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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.



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

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.... :)
 
But wait, there's more!

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:

torque_wrench_problem2.gif


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.
 
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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.
 
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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.
 
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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?
 
My Vote Funniest thread I have ever read!

Wow Guys,This is by far the funniest thread I have ever seen here. OK I will try!Snap -on makes two wrenches one is 12" on is 24". They are calibrated at the factory to read the same at the correct setting (any setting you the user wishes) at the 1/2" or 3/8" drive point. The 24" wrench uses 1/2 the force on the handle to reach this number then the 12" handle,the result is the same at the drive point 1/2" or 3/8". Now you add Allan's 3" extension to the 1/2" or 3/8"cut out in the tool you need to reduce the wrench setting to (desired number x .8 = new number with Allan's wrench.)This is true if Allan's wrench is straight inline with the torque wrench of any length. If Allan's wrench is 90 degrees to the torque wrench of any length the full value is applied. So 70ft/lb= 56ft/lb with Allan's wrench inline or 70ft/lb=70ft/lb with Allan's wrench at a 90 degree angle to the torque wrench of any length.In the end the nut gets 70 ft/lbs.
Bob
 
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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?

To the degree that the concept of consensus applies to what we're doing here, I get the sense that we're satisfied that the formulas in AC43.13 are generally applicable, and that the "0.8 correction" is only applicable in the one case where the length of the torque wrench arm is 12". I have only one torque wrench in that size range; it is calibrated in in-lbs and is what I used to use to snug up the 6mm bolts in Formula roadbike crankcases.

For reference, here's the figure in AC43.13 with the formulas for correcting for torque wrench extensions. If you work the math, you'll see it basically says the same thing as the free body diagrams I posted earlier:

torque_wrench_ac43.13.GIF


Edit add: Okay, let's use those formulas to work the math for the example I used earlier:

(T*L)/(L+E)=Y

Where:

T= desired torque = 100 ft-lbs
L= torque wrench length = 1.5 ft
E= extension length = 0.25 ft
Y= indicated torque or click setting

Y = (100 ft-lbs * 1.5 ft) / (1.5 ft + 0.25 ft)
= 150 ft^2-lbs / 1.75 ft (funny how we get ft^2 lb units, but it cleans up in the next step)
= 85.7 ft-lbs (see, the extra ft cancels out!)

Hey, that 85.7 ft-lbs sounds kinda familiar!

Edit add again: Just to be clear, I am not an engineer. I used to be a liberal studies dropout, but I relapsed. Even if I was an engineer, though, I could still be wrong. Your mileage has already varied.

Thanks, Bob K.
 
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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?

Not until Allan says uncle.

Allan? You still out there? Think of this all as tough love buddy. We do it because we care about you. Ok, and also so our props don't fall off.:)

Erich
 
the concept is very easy to verify.
all you need is a connecting piece (weld 2 bolts together?).


take 2 torque wrenches (that are known to be accurate),

join wrench A to the connecting piece.

join wrench B to Allen's 3" contraption, then to the connecting piece.


Pull the wrenches handles toward each other,

If the calculation is correct,
when wrench A reads 100lbs,
wrench B should read 80lbs.
 
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Guys,

You are forgetting the much, much simpler system used over here for years....

Forget the torque wrench.

Tighten it up until the nut strips, back it off a flat, safety wire - continue :D
 
O.K. I'M BACK!!!!!!!!!!!

Wow Guys,This is by far the funniest thread I have ever seen here. OK I will try!Snap -on makes two wrenches one is 12" on is 24". They are calibrated at the factory to read the same at the correct setting (any setting you the user wishes) at the 1/2" or 3/8" drive point. The 24" wrench uses 1/2 the force on the handle to reach this number then the 12" handle,the result is the same at the drive point 1/2" or 3/8". Now you add Allan's 3" extension to the 1/2" or 3/8"cut out in the tool you need to reduce the wrench setting to (desired number x .8 = new number with Allan's wrench.)This is true if Allan's wrench is straight inline with the torque wrench of any length. If Allan's wrench is 90 degrees to the torque wrench of any length the full value is applied. So 70ft/lb= 56ft/lb with Allan's wrench inline or 70ft/lb=70ft/lb with Allan's wrench at a 90 degree angle to the torque wrench of any length.In the end the nut gets 70 ft/lbs.
Bob
Bob has it exactly!!!!!!!!!
Now forget all you think you know about this and listen. Pretend for a moment you have no idea what a torque wrench is and have never seen one. The .8 multiplier is a constant that applies to our prop wrench and does not change. Our 3" tool is simply a torque multiplier period, nothing else. If you use this tool on your bolts, it and it alone boosts the power of what ever devise you use to tighten the bolts by a factor of plus 20%. That requires you to set your tool, be it your weight lifter mother in-law or a heavy rock on a rope to EXACTLY 80% of the desired value. THIS IS A FACT AND WILL NOT CHANGE under any circumstances. This is 7th grade math, not rocket science. If you are looking for 70ft.lb. on your bolts 70 X .8= 56ft.lb. setting on what ever you use to pull with.
... Now I need to search out a formula to re-calculate the 3.1416 number in pie as it will no longer hold true because I am using a longer pencil to do the math.:rolleyes:
 
ft/lbs is ft/lbs is it not?

(T*L)/(L+E)=Y

Where:

T= desired torque = 100 ft-lbs
L= torque wrench length = 1.5 ft
E= extension length = 0.25 ft
Y= indicated torque or click setting

Y = (100 ft-lbs * 1.5 ft) / (1.5 ft + 0.25 ft)
= 150/1.75
= 85.71 ft/lbs

Hey, that 85.7 ft-lbs
--------------------------------------------------------------------------------------------
(T*L)/(L+E)=Y

Where:

T= desired torque = 100 ft-lbs
L= torque wrench length = 4 ft
E= extension length = 0.25 ft
Y= indicated torque or click setting

Y = (100 ft-lbs * 4 ft) / (4 ft + 0.25 ft)
= 400/4.25
= 94.11 ft/lbs
-------------------------------------------------------------------------------------------
(T*L)/(L+E)=Y

Where:

T= desired torque = 100 ft-lbs
L= torque wrench length = .25 ft
E= extension length = 0.25 ft
Y= indicated torque or click setting

Y = (100 ft-lbs * .25 ft) / (.25 ft + 0.25 ft)
= 25/.5
=50 ft/lbs

I admit this is a bit of a brain teaser. The formula cited clearly yields different torque wrench settings for different wrench lengths (same extension). Why would the torque setting change? Wouldn't the force needed at the extension be the same regardless of how you generate it? Flame suit on!
 
Allan, you and your group are very good at making videos. The ones you have made so far are very logical and informative.

How about making one to prove your position, which I believe is regardless of the length of the torque wrench, when using your adapter the reading on the torque wrench will always be 80% of the desired torque on the nut.

All you need are a couple of torque wrenches of different lengths, some way to measure torque at the nut, and then we can see the results without, and with, your wrench.
 
L is not substituted for the length of your torque wrench. .8 was determined bu the length of the extension being 3 inches and it multiplying the torque by .25 (becasue it is one fourth of the length of a foot lb) thereby needing a reduction at the setting of whatever torque wrench to .8 of the desired torque product. The torque applied to the extension to get 100 lbs of torque will be 80 lbs set on ANY wrench no matter the length of said wrench.
 
Just when I had hopes this was reaching a resolution...

Its not like primer wars or the nose dragger/tail dragger arguments. One side is just plain wrong here.

Bob K's drawings and explanations seemed simple and straightforward. Can Allan or others in the "0.8 is a constant" camp explain succinctly where they are in error? Simply insisting that 0.8 is a constant as proven through 7th grade math isnt very convincing. There are some experienced, smart folks that made it through 7th grade and well beyond that say otherwise, and they have some convincing documentation that appears to back them up. I have yet to see ANY published documentation that makes the case for using a constant multiplier for a torque wrench extension of a given size regardless of the length of the torque wrench. Staying open minded however...

Somebody kidnap Dan Horton and refuse to release him until he takes a stand on this

erich
 
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Allan

The best compliment I can give you is to ask whether you are really Dilbert in this cartoon. http://www.youtube.com/watch?v=FlJsPa6UwcM

I love your intuitive answers to 'our' problems.

Thank you Steve for the compliment! I am beginning to feel like him (Delbert). This thread is actually beginning to scare me. The lack of understanding I am witnessing here could potentially have costly or dire consequences impacting safety. I am somewhat in disbelief!:eek:
 
...
Now forget all you think you know about this and listen. Pretend for a moment you have no idea what a torque wrench is and have never seen one. The .8 multiplier is a constant that applies to our prop wrench and does not change. Our 3" tool is simply a torque multiplier period, nothing else. If you use this tool on your bolts, it and it alone boosts the power of what ever devise you use to tighten the bolts by a factor of plus 20%. That requires you to set your tool, be it your weight lifter mother in-law or a heavy rock on a rope to EXACTLY 80% of the desired value. THIS IS A FACT AND WILL NOT CHANGE under any circumstances. This is 7th grade math, not rocket science. If you are looking for 70ft.lb. on your bolts 70 X .8= 56ft.lb. setting on what ever you use to pull with.
... Now I need to search out a formula to re-calculate the 3.1416 number in pie as it will no longer hold true because I am using a longer pencil to do the math.:rolleyes:

I'm sorry, but all the color, exclamation points, and emoticons in the world aren't going to make you right. Go back to your statics textbook and study the chapters on moments and rigid bodies. As I have shown, they are the basis for the simple applications of physics behind the formulas for torque wrench correction embodied in AC43.13.

Edit Add: Really, I encourage you to find a real mechanical engineer and have them explain this to you. Trust me, it will be better for you and for your ability to develop and promote your products.

Thanks, Bob K.
 
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Bob has it exactly!!!!!!!!!
Now forget all you think you know about this and listen. Pretend for a moment you have no idea what a torque wrench is and have never seen one. The .8 multiplier is a constant that applies to our prop wrench and does not change. Our 3" tool is simply a torque multiplier period, nothing else. <<SNIP>>:rolleyes:

If I forget all that I know about this then I guess I could be convinced that you are correct, but you are NOT correct. Your little device is NOT a torque multiplier. It is a lever. A lever is a FORCE MULTIPLIER. The math has been done here to the extent that I don't know why we are still arguing about it but let me try my hand.

Say I have a 2' long torque wrench. I grip the wrench on it's handle and apply 100 ft-lbs of torque to it. That'll be 50 lbs of force at the handle. Now I add your extension, and apply that same 50 lbs of force (the wrench will still read 100 ft-lbs because it has no way of knowing about the extension). Now I am applying 50 lbs of force at a distance of 2.25 ft. That is 112.5 ft-lbs. That is a multiplier of 0.8889 and 0.8889 is not equal to 0.8.

Guys seriously. This is not complicated. These are prop bolts that we are talking about. Let's get it right.
 
I'm sorry, but all the color, exclamation points, and emoticons in the world aren't going to make you right. Go back to your statics textbook and study the chapters on moments and rigid bodies. As I have shown, they are the basis for the simple applications of physics behind the formulas for torque wrench correction embodied in AC43.13.

I'd still like to see definate proof, beyond all doubt...........that the spring release mechanism of a "click" torque wrench, has to comply with the same physics that apply to a beam torque wrench. While the tables presented so far, make perfect sense for a "beam", how do they really relate to a compressed spring?

L.Adamson
 
I wonder...

if this is all a very clever marketing plan to keep the wrench at the top of the new posts lists for days on end... ;)
 
If you use this tool on your bolts, it and it alone boosts the power of what ever devise you use to tighten the bolts by a factor of plus 20%.
If you use this tool (inline) and a 0.8 correction factor on your bolts, and you use it with an 18" wrench, your bolts will be just under 7% undertorqued. 24" wrench? 10% undertorqued.
 
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I'd still like to see definate proof, beyond all doubt...........that the spring release mechanism of a "click" torque wrench, has to comply with the same physics that apply to a beam torque wrench. While the tables presented so far, make perfect sense for a "beam", how do they really relate to a compressed spring?

L.Adamson

This is an issue of proving a negative. I think you would find it more instructive to prove that the two types are fundamentally different than to try to disprove that they are the same.

if this is all a very clever marketing plan to keep the wrench at the top of the new posts lists for days on end... ;)

"...all sound and fury..."

Thanks, Bob K.
 
Yep

I'm sorry, but all the color, exclamation points, and emoticons in the world aren't going to make you right. Go back to your statics textbook and study the chapters on moments and rigid bodies. As I have shown, they are the basis for the simple applications of physics behind the formulas for torque wrench correction embodied in AC43.13.

Edit Add: Really, I encourage you to find a real mechanical engineer and have them explain this to you. Trust me, it will be better for you and for your ability to develop and promote your products.

Thanks, Bob K.


Yep,
The nice thing about this is that consensus has nothing to do with it. The right answer (not Allens, sorry) isn't a matter of opinion.

If you can't or won't follow the statics in Bob's postings, then just take Part 43 on faith and move on.

It has been really entertaining though. Just wish I find some way to introduce primer or tailwheels into it. :D:D
 
This is an issue of proving a negative. I think you would find it more instructive to prove that the two types are fundamentally different than to try to disprove that they are the same.

It's as simple as making some tests, between different wrenches. But of course, I have to think deeper (even if it's wrong) .

What would change the inherent characteristics of a spring/pawl system in a click wrench, if a cheater bar is added for length? Why would the spring compression figures & release mechanism.......suddenly change, because I added some length to the handle. I suppose, I'd like to see an engineering graph on this, also.
 
Edit Add: Really, I encourage you to find a real mechanical engineer and have them explain this to you. Trust me, it will be better for you and for your ability to develop and promote your products.

Thanks, Bob K.

Bob, I'm a "real" mechanical engineer, and your explainations are 100% correct and in my opinion very clear. I'm not sure how much more explaination you could give.

........This thread is actually beginning to scare me. The lack of understanding I am witnessing here could potentially have costly or dire consequences impacting safety. I am somewhat in disbelief!

Allan, I agree with this part of your statement. I'm sorry but maybe you should do an experiment to prove to yourself that Bob is correct. .8 is not a constant. Please study Bob's figures "G and H" in post #154. It shows why if you're using a longer than one foot torque wrench, you need more than 80 ft-lbs at the connection to the extension to get 100 ft-lbs at the bolt. So no matter what type of wrench (clicker or beam) you need to set it at the value given by the equation shown multiple times thru out this thread.
 
It's as simple as making some tests, between different wrenches. But of course, I have to think deeper (even if it's wrong) .

What would change the inherent characteristics of a spring/pawl system in a click wrench, if a cheater bar is added for length? Why would the spring compression figures & release mechanism.......suddenly change, because I added some length to the handle. I suppose, I'd like to see an engineering graph on this, also.

The cheater bar would not affect this at all as long as you don't put Allen's little extension on it. If you put the little extension on it then the cheater bar starts to matter. The torque wrench will still be measuring the same torque at the end of the torque wrench but that doesn't mean that the torque three inches away is simply 25% higher because the extension is 3" long.

The mechanism in the torque wrench is simply a method to measure torque. Granted a beam torque wrench measures torque in such a way that where you apply the force matters but other than that they are simply measuring the same torque.
 
The cheater bar would not affect this at all as long as you don't put Allen's little extension on it. If you put the little extension on it then the cheater bar starts to matter. The torque wrench will still be measuring the same torque at the end of the torque wrench but that doesn't mean that the torque three inches away is simply 25% higher because the extension is 3" long.

I'm visualizing that last partial revolution of the nut..........as it get's very close to the final torque setting. Let's say that it has 1/8" of travel to go...........but the 12" bar (along with the 3" extention), is suddenly changed to 24" with a cheater. Does that last 1/8", now become 1/4, 3/16, or whatever.............because the torque setting must now be 88.9?

edit: This example would be with a click wrench
 
All this noise can't be for real. Just can't be.

I know how smart some of the posters here are

"What would change the inherent characteristics of a spring/pawl system in a click wrench, if a cheater bar is added for length? Why would the spring compression figures & release mechanism.......suddenly change"

Nothing changes if you add leverage to the end of the handle. That's just making life easier by not requiring as much arm strength to get the torque wrench to ''click''. You wouldn't be stoking the fires here would you? I can't believe I'm even responding to this thread. Now, if you add the leverage extension between the torque wrench and the nut, now you have changed everything and you need to make the adjustment to the torque setting per the very correct formulas listed way back. Forget that .8 nonsense. That's one answer to a 12 inch wrench with a 3 inch extension between the torque wrench and the nut in a straight line, not 90 degrees. Jeeeesz.:eek:
 
I know how smart some of the posters here are

"What would change the inherent characteristics of a spring/pawl system in a click wrench, if a cheater bar is added for length? Why would the spring compression figures & release mechanism.......suddenly change"

Nothing changes if you add leverage to the end of the handle. That's just making life easier by not requiring as much arm strength to get the torque wrench to ''click''. You wouldn't be stoking the fires here would you? I can't believe I'm even responding to this thread. Now, if you add the leverage extension between the torque wrench and the nut, now you have changed everything and you need to make the adjustment to the torque setting per the very correct formulas listed way back. Forget that .8 nonsense. That's one answer to a 12 inch wrench with a 3 inch extension between the torque wrench and the nut in a straight line, not 90 degrees. Jeeeesz.:eek:

When you actually explain the compressed spring characteristics, instead of just repeating the laws of "beam" physics, then I'll be much more interested in what's said. Up to this point.................all I'm hearing is.. "because" it's written, and no actual explanation. I just figure that everyone here so far, doesn't really know...

I want to know, the exact workings of a compressed spring/pawl system, compared to physics and calculations derived for a beam type torque wrench. And yes, this applies to use with an extension.

edit: We already know that the relationship changes, when an extention is added, and that the setting will be less. Now, why does the spring loaded setting, change, because I just added some additional length, to the handle. The required force has changed, but what has changed within the compressed spring?
 
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I'm visualizing that last partial revolution of the nut..........as it get's very close to the final torque setting. Let's say that it has 1/8" of travel to go...........but the 12" bar (along with the 3" extention), is suddenly changed to 24" with a cheater. Does that last 1/8", now become 1/4, 3/16, or whatever.............because the torque setting must now be 88.9?

edit: This example would be with a click wrench

The 'movement' of the nut is irrelevant. As is the type of wrench. We are talking about torque. The nut will be in the same place after we are done with all this regardless of how we applied the torque. The question is how did we measure the torque on the nut. We are measuring the torque 3" away from the nut and that is what causes all the assumptions to change.
 
The 'movement' of the nut is irrelevant. As is the type of wrench. We are talking about torque. The nut will be in the same place after we are done with all this regardless of how we applied the torque. The question is how did we measure the torque on the nut. We are measuring the torque 3" away from the nut and that is what causes all the assumptions to change.

I must admit.............I'm ready to go out to the garage, rig up a three inch extension; torque the nut to 80 ft. lbs.; then add one foot, change the setting to 88. whatever.........and see if the nut is still at the "click" setting. If it so, then it's so.
 
When you actually explain the compressed spring characteristics, instead of just repeating the laws of "beam" physics, then I'll be much more interested in what's said. Up to this point.................all I'm hearing is.. "because" it's written, and no actual explanation. I just figure that everyone here so far, doesn't really know...

Seriously!? This is elementary, first year engineering school stuff. It's not complicated and the fact that we are still having the discussion is becoming laughable.

I want to know, the exact workings of a compressed spring/pawl system, compared to physics and calculations derived for a beam type torque wrench. And yes, this applies to use with an extension.

None of these physics are derived from a beam type torque wrench. The measurement method has no bearing on the physics. If your physics are dependent on how you measure the parameters then there is something seriously wrong with your assumptions.
 
Allan,

I think your wrench (and other products) are great innovations, and I wish you a lot of success. However, due to the safety factor of proper torque values, especially on a propeller, I would advise you to simply instruct people to follow the instructions in AC-43.13 for using a 3" extension on their torque wrench instead of telling them to use a constant multiplier of 0.8. Just my $.02.
 
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One last attempt

Larry,

I'll try one more way to explain it.

Let's assume the "clicker" torque wrench is 2 feet long and the extension is .25 feet long (3").

Now put the extension on the torque wrench and you have 2.25 feet from the handle to the bolt. If you want 99 foot lbs (to make the math cleaner) at the bolt you need 44 lbs applied normal to the handle (44lbs x 2.25ft). The wrench would click at 88 ft-lbs (44lbs x 2ft).

Now leave the wrench at 88 ft-lbs and move your hand to the middle of the wrench (1 foot from the handle). Now it takes 88 lbs to get the wrench to click but the torque on the bolt is now 88 lbs x 1.25 ft= 110 ft-lbs.

Leave the wrench at 88 ft-lbs and add a 1 foot cheater to the handle. Now, it takes 29.3lbs (88/3) normal to the cheater to get the wrench to click but the torque on the bolt is now 29.3lbs x 3.25 = 95.3 ft-lbs.

The setting on the torque wrench indirectly measures the force your applying at a distance from the end of the wrench. If you're using a socket or an extension at 90 degrees to a clicker wrench then where you grab on the wrench doesn't matter much (although in reality it does introduce small errors if you don't grab the handle). But with an extension in line with the wrench, where you grab is critical to get a known torque at the bolt.

That's it for me.......
 
That's it for me.......

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.
 
If we aren't careful here Delta-Romeo is going to change the posting rules, No Religion, Politics or Torque Wrenches. Thanks Guys, Allan
 
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