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Car Power to Weight

  1. Oct 5, 2005 #1
    I hope this is posted in the right section.

    If car A is identical to car B in every way, and both have 100 horsepower.

    Only car B weighs significantly less than car A, obviously car B will win a race on the 1/4 mile because being lighter.

    Is there a formula that can be worked from to ascertain how much extra horse power car A would need to beat car B??

    This is all in a theoretical sense so no extras are taken into account such as gearing, wind resistance, driver skill.

    eg: Car A = 1000kg with 100HP and Car B = 750kg and 100HP

    How much extra HP does car A require to win.
  2. jcsd
  3. Oct 5, 2005 #2


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    If no wind resistance, then only the power to weight ratio matters, IOW, car A needs 133 hp to keep up.
  4. Oct 6, 2005 #3
    Arh right..
    I was actually interested in the equation involved so I understand how it works.
  5. Oct 6, 2005 #4

    Physics Monkey

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    If a car of mass [tex] m [/tex] supplies a constant power [tex] P [/tex], then it is increasing it's kinetic energy at the rate [tex] P [/tex]. Starting from rest, the velocity of the car satisfies
    v = \sqrt{\frac{2 P t}{m}}
    It therefore travels a distance a
    x = \frac{2}{3}\sqrt{\frac{2 P t^3}{m}}
    in a time [tex] t [/tex]. If car A is heavier than car B then this formula makes it obvious that car A must supply a greater power to traverse the same distance in the same time. As we increase the power provided by car A, there is a threshold where they are able to cross the same distance in the same time. This leads immediately to
    \frac{P_A}{m_A} = \frac{P_B}{m_B}
    at threshold. If [tex] P_A [/tex] is further increased past this threshold value, then car A will beat car B. This of course reproduces the result of krab since the mass of car A is 4/3 the mass of car B in your example.
  6. Oct 16, 2005 #5

    So being that a cars power is delivered through a system of gears and is not delivered at a constant, the actual equation between two cars would be astoundingly difficult, wouldnt it?
  7. Oct 16, 2005 #6


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    Well if you could determine the rates and timings of the engine's pistons, it's not THAT hard (again, without taking anything else into consideration). But then again im pretty sure the combustions don't come anywhere near making a sine wave...
  8. Oct 16, 2005 #7


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    I think that se7en1976 wasn't referring to this much detail, but rather that an engine's (potential) output is highly dependent on rpm. That is in turn related to vehicle speed and gear selection.

    Anyway, the resulting equations are not "astoundingly difficult" if this is all that's modelled, but they certainly are much more complicated than Physics Monkey's. The big problems come in when you try to take into account traction properly. Any reasonably powerful car is going to be able to spin its tires, and this can be very difficult to model realistically if you want an accurate (few %) 1/4 mile prediction.
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