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="" />