Car Motion: Power and Distance: Air Drag
You use the term "Power" but what you are asking about is "energy" or "work done".
Power is an instantaneous measure of the rate of doing work
which is the same as the rate of using or supplying energy.
The standard relationship at constant Velocity V over a distance D in time T is:
D = V x T
which can be rearranged as
.... V = D/T
or
.... T = D / V
Two identical vehicles travelling at identical speeds and identical conditions on a level surface will require the same power input to overcome air drag.
They will also use the same energy in equal time periods.
They will also use the same energy over equal distances.
However:
If one travels for twice as long it will travel twice as far and use twice the same energy.
For a given Velocity V a vehicle will have
Air drag = k x V^2 ... (where k is a constant - see notes below for calculation of k)
Power = Drag x V = k x V^3
From above
Distance = Velocity x time
and
time = Distance / Velocity
Energy = Power x time
So Energy = Power x Distance / Velocity
You can use these equations with your available data to calculate what you require.
For the value of "k" in the air drag equation, see below.
Power & Energy requirements due to air drag:
Power is the instantaneous rate of doing work.
Over a distance work done = Energy = Power x time.
If power input just matches power taken to overcome air drag forces then
V = constant &
Energy = force x distance = force x velocity x time.
Air drag force F is given to a reasonable approximation by the classic drag equation.
This can be derived by calculating the energy required to accelerate a mass of air out of the path of a vehicle as it passes but you don't need to do this to use the equation.
Drag Force "F" = 0.5 x Rho x Cd x A x V^2
SI units:
F = force (Newton)
Rho = air density - ~= 1.3 kg/m^3 at sea level
Cd = drag coefficient (0 < Cd <= 1)
A =Area - m^2
V = Velocity m/s
Power in Watts = F.V
Power = 0.5 x Rho x Cd x A x V^3
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vidzi
Updated on March 16, 2020Comments
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vidzi over 3 years
I need to workout the amount of Power required by a car, in order to overcome air drag forces. I have figured out that, Power = c(Velocity^3). But, I am unable to understand Power's relation with distance. For example, if there are 2 cars, 1 covering 1000 m, and other 500 m. So, which car will be requiring more power to overcome drag forces? Can you please help me figure out, relationship between Power required and the distance traveled.
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danimal over 8 yearsdo you know how long they each take to travel their distance?
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vidzi over 8 yearsNo, but is it going to affect the power requirement? From the formula, its evident that it depends only on the car's velocity. I wanted to inquire, that if both of them are traveling at the same speed, are there power requirement going to vary with the distance traveled?
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Russell McMahon over 8 yearsI have updated my answer. You should be able to use the formulae provided to calculate what you want. I STRONGLY recommend that you work through what I have written and understand it. The only exception is the air drag equation (F=0.5 x Rho x Cd x A x V^2) - you can use this without understanding how it is derived (although knowing why it works is useful).
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vidzi over 8 yearsThanks for the toolkit. Lets assume that both car are traveling at the same velocity(v). So, according to the formula, their Power requirement should be same. But intuitively, shouldn't Distance traveled should also play a role in Power requirement? Please let me know, in case any other information is required.
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danimal over 8 yearsif they are going at the same speed, then they will use the same power. however, if one is using the same power for a longer time, then it will use more energy (do more work). Power is the rate of doing work, so the power doesn't care how long work is being done.
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WhatRoughBeast over 8 yearsIf it helps, think of power as the strength of the car's engine. More horsepower, higher top speed (since, as you've realized, drag goes up as the cube of the velocity). But the power of the engine has nothing to do with the size of the gas tank. Two cars with identical engines, running at the same speed - the one which goes farther needs more gas.