|Oct14-12, 08:01 PM||#1|
Power on a slope
Im trying to figure out how much power is needed to maintain a vehicles speed on a slope
I found out that my car weighing in at about 2000 lbs needs 15hp to maintain 75 mph on a flat road with no wind. now I'm trying to figure out how to take that and calculate the power needed on a 10% grade and a 20% grade. (more looking for the formula than the answer)
I know that a 10% grade = 5.7° and 20% = 11.3°
The formula I used to find power needed to maintain a given speed works by taking the time it takes to slow a car from one speed to another in neutral (80mph to 70mph) gives me the hp needed to maintain 75 mph.
Heres the formula I developed a few years back and has been surpassingly accurate as it account for most all the losses in the tires and drive line rear of the transmission.
( [-6*10-5]*m*[Vi+Vf]*[Vf-Vi) )/t = hp
m = mass in lbs
Vi = Initial Velocity (mph)
Vf = Final Velocity (mph)
t = time (sec)
Now, how do I apply a slope to this?
|Oct17-12, 12:40 AM||#2|
Instead of using lbs to measure weight, try using Kgs, it makes equations like that a lot easier. For example, you want to work out acceleration and the formula for that is end velocity - start velocity / time. So all you do is take how fast the object (car) was going before it started accelerating (70mph) and subtract how fast the car was going after it completed the acceleration (80mph). Then divide that by the time it took to complete the acceleration. The only problem is instead of mph, you need to measure velocity in m/s. So the steeper the slope is, the longer it's going take to complete the acceleration, therefore the acceleration is larger because there's more work (force x distance). And use Newtons instead of horsepower.
|Oct17-12, 10:11 AM||#3|
For a slope, you need the same power as on a flat road (air resistance is the same, rolling resistance is nearly the same), and in addition you need power to increase the potential energy: The car gains height with a rate of sin(α)*v against a force of m*g, this gives sin(α)*v*m*g as additional power.
And here is the power of (and in ;)) SI units: With m=1000kg, v=20m/s, sin(α)=0.10 (for 5.7°) and g≈10m/s^2, the required power is 0.10*1000kg*20m/s*10m/s^2 = 20000kg m^2/s^2 = 20kW (plus the power required on a flat road)
No conversion factors required.
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