Copyright © 2006 jsd

1  Car versus Airplane

Here is the question, as it came to me:

A car on a perfect level surface in still air goes 120 mph using 200 hp. measured at the rear wheels. An airplane flies directly over the car maintaining exactly the same speed. They both turn into a 20 mph headwind. What happens? (Give the speeds of both after the turn.) Assume: 1. The car and the plane apply constant power to the world, before and after the turn. The car’s transmission has to allow the same power to be developed regardless of speed. 2. The plane files low so IAS = TAS if that matters. 3. There are no interactions between car and plane. 4. Ignore turning effects and loss of energy by the plane in the actual turn, we are interested in steady-state results long after the turn. 5. The car has very very low drag tires so that all the drag on the car is aerodynamic.

Here is my analysis:

Assumption #1: Let’s take the coefficient of drag of the car to be independent of slip angle. This is probably a reasonable approximation if the slip angle is not very large.

Assumption #2: Let’s assume the drag is simply proportional to speed squared. This assumption is conventional in problems of this sort, and makes sense over a fairly wide range of reasonable Reynolds numbers.

Assumption #3: The wind does not change during the story; the change in headwind component is due solely to the change in direction of travel.

Note: Unless explicitly stated otherwise, all velocities and speeds (including wind speed) are measured relative to the ground.

Note: Some people find the phrase “apply constant power to the world” to be ambiguous or even misleading. In this case, “world” certainly does not mean “earth”. The airplane’s world is different from the car’s world. The airplane applies thrust to the air, while the car applies thrust to the ground.

We can write the drag force of the airplane as

Fa = a |w − v| (w − v)              (1)

where w is the wind vector, v is the groundspeed vector, and a combines the coefficient of drag with some size and geometry factors. You can check that F depends on speed squared, and that in the absence of wind F is directed opposite to the groundspeed – as it should. For a direct headwind, w is directed opposite to v.

The power dissipated by the airplane is

Pa = a |w − v| (w − v) · (w − v)
   = a |w − v|3

The drag force on the car is

Fc = c |w − v| (w − v)              (3)

Here c combines the car’s coefficient of drag along with size and geometry factors.

The power dissipated by the car is

Pc = c |w − v| (w − v) · (−v)
   = c (v2 − w · v) |w − v|

Everything above here gives F and P as a function of w and v, for arbitrary w and v.

We now specialize to the initial condition, where v = vi (the initial velocity). Both the car and the airplane have the same v at this point, since we are told that they are keeping pace. We also know that w is perpendicular to v, which means w and v add in quadrature.

The power dissipated by the airplane is

Pa,i = a (w2 + vi2)3/2

and the power dissipated by the car is

Pc,i = c vi2 (w2 + vi2)1/2

Now we turn into the wind. The airplane now has velocity va and the car has velocity vc. The wind is now a direct headwind, so the wind magnitudes simply add.

The power dissipated by the airplane is

Pa,f = a (|w| + |va|)3

and the power dissipated by the car is

Pc,f = c |vc| (|w| + |vc|)2


To simplify the calculation, we can convert to dimensionless units, i.e. we express all speeds as a multiple of |vi|. We define

u := w / |vi|
  ra := va / |vi|
  rc := vc / |vi|

and we note that for the task at hand, |u| = 1/6.

Equating the airplane power before and after, we find

(u2 + 1)3/2 = (|u| + |ra|)3              (10)

(u2 + 1) = u2 + 2|u||ra| + |ra|2              (11)

which can be solved for ra using the quadratic formula to obtain

|ra| = 1 − |u| + .5 u2 + terms of order u4              (12)

Similarly, we can equate the car power before and afterward to obtain

(u2 + 1)1/2 = |rc| (|u| + |rc|)2              (13)

which is a mess, but can be solved for rc approximately as

|rc| = 1 − (2/3)|u| + 0.28 u2 + higher order terms              (14)

Comparing equation 14 with equation 12 we find that the car pulls ahead, by the approximate amount

Δr = (1/3)|u| − 0.22 u2 + ⋯              (15)

or, converting back to dimensionful units, the groundspeed of the airplane is 101.7 mph while the groundspeed of the car is 107.6 mph for a difference of 5.9 mph.

2  Remarks

The first-order terms in these equations could have been obtained more simply by neglecting the shape of the wind triangles and just using power-law scaling arguments. However, the second-order terms were worth going after. They contribute about 10% to the differential velocity, and perhaps more to the point, there was no easy way to be sure how big these terms would be, except by working them out.

You may be wondering, how did I know the second-order term would be important? How did I even know it might be important? The general rule is that you should always be wary of problems that involve a small difference between large numbers.

This principle is well illustrated by the present example. Suppose you are doing an ordinary wind-triangle problem, for purposes of ordinary navigation, estimating time enroute, et cetera. In that case, the zeroth-order term is the leading term, and the first-order term is a valuable correction term. Second-order terms are usually not worth bothering with. (That’s in part because the wind velocity is never known exactly anyway; any error introduced by dropping the second-order terms is small compared to the uncertainty in the wind velocity.)

In contrast, though, the speed of the car relative to the airplane is a small difference between large numbers. The zeroth-order term drops out, the first-order term becomes the leading term, and the second-order term becomes valuable as a correction term.

Beware of small differences between large numbers.

In this case, even the crudest preliminary analysis was sufficient to warn me that the race between the car and the airplane would be a close race, so I knew that guessing wouldn’t suffice, and rough “rules of thumb” wouldn’t suffice. I knew early on that it would be worthwhile to do a systematic analysis, expressing the wind triangles using vectors.

Also: In any case: Once you have obtained an approximate solution (such as from equation 12 and equation 14), you should plug it back into the exact expressions (such as equation 2 and equation 4) and see how well it fits, how well it really solves the problem.

Note that I didn’t say “don’t make assumptions” ... and I didn’t even say “don’t make undocumented assumptions”. That’s because people make approximations and assumptions all the time. You couldn’t get out of bed in the morning if you didn’t make lots of approximations. You can’t afford not to make approximations. What’s more, you can’t even afford to document all the approximations you make.

It is often said that you need to know the difference between good approximations and bad approximations ... but really you need at least a three-way classification.

Copyright © 2006 jsd