convert shaft-torque to heatflow?

If you have a force vector and motion along some line other than this, then you can either use vector math or you can do it longhand, e.g. diagonal of the rhomboid and replace the vector with the scalar of d*f. Thus there is nothing absolutely requiring that we define force as acting along the line of motion, it's just a convention, or IOW, another method of solution. In the former case the force doesn't act along the line, only a component of it does. How is the operator thus distinct from the cross product in your r x f ? You are changing conventions from one equation to the other and thus comparing apples and oranges. Your argument about the difference between torque and energy is invalid. If the entire problem were worked out in vector format then the only difference, dimensionally, between the two terms would be the inclusion of radians in the torque expression to make them equal, i.e. to convert them into each other in a rotational system, as pointed out by Don Gilmore. More specifically some angular velocity reference must be supplied. It isn't the cross product that separates these terms from each other, it is the non-equivalence of the dimensionality of r and d. The radian should not be dimensionless.

hvacrmedic

Reply to
RP
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Incorrect. Torque is a vector, not a scalar. Let me repeat: TORQUE IS NOT A SCALAR, IT IS A VECTOR. As you stated above, you are treating it as a scalar. Your oversimplification of F*d eliminates all vector components, and thus treats torque as a scalar (which is incorrect).

See the following quote: "Vectors are not ordinary numbers so normal multiplication has no direct analog"

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What you are trying to do is multiply two vectors (R and F) and a cross product is not the same as scalar multiplication.

Dave

Reply to
dave.harper

Lots of engineering quantities are vectors. Force is a vector. That doesn't change its units at all. A vector has magnitude and direction. The magnitude is indicated with units, like newtons or lbs. The direction is never indicated in the magnitude units; this would be absurd.

Vector math does not explain why torque appears to have similar units to energy. Your cross-product will maintain proper vector direction for the torque, if dealing with vector equations for force and radius, but it does not reconcile the concept of energy, which is what we are talking about.

As I explained in a previous post, the answer is simple. Angular energy is torque multiplied by angle traveled. Since angles are measured in radians, which technically have no units, the units for the angular work look deceivingly the same as the torque alone.

The scalar product does not take any angle of rotation into account...but then neither does your cross-product. It will only return a equation of torque which has magnitude and direction. When you multiply it by an angle of rotation, it *will* be energy and it is no longer a vector. Work is a scalar quantity because it doesn't matter in what direction the work takes place, you have still expended energy.

Don Kansas City

Reply to
eromlignod

Vector math DOES expain the reason if you look carefully at the formula for torque and don't oversimplify things.

Question: How come torque, which is a VECTOR, has the units of a SCALAR? Answer: because it's difficult to express a cross-product in the units, so we simplify.

Example:

Volume=l * w * h, so the units are in * in * in (in^3 or equivalent). The units of each component (inches) are multiplied together to get a SCALAR.

Torque=r X F (that's r CROSS F). It's NOT using the same mathematical operation (multiplication) as for volume. Yet the units N-m imply that we DID use the same mathematical operation. So why do we want to relate units that imply we multiplied force and distance? As mentioned before, it's easier to say ft-lbs or N-m instead of worrying about how to express a cross-product in the units.

Dave

Reply to
dave.harper

The units *ARE* the scalar portion of a vector equation. When coupled with a real number they are literally the magnitude. Pounds are not "force"; pounds are a "magnitude" that describes force, irrespective of direction. A "force" has a directional component *added* to it. Pounds have no direction. For something to be treated as a vector, it must *be* a vector and have an innate directional component with i's, j's and k's. This is freshman physics.

All ordinary units are scalar. That's the whole point of having them--They are pure magnitudes. Which way is a ft/s? Which way is a newton?

Don Kansas City

Reply to
eromlignod

That's really what I'm getting at. The units are simplified down to a scalar quantity. If we wanted to be exactly right with the units, we'd have a dimensional formula beside all answers: force-dot-meter (aka Joule), newton-cross-meter, magnetism, etc. But that'd be a hassle and would scare away little kids from ever learning physics.

Reply to
dave.harper

Dear eromlignod:

Just as a sidelight... Note that voltage has units of work, but you don't actually get power until current flows, and you don't get work until you integrate that power. A similar argument to yours...

David A. Smith

Reply to
N:dlzc D:aol T:com (dlzc)

It is similar, but only when corrected :) V=E/Q But note that Torque=E/theta

Even though you missed it a bit, thanx for the suggestion, because this example does serve very well to illustrate my point in much simpler fashion than my previous attempts.

Now, do any of you see the logical inconsistency between the forms of these equations?

hvacrmedic BTW David, you've known me all along as Richard Perry over there in sci.physics land. How goes it lately in the nuthouse? SOS no doubt :)

Reply to
RP

..

I suspected.

SSDD (same sh*t, different day).

I have a problem with both the coulomb and the mole as well. Dimensional or non-dimensioned... They seem a whole lot like "a dozen" to me.

But radians turn radius into area, then into volume. Let's see if any "bright sparks" have been lit...

David A. Smith

Reply to
N:dlzc D:aol T:com (dlzc)

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