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L.Adamson said:
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.
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The spring will release when the torque *at the wrench head* reaches the value you have set (assuming the wrench is calibrated). This will be will be consistent regardless of the extension you have attached as long as the setting does not change. The torque at the wrench head will equal the force you've applied times the distance from the point of application of that force to the wrench head.

The torque at the end of the extension, however, doesn't simply scale as a function of torque and distance, it's a function of *force* and distance, the force being the force you've applied and the distance being the distance from the point of application of the force to the end of the extension. It does not matter that the torque in/on the system somewhere in between (say, at the wrench head) is less, the torque at the extension end is a function of total distance (wrench plus extension) times force applied.
 
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The torque on the nut is not being applied at the drive end of the torque wrench. It is being applied by the force on the handle.

If you applied a pure torque to the drive adaptor in the prop wrench (e.g. by using a pneumatic nut driver) you actually wouldn't apply any torque at the nut end of the prop wrench. (Imagine putting a screwdriver handle in the drive hole; it wouldn't matter how long the wrench was, you would just generate a linear force at the other end, not a torque)

I think this debate is a bit like the aircraft on a conveyor belt saga.
 
nauga said:
The spring will release when the torque *at the wrench head* reaches the value you have set (assuming the wrench is calibrated). This will be will be consistent regardless of the extension you have attached as long as the setting does not change. The torque at the wrench head will equal the force you've applied times the distance from the point of application of that force to the wrench head.

The torque at the end of the extension, however, doesn't simply scale as a function of torque and distance, it's a function of *force* and distance, the force being the force you've applied and the distance being the distance from the point of application of the force to the end of the extension. It does not matter that the torque in/on the system somewhere in between (say, at the wrench head) is less, the torque at the extension end is a function of total distance (wrench plus extension) times force applied.
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I appreciate the long explanation, but the "spring" has perhaps already clicked........before your potential force and distance ever got there.

Your correct about force and distance in relation to absolute torque. But it has nothing to do with the spring setting. The spring only cares about the correct torque applied to the nut. Not the potential torque that you can add to it, by the force of leverage.
 
L.Adamson said:
The spring only cares about the correct torque applied to the nut.
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No, the spring responds to, and is calibrated to, the torque at the wrench head. The torque is different at any other position in the system and is only a function of force and distance from that force to the point in question.
 
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This would be a really nice place for someone with the proper equipment to set up a demonstration. Easier to do here than with "the downwind turn."
 
The formulas that all of you are getting stuck on are only for simple levers and pivots.

Torque wrenches when you include the indicating mechanizm are complex machines not simple levers.

The key is the calibrated indicating method of the torque being applied. If you take a typical high quality aircraft torque wrench and set it to click at 100ft pounds and pull it on a nut in stock configuration, it should click when there is 100ft pounds on the nut. How much you have to pull is based on how long the wrench handle is. Remember the clicker is calibrated to a certain torque on the nut driver shaft.

Now take the same wrench and add a 3 ft extension to the handle. Pull the handle till the clicker clicks.....You still get 100ft pounds at the nut as long as you stop pulling when the clicker clicks. The amount you have to pull is reduced because of the longer handle. The calibrated clicker still clicks at 100ft lbs at the nut driver shaft.

If you take all the indicating mechanizms away and use a spring scale on the end of a simple lever, all of these textbook formulas work just fine with including the lever length.
 
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Well said Brian...

"If you take all the indicating mechanizms away and use a spring scale on the end of a simple lever, all of these textbook formulas work just fine with including the lever length."

I will add that if the attachment was welded on to the torque wrench then the over-all length would matter but it is not, the torque is all measured and applied to the attachmnet at a point 3 inches from the nut and this gives the additional leverage that must be compensated for.....
 
Not to ruin the fun, but I think it's way past time for the moderators to move this out of the classifieds section. I asked a question last week (on page 2) as to whether this works on a MT prop and there was so much talk about how to torque a prop that no one answered my question that could actually turn into a sale....

We are on page 11 now, and nothing in the last 8 pages has anything to do with the classifieds. Let's move this to the proper list.

And I'm still hoping someone can answer my question on whether this works on an MT 3 bladed RV-10 prop? The web site mentions Hartzell....

Thanks
-Mike Kraus
RV-10 Flying
 
mkraus01 said:
Not to ruin the fun, but I think it's way past time for the moderators to move this out of the classifieds section. I asked a question last week (on page 2) as to whether this works on a MT prop and there was so much talk about how to torque a prop that no one answered my question that could actually turn into a sale....

We are on page 11 now, and nothing in the last 8 pages has anything to do with the classifieds. Let's move this to the proper list.

And I'm still hoping someone can answer my question on whether this works on an MT 3 bladed RV-10 prop? The web site mentions Hartzell....

Thanks
-Mike Kraus
RV-10 Flying
Click to expand...

Nawwww, let's leave it in classifieds so it will be deleted in a couple of weeks. :D
 
Brantel said:
The formulas that all of you are getting stuck on are only for simple levers and pivots.

If you take all the indicating mechanizms away and use a spring scale on the end of a simple lever, all of these textbook formulas work just fine with including the lever length.
Click to expand...

I guess Hartzell in wrong too... have a look at page 3-6 of the Hartzell manual.

http://www.hartzellprop.com/public_dl.php?id=4

Again, the .8 factor is ONLY for a 1 ft torque wrench, for any other length you need to do the math for your wrench.
 
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Walt said:
I guess Hartzell in wrong too... have a look at page 3-6 of the Hartzell manual.

http://www.hartzellprop.com/public_dl.php?id=4

Again, the .8 factor is ONLY for a 1 ft torque wrench, for any other length you need to do the math for your wrench.
Click to expand...

Hartzell is saying exactly what Allan, myself, and some others are saying.

How many torque wrenches, that you would be using on a prop.........are just one foot long? I doubt there are any!

That one foot figure has to do with 12" meaning a foot pound. That's the distance that torque values are measured from. The drawing isn't even close to being scale.

L.Adamson


P.S.--- the extensions, crows foot, or whatever, are required, because you can't just place a torque wrench directly over the prop nuts. There is a flange in the way.
As Hartzell says, if the extension is at 90 degrees, just use the normal reading.
 
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Since so many here, are ignoring the mechanical aspects of a torque wrench, as Brantel mentioned.....

We'll just all go out, and take an ordinary wrench & stick 100 lbs of dumbbell weights at the 12" mark, or 50 lbs. at the 24" mark. If the 3" extension is used, we'll put only 80 lbs. at the 12" mark. Note: It doesn't matter how long your wrench actually is. It's those 12", 24" or any calculated mark in between that matters. Figure in a bit of weight, for the shaft, too.

If using an actual torque wrench, then ignore the above...

L.Adamson
 
L.Adamson said:
Hartzell is saying exactly what Allan, myself, and some others are saying.

How many torque wrenches, that you would be using on a prop.........are just one foot long? I doubt there are any!
Click to expand...

Not to take away from your comments because you are correct, but the torque wrench I use from my neighbor is exactly 12", so they do exist. I will try to get the brand from him as it is perfect for the job with a 3" extension (yes, you need to do the math) The formula's from Hartzell, and every manual I have ever seen are correct. Not sure why anybody would argue what should be considered common knowledge.
 
JonJay said:
Not to take away from your comments because you are correct, but the torque wrench I use from my neighbor is exactly 12", so they do exist. I will try to get the brand from him as it is perfect for the job with a 3" extension (yes, you need to do the math) The formula's from Hartzell, and every manual I have ever seen are correct. Not sure why anybody would argue what should be considered common knowledge.
Click to expand...

Is that wrench for inch pounds only? A wrench that's capable of 50 to 150 ft/lbs., would be a tough one to use, at only 12" long.
 
It seems that this whole discussion is about whether the 0.8 multiplier with a 3" extension is universal for any torque wrench. I propose it is not.

Let's use the Hartzell formula:

(actual torque required) X (torque wrench length) / (torque wrench length) + (length of adapter) = Torque wrench reading to achieve required actual
torque

Example 1:

Actual torque required- 100 ft-lbs
Torque wrench length- 12" (1.0')
Length of adapter- 3" (0.25')

Formula: (100 ft-lbs x 1 ft) / (1 ft + 0.25 ft) = 100 ft-lbs / 1.25 ft = 80 lbs torque wrench reading

Example 2:

Actual torque required- 100 ft-lbs
Torque wrench length- 18" (1.5')
Length of adapter- 3" (0.25')

Formula: (100 ft-lbs x 1.5 ft) / (1.5 ft + 0.25 ft) = 150 ft-lbs / 1.75 ft = 85.71 lbs torque wrench reading

So, using the math that has been shown before seems to indicate that a 0.80 factor is only valid when the torque wrench is 1' long...

There is a relatively easy way to confirm or debunk the 0.8 universal multiplier theory. Three beam type torque wrenches (of at least two different lengths) and some adapters are required.

Take one torque wrench and clamp the shaft in a vise such that a torque applied to the end can give a reading (the captured wrench). Using an adapter attach the second torque wrench (the free wrench) to the captured wrench. Press the free wrench to 100 ft-lbs. The captured wrench should also read 100 ft-lbs.

Do the same test with the third wrench, of a different length. Again, both readings should be the same, as both wrenches, regardless of their length, are calibrated to read the same torque at the head.

Now, do the same test with the 3" extension.

If at 100 ft-lbs on the captured wrench, both free wrenches read the same, then the 0.8 multiplier is proven.

If at 100 ft-lbs on the captured wrench, both free wrenches read different, then the 0.8 multiplier is busted.

Hey, maybe we could get Kari from MythBusters involved!! :)
 
L.Adamson said:
Is that wrench for inch pounds only? A wrench that's capable of 50 to 150 ft/lbs., would be a tough one to use, at only 12" long.
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I found it very easy to use for the 60-70 ft lbs on my prop with a 3" extension. I will find out the brand and what the top end is, but yes, at 150 lbs with no extension it would take some grunt.
 
My 2 cents

Wow, I just picked up on this thread and have not been involved in the flurry of opinions up to this point. Seeing the chaos I hesitated to chime in but after considering that this could have safety of flight implications for some relying on the information they get here I decided to weigh in. We all bring different strengths to the discussion on this forum. There is a lot I don?t know but I do have a master's degree in mechanical engineering, so whatever that is worth, here are the key points I would offer:

1) A torque wrench is designed to accurately measure and set torque at the head of the torque wrench. To that extent, it does not matter if the handle is 12 inches or 24 inches; whether you apply the force at the end of the handle or in the middle of the handle. The wrench "clicks" when the torque applied at the head reaches the set value.

2) If an extension (or crows foot) is attached the torque applied at the wrench head will be different than the torque applied at the end of the extension unless the extension is turned 90 degrees to the axis of the wrench handle such that the moment arm for the applied force is the same for the extension as it is for the wrench head.

3) If you use an extension on a torque wrench rotate it 90 degrees to the wrench axis and you will be fine. Set the wrench to the same torque value you would without the extension and the same torque will be applied at the bolt head on the extension.

4) If you must use an extension straight in line with the handle of a torque wrench you should be very careful. The torque applied to the bolt will be greater than the value you set on the torque wrench and the correction factor depends on both the length of the extension and the moment arm of the force application. It is not just the length of the torque wrench handle that counts, it is also dependent on where you apply the force on the handle. The correction formula is

C = L / (E + L)

If the extension length E is 3 inches and the distance from the wrench head to the point on the handle where you apply the force (L) is 12 inches then the correction factor is 0.8. But realize it is not always simply 0.8 for a 3 inch extension. Notice I said ?the point where you apply the force?. Using an extension this way makes the true torque applied to the bolt dependent on your ?technique?. Not so when a torque wrench is used without any kind of extension.

I happen to work in aerospace where we build spacecraft and other aerospace equipment and we follow these guidelines in torquing fasteners. As a rule we avoid using extensions or crows feet whenever possible. If we have use an extension we turn the extension 90 degrees to the torque wrench handle. If we can?t do that, we are very careful to calculate the correction factor accurately based on the specific wrench and the force application technique.

I hope this helps.:)
 
ccsmith51 said:
So, using the math that has been shown before seems to indicate that a 0.80 factor is only valid when the torque wrench is 1' long...
Click to expand...

Okay.............

Start looking through catalogs, and look for torque wrenches in the 100 ft/lb. category...............and see how many you'll find at just 12" long. I've looked all morning, and found none. That .80 factor applies to 12" (1') as it relates to ft/lb torque. Since the measurement of torque that we're using to set the prop nuts is in ft/lbs, then it all makes sense.
 
jpharrell said:
I happen to work in aerospace where we build spacecraft and other aerospace equipment and we follow these guidelines in torquing fasteners. As a rule we avoid using extensions or crows feet whenever possible. If we have use an extension we turn the extension 90 degrees to the torque wrench handle. If we can?t do that, we are very careful to calculate the correction factor accurately based on the specific wrench and the force application technique.
Click to expand...

I'll assume, you won't be torquing too many Hartzell props, where and extension or crows foot, is a must. :)
 
No amount of math and physics will "fix" this -- some just don't get it and it's more important to believe than to learn.

Either way, for those "wrenchers" -- RTFM for your torque wrench and follow instructions in it.. Once you realize you were wrong all along, you have an option of creating your own torque wrench that works the way you describe it.. you WILL make a lot of money selling those. Heck, I'll buy one.. cause all the torque wrenches I own work the same way Hartzell, McCauley, AC 43.13, and every other publication written says...
 
Here is a question. Is it possible that the tried and true formula applies only to old school, beam type, wrenches?
The beam type handle has a pivot to insure that you are applying the pressure at one specific point on the handle. The "click" type do not. I assume you can apply pressure anywhere on the handle and it will click at the correct torque regardless. That is an assumption, but if it is true, the length of your click type wrench would not matter and the .8 multiplier would work on any of them since they are all calibrated at the head for foot pounds.
 
Jon.. no.. Go back and read post #64 by Marc Hudson... then re-read it 10 more times until you understand the importance of point "B" on the drawing.
 
Radomir said:
Jon.. no.. Go back and read post #64 by Marc Hudson... then re-read it 10 more times until you understand the importance of point "B" on the drawing.
Click to expand...

You know a thread is going on too long when I have to go back from #122 to #64. Sorry, I will go read that one.
 
Brantel said:
The formulas that all of you are getting stuck on are only for simple levers and pivots.
Click to expand...

I follow the thought process that argues for a constant 0.8 multiplier when using a ft lbs torque wrench, but this quote just isnt true according to the documentation that has been presented in this thread. The formulas in the Service Bulletin, the Hartzell manual and the AC are all essentially the same and specifically indicate to input the length of the TORQUE WRENCH into the equation. They do not qualify the formula as applying only to simple levers without torque settings, and they most certainly DO NOT say to just enter 1 ft into the formula if your torque wrench reads in ft lbs.

If the "0.8 is a constant" folks are somehow correct here, it is disturbing to think that so many people have misinterpreted or been deceived by all the published documentation to date for so long. Makes me think that perhaps the documents have it right after all.

erich
 
True dat!! The thread would have been a lot shorter if everyone just went and read manual for their torque wrench :) and from what I gather they all read the same.. PS.. there are several other good explanations on this thread by several other folks... maybe diferent styles would help.

Either way, you actually don't need to "understand" any of this.. just follow instructions from the manual, or 43.13, or any other publication.. they all show the same formula... there really should be no 100+ posts about this :)
 
Radomir said:
Jon.. no.. Go back and read post #64 by Marc Hudson... then re-read it 10 more times until you understand the importance of point "B" on the drawing.
Click to expand...

Ok, I read that, works great for a simple beam, but not really sure it applies to click type wrenches. I just dont know that much about how they work internally. I will accept that they operate exactly the same as a beam type but they don't, otherwise they would have a pivoting handle to make sure you apply the force at a set point on the wrench. Click type don't have that so I assume they work differently.
PS - I dont have a dog in this fight, just want to know what is correct. I am lucky the wrench I use is exactly 1'. Thank goodness.
 
Radomir said:
Either way, you actually don't need to "understand" any of this.. just follow instructions from the manual, or 43.13, or any other publication.. they all show the same formula... there really should be no 100+ posts about this :)
Click to expand...

Perhaps, but the reason there are 120 posts about this is because the manufacturer of the tool that is being advertised at the start of this thread has instructed the users that the 0.8 multiplier is a constant for a torque wrench measuring in ft lbs. If that isnt true, then it needs to get sorted out in a convincing way so that the instructions will be corrected. I like to think that most of the posters are trying to contribute toward an effort of understanding and consensus, but perhaps that isnt achievable in a forum such as this.

erich
 
You make a good point Erich.. The key here is to understand that this is just another extension.. like any ohter you'd use on your torque wrench.. so I'm suggesting that we stick to what wrench mfgr says, not extensions mfgr :)
 
jpharrell said:
4) If you must use an extension straight in line with the handle of a torque wrench you should be very careful. The torque applied to the bolt will be greater than the value you set on the torque wrench and the correction factor depends on both the length of the extension and the moment arm of the force application. It is not just the length of the torque wrench handle that counts, it is also dependent on where you apply the force on the handle.
Click to expand...

This is a really important point. Some of us have been referring to "length of the torque wrench" to mean "how far from the head of the torque wrench you are applying the force", but to be clear it is that distance, not the manufactured length of the wrench, that is important here (assuming a click-type wrench anyway). If you have a 12" torque wrench and for some reason are choking up on it applying the force halfway at 6" from the head, you need use 6" as the torque wrench length in the formula for torque at the business end of the extension. If you add a 12" cheater so you are applying the force at 24" from the head of the wrench, use 24" as the torque wrench length in the formula. If you are applying the force at 12" from the head of the wrench, and using Allan's 3" extension, use a factor of 0.8 :). Beam-type wrenches are designed to be used by applying force at one particular distance from the head, and that's the length you should use.

Of course as Larry and various other folks have been pointing out at length (pun intended, I guess) where you apply the force on the wrench doesn't matter if you are using a click-type wrench in the usual way, without an extension. It will measure the torque at the head of the wrench and that's all you want in that case.

--Paul
 
Garage Guy said:
Beam-type wrenches are designed to be used by applying force at one particular distance from the head, and that's the length you should use.
Click to expand...

Recall that most of the old, linear scale, "beam-type" torque wrenches had pivoting handles. The proper procedure is to ensure neither end of the handle comes into contact with the beam so that all the force is applied at a specific location.
 
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OK, let me keep this straight then. We have seem to have

(1) some that say the 0.8 multiplier is a constant for torque wrenches reading in ft lbs

(2) others that say hogwash, it depends on the length of your torque wrench; you must do the math to figure out the proper multiplier per the published literature;

and now

(3) others that split the difference and suggest 0.8 is a constant for the click type of torque wrench but not the other kind.

Wow. We are really all over the map on this. Guess I will just use Allan's product with the extension at 90 degrees (no change in length) and a 0.8 multiplier and be done with it. Unless someone wants to disagree with that too.... :)

erich

P.S. WRONG I should have said "extension at 90 degress and NO multiplier"
 
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erich weaver said:
Guess I will just use Allan's product with the extension at 90 degrees (no change in length) and a 0.8 multiplier and be done with it. Unless someone wants to disagree with that too.... :)

erich
Click to expand...

No, if using the tool at 90*, forget the .8 factor. Just set it at the correct torque value, and that is all.

Actually this is not strictly correct, but it is close enough.

An extension normal to the wrench (at 90*) is actually on a line that is tangent to the torque wrench centerline, and therefore the distance is really the hypotenuse of a right triangle ------and will therefore require a correction factor. But in real life it is probably smaller than the error factor of the wrench to begin with.
( but I dont want to muddy the waters here, so dont worry about Pythagoras ) :D
 
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Radomir said:
Jon.. no.. Go back and read post #64 by Marc Hudson... then re-read it 10 more times until you understand the importance of point "B" on the drawing.
Click to expand...

Figures are correct, but wrong for the application. I've changed the force numbers to the correct values..........which will satisfy the "click" for 80 lbs at the lengths shown. This is simply dividing 100 by the wrench & extension length. For example, in #2, the wrench will never click, if only 40 lbs is applied. But we don't care. Apply 44.44 lbs, and it will. Wrench setting stays at 80 & extension remains at 3".

6qbrsk.jpg
 
Let me try to understand this:

First, let me say that my understanding is that a torque wrench indicates the moment at the end of the instrument where it is attached to a fastener, or an extension, or some other device or fixture. If that is incorrect, then the below means nothing.

So, you're saying in example two that if you apply a force of 44.4 pounds two feet from the torque wrench head, that will equal a moment of 80 ft-lbs. How is that possible?

I would think it would be 88.8 ft-lbs.

I'm an EE, not an ME, and the few statics and dynamics courses I had in college were over 40 years ago, but I just can't get my mind around a situation where M≠F*D...

All this is pertinent to me as I will be starting my condition inspection this weekend on my RV-4 (there, RV related!! :)), and there may be a situation where I need to use a torque wrench with an extender of some type and I'd sure like to know what I am doing...:confused:
 
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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.
 
ccsmith51 said:
So, you're saying in example two that if you apply a force of 44.4 pounds two feet from the torque wrench head, that will equal a moment of 80 ft-lbs. How is that possible?

I would think it would be 88.8 ft-lbs.
Click to expand...

Keep in mind, that the torque wrench head is not rotating against a fixed extension at B. 44.4 * 2'+.25' = 100 lbs. (approx)
 
L.Adamson said:
Keep in mind, that the torque wrench head is not rotating against a fixed extension at B. 44.4 * 2'+.25' = 100 lbs. (approx)
Click to expand...

Your comment above doesn't address your prior statement that a 44.4 lb force applied 2' from the end of the torque wrench will cause the instrument to indicate 80 ft-lbs. That is what I cannot understand.
 
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L.Adamson said:
Figures are correct, but wrong for the application. I've changed the force numbers to the correct values..........which will satisfy the "click" for 80 lbs at the lengths shown. This is simply dividing 100 by the wrench & extension length. For example, in #2, the wrench will never click, if only 40 lbs is applied. But we don't care. Apply 44.44 lbs, and it will. Wrench setting stays at 80 & extension remains at 3".

6qbrsk.jpg
Click to expand...

This is simply not correct. The original drawing is right. The fact is that the torque wrench will 'click' when the torque at point 'B' reaches it's set point. This torque does not care where along the wrench the force is applied. That is a simple lever arm calculation. But, if you move the point in space where you are applying the torque without also moving the point in space where the torque is being measured then you MUST take into account where in space the force is being applied.

So not only should you apply the calculation based on the length of the wrench but make sure that you pull on the wrench in the right place or you will incorrectly torque the bolt.
 
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ccsmith51 said:
Your comment doesn't above address your prior statement that a 44.4 lb force applied 2' from the end of the torque wrench will cause the instrument to indicate 80 ft-lbs. That is what I cannot understand.
Click to expand...

It will take at least 44.4 lbs of force, to torque the prop nut to 100 lbs, with 2.25' of wrench length. 2.25' at 40 lbs force will never get that nut to 100 lbs., with or without a torque wrench. It will stop at 90 lbs., just like in the drawing.

As soon as that prop nut, reaches 100 lbs. torque, the click type wrench will click at 80 ft/lbs. with a 3" extension.
 
Birkelbach said:
This is simply not correct. The original drawing is right. The fact is that the torque wrench will 'click' when the torque at point 'B' reaches it's set point. This torque does not care where along the wrench the force is applied. That is a simple lever arm calculation. But, if you move the point in space where you are applying the torque without also moving the point in space where the torque is being measured then you MUST take into account where in space the force is being applied.

So not only should you apply the calculation based on the length of the wrench but make sure that you pull on the wrench in the right place or you will incorrectly torque the bolt.
Click to expand...

As in example #2 --- It will never click, as it will never reach it's set point, if no more than 40lbs of force is applied.
 
L.Adamson said:
It will take at least 44.4 lbs of force, to torque the prop nut to 100 lbs, with 2.25' of wrench length. 2.25' at 40 lbs force will never get that nut to 100 lbs., with or without a torque wrench. It will stop at 90 lbs., just like in the drawing.

As soon as that prop nut, reaches 100 lbs. torque, the click type wrench will click at 80 ft/lbs. with a 3" extension.
Click to expand...

No, the wrench will click when there is 80 ft-lbs (not ft/lbs) of torque at point 'B'. The little mechanism in the wrench that measures torque doesn't know anything about the 3" extension. It will click when a force of 40 lbs is applied two feet away from point 'B' or when a force of 80 lbs is applied 1 foot away from point 'B'. Point 'A' is not relevant as far as when the wrench will click.
 
L.Adamson said:
As soon as that prop nut, reaches 100 lbs. torque, the click type wrench will click at 80 ft/lbs. with a 3" extension.
Click to expand...

In the context of a 2' torque wrench with 44.4 lbs applied, with a 0.25' extension, please show me the math that supports your above claim.
 
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Birkelbach said:
No, the wrench will click when there is 80 ft-lbs (not ft/lbs) of torque at point 'B'. The little mechanism in the wrench that measures torque doesn't know anything about the 3" extension. It will click when a force of 40 lbs is applied two feet away from point 'B' or when a force of 80 lbs is applied 1 foot away from point 'B'. Point 'A' is not relevant as far as when the wrench will click.
Click to expand...

A, along with the extention is moving also.
 
ccsmith51 said:
In the context of a 2' torque wrench with 44.4 lbs applied, with a 0.25' extension, please show me the math that supports your above claim.
Click to expand...

That drawing certainly doesn't support it....

edit: We're discussing a prop nut, torqued to 100 ft. lbs. That is the unit of measurment, based on 1 lb. force, at 1'. 2.25' at 44.4 lbs. force, will yield 99.9 lbs.
 
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Larry.. you're getting somewhere... and that somewhere is further and further away from your theory :) if you apply 44.4 lbs at teh end of the 2' wrench (with the splat extension), you need to set it to 88.8 (NOT 80!!!!) for it to click at right spot and give you 100 at the nut itself.

If you keep the clicker at 80, you'll never reach the desired 100... it'll click before you get there and will leave the nut undertorqued.

the formula given by others lets you figure out that 88.8 as the click setting on the wrench. it's clear from the example above that the 0.8 multiplier is not applicable to this scenario.
 
L.Adamson said:
We're discussing a prop nut, torqued to 100 ft. lbs.
Click to expand...

Well, we're also discussing what the torque wrench, with head at "B", will read. Assuming it's not broken, it should read the torque about that point, no?

May I suggest you go back and review my #82 .

And also, I asked you a question in my #98 which you haven't answered yet.
--Paul
 
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L.Adamson said:
That drawing certainly doesn't support it....
Click to expand...

I disagree; I think the drawing supports his point well. This is all simple statics well addressed in the textbook I mentioned in an earlier post.

Here's another set of free-body diagrams to dissect what is going on:

torque_wrench_problem.gif


A examines the geometry of a hypothetical torque wrench. For the sake of argument, it is set to click at 80 ft-lbs, and has a length of 18" (1.5 feet) between the point of application of force (the center of the handle) and the output lug.

B examines the geometry of our hypothetical extended crows-foot wrench. It has a length of 3" between the input square and the center of the nut.

C examines the condition under which the torque wrench clicks at 80 ft-lbs. We see that it requires a normal force at the handle of 53.3 lbs, since 53.3 lbs*1.5 feet=80 ft-lbs.

Very important: Note that the 53.3 lbs of force applied at the handle is countered by an equal and opposite force at the nut. Were that not the case, the wrench would continue to accelerate across your shop under the influence of the 53.3 lbs applied at the handle. I hate it when that happens!

D examines the conditions under which the 3" extension applies force to the nut. Note that it has two inputs: Both the 80 ft-lbs of torque applied by the torque wrench, and also the 53.3 lb normal force originally applied at the handle of the torque wrench. The resulting torque at the nut is therefore the addition of the torque plus the normal force times its arm, or 80 ft-lbs + (53.3 lbs * 0.25 feet) = 80 ft-lbs + 13.3 ft-lbs = 93.3 ft-lbs.

E examines the whole system together. We know that the wrench clicks when the torque at its output lug reaches 80 ft-lbs, because that is what it is set at. We know that the handle force required to make it click is 53.3 lbs, because 80 ft-lbs/1.5 ft = 53.3 lbs. We know that the arm of the whole system is 21" (1.75 feet) because the 3" extension and the 18" wrench add up to that. So the torque applied to the nut is 1.75 ft * 53.3 lbs = 93.3 ft-lbs.

F examines a counter-example: what happens when we put the extension on the other way. We know from the previous examples that the torque wrench clicks at 80 ft-lbs, and that happens when 53.3 lbs are applied at the handle. However, in this case the system length is 15" because the extension is actually a retraction of 3" and 18" - 3" = 15" or 1.25 ft. The torque applied at the nut is 1.25 ft * 53.3 lbs = 66.6 ft-lbs. Oh noes, it's the torque of the devil!

Thanks, Bob K.
 
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