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Author Topic: Control Line Shape Equation  (Read 5778 times)

Offline TomLaw

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Control Line Shape Equation
« on: June 28, 2012, 02:27:46 PM »
I took time off from my control system spreadsheet to dredge up something I did several decades ago. Starting with the well known equation of a vibrating string.


Offline Tim Wescott

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Re: Control Line Shape Equation
« Reply #1 on: June 28, 2012, 02:44:05 PM »
Didn't any of your math teachers ever mark you down for not spelling out what your variables are?

I figured out r, at least.

You're setting drag proportional to velocity?
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Offline Howard Rush

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Re: Control Line Shape Equation
« Reply #2 on: June 28, 2012, 03:37:46 PM »
It looks like it came out right.  I did this by iteration, too.  

Here's the authority: http://www.tulsacl.com/Linelll.html .  I hope this has Pete Soule's paper with it.  Does it?
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Offline TomLaw

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Re: Control Line Shape Equation
« Reply #3 on: June 28, 2012, 04:39:36 PM »
Unfortunately, I have a Mac. I have Soule's paper. What inspired me to do this originally was reading Netzeband describe Soule's work (okay, more than a couple of decades). I also assumed the handle is at the pivot. The drag reacted at the handle is 1/4 and at the plane 3/4 of the total.

steven yampolsky

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Re: Control Line Shape Equation
« Reply #4 on: June 28, 2012, 08:49:20 PM »
It looks like it came out right.  I did this by iteration, too.  

Here's the authority: http://www.tulsacl.com/Linelll.html .  I hope this has Pete Soule's paper with it.  Does it?

The install program on this page does not work on 64-bit windows. Is there source code for this? I'd like to give it a 2012 makeover.

Offline Howard Rush

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Re: Control Line Shape Equation
« Reply #5 on: June 28, 2012, 09:11:20 PM »
See Bob Reeves.
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Offline TomLaw

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Re: Control Line Shape Equation
« Reply #6 on: June 28, 2012, 09:23:40 PM »
I have been looking at them on and off for 40 years. I forgot I was the only one!

T(∂2Y/∂x2)dx = K(∂2u/∂t2)dx refers specifically to a vibrating string; I applied it to the control line

T = line tension = centripetal force

∂2Y/∂r = second partial derivative of Y (bow in line); dx should be dr, is a differential element of length measured along line

2T(∂2Y/∂r2)dr = Kv^2dr

the left side is a measure of the tension directed perpendicularly to the radius (opposing the local drag)

K is a constant

v is the speed of the line at radius r

Y(r) = [Cdρtr/(12mR)](r^3 - R^3)

Cd is the drag coefficient of a single line

ρ is the density of air

m is the mass of the plane

t is the diameter of the line

r is the distance measured from the handle out toward plane

R is the total line length

Drag = .5Cdρtrv^2 where tR is the cross sectional area of the line exposed to the wind.

This could be improved by including the speed and radius of the handle motion, and including the variation in drag to changing Reynolds number as you progress out the line.

« Last Edit: June 29, 2012, 10:30:15 AM by TomLaw »

Online Trostle

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Re: Control Line Shape Equation
« Reply #7 on: June 29, 2012, 01:36:47 PM »
this may not have much bearing on the subject here, but what I understand the work of Peter Soule and then what was picked up by Netzeband regarding line drag and leadout sweep, is that the shape of the lines between the model and the handle closely approximates a catenary curve.

You can actually see this when flying your model by sighting down the lines towards the model.  The portion nearest the handle is almost a straight line that will normally point at some point on rear part of the fuselage.  As the model rotates around the circle on a windy day, that point being pointd at on the aft part of the fuselage will move forward and back.

Keith

Offline TomLaw

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Re: Control Line Shape Equation
« Reply #8 on: June 29, 2012, 09:07:22 PM »
I believe it would be a catenary if the drag on each segment was the same. As it is, the drag increases as the square of the distance out along the line. A catenary has the maximum bow at the midpoint.

Online Trostle

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Re: Control Line Shape Equation
« Reply #9 on: June 29, 2012, 10:39:19 PM »

(Clip)

 A catenary has the maximum bow at the midpoint.

I did not bring up the matter of catenary curves to start an argument.  What I was trying to explain that Soule and Netzeband were able to generate material to determine desirable leadout rake angles (to a satisfactory level of accuracy) based on the line shape resembling that of a catenary curve.  This catenary similar curve is not the same as a catenary curve generated by say a cable suspended between two locations at the same elevation but more like a catenary curve produced by a chain (or flexible cable) suspended between two differing elevations.

Depending on the end points and the section of a catenary curve being examined, for a material forming a catenary curve, the "maximum bow" (watever that is) is not necessarily at the "midpoint" (whatever that is).

Keith
« Last Edit: June 30, 2012, 08:20:24 PM by Trostle »

Offline Tim Wescott

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Re: Control Line Shape Equation
« Reply #10 on: June 29, 2012, 11:14:10 PM »
When I was a wet-behind-the-ears engineer, stuffed with all sorts of exact solutions to carefully chosen problems, I thought that it was my job to find an exact solution to every problem that came my way.

Boy, I wasted a whole bunch of time on that.

Now I understand that you just need to approximate things well enough that the error from the approximation is lost in the noise of the error from everything that I don't and can't know, and to make things adjustable so that I can fix them after they're all built and pretty.

It's still fun to solve things like catenary curves, though.
AMA 64232

The problem with electric is that once you get the smoke generator and sound system installed, the plane is too heavy.

Offline TomLaw

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Re: Control Line Shape Equation
« Reply #11 on: June 29, 2012, 11:38:37 PM »
I hear you. I still try to do everything as exactly as possible. I only give up when the mathematical model begins to exceed my understanding of the physical world. I once wrote thirty pages on an airplane making climbing/descending turns. n~

Offline Bob Reeves

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Re: Control Line Shape Equation
« Reply #12 on: June 30, 2012, 03:11:20 PM »
The install program on this page does not work on 64-bit windows. Is there source code for this? I'd like to give it a 2012 makeover.


Just install it on a windows xp machine and copy it over to your win 7 machine, should run in compatibility mode and it's self contained. Haven't bothered to update the installation, haven't had any requests except for this and the above work-around will get it up and running.

And yes, it contains two of Pete's papers in PDF files...

Offline TomLaw

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Re: Control Line Shape Equation
« Reply #13 on: June 30, 2012, 04:48:52 PM »
I'm not arguing either. Your point about the catenary is well taken. I was just pointing out the difference between my four ;)th degree equation and the sinh and cosh of the catenary.  ;)

Online Trostle

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Re: Control Line Shape Equation
« Reply #14 on: July 01, 2012, 06:20:10 PM »
Just for the record, here is some material frm the Jul/Aug 66 American article where Bill Netzeband published his second of a three part series on "Control Line Aerodynamics Made Painless".  This is the article where he presented nomographs to determine the appropriate leadout rake.  These nomographs included using Reynolds Number as a factor and I believe the information to generate these nomographs have been the basis for several computer programs that have since appeared.

In his article, Netzebad explains line shape.  He wrote "Closest parallel in engineering mechanics is the Catenary, a curved line between two points similar to a slack flexible cord hanging between two poles."  His article goes on to discuss the catenary in more detail and details the evolution and the various people that over time contributed to a better uderstanding of line drag.  Makes for interesting reading. 

The page shown is from that article to show the line shape he discusses in detail.

We owe Bill Netzeband a great deal of gratitude for his many contributions to this activity of ours.  He is certainly missed.

Offline TomLaw

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Re: Control Line Shape Equation
« Reply #15 on: July 01, 2012, 08:07:10 PM »
Thanks for your insight. I read that article when it was new!

"In 1964, I got a great book on airplane aerodynamics by Dommasch, Sherby, and Connolly. They taught by writing equations in coefficient form, and I was finally able to adapt their equations to the funny things that happen to a Control Line plane." Netzeband

I was reading that book at the same time as he published articles. I still have the book, "Airplane Aerodynamics."

Online Trostle

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Re: Control Line Shape Equation
« Reply #16 on: July 01, 2012, 08:44:41 PM »
Thanks for your insight. I read that article when it was new!

"In 1964, I got a great book on airplane aerodynamics by Dommasch, Sherby, and Connolly. They taught by writing equations in coefficient form, and I was finally able to adapt their equations to the funny things that happen to a Control Line plane." Netzeband

I was reading that book at the same time as he published articles. I still have the book, "Airplane Aerodynamics."

Hi Tom,

I also still have that book by Dommasch, Sherby and Connolly.  It sits on my shelf right beside my Perkins and Hage Airplane Performance Stability and Control .

Offline TomLaw

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Re: Control Line Shape Equation
« Reply #17 on: July 01, 2012, 08:50:29 PM »
I have that also. Two of my other favorites are "Mechanics of Flight ", Phillips; "Theory of Flight", von Mises.

Offline Howard Rush

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Re: Control Line Shape Equation
« Reply #18 on: July 01, 2012, 10:44:30 PM »
Where do you live, Tom?
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Offline TomLaw

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Re: Control Line Shape Equation
« Reply #19 on: July 04, 2012, 10:07:16 PM »
Bakersfield, CA  ~^. In the glory days I was 10 minutes from the Sepulveda Basin.

Offline Howard Rush

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Re: Control Line Shape Equation
« Reply #20 on: July 04, 2012, 10:57:17 PM »
Not too handy, but maybe I'll see you at the Clovis contest.  I look forward to meeting you.
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Offline Brett Buck

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Re: Control Line Shape Equation
« Reply #21 on: July 07, 2012, 09:11:52 PM »
Here is the paper that has the corrected equations (after the nomographs where published, I believe). This paper is the basis of LINEII.

   Brett


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