Loading of Control Surfaces?
 

Is there a formula that will tell how much force is being put on a control surface?
I assume that it takes into account sq area, velocity, angular deflection, and in some way the airfoil or lack of that it is attached to.
The reason I am asking has to do with picking the proper servos for control.
What has enough torque and and is not over kill?
Thanks

RE: Loading of Control Surfaces?
 

There are formulas for it, and ther is even an on-line calculator for it. (I'm sure someone will remember the URL...)

The online calculator takes into account Area, aspect ratio of the control surface, (it asks for "span" and "chord" of the surface to get the area) expected maximum deflection angle and expected maximum airspeed. If I remember crrectly... it does not take aerodynamic balancing into account (which would reduce the servo power needed.)
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http://www.csd.net/~cgadd/eflight/calcs_servo.htm (found the calculator)

RE: Loading of Control Surfaces?
 

It's sort of a backwards solution in this case but quite elegant. You know how strong your servo needs to be but it's also
telling you how much load there is on the surface.... Cool [8D]

RE: Loading of Control Surfaces?
 

Now all I have to do is figure out what inch/ounces is and I'll be ready.
my reason for asking the original question is I would like to build a mock-up , with a servo and a weight
the then measure the current ( In Milliamps) that it takes for a full deflection surface movement with different servos as a Show and tell sort of thing for my local Club,
Maybe I'll be able to get some of these guys to think and then put the right servos in
instead of this is what I had , I hope it works.
bummer that the gears striped or the surface would not move at all.
Thanks for the help
we all can learn from others

RE: Loading of Control Surfaces?
 

The force required to deflect any control surface is determined by the dimension of the control surface, the angle of deflection and the velocity in relative speed. To calculate the required servo strength, the geometry {mechanical advantage / disadvantage} of the linkage must also be considered. This is a very complex series of calculations best done with a computer. There is a shareware program called RC calculator that will get you very close.

RE: Loading of Control Surfaces?
 

More calculations--
The chance of airloads destroying a servo -are about zero.
The chance of calculating the load -then finding a servo which actually meets that criteria - less than zero.
The chance of poor geometry wrecking the real work availble from the servo- about 99.99%
The chance of improper power supply affecting the output of the servo - about 75%.
The servo selected is important -but far less important than understanding the correct linkage and proper servo power supply
But read the calcs- THEN find out how servos really operate and what the linkage can affect.
In most cases, a less powerful servo (rated at less torque) will do the job.
The new digital servos provide much more control about neutral ( some call it holding power) and a low torque digital can oft do a much better job than a higher rated (more torque) servo.
If you don't understand why - then servo selection is just a blind guess.

RE: Loading of Control Surfaces?
 

An in-oz (inch-ounce) is a torque force measurement. 16 inch ounces = 1 in-lb. 12 in-lb = 1 ft-lb (the common SAE torque measurement.) 192 in-oz = 1 ft-lb

You can get 1 inch-ounce torque by hanging 1 ounce of fishing weights on a 1 inch servo arm (that is parallel to the ground, with the output shaft parallel to the ground and 90 deg to the servo arm). 42 in-oz would be the same as either haning 42 oz off the end of a 1 inch arm, or 1 ounce off a 42 inch arm.

In physics, all forces have vectors, or directions they are acting on. In discucssing servos or tightening bolts in mechanics, Torque forces have a "rotational vector" acting to rotate some object about a given axis.

RE: Loading of Control Surfaces?
 

The calculator in FH's link already takes all that into account. It looks at the loads on the surface and the angular deflection and then you enter the servo angular travel (+ and - 45 by default) and it tells you what the load is ON THE SERVO. So if it says 14 in-oz then you need at least a 16 in-oz servo. The assumption being that the servo will stall at 16 in-oz so you need some overhead.

The only thing the calculator does not take into consideration from what I can see is any friction in the system. So if you use tube in tube or wire in tube pushrods or if the clevis pins are a tight force fit in some of the holes then the need for more servo power will be there. You can test some of that friction by using a scale and pulling against the friction.

If you're trying to educate the guys I'd just do the calcs and print the screens.

RE: Loading of Control Surfaces?
 

Thanks For the Info
Great to have this kind of suport