I have the shaft torque... what I need is a half horsepower of
heatflow at 300 F.
I'm having visions of paddles inside of an insulated tank full of gear

oil. Bet it would work just fine. But it smells a bit klugey.
Any more *elegant* way to do it?

"RP" wrote: Torque is in units of energy, heat flow in units of
energy/second. You can't convert them.
^^^^^^^^^^^^^^^
If you're going to adopt a superior attitude, and criticize the OP, at least
try to get it right. Torque is NOT in units of energy. Lb-ft and ft-lbs
are not the same. I'll leave it to you to figure out what I am talking
about.

What about Newton Meters??
or Meter Newtons??
I think I will have another Fig newton :-)
On the serious side, the paddle wheel thing...
Seems to me that when dealing with fluid dynamics, there is no heat unless
there is friction.......but I could be wrong

unless
As long as you are able to do work on the system, heat will be
generated. It's not reversible though, That is, if he were to apply
heat, the paddle wheel would not turn...

*only* apply heat.
Because we apply heat to a turbine, allowing it to escape (along
with the heated gas), and the turbine turns.
You can apply heat to one "side" of this "dashpot", and cool the
other, and convective currents in the fluid can be made to turn
the shaft...
David A. Smith

Again, I have no idea what the original topic is-this thread was
crossposted to RCM where I found it. It looked like he's trying to
just heat up some oil from here...

True. But "lb-ft-radians" *is* a unit of energy and the radians, being
unitless, could have been neglected, as in most rotary calculations. In
other words, turning a shaft with a torque of 1 lb-ft, through an angle of
one radian, does 1 ft-lb of work.
But I know what you meant, Leo. I'm just being pedantic.
Don
Kansas City

Well hell, as long as you brought it up, why not turn this useless
thread into something interesting.
Question: What situation can you come up with in which allowing radians
as a dimensional quantity will lead to mathematical contradiction.
After studying dimensional analysis, and formal logic, I put the two
together and concluded that there really was no logical justification in
prescribing dimensionless factors, period. Now, let's allow for the
moment the radian as a dimension, immediately the anomaly that led to
your out of line reply to me would no longer exist. What are your
thoughts on this. BTW, I wasn't attempting to be superior, it was just
a bit of dry humor, there's too little of that on usenet, so no wonder
it's almost always misinterpreted. It wasn't directed toward the OP,
since he most likely didn't even understand what I said.
hvacrmedic

Torque is NOT energy. The units "newton-meter" or "ft-lb" is a
simplification of the real vector formula for torque: r x f (that's
radius CROSS PRODUCT force, not radius TIME force). Radius "times"
force is an oversimplification assuming that the radius and force are
perpendicular.
If you want to be dimensionally accurate with torque, call it
"ft-cross-lbs" or "newton-cross-meters".
Dave

The fundamental difference between rxf and f*d, even if we assume the
convention of perpendicular force in the former case such that r*f
(which is more consistent with our convention that the force is all
ready defined to act along the line of motion in f*d), is that r and d
are fundamentally different dimensions. d is displacement, which is of
course relative to the FoR, while r is invariant (barring
lorentz/fitzgerald contraction of course). There is a fundamental
difference between length and displacement, and thus in reality these
should be expressed with differing dimensions. IOW, you were sort of
frame jumping <sic> between conventions in order to make your case. If
we then adopt delta_s (or d) as a different dimensional quality than r,
then we would be forced to set radians as a dimensional quality in order
to complete the set of relationships between d and r in a rotating
system. That's more or less the gist of my initial speculation :)
hvacrmedic

<SNIP>
Bingo. That's just it. With work, force acts ALONG the line of
motion. With torque, it acts perpendicular. Back to HS physics: if a
block of ice is pushed along a flat plate of ice, how much work is
gravity doing on the block? Answer: none, because the force is
perpendicular to the direction of the gravity component. Same with
torque.
See the following quote:
"However, torque is not energy."
http://www.physics.uoguelph.ca/tutorials/torque/Q.torque.intro.html

The point is that F and r are in perpendicular DIRECTIONS.

Why are you bringing frame jumping into this? There's no frame jumping
needed to make my case. R CROSS F still doesn't equal R TIMES F. R
TIMES F is a simplification, and the vectors lay in perpendicular
directions.
Dave

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

<SNIP>
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"
http://www.ac.wwu.edu/~vawter/PhysicsNet/Topics/Vectors/VectorProducts.html
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

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

<SNIP>
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

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

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.

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

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