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Torgue vs BHP, who is right?

  1. Dec 21, 2007 #1

    I'm new to this forum but have come here to find out the answer to a debate raging on a Model Flying forum, it is to do with Torque and BHP and 2 people are haveing a very good disagreement and using some very good sums, but who is right PDR or CTB?



    Last edited by a moderator: Apr 23, 2017
  2. jcsd
  3. Dec 21, 2007 #2


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    Staff: Mentor

    Could you give us a summary of the question? I'd rather not read their whole thread...
  4. Dec 21, 2007 #3


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    How about a quick synopsis. I started reading the thread but became bored very quickly when two people start bragging about their big brains.

    It appears that there is some disagreement as to what term is to be used when comparing an engine, either BHP or torque. Is this the case?
  5. Dec 21, 2007 #4
    It won't take long to read :smile:

    Basically the important(I think) bits are ctb says:-

    "One day I'll find the perfect article that explains the difference between "torque" and "power" in terms of the dynamic behaviour of the vehicle under propulsion. And I'll bookmark it for sure and then publish it here!"

    PDR replies:-

    "Firstly let's look at torque, because it's very simple and very real. Torque is a physical thing - a twisting force. Torque can do work; it can cause a stationary mass to rotate (or make a rotating one stop). Simple enough.

    Power is different. It's an abstract concept invented by engineers to help us understand and measure things, but in reality it doesn't actually exist. You can't touch, see or feel "power", and that's part of the problem. "Power" is just a handy number that tells us how quickly we can deliver forces to do work. Why does this matter? Well the bottom line is that if you have two machines which both deliver 100lbf of force they can do the same work, but if one can deliver it twice as quickly as the other then it can have this speed reduced (with gears or levers fregsample) in ways that magnify the force. So the faster machine can produce more force if required, whereas the slower one can't. This abstract thing called "power" is a way of describing and quantifying that simple thing."

    CTB replies

    "Unfortunately not. For example "In general terms the bigger the prop diameter and the lower its speed the greater its efficiency (that is to say that it will make more thrust for the same power). " is just going to confuse people. Propeller efficiency is not the ratio between thrust and shaft power. You need to add in the words "at a given airspeed" or introduce the concept of "thrust power" or similar."
    Then some further argueing and then CTB puts in some fancy equations:-

    "OK - let's try some Physics O level stuff.

    Hopefully we all remember that the Kinetic Energy (KE) of a rigid body in translation (ie no rotation) is given by the expression

    KE = 1/2*m*v^2

    Where m = the mass of the body and v is the scaler value of its velocity (AKA speed).

    Consider the simplied case where the rigid body is moving horizontally (ie no chnages in gravitational Potential Energy) and friction is minimal.

    Then any power input to the body will cause a change in KE.

    The rate of change of KE WRT time will be equal to the power.

    rate of change of KE WRT = d(KE)/dt (ie the differential of KE by time).

    d(KE)/dt = d(0.5m*v*v)/dt.

    mass is a constant.......... therefore the above differentation equals

    d(KE)/dt = 0.5m*v*dv/dt + 0.5m*dv/dt*v
    d(KE)/dt = m*v*dv/dt

    dv/dt = rate of change of velocity wrt time = acceleration

    d(KE)/dt = m*v*a = m*a*v

    However m*a = Force (F).

    d(KE)/dt = F*v


    Power = Force * velocity QED!"

    PDR replies thus:-

    "This should be:

    d(KE)/dt = F*dv

    Which is just as well, because if it wasn't then it would imply that objects required power input to maintain a constant speed in the absence of friction, and that would mean that planets would require power to maintain an orbit, so without external power the planets would all decelkerate and spiral into the sun...

    so your final line would become:

    Power = Force * change of velocity

    WHich is rather the point. The absolute value of the velocity is irrelevent (as well as rather difficult to define, because you get tied up in the whole "frames of reference" paradox). The relevent parameter is the CHANGE of velocity, which is why any two objects of equal mass would require the same force, and the same power, to produce the same change of velocity - regardless of their starting velocities."

    Then CTB counters:-

    "Utter garbage.

    Go talk to a Physics teacher and see how much he laughs.

    Or alternatively post the KE of an object under constant acceleration and explain how the power applied to it must have been constant. NOT."

    There might be some finer points that you need to read the thread for, but I think I've got the jist of it.
  6. Dec 21, 2007 #5


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    I haven't read the whole thread, but in terms of the expression for power CTB is correct (P=Fv). As for engine comparisons, that is better left to an engineer, which I am most definitely not.
    Last edited: Dec 21, 2007
  7. Dec 21, 2007 #6


    Staff: Mentor

    Torque and power, in terms of measurements of an engine system's performance, are measuring completely different things. Basically, if you want to outrun the cops you want a powerful engine, if you want to pull your friend out of a ditch you want a high-torque engine.
  8. Dec 21, 2007 #7


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    To add to that, torque would be the rotational force the car can put out continuously, and torque multiplied by time gives the "pushing power" or momentum/impulse the car can deliver, whereas power is equivalent to the net energy per unit time the car can put out continuously. Higher power means more energy released in a shorter period of time. Both are related to acceleration but torque, if you will, is the push the car can deliver whereas power may mean your machine can run fast, but certainly not whether it can pull out of a ditch.

    To ease the above drivel, power is energy per second, and hence mass * integral(v,dt) / dT or KE divided by time.

    Torque is angular force and is related to power - dMdV / dT = force, dMdV = p = momentum and p^2/(dT * 2m) = power.
    Last edited: Dec 21, 2007
  9. Dec 21, 2007 #8


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    Torque is a force.
    Horsepower is how much work a given amount of torque will provide over a certain amount of time.
    But torque can be static. You could take an electric motor and weld the output shaft to the case so it can' turn. Then you just have a heater.
  10. Dec 22, 2007 #9


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    Normally, the only thing that matters is power, since gearing will multiply the torque on a higher revving, higher power, but lower torque engine. However, in this case, the assumption is a direct drive, without gearing. Instead the only variable is prop diameter and prop pitch. I know that a larger diameter prop is more efficient, except that ground clearance limits this, requiring more blades if two blades aren't enough to take advantage of the power available from an engine. More shorter blades are less efficient than fewer longer blades. I don't know if there's an optimal "pitch" for a propellor.

    So what is needed is a good combination of engine with an appropriate power and torque versus rpm curve and an appropriate propellor. Also there is the trade off of maximum thrust (for 3d manuevers), versus top speed.

    For electric motors, gearing is more commonly used, and here it's clear that peak power and the shape of the torque versus rpm curve are the key factors. The actual torque numbers don't matter. An option to gearing is to use outrunner electric motors, which produce peak power at much lower rpms, allowing direct drive.

    Power is not an abstract concept. Given a fixed amount of mass, it determines the rate of acceleration at given speeds, which is something that can be seen and felt, and can be measured in the results of elasped time and top speed in a drag race.

    Most chassis dynamometers measure power, not torque. Either a variable load is used, or the rate of angular acceleration of a heavy drum is used. In either case, there are two inputs, the measured force, and the speed at which the force is measured over a range of speeds. In semi English units, horsepower = force (lbs) times speed (mph) divided by 375 (conversion factor). In metric units, watts = force (newtons) time speed (meters / second). Without additional input, such as engine rpm, chassis dynamometers can't determine the angular speed of the engine, so they can't calculate the effective torque without the additional input.

    This is correct. It's not power times change in velocity; change in velocity is related to the force (force equals mass times acceleration). There's an obvious frame of reference, the air itself. A model has an airspeed, and the thrust times the airspeed equals the power generated. In the previous paragraph, I listed two examples of equations relating power to force times speed.
    Last edited: Dec 22, 2007
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