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Horsepower based on distance and speed |
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| Oct8-09, 08:32 PM | #1 |
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Horsepower based on distance and speed
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 |
| Oct8-09, 08:43 PM | #2 |
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Mentor
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No, you can't. You need to know the force.
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| Oct9-09, 08:29 AM | #3 |
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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)
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| Oct10-09, 01:16 AM | #4 |
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Recognitions:
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Horsepower based on distance and speed
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? |
| Oct10-09, 05:12 AM | #5 |
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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. |
| Oct10-09, 05:22 AM | #6 |
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| Oct10-09, 06:04 AM | #7 |
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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.
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| Oct10-09, 05:02 PM | #8 |
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Recognitions:
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http://www.dragtimes.com/horsepower-...calculator.php http://www.dsm.org/tools/calchp.htm |
| Oct10-09, 05:05 PM | #9 |
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