Power Demand for Motion of Vehicle on a Road and Plain in Air with Air Resistance

Question

Let's say we have an ideal motorcycle without frictional losses (no tire slip, ideal machine without losses). But we want to consider wind force which is proportional to the square of speed

$F = k \cdot v^2$

The required Power would be

$P = F \cdot v = k \cdot v^3$

Now consider tailwind of velocity w:

$F = k \cdot (v-w)^2$

and the required Power

$P = F \cdot v = k \cdot (v-w)^2v$

Now consider a plain, moving exactly with same speed v against ground and within the same wind above the cycle:

The pilot would (as usual in aviation) argue to move against the packet of air, so his frame of reference is a system, where wind is zero. The relative speed against this "wind system" is (v-w). Wind force is again

$F = k \cdot (v-w)^2$

but his speed against reference system is also v-w. Power demand of the plane would be therefore

$P = F \cdot (v-w) = k \cdot (v-w)^3$

Assuming both cw-values are the same, I would expect also same power requirements. But this is not the case.

Where is the problem in my chain of thoughts? I miss something relevant but cannot identify it.

Answer 1

For the no wind situation

$$P=kv^3$$

as derived, however there is a need to be careful what this means...

It means that an engine, that can operate at a power $P$ (i.e. deliver a certain amount of energy per second), could cause the vehicle to reach the maximum speed of $v$.

for both the motorcycle and plane this means that if $F$ is reduced to $$F=k(v-w)^2$$

they could obtain a new greater maximum speed $v'$, from an engine delivering the same power $$P=k(v-w)^2 v' = kv^3$$

so $$v' = \frac{v^3}{(v-w)^2}$$

The energy required for the journey would be lowered to the same value (for both vehicles with the tailwind) and could be found from power multiplied by time

$$E=P\times \frac{d}{v'} = k (v-w)^2d$$

where $d$ is the distance travelled.

Answer 2

You just discovered that the work, kinetic energy and power are frame dependent quantities. Your result has nothing to do with the motorcicle versus plane. You just analysed the same motion from two different reference frames and found different results. If you do the same thing for the motorcycle from a reference frame moving with the wind (so the wind velocity is zero) you will find the same formula as for the airplane. Even better, consider two motorcyclists, going next o each other, as they do sometime. From the frame associated with one motorcyclist the velocity of the other is zero so the power associated with any force acting on him is zero. So the bottom line is that the work, kinetic energy, power depend on the reference frame you use to calculate them. This is a well known fact and is nothing wrong with finding different values in different frames.

Edit after the comment. Of course the fuel used does not depend on the referenece frame you use to calculate things. But here you don't calculate the fuel used. You calculate the work (and power) done by the resistance force. This is not directly related to the fuel spent at all. Again, it is not plane versus cycle. If you analyse the cycle in the same reference frame as you use for the plane you get the same formula as you have for the plane. Think about this: you can use up a full thank of fuel by starting your engine and leave it in "park" position until all the fuel is burned. The work done by the air resistance is zero but you burned fuel. The same is true if you drive the car (or cycle) and you analyse it from a frame moving with the same speed (a second motorcyclist).