Scott Bair cylinder pressure data

LAST EDITED ON Feb-28-03 AT 10:28 PM (CDT)
 
Here are several plots from Scott Bair's report.
First is the cylinder pressure at minimal 2-stroking (just over the line to a constant 2-stroke). Note that it's firing every time, but the pressure only averages about 300 psi.

[photo not recovered: 3e60337c749d60e6.jpg]

............................Figure 1. Cylinder pressure while 2-stroking

Second is cylinder pressure during 4-stroking operation. Note that "non-firing" strokes are only about 200 psi, and the "firing" strokes are about 450.

[photo not recovered: 3e6033ca74ce71d8.jpg]

............................Figure 2. Cylinder pressure when 4-stroking

Third is the cylinder pressure coasting down after pinching off the fuel line and making the engine quit. Note that the pressure is about 200 psi, with is about what you get from simply multiplying the compression ratio x atmospheric pressure. Note also that the it's the same as the "non-firing" stroke in the first picture.

[photo not recovered: 3e60340874e21591.jpg]

............................Figure 3. Coasting cylinder pressure

These, put together, seem to conclusively prove the "non-firing" stroke is truly not firing anything. It also explains why you get very little power change when you go from 4 to 2.

Stock ST46.

Brett,
The information looks great. Any speculation about why the 4 cycle has higher compression (more fuel ignition?) when it does fire on every other stroke? How about the perceived increase in power as we go into a 2 cycle, or is it an increased power output related only to the increase in engine RPM?

LAST EDITED ON Mar-01-03 AT 00:33 AM (CDT)
 
>Brett,
> The information looks great. Any speculation about why
>the 4 cycle has higher compression (more fuel ignition?)
>when it does fire on every other stroke? How about the
>perceived increase in power as we go into a 2 cycle, or is
>it an increased power output related only to the increase in
>engine RPM?

It's not really more compression, of course - just higher *pressure* due to a more powerful firing.

Quoting the report (which I may just turn into a post, now that I know how to shrink the illustrations):

To summarize, when a stunt engine is four-cycling there is a misfire on alternate revolutions. However, the peak pressure when the cylinder fires while four-cycling is about 50% higher than when two-cycling. This should be interpreted as meaning that a good portion of the charge is retained (not ejected from the exhaust port) after the misfire and is added to the charge that will burn during the next revolution.

If only about half of the fuel and air charge is swept out after a misfire then it is logical that only about half of the exhaust is swept out after a successful ignition. This may be the explanation for why a two-stroke engine will four-cycle with a rich mixture. The rich mixture cools the head and glow plug and a fuel and air charge that is diluted by the exhaust that has not been swept away will not fire. But, on the next revolution more fuel and air mixture are added raising the concentration of combustible material to where ignition can occur. If the engine temperature increases, either by leaning the mixture or by loading the engine, even the diluted charge will fire and the engine two-cycles.

On re-reading, my earlier statement (in the other thread) are actually covered by Scott's report. So I had short-changed his analysis - my apologies.

The reason there's a power change is that while the pressure on the alternative firings is 50% higher, it only fires half the time. To get the same power, you would need the pressure to be *approximately* twice what it is on the two-stroke firings. I say approximately, because the timing is probably also different, meaning the transfer function from "peak pressure" to "torque" is far from 1:1 because the crank angle vs. pressure is probably not the same in both cases. That's one element in why changing the compression changes the ratio of 4 to 2 power.

It does explain why the power change can be managable - a factor of two would be vastly too much!

Brett,

Be interesting to see that same testing and data set applied to a Saito.

Hmm...

Randy

OK, I'm convinced. Can't get much clearer than this. I, too, think it's probably better exhaust scavenging than the retention of the earlier charge. Actually, come to think of it.. the part that takes the place of the unscavenged exhaust is part of the previous stroke's charge.

It's great to see some actual data, rather than idle (pun semi-intended; maybe a 4 cycling pun?) speculation. Thanks for supplying them.

I'm not surprised at all, except for the higher than normal pressures on the "power" strokes when it is in a 4-2-4.

where did you get this? Where was it published?
Dana Wall

>where did you get this? Where was it published?

It circulated in about 1980 when I understand that Scott was an engineering student at Ga. Tech. It was also a reference in Scott's "Stunfire" article.

Dixon re-published it a few years ago in Stunt News.

I would really like to see temperature and air/fuel ratio data included.

Brett, is there any information on the type of transducers used? I'm surprized at the low firing pressure, even in the 4 stroke mode. It implies we're not producing a whole lot of power. I won't dispute it, but thet little pressure seems unreal. An gasoline automobile engine runns a firing pressure in th 2000-3500psi range with compression running 200-275psi speaking very generally.

>I would really like to see temperature and air/fuel ratio
>data included.
>
>Brett, is there any information on the type of transducers
>used? I'm surprized at the low firing pressure, even in the
>4 stroke mode. It implies we're not producing a whole lot of
>power. I won't dispute it, but thet little pressure seems
>unreal. An gasoline automobile engine runns a firing
>pressure in th 2000-3500psi range with compression running
>200-275psi speaking very generally.

There was no information about the transducers or their bandwidth, but it's clear that it was sufficiently responsive - you can see the effects of the exhaust port cracking open, which suggests a pretty high resolutions.

When I calculated the torque from the pressure and the crank moment arm, and then used it to calculate the power, I got around 1/2 a HP (before internal losses), which sounds about right.

Assuming the pressure *was* much higher and it was getting filtered out by the response, you wouldn't expect the peak pressures to be all that different - in fact, the 2-stroke peaks would be expected to look higher if the transducer was acting like a lag filter.

All things considered, it's too consistent to be invalid.

LAST EDITED ON Mar-02-03 AT 01:20 PM (CDT)
 
Brett,

I have to agree with you on the consistancy, I would go nowhere near trying to call this data invalid. Its just not what I would have expected and its good to see some solid data.
A as usual, I nulls alot of the misconceptions we(I) sometimes get.

Could you tell me specifically which issue of SN this was in? I seem to remember it but I can't locate it.

Thanks,

>I would really like to see temperature and air/fuel ratio
>data included.
>
>Brett, is there any information on the type of transducers
>used? I'm surprized at the low firing pressure, even in the
>4 stroke mode. It implies we're not producing a whole lot of
>power. I won't dispute it, but thet little pressure seems
>unreal. An gasoline automobile engine runns a firing
>pressure in th 2000-3500psi range with compression running
>200-275psi speaking very generally.
>
>Randy Ryan
>AMA 8500
>SAM 36
>I fly 'em all and love it!

The pressures aren't as high in a model engine for the same reason that reynolds numbers are much lower on model airfoils. The air molecules are the same size in both instances, so the flow is not scaled to the smaller size.
Bare

>>I would really like to see temperature and air/fuel ratio
>>data included.
>>
>>Brett, is there any information on the type of transducers
>>used? I'm surprized at the low firing pressure, even in the
>>4 stroke mode. It implies we're not producing a whole lot of
>>power. I won't dispute it, but thet little pressure seems
>>unreal. An gasoline automobile engine runns a firing
>>pressure in th 2000-3500psi range with compression running
>>200-275psi speaking very generally.
>>
>>Randy Ryan
>>AMA 8500
>>SAM 36
>>I fly 'em all and love it!
>
>The pressures aren't as high in a model engine for the same
>reason that reynolds numbers are much lower on model
>airfoils. The air molecules are the same size in both
>instances, so the flow is not scaled to the smaller size.
>Bare

This is well out of my realm of "expertise" but isn't it also true that alcohol burns at a lower temp, slower and likely, therefore, produces less volume expansion when it combusts? Seems like I read somewhere one of the reasons the sparkers on gas are so much louder is in part because of these differences in combustion.

Just curious.

Ted,

What you say is true, I just thought that it was disproportionately low, methanol is after all used as a racing fuel. I have been away from home on vacation for the last week and a half, but an going to look into this when I get back to work. We have some extremely knowledgeable engine designers that have much more 2 stroke experience than I. This has certainly whetted my curiosity.

The real limit is volume of oxygen for one revolution. The methanol does not convert that limited amount to thermal energy better than petrol, but its advantage is amount of thermal energy absorbed to vaporize. It brings big advantage in fill efficiency – it mean you can burn more oxygen per one revolution just because you can keep it colder before compression.

Our engines use ignition by catalytic / thermal agent – the glow plug which is also slower than spark plugs. Small body of our engine also allows much more loses (compared to bigger engines).

But there is one very important effect. Methanol contains relatively (compared to the petrol) high amount of hydrogen. The trick is called dissociation. Oxides at some pressure and temperature (after ignition) dissociates. It absorbs thermal energy, which can be released back only at lower pressure at lower temperature – past TDC. The result is, that the pressure peak is not so high, but is wider (for longer time). It is reason of higher octane number of methanol. The dissociation is also reason of that >>>temperature of burning<<< mentioned by Ted - it is really related to shape of the pressure peak at TDC.

It is exactly the effect used on high power petrol engines with water injection at high load. The engine is designed to deliver stable power at say ¾ of throttle (high CR) – gives efficient run at moderate load. If the engine must deliver more, for short period of time, the throttle is wide open and water injection allows cooling down maximal pressure in cylinder by dissociation. Hydrogen burns back past TDC and gives that energy back. There are loses for vaporization, and not perfect burning, but power is present even at lower efficiency.

LAST EDITED ON Mar-03-03 AT 11:54 AM (CDT)
 
Alcohol has only about half ('hope I'm remembering this fraction right) the releasable energy content of gasoline. Thus, carburetors are jetted with larger holes to pass more alcohol for a richer mixture than gasoline. Methanol is used in racing engines because of its cooling properties, allowing higher compression ratios without detonation. The net result on my open class racing kart was a slight but significant increase in performance over stock configuration w/gasoline - and the need to "flush" the engine (run it a minute on gasoline) after every day of racing to minimize oxidation of the aluminum carburetor parts. When my piston-port 2-cycle engine was flushed at day's end, it ran a very rich 4-cycle, and care was taken to prevent "detonation" that would leave pock marks in the cumbustion chamber (watched CHT). This seems to indicate a faster flame front for gasoline. I'm sure someone on this forum knows the actual particulars.

SK

P.S. Sorry, Igor's superior reply was not on the board when I started mine.

>>>Alcohol has only about half ('hope I'm remembering this fraction right) the releasable energy content of gasoline.<<<

Yes, about. Exactly it is 46 000 and 29 000 kJ/kg (~0.6). But it has nothing to do with volume necessary to one cycle, or consumption. If you want to know it, you must calculate amount of fuel to burn with given amount of oxygen - or just compare atomic weight between two different fuels - this will give you weight ratio – or – you can convert it by specific weight to volume.

If you will do it, you will easily come to facts you wrote. The calorimetric volume of methanol is lower, but consumption is higher, and thus the difference of delivered energy is not so bad. Now if you improve filling efficiency of better cooling methanol you will come to little better result.

Dana: Did you get the Ringmaster plans yet?

Yes. A day or two ago. Thank you very much. Dana

>>> fuel and air charge that is diluted by the exhaust that has not been swept away will not fire. But, on the next revolution more fuel and air mixture are added raising the concentration of combustible material to where ignition can occur.<<<
This is very repetitive effect. It can be also explored on engine with connected battery on glow plug while quickly turning by hand (not flip by finger, not turned slowly, but quick and not to allow run). If the engine is rich enough and you turn it over, you will feel misfiring and firing (kick) every other compression 2 or 3 times followed (lean) by firing every till end with less and less energy.

BTW, somewhere in deep history of computer modeling, far before graphical visualization, I saw a computer model of large two-cycle spark engine for small river ships (done by friend teacher on our technical university and his students). I was surprised when a guy (r/c boat racer) told me it (the computer model) could also do fourcycling. Very important was, that the fresh fuel volume was so limited and DELAYED, that it was not able to reach the plug!

I think something similar happens also to our small engines. We know that not only timing, but also construction of intake ports can modify 4-2-4 character – especially existence of boost schnuerle port – it can be that such port is able better “deliver” also limited amount of fresh mixture to proper place – to the glow and it can be reason of different 4-2-4 character.

But Brett:
>>>The reason there's a power change is that while the pressure on the alternative firings is 50% higher, it only fires half the time.<<<
Isn’t it about TORQUE instead of POWER?

>>>To get the same power, you would need the pressure to be *approximately* twice what it is on the two-stroke firings.<<<
Yes – if you have THE SAME RPM.

>>>a factor of two would be vastly too much<<<

If the pressure peak on your example is 50% higher, we can expect 50% more torque in the first revolution. The second revolution does not bring any. So the torque of for cycling engine is 1.5/2 = 0.75 – may be less thanx friction. But engine on your example is two cycling at 9000rpm and fourcycling at 7500rpm. So the power of fourcycling engine is 0.75*7500/9000 = 0.625 of power of twocycling. Yes, I know this is only example and rpms could be probably closer, but there is still some rpm drop. So I think it is still question if >>>two is vastly too much<<<

… I think >>>two is too much<<< sounds better

>>>> fuel and air charge that is diluted by the exhaust that has not been swept away will not fire. But, on the next revolution more fuel and air mixture are added raising the concentration of combustible material to where ignition can occur.<<<
>This is very repetitive effect. It can be also explored on
>engine with connected battery on glow plug while quickly
>turning by hand (not flip by finger, not turned slowly, but
>quick and not to allow run). If the engine is rich enough
>and you turn it over, you will feel misfiring and firing
>(kick) every other compression 2 or 3 times followed (lean)
>by firing every till end with less and less energy.
>
>BTW, somewhere in deep history of computer modeling, far
>before graphical visualization, I saw a computer model of
>large two-cycle spark engine for small river ships (done by
>friend teacher on our technical university and his
>students). I was surprised when a guy (r/c boat racer) told
>me it (the computer model) could also do fourcycling. Very
>important was, that the fresh fuel volume was so limited and
>DELAYED, that it was not able to reach the plug!
>
>I think something similar happens also to our small engines.
>We know that not only timing, but also construction of
>intake ports can modify 4-2-4 character – especially
>existence of boost schnuerle port – it can be that such port
>is able better “deliver” also limited amount of fresh
>mixture to proper place – to the glow and it can be reason
>of different 4-2-4 character.
>
>
>But Brett:
>>>>The reason there's a power change is that while the pressure on the alternative firings is 50% higher, it only fires half the time.<<<
>Isn’t it about TORQUE instead of POWER?
>
>>>>To get the same power, you would need the pressure to be *approximately* twice what it is on the two-stroke firings.<<<
>Yes – if you have THE SAME RPM.

I was making the assumption that the same thing happened at closer revs - 9000 and 4-stroking, and 9000.1 and 2-stroking still had a similar pattern. That may or may not be true, but qualitatively,. I think the idea is OK.

>>> I think the idea is OK.<<<
Definitely, I just pointed the pressure condition in cylinder are about torque while power is about torque and RPM

>>>I was making the assumption that the same thing happened at closer revs - 9000 and 4-stroking, and 9000.1 and 2-stroking still had a similar pattern.<<<
This is not real example because if you have constant load or real prop with load related to RPM than the switch to 4 cycling will mean rpm and thus more power drop. This is exactly what your example shows – two cycling at 9000 and 4 cycling at 7500. If I try to calculate theoretical rpm drop, I get 1.5/2=0.75 torque drop from your estimated 50% more energy of every second cycle, than I get Sqrt(0.75)*9000 ~= 7800 rpm from rotating prop load. It is with match of your example and in range of estimated numbers.

But you are right, the engine can 2 cycle and also 4 cycle at the same rpm (also at 9000 and 9000.1 ) with matched load - or – making that 0.75 lower torque and thus also power because of the same rpm.

But the same is true for loaded engine 2 cycling at 7500 and unloaded 4 cycling at 9000 – I mean you can get more power from 4 cycling engine if you let it run at higher rpm – this is what we do with smaller low pitch props.

But I have another curiosity:

That computer model I wrote in previous message needed some external impulse to switch from 2 cycling to 4 cycling. It looks that both are stable enough not to switch so easy. The obvious impulse is misfiring. But that model was able to switch only by programmed noise source of some energy in intake, slowly but safely to 4 cycling. I mean that it looks like a small impulse to better or worse filling in one cycle can mean also little - just more - in upcoming cycle and after a number of revolutions the full 4 cycling developed. It matches your pictures – you can see that it is not absolute fire – not fire – you can see pressure peaks of different pressure from 200 to 400 PSI. I think those differences are related rather to amount of mixture than quality of ignition (in case that it fires).

This reminds me what GMA several times wrote me – it was about ports in sleeve aligned to ports in cast. He wrote me that he always got better results if the ports do NOT match each other. And best is if boost port in sleeve is lower than the port in cast – making sharp edge in path of fresh fuel. It can be source of such noise. It looks exactly like a device making noise on musical air pipes stacked to a resonation tube.

interesting thread. I wonder how it applies to my Brodak 40. It definitely LOSES power when it leans out and switches to a 2 cycle. It also starts to misfire and run erratically. But the 4 cycle is sweet and steady

>interesting thread. I wonder how it applies to my Brodak
>40. It definitely LOSES power when it leans out and
>switches to a 2 cycle. It also starts to misfire and run
>erratically. But the 4 cycle is sweet and steady

This is typically a sign of undercompression. It was certainly possible with the Big Jim ST 60 to make the 2 cycle less powerful than the 4 cycle. My guess is that the Brodak has a open hemi head and very little difference in the intake and exhaust port heights. Makes for a pretty sound but no guts.

I like to tune the compression until the 2 cycle is only a ***hair" stronger than the 4 cycle, lamost equal really. It is amazing how you can tune the switch with very thin gaskets. Nitro will also make the 2 cycle proportionally stronger than the 4 cycle. You might try a boost in nitro.

Ron B.
F4Fguy

Brett:

Re the Scott Bair data: the peak pressure is not a good indication of the torque. What you need is the mean effective pressure. To get this, you would have to have the area underneath the curve, integrate it, and find the mean. The reason two strokes have less torque than four strokes, is that the effective stroke is much shorter due to the porting. The four stroke also has much better scavenging than the two stroke, since it has an entire stroke to exhaust and an entire stroke for intake.

Everything else being equal, the 4 stroke will also have lower piston temperatures, lower cylinder temperatures, and lower charge temperature. This means that for the same compression from atmospheric the final pressure will be higher for the 4 stroke. Also as mentioned above the two-stroke has a shorter effective stroke so that for the same mechanical compression ratio, it has a lower effective compression ratio.

Alcohol fuels generally have fewer BTUs per lb than gasoline fuels but they will also run with much lower air to fuel ratios, so that the final mixture has more BTUs per lb giving more power. This is further enhanced by the fact that alcohol has a higher heat of vaporization the gasoline, cooling the mixture and making for higher density in the final mixture.

All of this is not to run down Scott's data! On the contrary! This is really wonderful work, the kind we have far too little of. As someone said earlier in this thread, we need data not guesswork, or opinion.

If you would like to read some very good texts on this subject try the following:

The Internal Combustion Engine by Taylor and Taylor

Combustion Engine Processes by Lester C.Lichty

Two-stroke Tuning by Gordon Jennings ( my Bible).

Ron B.
F4F Guy
Formerly: Director ,I.C. Engine Development and Testing
Reynolds Metals Company

Ron,

I think and I hope Brett compared ONE 2 cycle engine in different modes of work – 2 cycling and 4 cycling. It is not about 2 cycle and 4-cycle engine. I think the peak pressure is VERY good indication of the torque in such case. Being everything equal (timing, geometry, friction, ignition point …) the peak pressure, and thus area under, will be very close linearly related to the torque. Am I missing something?

BTW: Those books you wrote – are they available somewhere? Are some parts from them available online? (just to see what is it about)

Ron B.
F4Fguy

Igor:

Almost any good technical library should have the Lichty
or Taylor and Taylor or both.

The Two Stroke Tuner's Handbook is out of print and is
increasingly difficult to find.

Ron, you know, our SLOVAK technical libraries are focused to different kind of books

But thanx anyway. I will try it in Germany nexttime.

BTW does anyone know about some internet technical library having books like this one:

http://www.monmouth.com/~jsd/how/

LAST EDITED ON Mar-07-03 AT 01:11 AM (CDT)
 
>Re the Scott Bair data: the peak pressure is not a good
>indication of the torque. What you need is the mean
>effective pressure. To get this, you would have to have the
>area underneath the curve, integrate it, and find the mean.

Actually, if you have a continuous record of the pressure, you don't need the mean effective pressure. The way you calculate the torque (before internal losses) is to multiply the pressure by the piston area to get the *force* on the piston, and then by the effective moment arm (dependent on the crank angle, conrod length, and stroke). This gives a *continuous record* of the torque, instead of the overall average torque. Which was the goal in the first place. The mean pressure sort of falls out of this analysis. I'll leave the derivation to the reader for now.

Here is a MATLAB function for calculating the torque from the pressure (with data for an ST46 embedded):

***function = press2torq(crankang,pressure)
bore=.8635;
stroke=.779;
rodlength=1.338;
pistonarea=pi*(bore/2)^2;
alphacrank=crankang/57.296;

% pin offset from c/l

xpin=stroke*sin(alphacrank);

% rod angle from c/l
alpharod=asin(xpin/rodlength);
fpiston=pressure*pistonarea;
frod=fpiston*cos(alpharod);

arm=stroke*sin(alphacrank+alpharod);

torque=frod*arm;

***********

I'm sure most anyone can understand this and use it in another programming format. And, for engines that include it, add a little modification for DeSaxe offset.

I did in fact integrate the torque over time to get the average torque, and that's what I was comparing.

[photo not recovered: 3e68454c2452751a.gif]

...................... Instanteous torque, derived from pressure data

>The reason two strokes have less torque than four strokes,
>is that the effective stroke is much shorter due to the
>porting. The four stroke also has much better scavenging
>than the two stroke, since it has an entire stroke to
>exhaust and an entire stroke for intake.
>
>Everything else being equal, the 4 stroke will also have
>lower piston temperatures, lower cylinder temperatures, and
>lower charge temperature. This means that for the same
>compression from atmospheric the final pressure will be
>higher for the 4 stroke. Also as mentioned above the
>two-stroke has a shorter effective stroke so that for the
>same mechanical compression ratio, it has a lower effective
>compression ratio.

You misunderstand - these are the *SAME ENGINE*, a nominally 2-stroke engine set to minimally 2-stroke, and then so rich that it 4-strokes. Your explanation of how a nominally 4-stroke engine works is correct but not directly applicable to this situation.

The original question was whether or not a 2-stroke engine actually skip pulses when it's set rich enough to *sound* like it's 4-stroking.

Note that the peak pressure is higher for a similar reason as you describe - but the details you describe are not applicable.

>
>Alcohol fuels generally have fewer BTUs per lb than gasoline
>fuels but they will also run with much lower air to fuel
>ratios, so that the final mixture has more BTUs per lb
>giving more power. This is further enhanced by the fact
>that alcohol has a higher heat of vaporization the gasoline,
>cooling the mixture and making for higher density in the
>final mixture.
>
>All of this is not to run down Scott's data! On the
>contrary! This is really wonderful work, the kind we have
>far too little of. As someone said earlier in this thread,
>we need data not guesswork, or opinion.

Indeed.

Brett

Brett,
Just out of curiosity, how did you get the angle out from the time axis--for example, by assuming constant angular velocity?--or something more sophisticated?

>Brett,
>Just out of curiosity, how did you get the angle out from
>the time axis--for example, by assuming constant angular
>velocity?--or something more sophisticated?

The original data was done with a constant rate. Later, I went back and used the derived instantaneous rpm and used it to modify the rate, and let it iterate a few times. It didn't really make a lot of difference, since the rpm variation was only 400 rpm/5%. I figure the error in the overall phase is probably about the same order of magnitude.

Ron B.
F4Fguy

Brett:I was aware you were referring to 2-4-2 break.I intended
the four stroke reference as a partial explanation of what
happens during "four-stroking" of a two-stroke.

The pressure/crank angle data appear to bear this out.
Peak pressure seems to occur at about 28 to 30 deg.,indicating
either late firing ,or a slow burn or both.The former could be
caused by a lower initial charge temp,due to more time for
internal cooling and/or better scavenging of high temp.exhaust
gasses.The slow burn is typical of very rich mixtures.

In passing,this also explains the "hard"vs"soft break with
raising/lowering compression.Raising compression "advances
the spark" which,within limits,would raise power significant-
ly.

One other point,re pressure,whether peak or mean,the
pressure rise during the firing stroke vs the non firing
stroke is the operating factor.On this basis four-stro-
king produces considerably more than twice the power per
firing stroke than two-stroking.

Ron,

I think it is instant TORQUE, it cannot show pressure in TDC and thus also speed of burning or ignition point, close the TDC.

But if I take closer look, I see very important think – while expanding of firing cycle and after opening ports, the torque fall to “0” till upcoming compression past closing of ports.

[photo not recovered: 3e68454c2452751a.gif]

It is obvious. But why the same does not happen in dead cycle? The only clear interpretation is, that the pressure over piston is higher after dead expansion. I would personally expect that higher pressure – because of muffler which was probably used, so my question is why firing stroke shows that pressure drop after port opening – The only how I can explain it is shock wave and upcoming pressure drop after opening of exhaust and/or also intake. Any other explanation? …

If that over is true, than it means that the dead cycle filling is not only diluted by burned mixture, but also of less pressure, what is another reason of misfire or low energy of stroke IF eventual ignition appears. It well responds to lower negative peak of compression coming past firing expansion, while compression after dead expansion has higher peak. (week point is that it can be because of slowly propagating early ignition as we have slow GLOW ignition instead of quick and exact spark ignition). If this is true, than it make sense only if expanded burned gases after firing expansion runs also TO THE INAKE PORTS – if not, than the fresh fuel from intake will better fill combustion chamber and that pressure drop would not appear. Important is also little, but evident growing of torque around 180 after firing. It can be explained by continual lowering of pressure over piston (what is hard to believe) or growing pressure under piston – what can be because of shock wave reaching piston from backside.

It also in matches the fact that too low or even negative blow down angle does not allow 2 cycling at all.

What you all think about such a theory does it make sense? Or is it Sci-fi?

Brett, how exact is the chart? Can I believe such small details?

>But if I take closer look, I see very important think –
>while expanding of firing cycle and after opening ports, the
>torque fall to “0” till upcoming compression past closing of
>ports.
>It is obvious. But why the same does not happen in dead
>cycle? The only clear interpretation is, that the pressure
>over piston is higher after dead expansion.

You can see the pressure at the head growing slowly during the dead stroke (once you ignore the pressure spike from compression of the dead charge.

>I would
>personally expect that higher pressure – because of muffler
>which was probably used, so my question is why firing stroke
>shows that pressure drop after port opening – The only how I
>can explain it is shock wave and upcoming pressure drop
>after opening of exhaust and/or also intake. Any other
>explanation? …

I don't know for sure, but if you look at the integral of the torque, it's very clear that there is some net positive torque on the dead cycles. It's really small, and I suspect it's the result of compression of the dead charge by "stolen heat" (really enthalpy) from the firing charge.

>If that over is true, than it means that the dead cycle
>filling is not only diluted by burned mixture, but also of
>less pressure, what is another reason of misfire or low
>energy of stroke IF eventual ignition appears. It well
>responds to lower negative peak of compression coming past
>firing expansion, while compression after dead expansion has
>higher peak. (week point is that it can be because of slowly
>propagating early ignition as we have slow GLOW ignition
>instead of quick and exact spark ignition). If this is true,
>than it make sense only if expanded burned gases after
>firing expansion runs also TO THE INAKE PORTS – if not, than
>the fresh fuel from intake will better fill combustion
>chamber and that pressure drop would not appear. Important
>is also little, but evident growing of torque around 180
>after firing. It can be explained by continual lowering of
>pressure over piston (what is hard to believe) or growing
>pressure under piston – what can be because of shock wave
>reaching piston from backside.
>
>It also in matches the fact that too low or even negative
>blow down angle does not allow 2 cycling at all.
>
>What you all think about such a theory does it make sense?
>Or is it Sci-fi?

I'll have to get back to you on that one - I can't immediately explain it. I *don't* think it's an artifact of my processing. This might tend to make me think it has nothing to do with anything below the piston. A combination of restrictions in the exhaust, slight positive pressure in the crankcase that pressurizes the cylinder once the intake port opens, and heat transfer from the head to the atmosphere in the cylinder would explain it. Of course it could easily be an issue of heat transfer to the pressure transducer, too.

>Brett, how exact is the chart? Can I believe such small
>details?

The small features could be a function of the data processing. But the torque over a significant amount of time is almost certainly real.

The best determiner is to see if the features in the pressure seem to be consistent with the torque.

Brett,

You wrote:
>>>I don't know for sure, but if you look at the integral of the torque, it's very clear that there is some net positive torque on the dead cycles. It's really small, and I suspect it's the result of compression of the dead charge by "stolen heat" (really enthalpy) from the firing charge.<<<

It is not what I mean. Try to look to the picture below. Red circles show what I mean. I think you will agree that heat transfer cannot make such difference, it is close to BDC, so it must be much more than could come from stolen heat. And how it can be more than pressure past firing cycle?

LAST EDITED ON Mar-08-03 AT 12:03 PM (CDT)
 
>Brett,
>
>You wrote:
>>>>I don't know for sure, but if you look at the integral of the torque, it's very clear that there is some net positive torque on the dead cycles. It's really small, and I suspect it's the result of compression of the dead charge by "stolen heat" (really enthalpy) from the firing charge.<<<
>
>It is not what I mean. Try to look to the picture below. Red
>circles show what I mean. I think you will agree that heat
>transfer cannot make such difference, it is close to BDC, so
>it must be much more than could come from stolen heat. And
>how it can be more than pressure past firing cycle?

I understood what you meant, but that's a good question. But this is clearly visible in the pressure data:

[photo not recovered: 3e6a2ba222272de0.jpg]

..................... 4-Cycling Pressure with "zero" drawn in and apparent secular pressure rise illustrated

It's clear to me that the pressure data shows the pressure slowly rising the entire time it's not firing. That's why the effect you are seeing (illustrated in the circles) is happening. It doesn't explain why the pressure seems to be rising there, but it is clear it's not an artifact of the processing I did.

The stolen enthalpy could possibly heat the charge and expand it faster than it could go out - making the pressure rise simply from heat transfer. But with both ports open, it would have to be pretty fast. But you can see a very similar effect in the 2-stroke pressure data. Note also that there seems to be some low-frequency drift in the "bottom of stroke" pressure. This is either real, or a function of the test setup (beating between the pressure transducer response and the driving function?)

It would perhaps make more sense if the pressure was zero as you hit BDC on the *nonfiring stroke*, making the pressure at the opening of the exhaust stroke less than atmospheric. This could easily be the result of scavenging from the muffler (ST "flow-through") by the propwash and the departing shock wave (as opposed to the returning shock as you suggested). Or it could be some artifact of the test setup (like a shock wave down the tube going to the transducer).

Without Scott available to ask, I don't know how to determine which.

Brett

p.s. note that I thought of the idea that the anomalous pressure might be due to ram effects in the muffler (with the prop ramming air into the front of the "flow through" muffler), but the pressure is way too high for that. bb

That is GREAT! I did not see that slow charging before. However, you are right it is really too slow. It should look like a logarithmic curve. Actually it looks more like a line, or very slow charging (do you also call it “long time constant” in technical English?).

I saw that slow oscillation on 2 cycling, but I did not think it has what to do with the real effects in the engine. If it is not an error in measuring, than it must be something out of engine, muffler, tank or so. It is somewhere at 20Hz and it must be something large – could it be oscillation of test stand? But it could be, that the “charging” you pointed out can be somehow related to that 20Hz oscillation – the speed will be OK, but how the firing can reset it? … ???

I think it is too little data ... it will need more and better measured values – something like a doctor measuring EKG

>
> Brett:I was aware you were referring to 2-4-2 break.I
>intended
> the four stroke reference as a partial explanation of
>what
> happens during "four-stroking" of a two-stroke.

Sorry, I guess I didn't make that connection. The effect is clearly the same as a "true 4-stroke" - the 'dead' stroke functions to scavenge out the exhaust products much better than it does in one pass. Scott's explanation of "retained charge" is not really necessary - you don't need extra intake charge to get more pop, you really just need to get rid of the exhaust products.

> The pressure/crank angle data appear to bear this
>out.
> Peak pressure seems to occur at about 28 to 30
>deg.,indicating
> either late firing ,or a slow burn or both.The former
>could be
> caused by a lower initial charge temp,due to more time
>for
> internal cooling and/or better scavenging of high
>temp.exhaust
> gasses.The slow burn is typical of very rich mixtures.

Of course, if the pressure was symmetrical around TDC, it doesn't run because there's no net torque.

If you just look at the point that the "firing stroke" pressure matched the "dead-stroke" pressure, it looks like its firing almost exactly at TDC, or just slightly before. Just look at the pressure slope headed into the "firing stroke" - it takes a very steep spike right at TDC, and it's already above the PV=NRT pressure by TDC, implying there it starts burning (slowly) sooner.

So I would vote for propogation delay vice late ignition.

In fact, I could probably back out the pressure profile of the combustion alone by subtracting the "Charles Law" pressure from the firing stroke.

> In passing,this also explains the "hard"vs"soft
>break with
> raising/lowering compression.Raising compression
>"advances
> the spark" which,within limits,would raise power
>significant-
> ly.

It's not all that clear to me, at least from this data, that advancing it would really help. Maybe, but maybe not. Depends on the shape of the pressure - if it's really spikes before TDC, and it's really slow to bleed off after TDC, that would be the maximum torque configuration. I can't tell whether the 30 ATDC is best, worst, or somewhere in between.

LAST EDITED ON Mar-06-03 AT 11:58 AM (CDT)
 
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