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Horsepower based on distance and speed

  1. Oct 8, 2009 #1
    Hi,

    I was wondering if you can calculate horsepower given this information:
    Distance traveled: 180 Meters
    Weight of object: 1582 KG
    Speed at the 180 meters: 93 KPH


    Thanks,

    Steven
     
  2. jcsd
  3. Oct 8, 2009 #2

    russ_watters

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    No, you can't. You need to know the force.
     
  4. Oct 9, 2009 #3

    Andy Resnick

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    You can calculate the power required to maintain that speed only with some assumptions, since power = energy/time. Equivalent to what Russ_watters said, you have to assume the only energy is kinetic (level ground, the power required to maintain constant speed works against frictional losses, etc)
     
  5. Oct 10, 2009 #4

    rcgldr

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    Define:

    a = acceleration
    v = velocity
    x = distance
    c = constant
    f = force
    m = mass
    p = power

    Assuming constant power, lossless continuously variable transmission:

    a = c / v
    f = m a = m c / v
    p = f v = (m c / v) v = m c

    a = dv/dt = c/v
    v dv = c dt
    1/2 v2 = c t
    v = (2 c)1/2 t1/2

    a = c / ((2 c)1/2 t1/2)
    v = (2 c)1/2 t1/2
    x = 2/3 (2 c)1/2 t3/2

    x = 180 m
    v = 93 kph = 25.833333 m / s

    using equation for v:
    25.833333 = (2 c)1/2 t1/2
    t = (25.833333)2 / (2 c)
    t = 333.68055 / c

    using equation for x
    180 = 2/3 (2 c)1/2 (333.68055 / c)3/2
    180 = 2/3 (2)1/2 (333.68055)3/2 (c / c3)1/2
    180 = 2/3 (2)1/2 (333.68055)3/2 / c
    c = 2/3 (2)1/2 (333.68055)3/2 / 180
    c = 31.926226

    p = m c = 1582 * 31.926226 = 50507.29 watts = 67.7314 hp

    check

    t = 333.68055 / c = 10.451613

    v = (2 c)1/2 t1/2
    v = (2 * 31.926226)1/2 10.4516131/2
    v = 25.833333

    x = 2/3 (2 c)1/2 t3/2
    x = 2/3 (2 * 31.926226)1/2 10.4516133/2
    x = 180.00000

    I've never encountered this type of problem before, so it was a discovery process for me, perhaps someone could check my math?
     
    Last edited: Oct 10, 2009
  6. Oct 10, 2009 #5
    I had a blast at this and would like to post my working, even if just for somebody to tell me why it's incorrect,

    Use the equation of motion v^2 = u^2 + 2as and rearrange for a. Assume initial speed is 0, s=180m and v=25.8m/s

    This gives an acceleration of 1.849m/s^2

    Now F=ma with m=1582kg gives a force of 2925N

    Work done = force x distance = 2925 x 180 = 526500J

    Power = Work done/time = 526500/13.95 = 37742W (Time here is calculated using v = u +at)

    Since 1 horsepower = 745.7 watts

    Power = 37742/745.7 = 50.6 Horsepower.

    The only incorrect assumption I can see is that you're not starting from rest, but I think the question suggests this.
     
  7. Oct 10, 2009 #6

    rcgldr

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    This formula assumes constant acceleration and constant force, so the power is increasing linearly with speed. My assumption was that the original post was asking for the constant power required to accelerate to speed for a given distance, similar to an estimate of power given time to distance or speed achieve in a distance, in a drag race for a given weight vehicle. The constant power is also the minimum power. Constant acceleration requires double the power at the end of the run, or force x speed => 101.2 horse power.
     
    Last edited: Oct 10, 2009
  8. Oct 10, 2009 #7

    Pythagorean

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    Well as long as we're all being creative here, I assumed a person carrying the object and approximated the motion of the object as a sine wave as he jogged along, then calculated the work done against gravity in every cycle and then used the frequency to find out how many cycles there were in 180m.
     
  9. Oct 10, 2009 #8

    rcgldr

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    Not that creative on my part, do a web search for "horsepower estimator" and you'll find a lot of hits that based on 1/4 mile speed or time and weight, which is the approach I took, determine power given speed and distance from a standing start and weight of vehicle, although I simplified this assuming unlimited traction, and constant power. Examples of power estimators:

    http://www.dragtimes.com/horsepower-et-trap-speed-calculator.php

    http://www.dsm.org/tools/calchp.htm
     
  10. Oct 10, 2009 #9

    Pythagorean

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    Well, I didn't mean it as an insult!
     
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