Requesting help with torque measurement geometry

[hevansrv7a]

I am struggling with how to calibrate my test club. I won't bore you with all the things that did not work. And this may not, either.

The present problem is how to correct the reading on the scale for the geometry including the offset between the pivot on which the airplane rests and the centerline of the crankshaft.


I already understand there is an issue with the sideways force on the ball or rod on which the airplane rests, but the analog scale under the wing is where I can measure force and convert it to a torque reading.

I am considering taking a tangent off the arc, the center of which is the crankshaft and correcting the down-force reading with normal trig, assuming the actual torque is the tangent to the arc where the arc intersects the hardpoint that presses on the scale.

I already know there is a vibration issue, so don't even go there. I'm looking for a geometry solution at this time.

Thanks in advance

[rv7boy]

Hmmm...this is going to be an interesting thread...

[Mike S]

Dan Horton, calling Mr. Horton..................

[BobTurner]

I don't understand the picture, or what you're trying to do, at all. But the physics formula for torque is:


Torque = R x F where R is the length of a straight line from the pivot point about which you want to know the torque, to the point where the force is being applied (by your hand, or by the scale, or whatever). F is the amount of force (lbs) you are pushing with.
Here's the tricky part: "x" means you multiply these two number together, then multiply by the sine of the angle formed by the intersection of R and F (e.g., the directions in which R is running and the direction in which you push).

You may notice that with a torque wrench in normal use, torquing a bolt, the angle is 90 degrees and the sine of 90 is one.

Also in practice it is often easier to break down R into components parallel to and perpendicular to F. The cross product is easier, since the sine of 90 is one, and the sine of zero is zero.

[rzbill]

If my memory of statics class is correct, I believe the correction for the geometry you show is zero.

The torque on the prop shaft will be equal to the product of the horizontal distance from the craft centerline to the scale center and the net weight on the scale.

I built a club dyno for ultralight engines many years ago as a much younger engineer and seem to remember the prop cl being different from the torque arm cl. (since it did not matter)

*EDIT* Checked my MARKs Handbook this AM at the office. My memory is correct.

[gasman]

I know who to ask, but he is busy installing wheel bearings........

[krw5927]

Originally Posted by gasman

I know who to ask, but he is busy installing wheel bearings........

That's good stuff right there

[Rick_A]

I too found your drawing confusing, but torque calculations are easy.
The basic formula is:
Tw= (Ta * L) / (L +A)
Tw= Indicated torque
Ta=Actual (desired) torque
L= length of torque wrench (from center of handle to center of drive head)
A= added length

If you 9" torque wrench and put an 1" extension on in line with the wrench, you add the full lenght of the extension, so the total length is now 10".

If you put the extention at 90 deg to the wrench, you haven't added anything to the lenght so A=0, so total length is still 9".

If you put the extension at an angle between 0 & 90, the extention is a positive number less than 1. If you put the extension beyond 90 degrees, you now are subtracting from the total lenght. "A"= a negative number.

You could do some math, to get the lengths to add/subtract if you put the extention at something other than 0 or 90. Or you could just measure it.

[AaronG]

If I understand what you are trying to do, the correction is zero. The torque applied at the point of your arrow ( prop shaft?) equals the horizontal distance from the point of torque application to the location where the scale touches the beam, times the reading on the scale. The vertical distance is not important here, as long as the force applied to the scale is precisely vertical.

Also, there are no horizontal forces on your free body diagram provided. Only offsetting torque and vertical forces. The overall setup should not want to move right or left.

Aaron

[hevansrv7a]

I do appreciate the attempts to help, but they all seem off-target to me.

The drawing is a front view (looking aft) of the airplane, only one wing is shown, a scale is on the ground under the port wing. A rod connects the scale to the hard point on the wing. The airplane fuse is resting on a large dowel which is supported by a saw horse. The "prop" is a test club that produces only drag, no fore-aft thrust. The air is all moved radially outwards and the torque is the equal and opposite reaction.

1. I already understand the basics of torque - a force applied at right angles to the radius (a tangent to the arc) such that the radius times the force is the torque expressed, in this case, as pound-feet.

2. The engine applies the torque because it uses the club to push air sideways. The torque is being applied to the crankshaft. As shown.

3. Because the torque is at right angles to the radius there IS sideways force on the rod (shown as a circle) under the airplane and on which the airplane teeters.

4. The force applied to the scale (which is on the ground) will be vertical. However, the issue is how to determine the torque from the reading on the scale. If the airplane were suspended on a rotisserie exactly through the center of the crankshaft, then the torque would be applied to the hard point on the wing at a small angle and there would be an easily determined correction using simple trig.

5. The real problem, then, is that the airplane teeters on the rod underneath it and even though I will cage the rod so that it does not actually move sideways relative to either the airplane or the main support, there is still a sideways force. How is that to be figured in the total? If the force on the scale is 100 pounds and the vertical distance between the center of the rod and the center of the crank is 1.5 feet and the hardpoint is 3 feet out from the center of the rod, then what is the torque? Values for convenience only.