NEW! Ultimate C/S Prop Wrench Anti Splat Aero

[Bob Kuykendall]

Originally Posted by CMW

...The moment (or torque in our case) along the beam is given by the linear equation M(x) = P * (L-x) where M is the moment, P is the force applied at the handle, L is the overall length of the beam, and x is a distance from the cantilever fixed end (the bolt in our case)...

Very good, let's follow up on that.

We start by recognizing that the torque wrench plus the extension, represented here by the line between M and P, is what we call a rigid body. Yes, it pivots a tiny bit at the head (represented here as a point X inches to the right of the nut at M) in the process of measuring torque. But not much, so for our purposes we can assume that it is completely rigid. If we were evaluating its dynamic behavior it would be different, but we're not so it doesn't. When I'm flying I care about wing flutter, when I'm rock climbing I care about gate flutter, but I don't hardly ever care about wrench flutter.

Now, let's say we are using a torque wrench that is 36" from handle to drive lug. When we add the 3" extension to it, the system length L is 39". In feet, that is 39/12 = 3.25 feet.

Let's further say that we are looking for a torque at the nut of 100 ft-lbs. So let's solve our equation for P and find out how much force is required at the handle:

We start with:

M(x) = P * (L-x)

Solving for P and plugging in the numbers:

P = M(x)/(L - X)
= 100 /(3.25 - 0)
= 100 / 3.25
= 30.8 lbs

Let's run that back through the original equation to validate it:

M(x) = P * (L - x)
= 30.8 * (3.25 - 0)
= 30.8 * 3.25
= 100

So, we know that if we take a 39" bar and apply 30.8 lbs at the handle, we'll get our target torque at the nut.

Now, let's use the original equation again to see what the torque wrench reads when we are applying that 100 ft-lbs at the nut. Using our equation to find M(x), the moment 0.25 feet to the right of the nut where the head of the torque wrench is:

M(x) = P * (L - x )
= 30.8 * (3.25 - 0.25)
= 30.8 * 3
= 92.4

That tells us that with the 3" extension on the 36" torque wrench, the wrench will read or click at 92.4 ft-lbs when the torque at the nut is 100 ft-lbs.

That also tells us that the correction factor we would apply to the wrench reading or setting to get the correct 100 ft-lbs of torque when using the 3" extension is 92.4/100 = 0.924.

That is right in line with the equation in AC43.13 that is basically the rule of law among A&P and IAs and disregarded at their peril:

Y=T*L/(L+E)
=100*3/(3+0.25)
=92.3

Now, suppose that instead of the AC43.13 correction factor, we used the constant 0.8 correction factor that Allan dictates.

100 * 0.8 = 80

That would tell us that if we wanted 100 ft-lbs of torque and we believed Allan, we would either watch the wrench for 80 ft-lbs, or set the click for 80 ft-lbs.

Let's work this backwards a bit and use our original equation to see how much force that requires at the handle:

M(x)=P*(L-x)

We know that M(x) is the moment applied at the head of the torque wrench, which is 3" or 0.25 feet to the right of the nut. Solving for P and then plugging in the numbers, we get:

P = M(x)/(L - X)
= 80 /(3.25 - 0.25)
= 80 / 3
= 26.7 lbs

Now, let's use the original equation again and see what torque we get at the nut when we apply 26.7 lbs at the handle 3.25 feet away from the nut. Remember, the nut is 0 feet to the right of the nut (it is where it is, right?) so x=0:

M(x) = P * (L - x)
= 26.7 * (3.25 - 0)
= 26.7 * 3.25
= 86.8 ft-lbs

What that means is that if we apply the mythical "0.8 correction factor," when using a 36" torque wrench with a 3" extension we would get an actual torque at the nut of 86.8 ft-lbs. That is 13.2 ft-lbs or 13.2% less than our 100 ft-lb target, and in some circumstances may constitute a dangerously loose joint.

Thanks, Bob K.

[ccsmith51]

Reiley, I think that this discussion would have died a timely death a long time ago, except for the fact that it revolves around a very important, flight safety issue.

Allan states that there is a constant multiplier for his tool, that being 0.8. Others believe that this is incorrect. The difference might be the under-torquing of flight critical components.

I think that virtually all posts have been respectful, and I think that Allan has been accommodating and professional in his responses.

I hope that between the videos he suggests he will provide, and the others that have already been presented, we can find an answer that all can accept.

Respectfully,

[Bob Kuykendall]

Originally Posted by ccsmith51

Reiley, I think that this discussion would have died a timely death a long time ago, except for the fact that it revolves around a very important, flight safety issue...

That's the only reason I'm pursuing the topic. I've seen what happens when propellers fall off.

Thanks, Bob K.

[LAMPSguy]

I was taught that proper "counter torque" was critical when using the wrench. Bob, does the math prove that it makes a big difference? If so, where do you place the other hand for the counter force...at the nut or at the connection to the extension?

[Bob Kuykendall]

Originally Posted by LAMPSguy

I was taught that proper "counter torque" was critical when using the wrench. Bob, does the math prove that it makes a big difference? If so, where do you place the other hand for the counter force...at the nut or at the connection to the extension?

You raise an excellent point!

The counter force, which keeps the wrench from collapsing or capsizing, must be applied only at the nut. Applying force anywhere else along the torque wrench or extension will disrupt the torque reading.

Thanks, Bob K.

[scsmith]

Is really counter FORCE. (oops, Bob beat me to it!)
Go back to Bob Kuykendall's drawings in post #150, and you will see where he drew in the counter force in some of his examples. This force is necessary to maintain all the parts in static equilibrium. You know this from experience if you use a wrench with socket extension on it that raises the plane of the wrench up above the nut, that in order to keep the socket from tipping over to the side, you need to push against the top (back) of the wrench to keep it straight.

If you have a crow's foot type of wrench installed, or a very low profile socket, you don't really need to push, because the nut or bolt pushes back for you with no error, since its all flat at the plane of the nut or bolt. The added side force on the bolt doesn't change its response to getting torqued.

Strictly speaking, the only accurate place to apply this counter force is at the nut or bolt. Anywhere else, and the added force will change the applied torque slightly. Realistically, its not going to make much difference as long as you push pretty close to end above the nut or bolt head.

This question is one of many that are answered automatically when the technique of "free body diagrams" is used. No one has yet here drawn a completely correct free body diagram. I'm resisting the temptation to write a short introduction to static equilibrium analysis (statics).

[CMW]

Mike,

This takes the length of the arm of the torque creating device out of the equation completely, other than Allan's tool there is effectively no arm to figure

There is no torque without an arm. Calculate the force on the arm to react against the motor torque and you are right back to a simple cantilevered beam.

Bob,

The equation given in AC43.13 is the moment equation just in a form with two torques. It can be easily derived by substituting:

W=T(l)/(l+x)

Let T=PL
and L=l+x

so W=T(L-x)/((L-x)+x)
or
W=T(L-x)/L
or
W=PL(L-x)/L
or
W=P(L-x)

Once again I just want to say I think Allan's wrench is a great idea, and probably using .8 wouldn't cause any problems since most everyone's torque wrench is probably close to a foot long. But, I think it is important to understand the ramifications of using a crow's foot. I, for one, have used a crows foot to torque most of my fuel fittings. This is application where I think knowing what torque you're applying is critical.

[Mike S]

Originally Posted by CMW

Mike,

There is no torque without an arm. Calculate the force on the arm to react against the motor torque and you are right back to a simple cantilevered beam.

Bob,

Yes, but in this case the arm happens to have my right hand permanently attached to it.

The "arm" of the handle on a typical cordless is somewhere around 4" I would guess, and that would equate to 300 lb/ft going into my hand and wrist. (still using the original 100 lb/ft figure)

There is no way I can hold that much if it were a simple lever.

[Bob Kuykendall]

Originally Posted by CMW

...probably using .8 wouldn't cause any problems since most everyone's torque wrench is probably close to a foot long...

Sorry, but I don't agree with that. Get a one-foot ruler out and have a look at it; does that look like the length of the sort of torque wrench you'd use on 7/16" nuts? Remember, we are talking about a wrench specifically designed for propeller nuts.

Thanks, Bob K.

[erich weaver]

Bob K.:

Your analyses seem indisputable. However, I torqued the nuts on my Whirlwind constant speed prop today, and used a 1 foot long click type torque wrench that read out in foot lbs along with Alans extension. No debate here, since we are all in agreement for that case. My point is that 1 foot long torque wrenches using ft lbs are common place and work fine for prop bolts like mine and on the Hartzells.

Regards

Erich

Edit: the wrench seems longer than 12 inches because that measurement only goes to the center of the handle; the far end of the handle extens a few inches past that.