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I may be crazy (3 ex-wives all agree!) BUT I found the above funny. Can't wait till we get down in the dirt with integrals & differential equations!

As an aside, as a design engineer, I never integrated or differentiated a bloody thing after the 3rd yr of college! Once tried to establish a formula to calculate the internal volume of a Weatherby shaped cartridge, failed. Turned it over to a graduate mathematician who also utterly failed. Had to laugh!
 

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As far as I know, the word order is in fact only a method of drawing a verbal distinction because we don't have different words for a unit of distance that is being traversed as opposed to one that is used as a factor in force's angular analog. You can see this in the analogous forms of Newton's Second Law:

In linear action:

F=ma

In angular action:

τ=Iα

Where τ is torque, I is the moment of inertia, and α is the acceleration of spin rate. Thus, torque is the angular analog to linear force. It is not a description of an action, as work is.
 

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Discussion Starter · #24 ·
Foot pounds as in a bullet or car engine is foot-pound force (dot product). Weight is the pound mass. The torque of a torque wrench is Foot-pound mass (cross product).
Foot pounds as in a bullet is foot-pound force (dot product).
Foot pounds as in a car engine is foot-pound force (cross product).
Weight is the pound mass x "g", hence pound force.
The torque of a torque wrench is foot-pound force (cross product), same as a car engine.
 

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Discussion Starter · #25 · (Edited)
Thus, torque is the angular analog to linear force. It is not a description of an action, as work is.
Also, work = force x distance (f and d parallel), so f = w/d (work per unit displacement).
Analogously, work = torque x angular displacement, so torque = work per unit angular displacement.
 

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Discussion Starter · #26 · (Edited)
Why do the units of torque seem to be that of work or energy?

Put a wrench of length 'r' on a nut and apply a force 'f' perpendicular to the wrench. Maintaining the perpendicularity, rotate the wrench through a complete circle. The work done by the force will be 'f' times the distance moved, the circumference of the circle = 2(pi)r. Then, work = f x 2(pi)r. Rearranging, w = rf x 2(pi). Rename rf = torque. Then, torque = w/{2(pi)], where 2(pi) is the number of radians (dimensionless angular measure) in one revolution.

In another instance, with the same force and wrench, half a revolution would correspond to an angular displacement of pi radians. The work and displacement would each be half their previous values, but their ratio (torque) would be the same. Continuing in this vein, we may calculate the work done during one radian of angular displacement. Conventionally, torque is expressed as the work per radian.

The similarity of the units of torque to that of work is seen in the units expressed here, foot-pounds/radian. Being dimensionless, one may drop the unit of radians, hence the term foot-pound or pound-foot, if you prefer. The key is that the 'foot' in the foot-pounds of torque, in this energy approach, is the arc length. That is, the length of that part of the circle traversed by the end of the wrench, not the length of the lever.
 

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Unless I'm missing something, not work done per revolution, but work done per radian of revolution.
 

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Discussion Starter · #28 · (Edited)
Unless I'm missing something, not work done per revolution, but work done per radian of revolution.
True. See previous post. I was emphasizing, poorly as it turns out, the total work done in the example, not creating units for torque. I didn't want to introduce the concept of arc length.

Fixed it. See next post.
 

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Discussion Starter · #29 ·
Why do the units of torque seem to be that of work or energy?

Put a wrench of length 'r' on a nut and apply a force 'f' perpendicular to the wrench. Maintaining the perpendicularity, rotate the wrench through a complete circle. The work done by the force will be 'f' times the distance moved, the circumference of the circle = 2(pi)r. Then, work = f x 2(pi)r. Rearranging, w = rf x 2(pi). Rename rf = torque. Then, torque = w/{2(pi)], where 2(pi) is the number of radians (dimensionless angular measure) in one revolution.

In another instance, with the same force and wrench, half a revolution would correspond to an angular displacement of pi radians. The work and displacement would each be half their previous values, but their ratio (torque) would be the same. Continuing in this vein, we may calculate the work done during one radian of angular displacement. Conventionally, torque is expressed as the work per radian.

The similarity of the units of torque to that of work is seen in the units expressed here, foot-pounds/radian. Being dimensionless, one may drop the unit of radians, hence the term foot-pound or pound-foot, if you prefer. The key is that the 'foot' in the foot-pounds of torque, in this energy approach, is the arc length. That is, the length of that part of the circle traversed by the end of the wrench, not the length of the lever.
 

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Discussion Starter · #31 ·
Reading this thread makes me feel like the idiot, I now, know I am.
Torque is more simply the lever arm times the perpendicular force. The explanation I gave
is only 'out there' because I was trying to show the connection to energy units.
 

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Reading this thread makes me feel like the idiot, I now, know I am.
We could switch over to why Scotch is measured in 'fifths' if that helps ;)
 

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To-ma-to/ Te-mah-to, the only people it matters to is the linguists.
I often meet people from other parts of the world in my work, the English language as it's spoken here in America does an amazing job of communicating idea's and concepts, one of the reasons is it's not polluted with slang as it is in other places.
Proof of that is your having a discussion over something like this.
Half the people on here don't believe there's any correlation to it and killing power anyway!
Just my observation, I could be wrong.
 

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Discussion Starter · #34 ·
We could switch over to why Scotch is measured in 'fifths' if that helps ;)
If you think you are getting a "fifth", you are actually being shorted 7 ml.
 

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I'm sure that's the real reason they standardized even U.S. and U.K. wine and spirits on 750 ml instead of gallon fractions. It saves them 0.94% product on every bottle! If the conversion had favored giving a little extra, I'm sure we would still be sold gallon fractions.
 
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Discussion Starter · #36 · (Edited)
I recall gasoline being sold by the liter for a short period.
 

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Funny, my German grandfather said goodntight with his german accent and then would laugh when tightening bolts on tractors and equipment. I had totally forgotten about that, thanks for the memory jog jimboro.
 
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