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NEW! Ultimate C/S Prop Wrench Anti Splat Aero

nauga said:
*sigh*

If I have a 12" torque wrench and I apply a 80 lb force at the grip there will be a 80 ft-lb torque at the head. If I add a 3" extension (0.25 ft) straight out from the head that 80 lb will still produce 80 ft-lb at the head but it will produce 100 ft-lb at the end of the extension. That's what everyone wants. Problem is, if I use an 18" wrench, I will apply a 53.3 lb force at the handle to get 80 ft-lb at the wrench head. The torque at the bolt will be 53.3lb*(1.5+0.25 ft), or 93.3 ft-lb.

Might be within tolerances, might not. It's better to know than to guess.
Click to expand...

You're still incorrect. The difference is where the 3" extention is applied. If it's between the nut & wrench head, then the .80 applies. It doesn't matter how long the handle of the wrench is, even with cheater bars, if the 3" between the nut and wrench head stays at 3".

As I previously said, that nut is still going to turn the same distance
to satisfy it's torque setting, wheather you apply 80 lbs or 53.3 pounds at the end of the handle, depending on length. Additional length of handle, only changes the required amount of force, that you apply to turn that nut the same amount of degrees.............as long as the original 3" remains the same.
 
PCHunt said:
The reason length doesn't matter is that the torque wrench is indicating the torque on the nut, and not the force on the end of the wrench.

A wrench handle can be any length, and if the torque reading is calibrated at the nut end of the wrench, it will tell you the torque on the nut, which is what you want.

So, wrench #1 is 10 feet long, with a dial that indicates ft-lbs (or more correctly lb-ft) at the NUT. if you want 100 pounds of torque, you would apply a force of 10 lbs at the handle, and the dial would read 100. (10 feet X 10 lbs = 100 ft-lbs)

Wrench #2 is 1 foot long, with a dial that reads torque at the NUT. If you want 100 lbs of torque, you would apply a force of 100 lbs at the HANDLE, and the dial would read 100. (1 foot X 100 lbs = 100 ft-lbs)

You guys are confusing the nut torque with the force needed at the handle.

All torque wrenches read TORQUE AT THE NUT, not force on the end of the handle.

That's why the length of the handle doesn't matter for torque wrenches.

Taking all bets! :)
Click to expand...


HEY! I get it now, thanks to a few of the posts above. When you set the torque wrench for 100 ft lbs, by definition, you are going to have a force of 100 lbs applied 1 foot from the end of the wrench regardless of how long your wrench is, so that's where the 12 inches comes from. When you add the 3 inch extension, you are applying that 100 lbs to the close end of the extension, but more than 100 lbs to the nut at the far end of the extension. Hence the multiplier of 12/(12 + 3) = 0.8

WOO HOO! Now I can sleep.

Erich
 
I got an idea

Everyone agrees its a great tool. Lets go build some RV's!!!! Then we can use the tool.
 
physics 101, or mechanics

Let's assume I own a torque wrench which is two feet long. If I wanted to torque a fastener to 100 Ft-lbs. I would apply 50 lbs.

50lbs. * 2 ft. = 100 ft-lbs.

I buy the tool and follow the directions. I want 100 ft-lbs. of torque at the fastener, so I would set the wrench to 80 ft-lbs. (.8 * 100 = 80). That means I would apply 40 pounds of force to my end of the torque wrench, or 80 ft-lbs. to the drive end of the torque wrench.

40 lbs. * 2 ft. = 80 ft-lbs. which is the recommended torque wrench setting.

Now I am ready to proceed. I attach the tool which adds 3 inches, or 1/4 foot to the system. Now my system is 2.25 feet long. (2 for the wrench + .25 for the tool) I apply the 40 pounds force to the end of the tool and the wrench clicks. I now have applied 40 lbs. * 2.25 ft. = 90 ft-lbs.
 
I believe!

Now I am ready to proceed. I attach the tool which adds 3 inches, or 1/4 foot to the system. Now my system is 2.25 feet long. (2 for the wrench + .25 for the tool) I apply the 40 pounds force to the end of the tool and the wrench clicks. I now have applied 40 lbs. * 2.25 ft. = 90 ft-lbs.

No, because your wrench doubles your efforts first at the attach end of your torque wrench making it apply 80 lb feet there and the 3 inch extension adds 25 percent to that making 100 lbs.

It really works the way Allan says!!!
 
Red wrench, blue wrench...

redwrenchbluewrench.jpg


Apparently Allan is a marketing genius...
 
Science vs. Religion?

I believe!
Click to expand...
-Religion? ;)

Now I am ready to proceed. I attach the tool which adds 3 inches, or 1/4 foot to the system. Now my system is 2.25 feet long. (2 for the wrench + .25 for the tool) I apply the 40 pounds force to the end of the tool and the wrench clicks. I now have applied 40 lbs. * 2.25 ft. = 90 ft-lbs.
Click to expand...
-Science ;)

No, because your wrench doubles your efforts first at the attach end of your torque wrench making it apply 80 lb feet there and the 3 inch extension adds 25 percent to that making 100 lbs....It really works the way Allan says!!!
Click to expand...
-Religion? ;)

.25 is not 25 percent of 2 feet, but 25 percent of 1 foot
- Science :D
 
I've just had a quick look at the thread, and there seems to be quite a mix of fact and fiction here, so thanks to schristo for showing the free body diagram.

Torque is the product of a force and the perpendicular distance from that force to the point of application.

So if you're applying the force at 90 degrees to the wrench then the torque measured (at the wrench) is that force times distance to the end of wrench, and the torque applied to the nut is force times distance from wrench handle to the nut.

So the correction is the ratio of those two distances. Put the crowsfoot at 90 degrees and the distances are the same, so no correction, but otherwise the correction will depend on the length of wrench and length of crowsfoot offset.

schristo has it right.
 
Has everything to do with science, nothing about religion....

Any torque wrench is set to measure the torque where it attaches.... 6 inches long or 4 feet long matters not! It is set to measure the applied force at the attach end. So that measure is constant no matter how long Your wrench is (pun intended) the only thing that changes is the .25 foot long extension that you attach to the end of your torque wrench and that is why the correction is always the same!!

My opinion, I believe it, has not one thing to do with religion. ;)
 
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How to Test?

This discussion raises a question that has concerned me. I have two torque wrenches (cheap and expensive)

Is there a way that I can use wrench B to test the torque applied by wrench A?

Or, can I use my wrench that's newly calibrated to test the torque applied to nuts before calibration?

Or, can I use my basic torque wrench to test the torque applied when I use the device discussed in this thread to see if 0.8 really is the correct value?

Or in general, do I just trust my torque wrench or is there a way I can test it?
 
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The length of the wrench does matter.
Have a look at the formula for calculating crows foot extension.

Wrench torque = Torque Spec * (Wrench length / (Wrench length + Crows foot extension length))

You can't just multiply your desired torque by .8 and expect to get the value to place into your torque wrench. This .8 value works for a 12" / 1 foot torque wrench only.

The attached image might help clear things up a bit. It assumes that you want to apply 100 FT-LB of torque to a bolt using a 3" extension using three different length torque wrenches. A one, two, and ten foot torque wrench. If you just multiply the 100FT-LB value by .8 and dial in 80 FT-LBS in your torque wrench, you can calculate the different values you will obtain.

torque.jpg


As you can see above, you won't be getting 100 FT-LBS of desired torque unless you are using a 1ft torque wrench.
Stick with the formula and it will tell you what value to put into your torque wrench.
 
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Excellent drawing!!!

The change only happens between A and B because all the wrenches apply 80 lb ft of torque to the "B" location and B x 1.25 = 100 lb ft of torque....

I believe!.....:)
 
Genius Alan - pure genius! You've managed to keep your ad at the very top of the "New Posts" for over a day! ;)

I'm putting one of your wrenches on my wish list. Now, of course, you need to make one that fits the prop bolts on our WW 151 (I think they are 5/8" heads, but I might be wrong...) - they are different from our Hartzells.
 
PCHunt said:
The reason length doesn't matter is that the torque wrench is indicating the torque on the nut, and not the force on the end of the wrench.

Yep

A wrench handle can be any length, and if the torque reading is calibrated at the nut end of the wrench, it will tell you the torque on the nut, which is what you want.

Yep again

You guys are confusing the nut torque with the force needed at the handle.

All torque wrenches read TORQUE AT THE NUT, not force on the end of the handle.

That's why the length of the handle doesn't matter for torque wrenches.

Taking all bets! :)
Click to expand...

Only problem here is torque is a rotational force, around a center point.

Adding the extension changes the location of the center point of rotation, and the head on the torque wrench is now moving in an arc, not rotation.

All that said, I still think the .8 formula is correct, not too sure of the logic presented above-------Pete's or mine:eek:
 
Ironflight said:
Genius Alan - pure genius! You've managed to keep your ad at the very top of the "New Posts" for over a day! ;)

I'm putting one of your wrenches on my wish list. Now, of course, you need to make one that fits the prop bolts on our WW 151 (I think they are 5/8" heads, but I might be wrong...) - they are different from our Hartzells.
Click to expand...

"Thank You"!!!!!
It is slowing down a little on the posts so apparently I am going to have to stir it up with the announcement of the new coating we are thinking about changing over to. It's called "Skuzmolimide 151" and I am absolutely positive that we have many experts here with lots of experience in the use and application of this product ( any and all information can only help). Thanks, Allan:D
 
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Ironflight said:
Genius Alan - pure genius! You've managed to keep your ad at the very top of the "New Posts" for over a day! ;)

I'm putting one of your wrenches on my wish list. Now, of course, you need to make one that fits the prop bolts on our WW 151 (I think they are 5/8" heads, but I might be wrong...) - they are different from our Hartzells.
Click to expand...

Yeah, Alan, you don't say what Size the wrench is! I have a Hartzell C/S prop on my O320 that I'm certain has a different bolt size (5/8") than that on a O360 :). I would love one of your wrenches, but it needs to be the right size :).
 
5/8" C/S Prop Bolts

scard said:
Yeah, Alan, you don't say what Size the wrench is! I have a Hartzell C/S prop on my O320 that I'm certain has a different bolt size (5/8") than that on a O360 :). I would love one of your wrenches, but it needs to be the right size :).
Click to expand...

Sorry about that Scott, you are correct I somehow overlooked that. The wrench is 3/4" as that is the only size we have seen on C/S props. If we receive the requests for a 5/8" we will make those also. I will contact the propeller manufacturers again today and find out the approximate ratio of 5/8" bolt use. Thank you for pointing this out to us. Allan:)
 
The Paintman said:
He has done the math for us
Click to expand...

So he says, but he hasn't showed his math to us. Meanwhile nauga and schristo and others have shown math that indicates that the 0.8 correction factor is a special case, and depends on the force on the torque wrench being applied one foot from the torque wrench head.

I suppose Allan is too busy filling orders to do it, but if anyone else wants to step up and show your derivation of a constant 0.8 correction factor, have at it. And while you are at it, show the derivation for the correction factor for torque wrenches that read in-lbs and newton-meters, as well as a ft-lbs one.

--Paul
 
Garage Guy said:
So he says, but he hasn't showed his math to us. Meanwhile nauga and schristo and others have shown math that indicates that the 0.8 correction factor is a special case, and depends on the force on the torque wrench being applied one foot from the torque wrench head.
Click to expand...

The math is incorrectly applied. That click you hear and feel is from pre-set spring tension, depending on the torque setting. It won't make a difference if the handle is 12, 24, or 36 inches..........as long as the same torque number is used (such as 100 foot pounds), and the "tool" remains at 3" length.

What you "feel" while pushing the handle will be different. But the nut won't be turning any more, or any less.........just because the handle is a different length.
 
erich weaver said:
When you set the torque wrench for 100 ft lbs, by definition, you are going to have a force of 100 lbs applied 1 foot from the end of the wrench regardless of how long your wrench is
Click to expand...

Or, you might have a force of 50 lbs applied 2 foot from the head of the torque wrench. Either way the wrench reads 100 ft-lbs of torque at its head.

But in the first case the torque at the nut is 100 lbs through a 1.25 foot arm, 125 ft lbs, for a conversion factor of 0.8. In the second case that torque at the nut is 50 lbs through a 2.25 foot arm, 112.5 ft lbs, for a conversion factor of 0.89.

so that's where the 12 inches comes from. When you add the 3 inch extension, you are applying that 100 lbs to the close end of the extension, but more than 100 lbs to the nut at the far end of the extension. Hence the multiplier of 12/(12 + 3) = 0.8
Click to expand...

Follow your logic for a torque wrench that reads in in-lbs. Set it for 1200 in-lbs; by definition that is 1200 lbs 1 inch from the torque wrench head. That's where the 1 inch comes from. When you add the 3 inch extension, you are applying 1200 in-lbs to the close end of the extension, but more at the nut. Hence a multiplier of 1/(1+3) = 0.25.

Something is wrong with that logic! 1200 in-lbs just is 100 ft-lbs. Changing the labels on the torque wrench certainly shouldn't change the multiplier...

WOO HOO! Now I can sleep.
Click to expand...

Sorry!

--Paul
 
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Another way of demonstrating.

A foot lb. is the force of 1 lb., 12" perpendicular to the pivot point.

If we take 100 lbs, and set it 12" from the pivot point, then it's 100 ft. lbs torque against the pivot point.

If we use a 24" or 36" wrench, we are NOT moving the 100 lbs past the 12" spot on the wrench. It remains at the 12" point. By adding a longer handle, we push with less force, to move that 100 lbs.

With the addition of the 3" tool, the 12" point (which is used for the "foot" in ft/lbs. is now 15". And that's why a fraction is used. 3" & 12" remain as constants.
 
Science

So he says, but he hasn't showed his math to us. Meanwhile nauga and schristo and others have shown math that indicates that the 0.8 correction factor is a special case, and depends on the force on the torque wrench being applied one foot from the torque wrench head.

I suppose Allan is too busy filling orders to do it, but if anyone else wants to step up and show your derivation of a constant 0.8 correction factor, have at it. And while you are at it, show the derivation for the correction factor for torque wrenches that read in-lbs and newton-meters, as well as a ft-lbs one.
- Paul
Click to expand...

It would be nice if everyone saw the math and agreed, but somehow I don't foresee that outcome. I could devise an experiment to prove that the torque wrench length matters. That would be science and perhaps wouldn't satisfy everyone either? Perhaps it could settle a bet? anyone?
 
To put it in perspective

My torque wrench Proto model J6014C has instructions and a formula
They use 22.57 for the handle length . If you used there formula the multiplier would be .882 (for that wrench only)

If you use a 3 " extension and want to apply 80ft lbs of torque to the bolt
you would set the wrench at 70.56 lbs

If the wrong multiplier was used for that wrench .8 it would set at 64 lbs

They would be under torqued by 9.6% if .8 was used with this wrench

The best thing to do is check with the manufacture of the torque wrench used and you should have the required information.

Be Safe

Tim
 
Easy Demo

An easy way to demo the induced torque would be to use the tool and a linear (non-settable) torque wrench. Torque a bolt with the extension, then read the required torque to loosen without it.
 
Lux Wrangler said:
It would be nice if everyone saw the math and agreed, but somehow I don't foresee that outcome. I could devise an experiment to prove that the torque wrench length matters. That would be science and perhaps wouldn't satisfy everyone either? Perhaps it could settle a bet? anyone?
Click to expand...

Sure, try it.

The wrench will still click at the exact same spot, and the nut rotation will still stop at the exact same spot. I don't care if the handle is 3' or 30'.
It's still based on that 12" (for ft/lb.) & 3". Those numbers don't change.
 
380mxc said:
My torque wrench Proto model J6014C has instructions and a formula
They use 22.57 for the handle length . If you used there formula the multiplier would be .882 (for that wrench only)

If you use a 3 " extension and want to apply 80ft lbs of torque to the bolt
you would set the wrench at 70.56 lbs

If the wrong multiplier was used for that wrench .8 it would set at 64 lbs

They would be under torqued by 9.6% if .8 was used with this wrench
Click to expand...

80 * .08 = 64 lbs. at 12"

3 more inches is .25 of 12"

64 * the additional .25 = 16

64 + 16 = 80
 
Larry

You are correct. Using your example, at 12" the applied force required is 64lbs. The three inch extension will experience 80 FT-LBS of torque. Your math example assumes a torque wrench that is calibrated at 12 inches.

The problem arrises when Tim (380mxc) has a torque wrench that calibrated at 22.57 inches.

Torque = force X distance

For a 12" torque wrench with a 3" extension, setting 64 FT-LBS results in 80 FT-LBS at the bolt. 64LBS x 1.25feet = 80 FT-LBS

For the 22.57" torque wrench with a 3" extension, setting 70.56 FT-LBS results in 80 FT-LBS at the bolt.

A force of 37.23LBS at 22.57 inches (1.88feet) = a torque of 70.56 FT-LBS
Adding a 3" extension to the equation. 37.23LBS x 25.57 inches (2.13feet) = 80 FT-LBS

Tim can't just multiply torque required by .8 and plug that into his torque wrench. His torque wrench manual values line up with the extension formula.

Wrench torque = Torque Spec * (Wrench length / (Wrench length + extension length))
 
Mike,

Adding the extension changes the location of the center point of rotation
Click to expand...

With the extension, now there are two center points of rotation, one at the torque wrench head, the other at the nut. This whole discussion is about the multiplier that relates the torques applied around these two axes.

and the head on the torque wrench is now moving in an arc, not rotation.
Click to expand...

Well if we're measuring static torque, nothing is moving at all, but you are getting at a key point. The head on the torque wrench has both a torque moment around the head axis, and is applying a force perpendicular to the extension wrench arm. The torque wrench is directly measuring the first of these (that's what torque wrenches do) and is not measuring the second one directly at all. (But you can compute the second one if you know the torque wrench and extension wrench lengths.) I think a big part of the confusion is folks confusing the two.

--Paul
 
Question

In the video the wrench appears to be made of 2 thinner pieces based on the appearance. is it? Or is it one piece with a line around it? Thanks Marty
 
nauga said:
This is supposed to be a prop wrench, how 'bout a prop service bulletin for reference?

http://www.mccauley.textron.com/sb227B.pdf, see the last page.
Click to expand...

To save everyone the trouble of an extra click, the McCauley Service Bulletin above says that when using a torque wrench with an extension:

dial reading = torque wrench length (ft) x desired torque (ft lbs) / [torque wrench length + extension length (ft)]


Clearly this falls into "the 0.8 multiplier is not a constant" camp. Im back to not knowing what to think. Could this SB really be wrong?

erich
 
erich weaver said:
To save everyone the trouble of an extra click, the McCauley Service Bulletin above says that when using a torque wrench with an extension:

dial reading = torque wrench length (ft) x desired torque (ft lbs) / [torque wrench length + extension length (ft)]


Clearly this falls into "the 0.8 multiplier is not a constant" camp. Im back to not knowing what to think. Could this SB really be wrong?

erich
Click to expand...

OMG... this is the same thing that AC 43-13 says, could it actually be that all these documents have been wrong all these years and remarkably the brilliant minds on VAF have finally discovered the truth :eek:

As I said before, show me the math and I'll be a believer, but I still haven't seen it.

All ft pound wrenches are not 1 ft long which is what this "theory" is based on. If every torque wrench was 1 ft long then Allen would be right, end of story.
 
Allan,

Looks like the Retail Price of your wrench has gone up... include a 12" 3/8" drive Snap On high quality torque wrench in the wrench kit and then nobody will have the need to do the math. :D
 
Take 3 torque wrenchs, one 1 ft long one 2 ft long and ine 3 ft long. All are calibrated for 100 ft lbs torque.
So even though they are different lengths, they all put exactly the same 100 ft lbs of torque at the head.
So all three torque wrenches put exactly 100 ft lbs of torque at the head and then you add the exact same 3 " extension to them. Wouldnt the 3 " extension have the same effect on the 100 ft lbs of torque at the head on all of them?

Just saying...:D
 
rvator51 said:
Take 3 torque wrenchs, one 1 ft long one 2 ft long and ine 3 ft long. All are calibrated for 100 ft lbs torque.
So even though they are different lengths, they all put exactly the same 100 ft lbs of torque at the head.
So all three torque wrenches put exactly 100 ft lbs of torque at the head and then you add the exact same 3 " extension to them. Wouldnt the 3 " extension have the same effect on the 100 ft lbs of torque at the head on all of them?

Just saying...:D
Click to expand...

.
EXACTLY!!!!!!!!:D
You got it perfect, Thanks, Allan
 
rvator51 said:
Wouldnt the 3 " extension have the same effect on the 100 ft lbs of torque at the head on all of them?
Click to expand...
No, because it's the force you put on the end of the wrench times the distance from the end that generates that torque, it doesn't just magically appear. As the arm gets longer the force to produce 100 lb at the head goes down, and it's this force applied over the total length, wrench plus extension, that produces a torque at the extension end.
 
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There are many publications out there that have printed it wrong I guess. It sure would be easy to prove.
After reading the last 10 pages, the only way I would use this product would be with the wrench at 90 degrees from the adapter..... easy to do with this application.

At least it's not a primer or F/P issue......................:D
 
rvator51 said:
Take 3 torque wrenchs, one 1 ft long one 2 ft long and ine 3 ft long. All are calibrated for 100 ft lbs torque.
So even though they are different lengths, they all put exactly the same 100 ft lbs of torque at the head.
So all three torque wrenches put exactly 100 ft lbs of torque at the head and then you add the exact same 3 " extension to them. Wouldnt the 3 " extension have the same effect on the 100 ft lbs of torque at the head on all of them?

Just saying...:D
Click to expand...
WRONG!!!!!!!:D:D

You got it wrong, sorry
It's already been explained correctly a few times.

The nut sees a torque of applied force * distance from handle to nut.
The wrench gauge sees a torque of applied force * distance from handle to drive end of wrench.
So correction is the ratio of those lengths.
Put the crowsfoot at 90 degrees; the lengths are the same so the correction is zero.
 
she's a witch!

Many of these posts are pretty humorous, I wonder if I use a large red font if it will have more impact...

It is interesting to see various responses and who makes them; I must be the dumbest guy in the forum; perhaps PI really is 3.

"Who are you, who are so wise in the ways of science?"
http://youtu.be/zrzMhU_4m-g
 
Larry, this is great, this is the closest I've seen to someone actually laying out a mathematical argument for a constant conversion factor.

Now my question is, if this is a good argument, why not apply to a torque wrench that reads inch-pounds. Just change foot to inch:

L.Adamson said:
Another way of demonstrating.

A inch lb. is the force of 1 lb., 1" perpendicular to the pivot point.

If we take 100 lbs, and set it 1" from the pivot point, then it's 100 in. lbs torque against the pivot point.

If we use a 24" or 36" wrench, we are NOT moving the 100 lbs past the 1" spot on the wrench. It remains at the 1" point. By adding a longer handle, we push with less force, to move that 100 lbs.

With the addition of the 3" tool, the 1" point (which is used for the "inch" in in/lbs. is now 4". And that's why a fraction is used. 3" & 1" remain as constants.
Click to expand...

If I'm following you, now you have a conversion factor of 1/(1+3), or 0.25. That is, if you want a certain torque X in-lbs at the nut, you would set your torque wrench to 0.25 X.

OK so far?

--Paul
 
Allan,

New Service Bullet
SB 2012-2-14

For everyone purchasing this wrench, do your own math calculations for torque value. :D
 
So tell me, wise ones...

A click wrench uses a spring and pawl, which releases when the desired torque setting is hit. This compressed spring is calibrated to match ft/lbs., which is based on one pound of force, at one foot from the pivot point. The one foot is only a measurement of force. The wrench does not have to be a foot long.

How is this spring going to know, if you're grasping the wrench at the exact same spot on the shaft. Does it care? No it doesn't! The force you apply at the handle depending on length, varies. But the spring just doesn't care. It's going to release at the setting it's set to...........which is based on one pound at one foot, for ft/lb settings.

edit: And I'm going to add a 100' extension, pulled by block and tackle, connected to a cement truck.................and the spring still won't care, or notice the difference!!!!
It will just merely click, when it reaches the set point..

L.Adamson
 
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