On a beam style torque wrench, you'll notice that there's a pivot in the handle. As long as when you apply a load through your extension and the handle is not cocked over (still pivoting about that point), you haven't affected the reading. If you have, then the reading isn't accurate.

Also, an extension that is used in line with the socket and the torque wrench doesn't affect the torque value.



This is easy, yes. That's why you can have two different length torque wrenches still give the same torque reading. You'll have to pull harder with a shorter torque wrench to achieve the same torque reading than with a longer wrench.

Frank,
I'm not sure I read correctly or not, but the mechanical advantage, or lever arm, is from the center of rotation to where the load is applied. So, if you have a breaker bar, the distance of the lever arm is from the center of the 3/8" or 1/2" square drive to where you apply the load. The radius of the nut doesn't get involved in the calculations. That's why when you torque a bolt, it doesn't matter whether it's a 1/4" nut or a 3/4" nut.

Hi Lee -

That's exactly why I said I wasn't certain about how the mechanical advantage should be determined. I see your point, but it does seem to me that the diameter of the nut should be used in the mechanical advantage relationship, because the rotational force is applied in its own radius - or so it seems to me. I might be way off base on it, though. Interesting question and one I've certainly never before considered.... that I remember. But then, school was a looong time ago.

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

Lee -

From this definition I'd say that you are correct:



http://www.pha.jhu.edu/~broholm/l18/node2.html

Thanks for the correction! <img src="/forums/images/graemlins/cheers.gif" alt="" />

Frank

Lee -

For the sake of curiousity, let's back up a second on that theory. Granted, the torque is generated on the axis of rotation, but if you don't use the diameter of the nut as a base for calculation, how would you find the mechanical advantage so you could determine the amount of force to apply at a fixed distance from the center of rotation?

IOW, using the absolute center of rotation - "0" - in relation to the length of lever arm, the mechanical advavantage would be "infinite" no matter how long the lever. Isn't that correct - or am I confused?

In order to determine mechanical advantage, 2 real values must be known. What am I missing? <img src="/forums/images/graemlins/confused.gif" alt="" />

Frank

All you want to know is what torque to apply. Torque = force X distance (or lever arm), or, using some algebra, Force = Torque/distance. From your definition, the axis of rotation is the centerline of the bolt or nut. This is the same as the centerline of the socket/extension/square driver. So, the distance (in our formula above) is the distance from the centerline of the square driver to where you apply the load on the handle of the wrench.

What you are confusing this with is the resultant force that gets applied to the nut. In engineering, we do a lot of summation of moments. A moment is an engineering term for torque. Summation of moments means that if nothing is moving (or is static) then there has to be a reaction moment (or torque) equal to what you apply. This is the force that the nut resists with on the socket acting through some lever arm.




Remember from the first post that a 3 lb force applied using a 2 foot bar gives 6 ft-lbs torque, but a 12 lb force using a 1/2 foot bar also gives 6 ft-lbs torque. Clear as mud?

ok for guy that only have a ged.bar type torqe wrench,79 ft of torqe for the head bolts and a 3"ext.was my torqe still the same or do i need to fix it?

Lee -

Understood - I spent my undergraduate years in e-school, but never worked a day as an engineer (other than summers between years), so have forgotten much. I think what you're referring to in the example of "nothing moving" is called "net torque".

What I was attempting to determine was not where the resulting net torque was applied (or even how much was applied), but what mechanical advantage would be at a given lever length without benefit of a torque wrench. Obviously there's a difference in MA when one grabs the lever at varying distances, but there also has to be a base-line for mechanical advantage determination and to avoid MA values of infinity there has to be a finite number to work with, doesn't it?

If you put a small diameter nut on a thread and then replace it with a larger diameter nut, mechanical advantage has increased in the ability to apply torque to the fastener. So, it seems that some starting point must be identified in order to calculate the mechanical advantage of one over the other. I can see where the nut rotational diameter would not be the base line - would it be the mean thread diameter since that is the only element that doesn't change?

If you use the mean thread diameter circumference as one element and the diameter of the lever length where force is applied as the second element, would division of the mean thread diameter into the lever rotational diameter not yield a mechanical advantage? If so, it would answer the original question of how much force to apply - at what distance - to the lever to gain a torque value without the use of a torque wrench. If that formula won't work, then the only answer lies in the torque angle tables, I guess.

Man, this is like going back to school.... <img src="/forums/images/graemlins/lol.gif" alt="" />

Frank

OK, let's see if I've got this.

If you have an extension like a very elongated crowsfoot, i.e., a bar with a female 1/2" drive socket on one end, and a male on the other end, both perpindicular to the shaft of the bar, like a breaker bar with a female socket on the handle end. I insert a torque wrench on the female end, and a socket to the work on the male end. The bar is 1' from center to center of the drives. I put the t/wrench on in line with the ext. bar and apply 50 ftlbs to a .5" radius to contact fastener. How much torque did I apply to the fastener? Now I put the t/wrench on at close to a right angle to the ex. bar (actually at a smaller angle to put the handle pivot of the t/wrench at the same distance from the center of the fastener as the female extension bar end) and apply 50 ftlbs on the wrench scale. What did I do to the fastener this time? Same thing? Different? I think it's the same, and I think it's 600ftlbs, but I get confused easy when I do the math in my head.

Confusing, isn't it! But, whether you have a large nut or a smaller nut, you still apply the same torque using a torque wrench or pulling at the correct force/distance on a breaker bar. The chance of rounding corners on a larger nut are less than on the small one but you still apply the same torque. That torque you apply with the wrench gets converted to forces acting on the nut, but this will still convert, when acting through the diameter of the nut, back into the same torque being applied. <img src="/forums/images/graemlins/confused.gif" alt="" /> Still clear as mud, isn't it!

Very good gentlemen. So any body have a good formula?

I came up with this one, but is really difficult to post hear so I will just go through verbally.

I took extension length c/c and divided that into twelve, (ratio for ft lbs.) then
inverted and multiplied that times the specified torque value. I then took that and subtracted it from the specified value to get my actual torque wrench setting. I think I said that right, just doesn't look the same as my formula on paper.

Theory was to figure the ratio advantage of the extension. and subtract that from the torque value, Now I have successfully eliminated the uneccessary variable right?

If the torque value was 50 ftlbs to be applied, and I had a 6" ext. My calc would be 12 divided by 6 = 2. invert and multiply times 50 and I would get 25 50 minus 25 is the corrected value right? So that was a round number. I need 45 ftlbs and have a 2.5 in ext. 12divided by 2.5=4.8 invert and multiply times 45 =9.375 subtract that from 45 for a crrected value of 35.625 This would be the striaght line calculation for the extension. What would it be for various angles? My calculator has the sin cos stuff so if neccesary for the formula so be it.

Did I over simplify this? My number came out only slightly different than using the formula with torque wrench length, but I beleive mine to be more accurate in the straight line use of the extension.

What would the formula be if I had to reach my fastener at an angle other than straight line? or for that the formula using any degree of angle of the extension. We must keep in mind the proper angle of pull in respect to the angle of extension also right. I mean if the extension is at 90 and you pull in a direction parrallel to the extension, the advantage would change if you pulled directly toward the center of rotation of the fastener right?

I believe the mud is getting clearer <img src="/forums/images/graemlins/confused.gif" alt="" />

Im not too sure on the moments of the dia of the bolt as in the diagram above, but will contemplate. But as rebutted, the force on the torque wrench wouldn't neccissarily know what size bolt head or socket head was being used right? Yet for some reason I beleive there is more to it than the stated obvious. Maybe not enough for us car mechanics to concern urselves about, but then again that would apply to my first question of the formula with the handle length.

Just hoping to get a better understanding of the applied torque and proper, methods of use, and calculations.