Servo torque (physics question)
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Servo torque (physics question)
Originally posted by David Cutler
True, slop is slop, but if the arms are, say, twice the length, then the slop (which doesn't change) is only half the amount it was in relation to the size of the movement of the arms.
Or, to put it another way, suppose your slop is 1/100 of an inch and the arms move 1 inch. If you double the movement of the arms to 2 inches the percentage slop is not now 1 in 100 but 1 in 200 so the angle the control horn makes as it slops is half what it was before, therefore half the control surface movement,
-David C.
True, slop is slop, but if the arms are, say, twice the length, then the slop (which doesn't change) is only half the amount it was in relation to the size of the movement of the arms.
Or, to put it another way, suppose your slop is 1/100 of an inch and the arms move 1 inch. If you double the movement of the arms to 2 inches the percentage slop is not now 1 in 100 but 1 in 200 so the angle the control horn makes as it slops is half what it was before, therefore half the control surface movement,
-David C.
I agree with your analogy, but my point was the slop remains the same irregardless of the arm length. The relationship to arm length is the other story. The thing is were not talking about inches in our application. I suspect the differences are barely notable in our scenario.
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Servo torque (physics question)
I suspect the differences are barely notable in our scenario.
Yes, that's probably true, although, if the arms get particularly short in a small model to allow free movement of the arms in the fuselage, the slop / arm length might become critical.
What I do know is, I always put the pushrod as far out as possible to minimize the effects of slop, because, well, I might as well, if the arm length is there to be used!
-David C.
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Servo torque (physics question)
Originally posted by mglavin
David
I agree with your analogy, but my point was the slop remains the same irregardless of the arm length. The relationship to arm length is the other story. The thing is were not talking about inches in our application. I suspect the differences are barely notable in our scenario.
David
I agree with your analogy, but my point was the slop remains the same irregardless of the arm length. The relationship to arm length is the other story. The thing is were not talking about inches in our application. I suspect the differences are barely notable in our scenario.
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Servo torque (physics question)
I don't think I have heard anyone mention yet the fact that the mechanical advantage increases with the 1/cos() of the angle through which the servo arm turns. At the farthest most travel limits of the servo arm, the pushrod is moving less and less actual distance per degree of servo arm rotation. The region that is most linear is at 90 degrees to the pushrod. Once the servo arm has rotated 60 degrees, the mechanical advantage has increased from 1.00 to 2.00. When it has rotated 70 degrees, the mecahnical advantage goes up to 2.92. At 80 degrees, it is 5.75!
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Servo torque (physics question)
Originally posted by mtthomps
Once the servo arm has rotated 60 degrees, the mechanical advantage has increased from 1.00 to 2.00. When it has rotated 70 degrees, the mecahnical advantage goes up to 2.92. At 80 degrees, it is 5.75!
Once the servo arm has rotated 60 degrees, the mechanical advantage has increased from 1.00 to 2.00. When it has rotated 70 degrees, the mecahnical advantage goes up to 2.92. At 80 degrees, it is 5.75!
Why some setups restrict you to short servo arm travel when just a few minor changes make major differences.
Ed M.
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Servo torque (physics question)
Originally posted by mtthomps
I don't think I have heard anyone mention yet the fact that the mechanical advantage increases with the 1/cos() of the angle through which the servo arm turns. At the farthest most travel limits of the servo arm, the pushrod is moving less and less actual distance per degree of servo arm rotation. The region that is most linear is at 90 degrees to the pushrod. Once the servo arm has rotated 60 degrees, the mechanical advantage has increased from 1.00 to 2.00. When it has rotated 70 degrees, the mecahnical advantage goes up to 2.92. At 80 degrees, it is 5.75!
I don't think I have heard anyone mention yet the fact that the mechanical advantage increases with the 1/cos() of the angle through which the servo arm turns. At the farthest most travel limits of the servo arm, the pushrod is moving less and less actual distance per degree of servo arm rotation. The region that is most linear is at 90 degrees to the pushrod. Once the servo arm has rotated 60 degrees, the mechanical advantage has increased from 1.00 to 2.00. When it has rotated 70 degrees, the mecahnical advantage goes up to 2.92. At 80 degrees, it is 5.75!
Hence, use the spreadsheet. It takes all of this into consideration and displays all of this information graphically.
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Thinking again
Using the same example: minimum 75 oz of torque needed at the surface, 100 oz/in servo, desired 50 degree surface deflection is met with ATV at 100%:
Say I want to increase the speed of the control surface so I move the horn in on the control arm to .75" with the servo arm at 1" for a mechanical disadvantage (we all did this before at one time right?).
If I did my math right, a 1" servo horn at 100 oz coupled with the .75 control horn will net 75 oz/in. So far I am meeting the torque requirements of the surface. But I will have to dial down my ATV to reduce the now increased surface deflection.
I can see that I lose resolution by dialling down my endpoints (ATV). Also I guess with less throw I won't be getting the full range of available mechanical advantage (re: what mtthomps posted)?
Did I just answer my own question or are there any other negatives of working at a mechanical disadvantage (even when the torque requirements are met)?
Say I want to increase the speed of the control surface so I move the horn in on the control arm to .75" with the servo arm at 1" for a mechanical disadvantage (we all did this before at one time right?).
If I did my math right, a 1" servo horn at 100 oz coupled with the .75 control horn will net 75 oz/in. So far I am meeting the torque requirements of the surface. But I will have to dial down my ATV to reduce the now increased surface deflection.
I can see that I lose resolution by dialling down my endpoints (ATV). Also I guess with less throw I won't be getting the full range of available mechanical advantage (re: what mtthomps posted)?
Did I just answer my own question or are there any other negatives of working at a mechanical disadvantage (even when the torque requirements are met)?
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Servo torque (physics question)
Basically yes you did answer your own question.
However I would allways go for 100% ATV to get max force at the control surface, having too much is better than too little.
To increase speed, I buy new servos.
However I would allways go for 100% ATV to get max force at the control surface, having too much is better than too little.
To increase speed, I buy new servos.
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RE: Servo torque (physics question)
ORIGINAL: rino
I thought this would reduce the amount of torque since the servo arm force vector is no longer 90 degrees to the pushrod.
I thought this would reduce the amount of torque since the servo arm force vector is no longer 90 degrees to the pushrod.
You have bigger angles due to the section of the wing than you will get going to a different hole on the control arm. It is not an issue.