Formulas for Airspeed and Line Tension

LAST EDITED ON Aug-28-03 AT 03:32 PM (CDT)
 
I posted this a little while back, but it got buried pretty deep in the "Line Length" thread. I'm repeating it to ensure that everybody who might find it useful sees it. With these simple formulas, a stopwatch, and a tape measure, you can take a quantitative approach to tuning the lap time/line length/airspeed parameters for best performance. I hope this proves useful.

Andrew Tomasch

Original Post (with a couple of additional comments):

>Is there a formula to work out the best (sweetest)
>line length for various designs?
>
>I'd like some feedback on the line length
>recommendations for a "Nobler" around 40oz
>in flying weight?

While there is no "magic formula" which will tell you what line length to use, consideration of the physics for uniform circular motion yields some useful simple formulas, which can help guide and inform your quest for the optimal line length for a given model.
The formulas are very easy to derive, and can be found in any high school or first year undergraduate physics text book. The only bother is in converting from "physics units" (meters, kilograms, seconds, Newtons of force, meters/second for speeds) to "standard control line units" (feet for line length, ounces for model weight, pounds of line tension, seconds for lap times, MPH for speeds). This I have done, so that the "experimenter's units" used by control line flyers can be plugged into the formulas directly, and the results obtained are in familiar units as well:

Brackets {} contain the units in which each quantity is expressed.

T{pounds}=Total tension for *both* lines. Each line carries a load of T/2
W{ounces}=Model weight
R{feet}=line length, from center of circle to center line of model
V{MPH}=Flight speed of model.
t{seconds}=Lap time of model

I've carried all the decimal places on my calculator. The constants can be rounded as desired. I use "*" for multiplication and "/" for division.
Here are three useful relationships:

(I) V{MPH}=(4.283989982)*(R{feet}/t{seconds})

(II) T{pounds}=(0.004181496)*(V{MPH}*V{MPH})*(W{ounces})/R{feet}

(III) T{pounds}=(0.076741207)*W{ounces}*R{feet}/(t{sec}*t{sec})

These relationships are for level flight. An elementary application of Newton's second law yields a simple relationship for the reduction in line tension during overhead flight:

(IV) T(overhead)=T(level flight)- Weight of Model
T(overhead){pounds}=T(level flight){pounds}-W{ounces}/16

That is, the tension for the model directly overhead is the reduced from that in level flight by the model's weight. This is because directly overhead, the model's weight is now providing part of the force needed to keep the model accelerating in a circular path, and the line tension is reduced accordingly so that the *sum* of the weight and tension produce the same acceleration toward the circle center as in level flight.

Now let's look at our generic example: a 40 ounce airplane on 60' lines turning 5 second laps:

Formula (I) yields V=51.408 MPH

Formula (II) yields T=7.367 pounds

Formula (III) also yields T=7.367 pounds, as it must.

Formula (IV) indicates that the tension directly overhead will be reduced by the model's weight of 40 ounces (2.5 pounds) and therefore will fall to 4.867 pounds.

Probably the most important lesson is to note that the tension in the lines increases as the *square* of the speed, or equivalently, as the *inverse square* of the lap time. So the most effective method for improving line tension is to fly faster, at a given line length. Line tension will also increase inversely with line length, and in proportion to model weight, provided that the flying speed remains approximately constant.

The general rule of thumb is that small changes in line length will not change the flying speed much, that is, we neglect the additional drag due to longer lines, or the drag reduction due to shorter lines, both of which will change the flying speed. You can always *measure* the actual flying speed by measuring the lap time and using (I). It is also important to note that these formulas ignore lift generated by the fuselage which will produce additional tension in the lines. That's why the amount and distribution of side area can make such a pronounced difference in line tension, particularly in the overheads. And has been pointed out many times, you cannot dictate the flying speed to the airplane. Heavier airplanes must fly faster to fly at all.

Once you determine at what speed the airplane is flying reasonably, (I)-(IV) can provide some guidance in tuning the line length to get reasonable lap times and line tension. Once you have determined the actual flying speed with (I) you can solve for the radius R and plug in the measured airspeed V and desired lap time t to get an estimate of the line length R required to give the desired lap time.

Finally, it should be easy to pop (I)-(IV) into a spread sheet to generate tables of flying speeds and line tensions given lap times, line lengths, and model weight.

Have Fun!
Andrew Tomasch

Scientists!

One problem with all this, Andrew. As I read it, the computations relate to a rock on a string. But there is no relationship to the flying airplane, nor to the thrust package or the vectoring thereof (sounds kind of lawyerish, doesn't it?)

As to the first point, you could trim the airplane to make a constant left turn of the same radius as the length of the lines and totally eliminate any line tension. Yet the lap speed, weight, line lengths as used in the forumula would remain the same. Or, you could trim it to turn the other direction and...you get the idea.

Also, the power package does, indeed, affect the line tension. Just changing from a smaller to a larger prop, but keeping the same rpm, pitch and flying speed will increase tension (if there was any to begin with.)

In other words, what I am saying is, it is not all according to formula. Only if it is a rock on a string.

Len,
At least my current understanding of Stunt Nirvana trim involves basically zero outward thrust and no rudder offset--so very little outward yaw, Andrew's analysis gives basically the first order result for line tension. As he points out the analysis is ignoring any lift that the fuselage produces (which if outward would have to come from an outward yaw assuming the fuse is symmetrical).

My guess is that his obsevation that the planes tangential velocity (or lack thereof at any location on the flying hemisphere) is where most of our line tension problems come from. For the most part this is a result of engine thrust (engine +prop).

>My guess is that his observation that the planes tangential
>velocity (or lack thereof at any location on the flying
>hemisphere) is where most of our line tension problems come
>from. For the most part this is a result of engine thrust
>(engine +prop).

An excellent point, which again I probably didn't emphasize enough: slow down and lose tension as the *square* of the speed. So getting slow in the overheads is a killer. You lose tension due to the model's weight, and additionally, as the square of the speed if you slow down. Here the powerplant can make a big difference.

I had come to the conclusion that good trim emphasized little if any radial force due to engine offset or other aerodynamic effects like fuselage lift (say from rudder offset or outward yaw). So I figured that the simple rock on a string would be useful for models trimmed that way. Good of you to point that out explicitly.

Thanks Alan!

Andrew Tomasch

>
>In other words, what I am saying is, it is not all according
>to formula. Only if it is a rock on a string.

Leonard,

Correct. The tension formulas are for a rock on a string. The airspeed formula is of course always valid, since it just divides distance traveled by time to travel the distance.

I did state explicitly that any additional force due to aerodynamics would change the line tension. I should have added thrust vectoring (engine offset) as well. The tension reported is that demanded by physics to keep the model of a given weight flying in a circle of a given radius, at a certain speed. The rock on a string (ROS) case is the simplest, where the line tension is the *only* force available to accelerate the model toward the circle center. The model *must* experience a force of that magnitude, but other forces can contribute as well, reducing the line tension required. I gave the example of overhead flight, where the weight of the model contributes, and thus reduces the tension. And additional forces can be directed radially outward, demanding more line tension too. Engine offset and lift from the fuselage are examples of this. All the physics tells you is there must be a final resultant force on the model radially inward to keep it flying in the specified circle at the specified speed. The ROS is the simplest example where that force is only the line tension.

Bottom line, I completely agree with you.

I would contend that it is useful to understand the ROS, so as to be able to study the contributions of thrust offset, fuselage lift etc. To really understand what goes on, you need to *measure* line tension, say hypothetically with a scale built into the handle (spring balance? strain gauge?), to understand how much the actual line tension deviates from ROS. Then you have a measure of how important other forces are in contributing to the final line tension, and you can make changes and measure the effect. Without measurements of this type, there is little that can be done quantitatively to understand line tension, IMHO. But the ROS analysis at least lets you get ball park ideas as to how much line tension you might expect from a given setup in the absence of major contributions from other effects.

Thanks for the thoughtful reply. This was just the sort of discussion I had hoped for.

Andrew Tomasch

And what you are contributing here, Andrew, also helps others see the importance of all factors. A lighter airplane, all else being equal, will have less "pull" on the lines than a heavier airplane. A faster flying plane will have more "pull" than a slower airplane. And, if you slow down too much in the maneuvers, you can (and will) lose tension. Precession is also a factor, and this is a product of prop (and spinner) weight and engine speed minus any aerodynamics that have been applied to counter this.

Where are the weak spots in the pattern? How about the third corner of the hourglass? You are slowing down to "take it up". You are flying overhead after just having turned a corner (another slow down point). And now you have to make another dreaded outside corner which further decreases speed and tension while adding precession to the picture.

All of this shows that you need a certain air speed to fly the airplane and a certain lap speed to maintain tension plus the proper aerodynamics and trim to offset some of these forces while providing a good power package to keep things moving.

Complicated little buggers, these ariplanes of ours, aren't they?

Hey, This should work for me as Ive flown a few Rocks on Strings!

I'm more a boulders on cables man myself.

> A lighter
>airplane, all else being equal, will have less "pull" on the
>lines than a heavier airplane. A faster flying plane will
>have more "pull" than a slower airplane. And, if you slow
>down too much in the maneuvers, you can (and will) lose
>tension. Precession is also a factor, and this is a product
>of prop (and spinner) weight and engine speed minus any
>aerodynamics that have been applied to counter this.
>
>Where are the weak spots in the pattern? How about the
>third corner of the hourglass? You are slowing down to
>"take it up". You are flying overhead after just having
>turned a corner (another slow down point). And now you have
>to make another dreaded outside corner which further
>decreases speed and tension while adding precession to the
>picture.
>
>All of this shows that you need a certain air speed to fly
>the airplane and a certain lap speed to maintain tension
>plus the proper aerodynamics and trim to offset some of
>these forces while providing a good power package to keep
>things moving.
>
>Complicated little buggers, these ariplanes of ours, aren't
>they?

Ron B.
F4Fguy
All true.All too true!Which ,paradoxically,is why a lighter model is a better model.While it may have less tension,it will,all other things equal,have more uniform"pull"throughout the pattern,due to it's ability to accellerate faster after a slowdown.In addition,it won't decellerate as much,since lower weight allows lower incidence for a given turn radius,hence lower induced drag and less speed lost,allowing a tighter corner...............etc.

As you correctly point out,the power package is a big(maybe the biggest)contributor to overall line tension.In-flight accelleration is the name of the game,i.e. the ability to get back to(or as near as possible to)level flight speed after a speed loss from any decelleration.This is greatly aided by working on the"back"side of the torque curve.With a loss of RPM torque goes up allowing rapid accelleration back to"set"speed.The Fox.35 and S.T.46 are prime examples,both have almost straight line increasing torque with falling RPM within the useable range.Pipes,as presently used,operate the same way,albeit at higher revs/power,and over narrower ranges.

Props also contribute.The ability of modern stunt engines to turn larger diameters has greatly enhanced available thrust,again increasing the ability to return to speed.

The ROS analogy is apt.It doesn't matter where you get the energy,the only way to get line tension is lap speed(I'm not going to go into hovering with a Bi-Slob).To get speed you can either pull with a prop or pull on the lines,as in whipping,but you've got to keep the energy flowing.

Ron B.

PS:the Bi-Slob isn't flying on the wing in hover,in that regime it's a vectored thrust machine.