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-   -   NEW! Ultimate C/S Prop Wrench Anti Splat Aero (https://vansairforce.net/community/showthread.php?t=82592)

9GT 02-16-2012 04:47 PM

Quote:

Originally Posted by PerfTech (Post 629932)
Bravo Mike!!!!
This is the best post so far!

Actually the best post so far Allan is the first post of this thread where you offer this wonderful tool to those of us who have fought the battle too many times taking a C/S prop on and off.

scsmith 02-16-2012 06:07 PM

Larry is raising an interesting point....
 
In Larry's defense, the question he is asking, specifically about the behavior of a click-type torque wrench, is worth pursuing.
This is entirely separate from the whole statics issues of the resulting torque obtained at the bolt using an extension. (A discussion which I have excercised an uncharacteristic level of impulse control in not wading into...plenty of others have done a good job there, including my good friend Bob Kuykendall)

The question is this:

Does a click-type torque wrench actually measure the true moment at the head of the wrench, or, does it do this approximately by responding to the combination of moment and shear force at the head that results from pulling at the nominal handle end?

Evidence that suggests that it does respond only to the moment:
The fact that click-type torque wrenches do not use a teetering handle to fix the point of application suggests that one benefit of click-type torque wrenches is that they are at least somewhat insensitive to the point of application of the force, and therefore insensitive to the shear force at the head of the wrench. So, if you grab it an inch or two away from the nominal handle posittion, it shouldn't affect the torque.

But.....

Is the internal design of the wrench such that this is REALLY true, or only approximately true? If we double the length of the handle with a cheater, and thus cut in half the shear force at the wrench head, does the wrench REALLY still click at exactly the right torque, or does it click at a somewhat different torque because it is influenced by the shear force and moment in combination?

Evidence that suggests that it may respond to a combination of moment and shear:
Looking at the outside of a click-type wrench and imagining what is inside, it appears there is some kind of pivot point a finite distance away from the actual wrench head, and the part that the head is attached to must move when the part snaps past a sear or cam that is spring loaded to a certain force. So, the fact that this pivot is a finite distance away from the wrench head means that it is really only at that pivot that a pure moment is measured. The moment change from the pivot to the head is dependent on the ratio of the distance from the pivot to the head and the distance from the pivot to the handle. Since the pivot is much closer to the head, the moment gradient (from the shear force) doesn't cause much change in moment between the two points....but it would cause SOME change. So, for large changes in handle length (and thus large changes in shear force) there would be a significant change in the wrench calibration, i.e. it would click at the wrong torque at the head.

Honestly, I don't know the answer to this. I think a lot of us have assumed that it must respond purely and only to the torque, or it wouldn't be a very good concept for a torque wrench. But, as Larry is trying to get to, the devil is in the details. What is the internal design such that it would have no sensitivity at all to the shear force at the wrench head?

This would be an easy test to do with two torque wrenches. No need to weld anything, just find a 12-point socket that fits the drive lug of the second wrench. Do they both click at the same time when both are pulled on their intended handles? If so, do they also both click at the same time if one is pulled at its intended point and the other pulled from a long cheater bar? Or, just compare a clicker to a beam type wrench.

PerfTech 02-16-2012 06:40 PM

Explanation With The Math
 
Quote:

Originally Posted by L.Adamson (Post 629935)
Good...

Because I'm aware of everything you just said. But, it doesn't explain, what I'm looking for.

I've spent the last two days, scouring the Internet for the exact info I want. I've linked to numerous calculators. I don't need to hear anymore ...re quoting of the same numbers.

OK Larry, You are correct and I feel like it is time to explain this with the true numbers and set everyone's mind at ease. This discussion has been absolutely wonderful for us and since Sunday PM we have sold all but nine from our first run of 400 wrenches. Thank you to everyone for your support.
So here goes! First lets get the numbers established we are working with. To make this easier we will use 100 ft lb as the called out torque (ft lb is based at 12"). So 12" = 100% without our wrench. Our prop wrench is exactly 3" center distance. You take the 12" divided by 3" = 4 times or 25%. Now you add the 12"+3"=15" or 125% with no correction you would be over torquing to 125 ft lb with our adapter. So you must divide the 100 ft lb call out by 1.25. 100 / 1.25 =80.
thus the .8 multiplier. Using a 3" adapter you must set your ft lb torque devise (what ever it may be) to 80 ft lb to achieve 100 ft lb at the bolt. This is the math and nothing will change this fact. I hope I haven't offended anyone by presenting this in it's simplest form. I just want to clarify this to everyone. Love this forum and all who use it. Allan
.
PS Keep an eye on our web site as we will be introducing several new products in the coming weeks.
.

schristo@mac.com 02-16-2012 06:53 PM

How are the halves bonded together?
 
Quote:

Originally Posted by PerfTech (Post 628472)
Ultimate C/S Prop Wrench
We have completed our testing and development on this product and now are offering them for sale. Please have a look at the demonstration video below. If you have a constant speed propeller or work on airplanes you will definitely want one of these. Let us know your opinions. Regards all, Allan
.
http://antisplataero.com/Videos.html
.
http://antisplataero.com/Products.html

The wrench appears to be made from two pieces bonded together to double the thickness... how are the two halves bonded together?

Mike S 02-16-2012 07:01 PM

One more try:eek:

I will use the same 100 #/ft as before to keep the math easy

Torque is properly expressed as "pounds force foot", not "foot pounds"

It does not matter is the torque is generated by by a hundred pounds of force on a one foot lever arm, or a fifty pound force on a two foot arm or a one pound force on a hundred foot arm, or two hundred pounds of force on a half foot arm, it will all equal a torque of 100 pounds force foot.

Remember torque and force are not the same thing.

Torque is a form of force, as related to moving something around a center.

Force is a straight line effort to move something.

Going back to the problem at hand, the end point of whatever kind, or size of torque wrench is seeing a torque of 100 pound force feet. It does not matter what kind or size wrench generated this torque, or if it was even a wrench---could be an electric motor, a spring, or pixies dancing on a toothpick, just forget about how the torque is created. If we add an extension of .three inches (25 foot) to the arm, what does that do to the to the torque????

It is no longer 100 pounds force foot, it is now 125 pounds force foot, at the end of the added arm.

To get the torque back to our target of 100 pounds force foot, we need to reduce the force side of the equation by 100/125, or .8.

All the diagrams with pivot points and such are not showing torque, they are showing force-------they are lever diagrams.

At least that is how I see things, YMMV.

Like you-know-who always says, my $0.02

And yes, it is a pretty strong toothpick.

PerfTech 02-16-2012 07:10 PM

Quote:

Originally Posted by schristo@mac.com (Post 630006)
The wrench appears to be made from two pieces bonded together to double the thickness... how are the two halves bonded together?

You are correct! They are laminated together and plug welded two places in the beam. This is done to give it extra strength as the grain structure is omnidirectional and the heat treating processes are done before the two are joined together.This process also enables us to hold much tighter tolerances with our laser cutter due to the thinner material. Thanks, Allan

CMW 02-16-2012 07:20 PM

Ok one more
 
Here's a video I just made answering one of the questions that I think Larry was asking and Steve clarified. Does it matter where you hold a clicker type torque wrench. Intuitively you would think it shouldn't matter much. But as you can see using a Snap-On 100 ft-lbs clicker torque wrench set to 25 ft-lbs (300 in-lbs) and pushing against a Snap-On 600 in-lbs dial torque wrench, there is a significant error when pushing in the middle of the wrench instead of the handle. I don't know exactly how the mechanism works but I know it's calibrated using the handle.

http://youtu.be/_1LZk0jnQrA

By the way, this clicker wrench just happens to have an effective length of 12 inches so it should work perfectly with Allan's wrench and a .8 multiplier.

PerfTech 02-16-2012 07:30 PM

Quote:

Originally Posted by CMW (Post 630018)
Here's a video I just made answering one of the questions that I think Larry was asking and Steve clarified. Does it matter where you hold a clicker type torque wrench. Intuitively you would think it shouldn't matter much. But as you can see using a Snap-On 100 ft-lbs clicker torque wrench set to 25 ft-lbs (300 in-lbs) and pushing against a Snap-On 600 in-lbs dial torque wrench, there is a significant error when pushing in the middle of the wrench instead of the handle. I don't know exactly how the mechanism works but I know it's calibrated using the handle.

http://youtu.be/_1LZk0jnQrA

By the way, this clicker wrench just happens to have an effective length of 12 inches so it should work perfectly with Allan's wrench and a .8 multiplier.

Yes, you should always use all torque wrenches as designed. With our tool you need the .8 multiplier, always. Please read my post #203.Thanks, Allan

Mike S 02-16-2012 07:35 PM

OOPS
 
I just realized my last post left out a bit of information that may be caught by some of the sharp eyed folks here ...........

When the 3" tool is added, the rotational point is displaced a bit, and will probably cause a very minor error to be induced into the formula.

Probably some global pi r cubed thing, factored at arc sine zero, or thereabouts.

Any math majors out there want to clarify???

scsmith 02-16-2012 07:40 PM

Allan, your math is incomplete
 
I'd like to suggest that you should really open your mind to the possibility that you are seeing this incompletely and have something to learn from the several educated sources here that are trying to help.

Look what happens to your math if we agree to use metric units for your equation? Nature doesn't know anything about measurement units. There's nothing magic about a 1 ft length basis for a torque measurement.
What if we use Newton-Meters? Your extenstion wrench is 7.6% of a meter long. So why wouldn't the torque increase be 1.076?

What is missing from your math is the added torque from the shear force at the head of the wrench acting over the length of the extension. Study Bob Kuykendall's figures, or the figures in AC43.13. They are correct.

This is more than an academic arguement because people need to be able to use your great product in a way that insures that they are safe.

By the way, we have far exceeded the expected number of posts for Goodwin's Law to kick in. I applaud everyone's patience.

Quote:

Originally Posted by PerfTech (Post 630002)
OK Larry, You are correct and I feel like it is time to explain this with the true numbers and set everyone's mind at ease. This discussion has been absolutely wonderful for us and since Sunday PM we have sold all but nine from our first run of 400 wrenches. Thank you to everyone for your support.
So here goes! First lets get the numbers established we are working with. To make this easier we will use 100 ft lb as the called out torque (ft lb is based at 12"). So 12" = 100% without our wrench. Our prop wrench is exactly 3" center distance. You take the 12" divided by 3" = 4 times or 25%. Now you add the 12"+3"=15" or 125% with no correction you would be over torquing to 125 ft lb with our adapter. So you must divide the 100 ft lb call out by 1.25. 100 / 1.25 =80.
thus the .8 multiplier. Using a 3" adapter you must set your ft lb torque devise (what ever it may be) to 80 ft lb to achieve 100 ft lb at the bolt. This is the math and nothing will change this fact. I hope I haven't offended anyone by presenting this in it's simplest form. I just want to clarify this to everyone. Love this forum and all who use it. Allan
.
PS Keep an eye on our web site as we will be introducing several new products in the coming weeks.
.



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