mag00,

Your example of a cross lug wrench with an extension on the forward axis arm so it can reach further into a recess and then use the 2 remaining levers to apply force in a rotational manner around the axis of the bolt while reading a longitudinal torque indicator on the shaft of the rotational force generator, appears to be the same principle as placing an extension on a torque wrench where the socket normally resides and then attaching the socket to the extension. IOW, the extension is on the rotational axis and at a 90 degree angle to the wrench handle (lever). Is that correct? If so, you've certainly led everyone here on a wild goose chase by not clearly stating your needs - or - we have gotten so wrapped around the axle that we missed your point entirely.

Read this and you'll understand a bit more about torque:

http://www.ship.edu/~sagoul/Torque_rotation.htm

If all you're doing is extending the right angle socket by use of an extension, then the extension has virtually no effect, so long as you provide reasonable alignment support at the 90 degree angle at the end of the torque wrench. In reality, there probably is a slight difference because of your inability to exactly align and support the mechanism, but if you can keep it relatively aligned, you'll get close enough for horseshoes and hand grenades.

I'm sorry I don't have a formula, but does that help?

Frank

I just went back and reviewed the first page of this thread. If I now understand your question, TobyB answered it there in about the 6th response on page 1:



Yikes!

Frank <img src="/forums/images/graemlins/lol.gif" alt="" /> <img src="/forums/images/graemlins/lol.gif" alt="" />

I told you we were having fun. Scotty, wish I had a nickel for everytime I've been wrong, but you learn more by being wrong than by being right, right?

"Logic is just a way of going wrong with confidence", thanks to R. A. Heinlein....

I don't know how long anybody wants to stay after this, but I still don't have my formula.

@ Frank, I'm not talking about putting an extinsion in axis with the drive and socket, that was to answer the head bolt torque question quoted.

I will not except any formula with the use of the length of the torque wrench as an accurate formula, those formula's are for those who care not to be exact and have no trig or math comprehension. Those formulas using handle length only serve for quick and easy torkin, and are typically close enough for mechanics.

Being the perfectionists we all are, I take and plug different numbers in the above mentioned formulas and keep getting different adjusted torque settings for the same extension length in a straight line, this is why handle lenth is confusing everyone hear.

Try this formula (TxL)/(T+L) first use a 14 inch torque wrench and then us a 18 in length. Keep the extension straight and same ft lb desired. You will get two separate answers for the same extension and applied torque.

Also take the same torque wrench and change the extension length from 6" down to 3" and the adjusted torque value does not double.

Therefore a more logical formula may contain some form of trig, or even calc, and no variable other than the extension length and torque needed. That may be why no-one is doing it. And yet it may be as simple as the one Scotty provided.


The one link above contained great info about torque but with all the 's I couldn't see what they were explaining.

Someone must have derived the correct formula right? It's not like this question has never been asked before, and I just can't believe that everone has adopted those other formulae with handle length in the equation as accurate when in fact they are only close.

mag00,

As far as I can determine, the formula is correct as posted.

What I haven't explained as well as I should (not a very good teacher) and what you don't seem to yet grasp is this:

The torque wrench is calibrated with the center of rotation at the bolt center and a given force applied at a given distance along the handle.

When an extension is added, the square-drive end of the torque wrench is no longer operating in the same swing radius relative to the center of applied torque and the calibration of the wrench is now altered.

Furthermore, the extension has increased the total effective distance from applied force to the point of application and the relationship between the 2 changed factors must be resolved to regain accuracy.



The formula is not just "close enough for mechanics". Mechanics are the very folks who need, want and expect the most accuracy. The formula was developed to give mechanics the exact benefit of accuracy in work that most are proud to do and damned good at doing and no amount of math or trig comprehension will change the existing facts.

I'd suggest that you find someone who'll give you the answer you want to hear. Being a "perfectionist" is one thing, but it seems an exercise in futility to attempt to prove experts wrong at their own business.... but hey - if you can do it, go ahead.... you might find it "up there somewhere".

Now on the off-chance that I still haven't understood what you want to know, there are "other" factors that affect torque values - i.e. friction loading, thread pitch, etc. If the adjustment corrections you're looking for fall into those types of "perfectionist" areas, I can't help you - that's beyond my "need to know" or desire to understand - much less try to explain.

Frank <img src="/forums/images/graemlins/cheers.gif" alt="" />

mag00 -

I'm still confused as to what you're trying to do and with what type of equipment you're using, so maybe there's still some basic misunderstanding that's getting lost in the communications. Try this and let's see if you can get your answer -

Clearly state:

1) what type of torque wrench you're using.
2) which end of the wrench will have the extension
3) what type of extension you're adding.
4) the length of the extension
5) what angle to applied rotational force (handle) the extension is set.
6) the desired torque value on the fastener

Maybe with this information someone can help. <img src="/forums/images/graemlins/cheers.gif" alt="" />

Frank

Project is done, but in the process of trying to calculate proper torque setting too many discrepencies of formula has occured. All the formulas with the length of handle of torque wrench have errors when different values used.

To keep it simple I have 1/2 bolt head which needs 20 ft lbs of torque exactly. It cannot be reached directly with a socket in a conventional manner. There was just enough room between to reach in and get the bolt in. If the box end of a 1/2" combination wrench is placed over the bolt head to be tightened the open end of the wrench is accessable. The combination wrench is 6" long. A half in drive fits nicely in the open end of the wrench and is 6" c/c.

Question is: What is the formula to figure the amount of torque (rotating force) needed to give 20 ft lbs at the bolt to be tightened, with the torque applied at the open end of the combination wrench.

I=A/(1+E) is the best I have so far. I=20/(1+ .5) I=13 1/3 ft lbs.

I=(AxL)/(E+L) The accepted formula, the adjusted torque value changes as the length of the handle changes using this formula.

example I =(20x14)/(6+14) with 14" torque wrench I=14 ft lbs

and for the 16" torque wrench I=(20x16)/(6+16) I=14.8

Now that's a pretty big difference using a different length torque wrench on the same exact bolt to be tightened with the same exact extension at the same point of contact on the extension. That is not close enough for my liking.
The torque delivered should be the same at the drive end of each of the torque wrenches delivered to the open end of the 1/2" combination wrench.

Now if I were to put 40 lbs on the 1/2" combination wrench which is 6" long I would achieve 20 ft lbs at the bolt to be tightened. If I slip a pipe over the combination wrench and now place 20 lbs at the 1 ft mark I will torque to 20 ft lbs (given the pipe wieght nothing). I now place 10 lbs on the pipe which is over the combination wrench at the 2 ft mark and I still have 20 ft lbs of torque on the bolt to be tightened.

Lets go back now to the torque wrenches. I apply 13.3 ft lbs of torque to the open end of the 1/2 inch combination wrench to acheive 20 ft lbs of torque on the bolt to be tightened.

Now I apply 14 ft lbs of torque at the open end of the combination wrench, is the torque applied to the bolt to be tightened the same?

Now I apply 14.8 ft lbs to the open end of the 1/2 combination wrench is that still the same 20 ft lbs of desired torque? I think not, therefore length of torque wrench is not a correct way to calculate extension advantage in torque application.

Thus Scotty's formula I=A/(1+E) is the most consistant formula given in the whole thread. It stands as a good formula still. His own claim denouncing his formula was without merrit and that can be revisited if needed.

Handle length formula has just been proven to be inconsistant. I got the second formula using the handle length in the calculations from this
web page which comes up on the google search highly ranked.

So inquiring minds want to know... What's the stinkin formula if the formula scotty presented isn't accurate? <img src="/forums/images/graemlins/lol.gif" alt="" /> <img src="/forums/images/graemlins/lol.gif" alt="" /> <img src="/forums/images/graemlins/lol.gif" alt="" />

Finally figured out the descrepencies and scotty don't go running to your wife yet.
What the other formulas didn't take into consideration is that 30 ft lbs = 30 lbs on the end of a 1 foot lever. The longer handle of a torque wrench is only for ease of applting the desired torque value. If you were apply 150 ft lbs of torque from a 1 ft bar you would work much harder to get to the 150. By making a torque wrench manageable, it is easer and faster to do the work, but still must be delivering the equivelent of 150 lbs on the 1 foot lever. Thus the constant of 1 in scottys formula.

Now take their formula y=(AxL)/(E+L)and plug in the numbers. The reduction of ft lbs has already been made to reflect the amount of weight on a 1 ft lever, thus the constant length to be 1 ft.

y=(30x1)/(1+6)or Y=(AxL)/(E+L) where L=the one foot lever giving the lbs/ft nomenclature. This will be a constant with any type torque wrench as long as it reads in lbs/ft.

Simply stated[color:"blue"] Y=A/(1+E) [/color]

where E is the length in feet of the extension.
and A is the desired torque and Y is the adjusted value setting for the torque wrench.

You've been given the "stinkin" formula numerous times - and you still haven't supplied the information wrt the type of torque wrench or the length of the torque wrench handle you're using - but I think you've given me enough information to understand what's confusing you. It's easy to see why - it's even more difficult to explain, I think.

Again, torque wrenches are calibrated using the length from the drive axis to the center point of the handle. Because of the mechanical advantage of a longer handle torque wrench, less effort is required on the handle to deliver the same torque at the fastener. You are correct that the same force can be applied at the end of the two torque wrenches, but different input forces (you) are required to accomplish it. However, the force you want to apply is at the axis of the bolt and the mechanical advantage comes back into consideration in handle length as it relates to calibration of the wrench indicator (gauge). When the extension is added as a torque multiplier, it becomes a factor in the overall equation and although it is as one with the torque wrench, the reading must be changed to derive applied torque at the fastener. IOW, you are using the formula to recalibrate the torque wrench dial to take into account the change in total length and applied force. The change is less of a factor with a longer handle than with a short one because it represents a smaller percentage of total deviation from overall change (length of swing arc determined mechanical advantage) in relationship to the original calibration of the torque wrench.

In your example of the 14" vs. 16" handles, your answers are correct because of the above (admittedly confusing)explanation. If you extend the handle out to infinity (AND CALIBRATE IT THERE BEFORE ADDING THE EXTENSION), then add the extension, the 6" extension would be such a small overall effect as to not require an adjustment - you would simply set the dial at 20 ft/lbs.

For example - look at a handle of 100" (that is calibrated at that length) - the formula tells you to set the torque wrench to 18.9 ft/lbs. That is the ADJUSTED CALIBRATION of the wrench needed to provide 20 ft/lbs on the fastener with a 6" extension.

What you're not grasping - and perhaps I'm not explaining well - is that a longer torque wrench has more mechanical advantage than a short one, but the fixed length of the extension has less effect on the overall calibration of the longer handle wrench than on the short one.

Does that help? If not, it's the best I can explain it - perhaps someone else wouild like to take a try.... or you can use an incorrect formula if it better suits you needs.

Now inquiring minds should know....... <img src="/forums/images/graemlins/kewl.gif" alt="" />

Frank <img src="/forums/images/graemlins/cheers.gif" alt="" />

Okay Magoo, let's work it backward.You want 20 ft/lb on the bolt, right? There's only 2 variables left: force and distance. Now if my extension is .5 feet and t-wrench effective lentgh is 2 foot, then the lever is 2.5 feet long. How hard do I pull on a 2.5 foot lever to get 20 ft/lbs on bolt? 8 lbs, right? 2.5 X 8 = 20. Now I need to figure out how to get an accurate 8 lbs pull (NOT ft/lbs) on the end of my 2.5 foot lever (t-wrench + extension). The only gauge available is the torque scale on the t-wrench (handy eh?). 8 lbs pull on a 2 foot t-wrench yields a reading of 16 ft/lbs on the scale, so, pull the torque wrench with the extension on till it reads 16 ft/lbs and your bolt will be torqued accurately to 20 ft/lbs. Let's try it with a 1 foot effective length t-wrench + .5 foot extension. Lever length is 1.5 feet, so,1.5 X 13.33 = 20 . I need 13.33 lbs pull on a 1.5 foot lever to get 20 ft/lbs, right? Okay it's easy this time,to get 13.33 lbs pull on a 1 foot effective length t-wrench, you pull till the gauge reads 13.33 ft/lbs. So put the extension on your 1 foot t-wrench and pull till the gauge reads 13.33 ft/lbs and your bolt will be torqued accurately to 20 ft/lbs. Just think, pull, in lbs times lever length, in feet gives you accurate torque in ft/lbs on your bolt. I guarant....aw heck just check it out. <img src="/forums/images/graemlins/lol.gif" alt="" /> <img src="/forums/images/graemlins/lol.gif" alt="" /> <img src="/forums/images/graemlins/lol.gif" alt="" />

Scotty