease.

And, by the way, on the same subject, we all know of the engine richening when we fly into the wind. The real problem here happens not just from a normal "slowing down" of the airplane as happens when we turn a corner and start off in a new direction. (Level flight to flying up, for instance)

The problem is that we are flying, say, at 50 miles per hour, and the prop is turning at perhaps 55 miles per hour (10% difference, or "slippage") in order to generate sufficient thrust to overcome the drag of the airplane (drag and thrust produced are equal.) If the airplane and engine are thus flying through this air at 50 miles per hour, and it turns into a 20 mile per hour head wind, the air the engine now sees is moving 70 miles per hour while the engine still is only turning 55. The result is that the air speed alone is capable of turning the prop (which is now acting as a brake) and the engine is totally unloaded. In this case we do have a case where the airplane is slowing and the fuel is moving forward, causing the engine to richen and worsening the situation. It still produces a fuel problem that a constantly regulated source of fuel would help, but this is totally unrelated to the present discussion on gravity.

Now, just as I said the engine is totally unaware of the airplane behind or beneath it and the load on the engine is related to air speed only, so the fuel inside the tank is totally unaware of what the engine is doing. It is affected only by acceleration, deceleration, gravity, and suction (which affects only that portion being pulled "up" the feed tube).

We agree, I think, on the affects of acceleratin and deceleration.

If the airplane is flying at a constant speed while climbing, then there is no difference from what happens when the plane is at rest. Point the nose up, and it leans. It is the effect of gravity. (See my elevator comments in reply 129).

If the airplane is accelerating while climbing, then the engine must draw fuel against both the force of gravity and the force (or effects) of the acceleration. In effect the engine is moving away from the fuel. So here it is acceleration and gravity that would be the source causes of (even greater) leaning. (Whip up in the wind?)

Now if the airplane is slowing down because of insufficient thrust to maintainn the speed up hill, then the fuel knows nothing of what the engine is doing. The airplane slows down simply because of the added effects of gravity and a lack of sufficient thrust to maintin the speed. And the fuel in the airplane slows down along with it. And, here is my point, the fuel slows down at the same rate as the airplane. There is no braking effect here, just a normal increase in drag on both the plane, engine, and fuel because of the increased effect of gravity (the airplane is moving away from it) and insufficient thrust to maintain the former speed. At some point thrust and drag will equalize again, but it makes no difference. The same forces of gravity are acting on all portions of the equation, and there is no leaning effect other than by gravity alone. (And this does not count the slight increase in load on the prop which is independent of this point in the argument.)

>I do not say there is NO gravity I say that if the engine
>pulls somehow, the fuel pressure is some, but if you change
>gravity (do not ask me how) and you do not change engine
>pull; the pressure is still the same. …. Even I agree that
>the engine pull will probably change, but still – the
>pressure is linear to PULL.

You keep mentioning pressure. I keep mentioning gravity (along with acceleration or deceleration which are independent but adding or subtracting from each other). Are we using different terms or are we talking different points?

>One example: imagine a rocket. Its engine pulls exactly the
>weight of the rocket on ground and let us assume, that it
>does not change mass of fuel with burning (it is not true,
>but let us expect it). So if you launch it, it will hang on
>one place.
>
>You will say that the pressure in tank is still the same,
>because of gravity. OK I will not try to argue against. It
>is really exactly like on ground.
>
>But now - imagine you give it very small short impulse up,
>so the rocket will slowly move up by constant speed. The
>engine pull is still constant, the speed is constant so the
>pressure will be still the same. OK.

Here we are in agreement and I understand your terms. It is the same as with the engine and its fuel component. As you point the nose of the plane up, you lower the level of the fuel and it causes leaning of the engine. True whether at rest or in motion.

>But now – as rocket slowly but safely goes out of Earth, the
>gravity is descending. But the engine still pulls the same
>way. Do you still think that the pressure in tank is lower
>because the gravity is lower? I say not, because the engine
>pull is still the same.

OK, now I know we are talking about two different things. I am talking about gravity affecting the engine run, that when you lower the tank by 3 inches, 6 inches, whatever, and the engine has to draw fuel from that level, that it causes the engine to lean. If we were to do this same experiment inside this rocket that you have mentioned instead of the elevator that I used earlier, then there were be a definite change because of gravity and the engine would gradually richen as it climbed because the fuel draw would become easier. So the fuel draw because of gravity is there, unless we repeal the law of gravity (as in taking us into outer space). Fortunately, we do not fly that high with our airplanes.

>The smaller gravity just allows better mechanical
>acceleration, but inside the rocket you feel still the same
>force as you would expect from gravity – even far away from
>earth even out of solar system, and INDEPENDENTLY from real
>actual gravity.
>
>igor

Here I think you missed the point, Igor. Here I believe we would suddenly find ourselves "weightless" even though the rocket was moving. Yes, we would be moving at the same speed as the rocket, but there would be nothing pulling us in any direction--unless the rocket suddenly slowed or accelerated.

And that is my point. I accept the affects of acceleration and deceleration on the fuel and the fuel draw to the engine. But we cannot eliminate the continual effects of gravity unless we take our planes into outer space (or watch Star Trek with their "artificial gravity" Hey, maybe that is a good idea. that will give us a break from this discussion).

As always, thanks.

I GOT IT ALREADY!!!

You say:

>>>Here I think you missed the point, Igor. Here I believe we would suddenly find ourselves "weightless" even though the rocket was moving. Yes, we would be moving at the same speed as the rocket, but there would be nothing pulling us in any direction--unless the rocket suddenly slowed or accelerated.<<<

You are NOT weightless. You DO feel weight. The weight exactly as the weight on Earth in gravity field in static position. It is so not because >>> the rocket was moving<<<, it is because of engine still pulling.

The Earth acceleration is 10m/s/s. Until rocket hangs close to ground, you feel that as gravity. If the rocket is far away from Earth where the gravity has only its half: 5 m/s/s, its effect on you inside the rocket is only half, but your engine is pulling the same way so its excessive pull mechanically accelerates the rocket by that missing half of gravity – it is 5m/s/s. It is mechanical acceleration, which adds its effect to the gravity, so you have still the same weight or better said force pressing you to bottom of rocket, so you still feel the same weight.

If the rocket is completely out of gravity, nothing brakes you, engine is pulling still the same and all its force is converted to the mechanical acceleration. You do not have gravity effect, but mechanical acceleration is exactly 10m/s/s and that makes exactly effect of gravity, which you can feel inside the rocked.

, (snip) so you still feel the same weight.
>
>If the rocket is completely out of gravity, nothing brakes
>you, engine is pulling still the same and all its force is
>converted to the mechanical acceleration. You do not have
>gravity effect, but mechanical acceleration is exactly
>10m/s/s and that makes exactly effect of gravity, which you
>can feel inside the rocked.
>
>
>
>igor

OK, Igor, you are talking acceleration. I missed that. You started out saying constant speed. (It is that "term" thing again.) Here is how you began.

>But now - imagine you give it very small short impulse up,
>so the rocket will slowly move up by constant speed. The
>engine pull is still constant, the speed is constant so the >pressure will be still the same. OK.

You said, "constant speed". That was what I was going by. Somehow, I guess, you moved from constant speed to acceleration. I was still looking at the hypothetical "constant speed" because that is what we are striving for in our airplane.

>But now – as rocket slowly but safely goes out of Earth, the >gravity is descending. But the engine still pulls the same way.
>Do you still think that the pressure in tank is lower because
>the gravity is lower? I say not, because the engine pull is
>still the same.

Here is where you sneaked that acceleration thing in again. Nobody (certainly not me) is denying that acceleration forces are felt, and that the effect is very similar to gravity. If the rocket accelerates, fine, then that force will be there. If not, and you are out of the gratitational pull, then you will feel nothing. And if you are in the gravitational pull, then you will feel gravity.

But what does all this have to do with the argument? We grant that acceleration is a force. But we are not constantly accelerating with our airplanes. If we are, then acceleration adds to gravity and deceleration, if it is a braking force other than gravity, subtracts. If it is just gravity that causes it to decelerate, then no. Gravity is still there in all its full force.

I do not understand what all of the rest of this has to do with it. Does gravity still exist with our airplanes? Yes. Does it affect the run of the engine when we point the nose up? Yes. Does it do so in flight? Yes. Other forces exist. I grant that. But this is the only one we were talking about.

>The smaller gravity just allows better mechanical acceleration,
>but inside the rocket you feel still the same force as you would >expect from gravity – even far away from earth even out of solar >system, and INDEPENDENTLY from real actual gravity.

If you could keep accelerating forever, that would be true. But at some point acceleration stops. And since we do not escape the gravitational pull with our airplanes, if the airplane in exerting a constant speed in the climb, then it is all gravity that will affect the engine run. If the airplane is accelerating, then it is gravity and acceleration together. And if the plane is slowing down because of insufficient thrust to overcome the added pull of gravity, then it is all gravity again. Whatever else you want to add, subtract, multiply or divide is meaningless to the argument concerning gravity because gravity is still there. And if we have lowered the tank below the engine, the effects of gravity are going to cause the engine to run leaner (as will acceleration, but that would be in addition to gravity).

Len,

In that example with rocket - do not you see that the force inside is not changing? Even the gravity changes or goes away at all?

I do not say anything else, only that the pressure in liquid column is

height * density * pull / mass

and thus determining values for the pressure is height and pull, because density and mass is constant.

I do not say anything else, like effect to the speed and effect of speed to the pull. I did not say that gravity disapear or that has no effect to the model and its speed. I fully agree that it makes effect to speed, it is counter force to prop pull (and then somehow mixed to pull thanx action and reaction), the speed has effect to prop AoA and so and so and so … I thing you are arguing to something what I did not say.

You say that if you fly at constant speed then the pressure is like on ground in your hands, I completely agree and it is completely compatible with what I say – just ask yourself:

Isn’t the thrust in that moment equivalent to the weight of model?

Isn’t the pull/mass component equal to the G in that case?

So is that equation compatible what you say?

And isn’t that constant speed dependent on that pull?

If you answered everytime yes, then we mean the same

LAST EDITED ON Dec-03-03 AT 10:17 AM (CST)
 
>>The smaller gravity just allows better mechanical
>>acceleration, but inside the rocket you feel still the same
>>force as you would expect from gravity – even far away from
>>earth even out of solar system, and INDEPENDENTLY from real
>>actual gravity.
>>
>>igor

>Here I think you missed the point, Igor. Here I believe we would suddenly find ourselves
>"weightless" even though the rocket was moving.

Because of decreasing gravity the constant thrust accelerates the rocket with increased acceleration. Inside the rocket, occupants cannot tell the difference between forces arising from gravity and acceleration. Their sum does not change. Using Newton's second law, with...

a = acceleration of rocket
g = acceleration of gravity (positive numerical value only here)
m = mass of rocket, occupants, and hypothetically undiminished fuel (constant sum)
F = upward force = thrust (hypothetically unchanging)

Then, since thrust = force to accelerate plus weight, and Weight = mg...

1) F = ma + mg
2) F = m(a + g)
3) a + g = F/m (constant)
4) a = F/m - g (a increases as g decreases)

From #4, ultimately "a" will equal the original acceleration of gravity when g has diminished to zero. The rocket accelerates its occupants, who must feel the force causing their acceleration. Even after g has disappeared, they will feel a force equal to their own earthbound weights.

SK

P.S. I hope the "rocket scientists" will forgive my choice of a positive "g", rather than just setting up a vector equation like "ma = F + mg", with "g" negative. 'thought it would be clearer this way.

P.P.S. Edit: I guess Igor typed faster than I.

Hej, I like the point 3:

3) a + g = F/m (constant)

It clearly says that whatever gravity is, at constant force there must be another replacing acceleration. Or that what evever gravity is, you feel the same force inside. By other word the determinig is thrust, while gravity has no real effect to forces inside That is it.

BTW, I am laughing

I know Einstein did lot of such examples in his work, but I did not knew he did exactly this one, much earlier then we did it – it is from his book from 1920

http://www.bartleby.com/173/20.html

Yes, I remember reading his little book *Relativity* and other examples of his "thought experiments" - great stuff.

SK

This is response to Doug’s #122
>>>But what I was wondering is a constant speed what we really want? <<<
It was here several times, I personally think little loosing is not bad, but I wanted something else what is related here:

>>>Maybe so but I am thinking of a vert 8 in some pretty strong winds and that little boost is pretty nice and the slowing effect on the way down seems to help kill wind up.<<<
If you believe me that leaning is caused by prop pull, than the answer is that it does not help too much in wind. If you fly against the wind (in level just before upwind side, or over head in uphill parts like first leg of hourglass), the prop pull descends because of unloading (upwind, AoA and so) and thus it does not make leaning effect.

And there is another problem, if you count with leaning too much, and leaning is by pull, then you see what is happening on top of loops, engine does not switch from 4 to 2 cycling immediately when you point nose up, it must first little slow down, only then comes boost and that is happening also on end of it – switching 2 to 4, it happens later than on top of loop, because it must little overspeed, only then it unloads and richen back. That is reason for that 2cycling on top or even after the top.

But anyway that is good argument for that several time mentioned small model. Just imagine: smaller model with smaller engine running close its max (relatively lean – only slightly rich – just not to sag) which does not accelerate too much, flying on shorter lines giving good feel and controllability AND keeping figures smaller and thus not giving too much speed difference even with stronger deceleration (smaller distance at the same decceleartion -> less absolute speed difference) – AND thanx only little leaning / richening better performing against the wind. It really makes sense in my eyes. I always feel advantage in strong wind with my smaller .46 piped model. I believe it will be at least a fraction of success of .40VF.

>>>>
>We have a liquid column. The height is H, the density is R,
>the gravity acceleration is G and column is fixed on one
>place on ground.
>And now, my question is, what is the pressure in the same
>column if it is in M heavy vertically flying vehicle, if
>engine thrust is T drag of vehicle is D.
><<<
>
>I see no one want to post the solution so I do it, I hope it
>will say what I mean:

I would've answered sooner, I wasn't online at the time.. ^_^;;

>1/ I would like to go the way my opponents:
>If the wehicle will be static on one place, there will be
>static pressure coming from gravity:
>
>P = H * R * G
>
>2/ But model moves, the acceleration / decceleartion
>changes the pressure also, so let us analyze speed and
>acceleration:
>
>The force acting to the model is:
>
>F = Fthrust - Fgravity
>
>Fthrust = T – D and Fgravity = M * G
>
>So the force acting to the model is:
>
>F = T – D - M * G

Interesting.. if you have downward flight,

F = T-D + MG

>So thanx Newton’s laws the acceleration is:
>
>a = F / M = (T – D - M * G) / M = (T – D) / M - G

This applies to upward flight only.

a = F/M = (T - D + M*G)/M = ((T-D)/M) + G for downward flight.

And

a = T-D/M for level flight.

>quite clear I think, so thus the pressure difference thanx
>acceleration is:
>
>P = H * R * a = H * R * ( (T – D) / M – G) =
>= H * R * (T – D) / M – H * R * G

Let's rewrite this: P=HR((T-D)/M)-HRG for upward flight

>So now we know acceleration of the model movement and so we
>can apply it to the column also.

Also, since this applies to upward flight, it doesn't apply to downward flight.

P = H*R*a = H*R*(((T-D)/M)+G) = HR((T-D)/M) + HRG

for downward flight.

>Now take both components together and the residual pressure
>is:
>
>P = H * R * G + H * R * (T – D) / M – H * R * G =
>= H * R * (T – D) / M

Ah.. I see what you're doing. The pressure that's attributable to gravity is

P1=HRG

and that attributable to acceleration of the airplane is

P2=HR((T-D)/M) - HRG

and so P1+P2 = HR((T-D)/M) because +HRG and -HRG cancel out.

>If you take closer look, the (T – D) / M is exactly the
>acceleration of the engine thrust minus drag on the model,
>so it clearly shows that the pressure depends ONLY on engine
>thrust and is independent on gravity:
>
>P = H * R * pull / M
>
>OK?

OK, and in downward flight, P1 is now in the opposite direction so you end up with the same results - the two HRG terms still cancel out.

The MAIN problem with this analysis is that, in Stunt planes, thrust is NOT the same going up and going down. Going up, thrust has to at least equal D + MG to maintain the same upwards velocity. OK, maybe in many cases it doesn't quite equal D + MG, in which case the plane slows down going up. Going down, it has to be equal to D - MG. If it exceeds this, it speeds up going downhill. On a slick, heavy model, thrust would even have to be negative (and this is achieved through using a low pitch propeller and an engine on the back side of the torque curve).

As I've said before, we're trying to determine whether flying nose up or nose down will affect fuel pressure at the needle. The simplest model to calculate that has some semblance to reality in the Stunt situation would be a constant speed setup - i.e. NO acceleration going up or going down, in which case the difference in fuel pressure between up and down legs is on the order of 2G*RH.

Nice, so now you agree, that the pressure is linear to the thrust, and I fully agree that the thrust is related to speed and the speed if affected by gravity.

So you are right if you say that if the engine maintain perfectly constant speed, then the fuel pressure is exactly like on ground nose up or nose down. And I note that also prop thrust makes that thrust exactly canceling gravity (may be with drag somehow mixed to the thrust).

What I wrote in original message was, that we cannot secure that perfect stable speed, and if the engine does not pull itself, the leaning does not come and does not help to accelerate – and – if the engine sags, the leaning is cancelled.

Are we on one wave now?

>Nice, so now you agree, that the pressure is linear to the
>thrust, and I fully agree that the thrust is related to
>speed and the speed if affected by gravity.
>
>So you are right if you say that if the engine maintain
>perfectly constant speed, then the fuel pressure is exactly
>like on ground nose up or nose down. And I note that also
>prop thrust makes that thrust exactly canceling gravity (may
>be with drag somehow mixed to the thrust).

>
>igor

I think this is where we really differ in terminology, Igor. As I understand you, you are saying the effects of gravity are canceled because the body is in motion. (It is this thrust equals drag thing.) But I say I still feel the gravity, even when I am in an elevator because it is still holding me to the floor. And this is what is happening to our fuel. Even though the fuel is being put into motion, the force of gravity is still "pulling" on it, and the effects remain the same on the fuel draw whether in flight or on the ground. Check my reply number 135.

Igor wrote:
>So you are right if you say that if the engine maintain
>perfectly constant speed, then the fuel pressure is exactly
>like on ground nose up or nose down. And I note that also
>prop thrust makes that thrust exactly canceling gravity (may
>be with drag somehow mixed to the thrust).
>
>snip
>
>Are we on one wave now?
>
>
>
>igor

Still don't think so.

Per your statement above, assume a constant speed of 0 (zero, also T=D here). Your formula would then say that plane level, plane nose up, or plane nose down would all be the same. I don't agree. They would all be different by gravity, + on one, nothing on two and - on three. The gravity term must remain in your equation as you indicated by a previous formula: P= H * R * (T – D) / M – H * R * G

P= H * R * <((T-D)/M)-G> (this is in a nose up/down orientation)

T-D accounts for the acceleration the plane sees due to imbalances in a steady state condition (where a=0), and the -G term account for the ever present gravity we live in. Applying the above equation to the standing on ground scenario above (where V=0), level on ground, delta P =0 as H=0, nose up, delta P = -H*R*G, and nose down, delta P= H*R*G.

This formula does indicate that nose up is leaner and nose down is richer.

LAST EDITED ON Dec-03-03 AT 02:15 PM (CST)
 
>>>Per your statement above, assume a constant speed of 0 (zero, also T=D here).<<<

My formula does not count with SPEED itself, only with acceleration ( and therefore force – either thrust or drag). But I say the pressure is invariant with speed so I agree, we can presume the speed is zero. In reality does not matter what is speed. Only acceleration counts.

>>>Your formula would then say that plane level, plane nose up, or plane nose down would all be the same.<<<

Definitely. If speed is zero, also the drag is zero and thus also thrust is zero – that was your condition to the example. In gravity field it is begin of free fall. And in free fall does not matter what is direction of nose – up, down or horizontal – every time the result pressure is ZERO.

>>>I don't agree.<<<

I am sorry

I haven't read all the above, but here's what I think happens:
Suppose a model is cruising along level in equilibrium (T=D)at 60 mph. Let D=D1+D2, where D1 is the drag the airplane would have at 60 mph if it didn't have to make any lift and D2 is the component of drag, induced and other, that it takes to make 1 g of lift in level flight. D1 is a whole lot bigger than D2. T=D1+D2. Now suppose it starts into a wingover. To simplify the problem, suppose that when it finishes the corner, it is still going 60 mph. Now it's pointed straight up, going 60 mph: the initial condition. Gravity has changed direction in the airplane axis system by 90 degrees and lift is zero. This causes a net force acting in the direction opposite that of the airplane's velocity: (mg + D1) > (T + D2). So the airplane will initially accelerate backward at approximately 1g. A longitudinal accelerometer on the airplane would read close to zero. As a consequence of accelerating backward, the airplane will slow down. As the airplane slows down, propeller angle of attack increases, so there's more thrust. There is also less drag, which is pretty much proportional to the square of airspeed. The airplane fairly rapidly reaches a new equilibrium, where new thrust = weight + new drag. It is no longer accelerating backward, so an onboard longitudinal accelerometer would read 1g toward the rear (x the cosine of whatever angle it's gone through since it started the wingover). A typical stunt engine fuel system would run leaner with the weight of the fuel pulling it away from the engine. This effect usually causes more power, hence more thrust, so the new equilibrium speed is not quite as low as it would have been at constant engine power.

I was pretty sloppy with vectors and scalars above. If it's not clear, assume D2=0, which is as reasonable an assumption as the one that has the airplane losing no speed in the entry corner. I'll bet that the airspeed the airplane has as it comes out of the corner is pretty close to the equilibrium speed going uphill. Hence, the full force of gravity acts on the fuel as soon as the airplane is going straight up.

It would be fun to simulate this. I have prop performance data for APC 9-4s and 9-5s, and Preston has inflight RPM data for The Bomber, so I should be able to make a fairly good simulation. I don't have time to do that now, and I hope Preston doesn't read this, lest we go off on yet another tangent.

Like Howard, I have not read the whole thread, but here is my 2¢.
The longitudinal acceleration due to more drag is not the answer, just a distraction. There is a steady state explaination:
In the case of a 10G loop I would imagine the angle of attack would be at least 5 degrees. Lift is
perpendicular to direction of motion, not perpendicular to the chord line of the wing. This would
result in an acceleration component of 10*sin(5)= .87g longitudinally. Pretty close to the extra .87g
you get putting the nose up 60 degrees when you are on the ground. When going straight up in the
loop 1.87g affecting the fuel head, down would be -.13g. Leaner going up, richer coming downhill.
A hard corner (even the first corner of the outside) would always have the tendency to reduce
pressure at the venturi.
I am pretty convinced that it is the change in fuel head as a result of angle of attack in
manoeuvres that causes the 4-2 break, not the change in load.

Pat MacKenzie

P.S. the 10g loop at 5 degrees AOA is a guess, but a reasonable one I think. Perhaps someone could
give more accurate estimates. Oh and Leonard, there is no such thing as centrifugal force. It is a
fiction, though a very convenient one that I will let slip out if I am not being careful! There is only
centripetal acceleration.

LAST EDITED ON Dec-04-03 AT 09:14 AM (CST)
 
Pat-

>Oh and Leonard, there is no such thing as centrifugal force. It is a fiction, though a very
>convenient one that I will let slip out if I am not being careful! There is only centripetal
>acceleration.

I hope I'm not starting last year's argument all over again, BUT...I have to disagree. It is "fictional" only if the term is applied wrong - as in applying it to the wrong object, confusing it with centripetal force or inertial forces exerted in turn by objects being accelerated by the centripetally accelerated object - or when it is used conveniently to name "outwardly" directed forces in a centripetally accelerated reference frame. It is IMO sensibly defined in Halliday and Resnick, my old Handbook of Chemistry and Physics, and several other reputable texts as the reaction force on the object exerting the centripetal force. For instance, in a CL application, centrifugal force is the outwardly directed force exerted on the pilot at the center of the flight circle (via lines) by the model plane, while the centripetal force is the force exerted by the pilot (through the lines) on the model plane to keep it accelerating centripetally in its circular path. Pilot and plane, as "sources" of these forces, are a "reaction pair", in accordance with Newton's Third Law of Motion. Since the plane does pull radially outward on the pilot, I believe that these authors are sensibly applying the word "centrifugal". There can be "centrifugal" acceleration of the pilot, if the model is massive or fast enough to dislodge him/her.

I'm hoping that last year's debate doesn't get repeated, but if it does, it should be on a new thread, where preferably, the participants actually read each others' posts before replying - just my preference. I won't start such a thread.

SK

Edit: I see that I made one mistake writing this late at night. The "inertial forces exerted in turn by objects being accelerated by the centripetally accelerated object" are indeed "centrifugal" forces, since the "centripetally accelerated object" has come in its turn to exert a centripetal force on, say, its contents, which in turn react outwardly against it.

I agree with your example, but the flame was about another thing. And that is if the gravity effect in fuel pressure depeneds only on prop thrust or there is gravity on top of that. But you wrote:

>>> Now it's pointed straight up, going 60 mph: the initial condition. Gravity has changed direction in the airplane axis system by 90 degrees and lift is zero. This causes a net force acting in the direction opposite that of the airplane's velocity: (mg + D1) > (T + D2). So the airplane will initially accelerate backward at approximately 1g. A longitudinal accelerometer on the airplane would read close to zero.<<<

And that is what I say. You do not feel the gravity if you do not pull agains it. That is all.

But another thing:
You say that the stable speed is reached early. I do not think so (at least not “early”).

Look:
I presume that the drag is too low to change the thing too much (it is not, and it will make my derivation worse, but it can be 5N compared to 15N of model weight so it is really not so big). So I presume it is zero, it will make thing better to understand. I assume that I have constant RPM engine and I also borrow Lens idea of perfect prop with linear response of thrust to speed. We are not far from that with our low pitch props. Now if you have very strong engine giving thrust 2 * model weight, the constant speed will be at half of level speed (I know, the drag little deforms it). That is because the prop pull is equal zero in level normal flight and it’s pull is 2*weight at 0 speed. Thus the half speed gives exactly pull of the model and that is at balance, but it is too slow for us.

And another point of view: If you fly 20m/s and gravity acceleration is 10m/s/s, and the square loop vertical leg takes say 0.5s, then you lose only 1/4 of normal speed, and thus I think you do upper corner BEFORE stabilizing. (it is without engine trust!!!)

So I think the only question is what Len already mentioned before also, and it is losing of speed in corner. As far as I know from my sheets the drag is somewhere at 15N and it is close to model weight and thus similar to gravity effect in vertical flight, so in reality the slowing down starts BEFORE pointing nose up, but even if it takes full second, it still loses only 10m/s – it is half of normal speed. So I really think that the model slows down all the way in vertical parts – at least little bit – and so I think it never gets FULL ground leaning.

AND if the engine sags then it is really surviving factor for lean engine, because it will not lean out so much.

>>>>Your formula would then say that plane level, plane nose up, or plane nose down would all be the same.<<<
>
>Definitely. If speed is zero, also the drag is zero and thus
>also thrust is zero – that was your condition to the
>example. In gravity field it is begin of free fall. And in
>free fall does not matter what is direction of nose – up,
>down or horizontal – every time the result pressure is ZERO.

Hmmmmmmmm.... true, very true. And very insightful.

And also, speed could be zero, but with thrust = g. With the nose up, this would hold the airplane in one position, and the acceleration on the fuel is g. Which also is equal to the thrust. Nose down, you'd have to have a reverse pitch prop providing a thrust of -g, which also leads to acceleration on the fuel of -1g.

:-)

or:

Nose down, still the same thrust of +1g, it accelerates down with 2g but it also leads to acceleration on the fuel of +!!!1g even nose down.

LAST EDITED ON Dec-04-03 AT 08:05 AM (CST)
 
See...

Wouldn't it be cool to just ignore the fuel head question?

Can we all agree the fuel head is changing? Which is (in turn) affecting the power curve? Can ANYONE agree that there is no true way to follow a power ALONE as long as the mixture at the venturi is constently changing?

I do agree that fuel head can change, and I'm quite sure it affects the power output of the engine, but I'm not convinced it necessarily makes the engine jump off the power/torque vs. RPM curve.

>I do agree that fuel head can change, and I'm quite sure it
>affects the power output of the engine, but I'm not
>convinced it necessarily makes the engine jump off the
>power/torque vs. RPM curve.

All power curves (at least the ones generated by Frank Williams) are generated at a constant mixture, with a constant fuel head. Any change in the mixture will change the curve. So, in effect, you have a new curve. Simple.

You can't generate entire torque-RPM curves on one "mixture" (i.e. needle setting), because you need a bunch of data points at different RPMs. You can't get get this using one prop unless you a) play with the throttle or b) change the mixture, so you need to use a variety of props. If you use prop A, and adjust the mixture for maximum available power, you'll get one RPM and torque reading. To get a second one, you change to prop B. Prop B is larger than prop A, so when you start the engine, you'll find it going lean and sagging, unless you open up the needle to give it more fuel. So by definition, you've already changed the "mixture" (it's running fewer RPM, and needs more fuel, so there's more fuel going in with less air, and it's by definition richer).

>You can't generate entire torque-RPM curves on one "mixture"
>(i.e. needle setting), because you need a bunch of data
>points at different RPMs. You can't get get this using one
>prop unless you a) play with the throttle or b) change the
>mixture, so you need to use a variety of props. If you use
>prop A, and adjust the mixture for maximum available power,
>you'll get one RPM and torque reading. To get a second one,
>you change to prop B. Prop B is larger than prop A, so when
>you start the engine, you'll find it going lean and sagging,
>unless you open up the needle to give it more fuel. So by
>definition, you've already changed the "mixture" (it's
>running fewer RPM, and needs more fuel, so there's more fuel
>going in with less air, and it's by definition richer).

That is not how Frank does it.

What if you did it with a real dyno? Real dynos don't go through all that. I believe Frank described a brake dyno that got the whole curve in about 5 seconds.

I just can't believe that we are discussing that it would be good practice to change the carb mixture in the middle of a dyno test. To suggest such a thing on a car engine would be ridiculous. I guess in stunt terms it is somehow reasonable.

If you change the mixture the output changes. Typically you will lose RPM with any less than perfect mixture.