NEW! Ultimate C/S Prop Wrench Anti Splat Aero

[dedgemon]

Originally Posted by Bob Kuykendall

Can I trade my old Statics textbook for one of these wrenches? Judging by the length and breadth of this thread, it could be a good deal.

Thanks, Bob K.

PS-I'm keeping my Beer and Johnson Ninth Edition. I may be a relapsed liberal studies dropout, but I do needs my vectorz on th' occasional.

Bob, Keep the statics book, most of these guys should go read it. Besides, have you priced textbooks thse days?

[ccsmith51]

Originally Posted by dedgemon

Bob, Keep the statics book, most of these guys should go read it. Besides, have you priced textbooks thse days?

I'd be afraid to. I remember back in the late 60's and early 70's that textbooks cost about as much as tuition!!

Anyway, thanks to Bob for the great diagrams and examples. I expect that post will put things to bed. I hope....

[Sam Buchanan]

Originally Posted by Sam Buchanan;
Nawwww, let's leave it in classifieds so it will be deleted in a couple of weeks.

Originally Posted by dedgemon

Aww come on Sam. Math done in public is really entertaining. At least to some of us geeks!

If I take off into a 70kt headwind then turn 180 will my plane stall??

[Bob Kuykendall]

Now, suppose you wanted to start with the target nut torque and work backwards through the extension and torque wrench lengths to find the appropriate torque wrench setting. This is what the formulas in AC43.13 do for you:



G shows the whole system. You know that the total arm is 21" or 1.75 feet, and you know that you are looking for 100 ft-lbs at the nut. So you take 100 ft-lbs / 1.75 ft = 57.1 lbs. That's the force you need to apply to the handle of the torque wrench.

That's all well and good, but you're not going to use a scale to measure out 57.1 lbs to apply to the handle. You want the torque wrench to measure it for you. So you take that 57.1 lbs, multiply it by the length of the torque wrench from the handle to the output lug, which we know is 18" or 1.5 ft. 57.1 lbs * 1.5 ft = 85.6 ft-lbs. But is it really?

H shows how we confirm our math by adding up the forces and torques on the extension. First, we know that we are plugging 85.6 ft-lbs of torque into its drive square. And we also know that we are applying a normal force of 57.1 lbs at the nut and of course a countering force at the input square. We know the normal forces are equal and opposite because the thing isn't flying across the hangar. The 3" (0.25 ft) distance between the application lines of the normal forces yields a torque of .25 ft * 57.1 lbs = 14.3 ft-lbs.

So the total torque is 85.6 ft-lbs + 14.3 ft-lbs = 99.9 ft-lbs. That's within .1%, well within the fuzzband for standard hardware.

Thanks, Bob K.

[Brantel]

While my argument really was not focused on the whole extension thing...

I would say that Bob wins since he is the one that accounted for the normal/countering force being applied to the machine in his illustration.

[nauga]

Originally Posted by Bob Kuykendall

Oh noes, it's the torque of the devil!

Bob, I don't know if you recall the good old days of RAH but you're still my hero.

[nauga]

Originally Posted by Brantel

I would say that Bob wins since he is the one that accounted for the normal/countering force being applied to the machine in his illustration.

Bob is saying the same thing the 'size matters' folks have been saying all along.

Setting = Bolt Torque*wrench length/[wrench + extension length]

From his example:
100*1.5/(1.75) = 85.7 ft-lbs, within roundoff error of Bob's results in post 154.

[Brantel]

Yes he is but he included a very good example illustration showing why....

[dedgemon]

Originally Posted by Sam Buchanan

If I take off into a 70kt headwind then turn 180 will my plane stall??

Depends on how you primed it.

Maybe a new thread?

[jaredyates]

Do we have a consensus here on what I need to set my torque wrench to to get 70 foot-pounds of final nut torque with this tool?