What are the risks of using 44 oz-in vs. 54 servos
02-19-2010 | 04:08 PM
#28
Senior Member
Joined: Sep 2006
Posts: 497
Likes: 0
Received 0 Likes
on
0 Posts
From: Mayking,
KY
RE: What are the risks of using 44 oz-in vs. 54 servos
ORIGINAL: fcomer84
As far as I know, mfg. servo torque is with a 1'' arm on the servo. Think leverage here. A long cheater bar works way better than a short one. Use the longest one you can fit in the space you have.
And, yes, most applications do not need as much torque as most folks are led to believe.
As far as I know, mfg. servo torque is with a 1'' arm on the servo. Think leverage here. A long cheater bar works way better than a short one. Use the longest one you can fit in the space you have.
And, yes, most applications do not need as much torque as most folks are led to believe.
02-19-2010 | 04:25 PM
#29
Member
Joined: Jul 2006
Posts: 76
Likes: 0
Received 0 Likes
on
0 Posts
From:
RE: What are the risks of using 44 oz-in vs. 54 servos
I think you might want to consider a retract servo similar to the hi-tec hs-75, jr rt-88, or jr 791. These are all servo's designed for retracts. JMHO. I have at least one of each, of course the 791 is the best and has 260 oz of torque, but will only work with 4.8 volts.
And you might want to close your eyes when you write the check, it does leave a mark when you buy it.
And you might want to close your eyes when you write the check, it does leave a mark when you buy it.
02-19-2010 | 07:16 PM
#30
My Feedback: (20)
RE: What are the risks of using 44 oz-in vs. 54 servos
R2champion
torque is the product of force multiplied by distance (actually the distance perpendicular to the force)
The servo can always make hte same amount of torque. The FORCE measured at the pushrod will vary with the length of the arm. A 50 ounce-inch servo with the arm at a right angle to the pushrod will stall under these conditions:
1 inch arm, 50 ounces of force, 5 inch arm 10 ounces of force.
Of course the total available travel of the pushrod also varies with the length of the arm.
torque is the product of force multiplied by distance (actually the distance perpendicular to the force)
The servo can always make hte same amount of torque. The FORCE measured at the pushrod will vary with the length of the arm. A 50 ounce-inch servo with the arm at a right angle to the pushrod will stall under these conditions:
1 inch arm, 50 ounces of force, 5 inch arm 10 ounces of force.
Of course the total available travel of the pushrod also varies with the length of the arm.
02-20-2010 | 07:29 AM
#31
RE: What are the risks of using 44 oz-in vs. 54 servos
ORIGINAL: r2champion
Wouldn't you gain torque using a smaller servo arm, and a longer control surface horn rather than a longer servo arm?
ORIGINAL: fcomer84
As far as I know, mfg. servo torque is with a 1'' arm on the servo. Think leverage here. A long cheater bar works way better than a short one. Use the longest one you can fit in the space you have.
And, yes, most applications do not need as much torque as most folks are led to believe.
As far as I know, mfg. servo torque is with a 1'' arm on the servo. Think leverage here. A long cheater bar works way better than a short one. Use the longest one you can fit in the space you have.
And, yes, most applications do not need as much torque as most folks are led to believe.
Torque is just a concept for twisting or rotational forces.
Rotational forces are considered to be applied to a shaft with no lever; hence, it is not correct to speak of torque with certain length of arm or lever.
On the other hand, once a motor generates a torque for a mechanism, that remains constant; hence, beyond that point, torque is not gained or lost (in reality, some is lost due to friction in the parts of the mechanism).
For practical use, most of these rotational forces need to be converted to linear forces; then, a lever or arm attached to that shaft is used.
At the end of that arm, there is a linear force.
The magnitude of that linear force depends of the torque and the length of the arm.
Linear force = Torque / Arm length
In that mechanism, any force we gain, is at the expense of displacement of the point at which that force is applied; and any displacement we gain, is at the expense of the force generated at the end of the arm.
Just a schematic to support R8893's post:
http://www.angelfire.com/indie/aeros...eometry-01.jpg
