Simple explaination of HP vs Torque, using an FRS as an example..

[switchlanez]

Originally Posted by ZDan

If power is a function of time, then so must be either torque and/or rpm.

If torque and engine rotational speed are constant, not varying with time, then power P = T*w (should be a tau and an omega there, but no greek symbol font available!), torque multiplied by angular rotation speed, not a function of time.

You could have an expression for power varying over time shown as P(t), but then you would have an equation in terms of time "t".
Saying P(t) = T*w doesn't make sense, as there's no "t" in there, and if T and w are fixed, so is P.

I figured out why the right side of the equation is written without (t). Because it is a dot product of torque and angular velocity in vector notation, not scalar notation (where only magnitude matters). In vector notation, you get the full package of information (magnitude and direction as functions of time) and is denoted using italics and boldface.

w = <wx(t), wy(t), wz(t)>
T = <Tx(t), Ty(t), Tz(t)>

where x, y, and z are supposed to be subscripts denoting the x, y, and z directions. Vector notation is implied to be a function of time considering its x, y, and z scalar components are functions of time.

So the original power equation was correctly written as

P(t) = T*w

and the vectors T and w can be expanded in longhand so

P(t) = <Tx(t), Ty(t), Tz(t)>*<wx(t), wy(t), wz(t)>

where "*" means dot product of two vectors (which is not straight multiplication). Note the dot product operation results in a scalar value P(t), not vector P.

[TurtleFactory]

Thanks!

[ZDan]

No need for vector notation, we know that the axis of rotation and the axis of torque are the same. Simple scalar arithmetic.

Power P, torque T, and angular rotation w (omega) can be fixed or varying over time.

Rotational speed w, while being time-related, isn't necessarily a *function* of time.
If revs are held at, say, 5000rpm (524 radians/sec), that's not a *function* of time, it's just w = 524 rad/sec.

Usually, we're talking about constant, instantaneous values. P = T*w. None as a function of time.

If you have constant T and constant w, P is constant. If P is changing with time, then torque and/or rpm must be changing as well.

Calling P a function of time when T and w are fixed is incorrect, and reinforces the INCORRECT interpretation of power as "torque over time". It isn't. Power is the *instantaneous* rate of doing work. At a given torque and rpm, power is constant, not a function of time.

[switchlanez]

Originally Posted by ZDan

No need for vector notation, we know that the axis of rotation and the axis of torque are the same. Simple scalar arithmetic.

Power P, torque T, and angular rotation w (omega) can be fixed or varying over time.

Rotational speed w, while being time-related, isn't necessarily a *function* of time.
If revs are held at, say, 5000rpm (524 radians/sec), that's not a *function* of time, it's just w = 524 rad/sec.

Usually, we're talking about constant, instantaneous values. P = T*w. None as a function of time.

If you have constant T and constant w, P is constant. If P is changing with time, then torque and/or rpm must be changing as well.

Calling P a function of time when T and w are fixed is incorrect, and reinforces the INCORRECT interpretation of power as "torque over time". It isn't. Power is the *instantaneous* rate of doing work. At a given torque and rpm, power is constant, not a function of time.

I agree, scalar is all that's needed for our purposes. I was just explaining the image I copy/pasted which happened to be in vector notation (I didn't find one written with constants independent of time). That function written with vectors as a function of time is NOT incorrect, just extraneous for our purposes.

[Xanatos]

Originally Posted by ZDan

No need for vector notation, we know that the axis of rotation and the axis of torque are the same. Simple scalar arithmetic.

Power P, torque T, and angular rotation w (omega) can be fixed or varying over time.

Rotational speed w, while being time-related, isn't necessarily a *function* of time.
If revs are held at, say, 5000rpm (524 radians/sec), that's not a *function* of time, it's just w = 524 rad/sec.

Usually, we're talking about constant, instantaneous values. P = T*w. None as a function of time.

If you have constant T and constant w, P is constant. If P is changing with time, then torque and/or rpm must be changing as well.

Calling P a function of time when T and w are fixed is incorrect, and reinforces the INCORRECT interpretation of power as "torque over time". It isn't. Power is the *instantaneous* rate of doing work. At a given torque and rpm, power is constant, not a function of time.

Angular Velocity (w) is a function of time. "Holding a RPM" is just stretching time. Remember the key words and definitions at play: Rotations Per Minute. Without the relation of a scalar value to time that scalar value would be useless. Take for example the speed of a space shuttle is roughly 17,000 miles per hour. By dropping the "function of time" you drop per hour. Well in relation 17,000 mph is f'in fast. Without knowing the function of time its plain 17,000 miles. It could be miles per second, miles per minute, year, decade and so on. There is a direct relation between the speed and the delta time at play in all velocities. Thats why its a function of time.

[ZDan]

Originally Posted by Xanatos

Angular Velocity (w) is a function of time. "Holding a RPM" is just stretching time.

Time is involved, but constant rpm => rpm is not a *function* of time.

[Xanatos]

Your definition of a function must be different from everybody else's.

Say we have 5000 RPM. RPM = rotations / minute.

5000 RPM could mean 10000 rotations / 2 minutes or 20000 rotations / 4 seconds. The math is the same when you maintain the same RPM but when you change one value another must change to keep the constant RPM. Thats why its a function of time. It can also be stated that its a function of angular speed as well.

When looking at instantaneous power you use the limit of t(time) as it approaches zero. Meaning that without time power cannot be calculated. Otherwise you would just factor zero as time and power would be infinite in all cases.

http://www.lmgtfy.com/?q=power+function+of+time

[ZDan]

Obviously you have to keep your units straight. For most calcs you'll want to stick with radians per second.

You don't need to do any limits as dt approaches zero. You *have* the constant rotational speed.

You absolutely can calculate power without a value for time.

5000rpm (524 radians/sec), 200 lb-ft torque:

Power = 200 lb-ft * 524 rad/sec = 104,800 ft-lb/s
1hp = 550 ft-lb/s

Power = 190hp

I just calculated power without time.

If something is a "function of time", that means it is variable with time. If we're talking about constant rpm and constant torque, power is also constant, none is a "function of time".

[Xanatos]

Originally Posted by ZDan

Obviously you have to keep your units straight. For most calcs you'll want to stick with radians per second.

You don't need to do any limits as dt approaches zero. You *have* the constant rotational speed.

You absolutely can calculate power without a value for time.

5000rpm (524 radians/sec), 200 lb-ft torque:

Power = 200 lb-ft * 524 rad/sec = 104,800 ft-lb/s
1hp = 550 ft-lb/s

Power = 190hp

I just calculated power without time.

If something is a "function of time", that means it is variable with time. If we're talking about constant rpm and constant torque, power is also constant, none is a "function of time".

You sure did! When you restrict RPM to a constant you lock time as a ratio. So whatever calculations you do power will always use time. Your formulas assume 1sec interval right now. Plug in .5 sec and your power will change. Thats power as a function of time.

[ZDan]

I'm not assuming "1 second interval". I *have* the rotational speed: 5000rpm (or 524 rad/s). I don't have any need of a time interval.

There is no "interval". The calculations are outside of time, instantaneous.

[Want.FR-S]

^ w.r.t. the term "function of time", one can define a function F of time by denoting F(t). So if we say Y(t) =F(t), that means the value Y changes over time T based on the function F. It has nothing to do with the unit of F.

Example: suppose I am calculating the rain drops into a bucket. At Time T = 0 second, no drops. At T = 1 second, 2 drops into bucket. At T = 2 seconds, 3 more drops into the bucket and I have 5 drops in the bucket. Therefore, we can say:

Y = total drops in the bucket
Y(0) = F(0) = 0;
Y(1) = F(1) = 2;
Y(2) = F(2) = 5;

Since Y is the sum of drops in the bucket.

Yes, the values of F and Y changes over time, so they are *function of time*. However, their *units* are drops, not drops/second. The unit drops/second is used as the *rate of change* for the function Y, and in Calculus term, that is the first derivative of function Y over time. This is just like the term distance and velocity of a moving object.

So back to the discussion, the original equation P(t) = T*W/5252 is incorrect because if the power is a function of time, both Torque and W (or RPM) should also be the function of time. In real life, Torque and W changes over time. It is rare that you can hold constant torque and RPM over a period of time. The correct formula, if we want to represent it as a function of time, should be

P(t) = T(t) * W (t)/5252

1. I just use the English unit system (HP/lbf-ft) for simplicity.
2. The term t does not mean something divided by time. The term t means at time equal to T at this observation. So you can say at this moment 16:38:40 March 4, 2013 I have an engine generating 200 lb-ft of torque @ 5000 RPM, and that engine produces 190 HP horsepower work at this moment.

Hopefully that clarifies the mathematical terms and things.

[Yruyur]

After reading this thread I just realized I am dumb.

Sent from my SCH-I535 using Tapatalk 2

[Speedy000]

100% i dropped physics for a reason.

[switchlanez]

I blame myself for copying the image from wikipedia and confusing everyone lol. But that equation is correct! And I already explained why. Tau and omega are time function vectors!

P(t) = T*w (where T and w are vectors, NOT scalar constants)
w = <wx(t), wy(t), wz(t)> <-- omega vector is a function of time
T = <Tx(t), Ty(t), Tz(t)> <-- Tau vector is a function of time

Substitute time vectors in for w, T and we get

P(t) = <Tx(t), Ty(t), Tz(t)>*<wx(t), wy(t), wz(t)>

But we are not concerned with 3 dimensions since torque is applied in 1 dimension and angular velocity is spinning on 1 axis so we can rewrite the equation in 1 dimension as

P(t) = T(t)*w(t)

As @ZDan mentioned, we are plugging in constants for torque and angular velocity that don't change w.r.t. time:
T(t) = T(-infinity) = T(0) = T(+infinity) = T (constant such as 151 ft-lbs)
w(t) = w(-infinity) = w(0) = w(+infinity) = w (constant such as 7400 rpm)

Substitute those in and we get

P(t) = T*w = P

You need to understand what vectors are for any of this to make sense. Again, the original P(t) = T*w is correct; it just carries a lot of uncecessary information writing the torque and angular velocity variables as 3-dimensional (x,y,z) vectors which are implied functions of time.