Philostang

Hey Texotic,

In this instance, you're probably right as a practical matter, since the torque values are relatively small for the bolt/stud in question.

However, if you look at the calculator on the diagram, you'll note that any extension of the pivot point beyond the torque wrench's original design will yield an increase in torque over the value either read or set on the wrench. A crow's foot by design will extend the pivot point, albeit a small amount (say an inch or so). That is, the center of the bolt or nut turned by the crow's foot will be 1" or so further than the ratchet pivot on the wrench. This will change the value of actual torque being applied as a matter of basic physics. Of course, as I opened, the change in this case is likely to be so small as to not matter much (perhaps others disagree?).

Best,
-j

Ato

Although not really the right way to do things I just tightened it down as far as I could w/o any torque specs and marked the nut and body w/ nail polish. Considering we are talking about 35-48 lbs tq depending on the the front or back I figured I would be hard pressed to go overboard w/ 2 wrenches in my hand and would just keep an eye on the nut w/ the polish, so far no movement.

Norm Peterson

Since torque values are generally given as a range of acceptable torque, and since you know the effect will be small, just shooting for the middle of the range instead of the high end will be good enough.


Norm

Philostang

Thanks Norm, I was hoping to hear from you or some others to confirm.

Playing around with the calculator, my 18" torque wrench with an additional 1" added by the crowfoot produced 42 lb-ft from a 40 lb-ft setting. Not much to worry about at all.

And...had I known all this (or thought it through) at the time, I would have just gone with purchasing the crow's foot rather than fab-up a couple of sockets. Live and learn.

Best,
-j

Texotic

That's why I said if the crow's foot was perpindicular to the handle. Like if you were looking over the pivot point of the wrench with the handle at 6 o'clock, the crow's foot would be at 9.

Torque is just the dot product of the radius vector and force vector. Therefore, if you just offset the torque wrench to the right along the axis of the force, it doesn't change the torque value since the distance of the force vector from the axis of rotation hasn't changed.

If you put the crow's foot on the end pointing in the direction of the handle (12 o'clock), then it would increase the distance of the radius vector and you need to make the necessary adjustments.

Unless I just wasn't paying attention in Physics when we did torque...

Philostang

Hah! Texotic, I must have been low on some very vital level of caffeine when I read your post. Sorry, I didn't catch the ever-so-subtle "perpendicular" point. My bad. But now I'm just curious...like way more than is good for me while sitting here allegedly at work...

I think I see what you mean about the force and radius vector lengths being identical in both cases (no foot vs. perp. foot). But don't both of these actually change under the "L" configuration of the crow foot being perpendicular to the torque wrench?

Take the radius vector first. Doesn't the "L" configuration derive its radius from the hypotenuse of the triangle in which the wrench length and the crow foot form the legs of the triangle? If so, then the radius vector is longer with the crow foot extension even in the perpendicular position. The hypotenuse is always the longest side of a right triangle, so it will be longer than the length of the torque wrench (our original radius vector length).

Moving on to the force vector, assuming these are both pulled on the same amount, the force would seem to be equal. My concern is that they are not brute forces, but applied vectors, so their direction matters. In the case of the non-crow torquing, the force is applied at 90 degrees to the wrench handle. The same is true in the "L" configuration, but in this configuration 90 degrees to the handle (the longer leg of the right triangle) is not 90 degrees to the radius vector (hypotenuse). This angle is less than 90, and so less force is being applied directly perpendicular to the radius. Doesn't this mean that less torque is in effect being applied? Of course, if the crow-foot leg was really long, this angle would be substantially less than 90. Then again, would this cancel the increase torque of the longer radius?

Anyway! We don't apply a constant pull to a torque wrench (after all, that's sort of the point isn't it...). The force will be read or set off of a pull perpendicular to the handle, which means that even if the angle is smaller we'll just keep pulling harder. But this doesn't address the problem now that the radius has gotten longer; we will be applying more force than our reading or set indicates. So we're back to possible changes in the radius vector given the "L" configuration...that's my real uncertainty.

Well, my physics is rusty, so you see why all the questions come up. Mostly I think all of this is just a matter of curiosity (the struts can be safely torqued with just about any of the above suggestions), but I'm a curious guy and I really do appreciate others who will engage in cat-like discussion with me. Thanks Texotic, you woke me up today.

Best,
-j

Texotic

I'm pretty sure that you don't use the hypotenuse as your radius. When you measure torque, the radius and force vectors have to be perpendicular. As long as you are pulling directly perpendicular to the handle, then the orthogonal radius vector would be equal to the length of the handle. If you were assuming the radius vector was the length from the bolt to the handle (hypotenuse), then the force vector wouldn't be how hard you are pulling, it would be how hard you are pulling multiplied by the sine of the angle between the crow's foot and hypotenuse (I think... don't really feel like doing trig right now).

Although, it just occurred to me that the torque isn't being measured at the crow's foot, it's being measured at the joint between the wrench and crow's foot, so I'm sure that screws everything up. I'm on Spring Break, so I don't feel like thinking about it any more; I think the percentage difference isn't a big enough deal to lose sleep over.

Philostang

Cool. We're on Spring Break next week, so I understand (sadly, I'll be with my in-laws the entire week...may prefer working on torque problems).

Anyway, I think I was also overlooking the fact that the torque is not being measured at the crow's foot, so this has helped (at least ferret out a mistake). Have a great break w/what's left of it! I should get back to grading.

Best,
-j

Norm Peterson

Easy to get confused with this.

You'd use the hypotenuse only if the direction of your force was perpendicular to it instead of to the torque wrench's beam. Even a pin-connected handle will support a component of force that's axial to the wrench beam.

Normally, the moment (torque reading) that you obtain at the center of the wrench's drive square simply remains constant at that value along the crowsfoot (assuming perfect perpendicularity).

I imagine that a certain amount of uncertainty regarding the direction of the force that you apply to get to a reading is taken into account in the permissible torque spec range.


Norm

Texotic

So are you saying that if the crow's foot was perpendicular and you were pulling perpendicular, you wouldn't have to make any adjustment?