I know that the drag force on an airplane scales as (velocity)^2 which implies that doubling the airspeed will produce a 4x increase in the drag force. So my question is:

Imagine we have an airplane that flies 100MPH on 100HP. If we replace the engine with one that produces twice the power (200HP) I would anticipate that, everything else being equal, the most speed increase I could possibly expect would be (100MPH) x sqrt(2) or about 140MPH- and that if I wanted to double the airspeed I would need a 400HP engine. Is this correct?


2 Answers 2


In theory, yes. However, everything else isn't equal.

When you double the speed, you probably exceed the design parameters of the airframe. You'll definitely exceed the VNE of a 100mph airframe if you operate it at 200mph, and then all sorts of ugly things can start happening, from weird handling characteristics to large pieces, like the tail, coming off inflight. Remember that 4x drag means 4x the strain on the airframe. Is your airframe up to the task?

Most 100mph aircraft have a design optimized for short takeoff and landing, wings that are very efficient at slow speed, but not efficient at higher speeds. So the simple drag calculation may not hold true, when you take wings designed for slow speed and operate them at a much higher speed. The actual drag will probably be higher, if you consider the design of the airframe.

Compare a 100mph aircraft, say a Cessna 150, to a Mooney 201, that can go 200mph. The Mooney has much thinner wings, that result in a high landing speed, but lower drag at higher speeds, while the 150 has a fatter wing that gives a lower landing speed and better low speed handling, but may not fare so well at a much higher speed than it was designed to handle. The other tradeoff, aside from speed, is the 150 is far more forgiving of a clumsy pilot, while the Mooney is probably not the best trainer for first time students.

Note that the Cessna goes 100mph on 100hp, while the Mooney goes 200mph on 200hp, so the advantages of designing for the target speed can be seen in not just upping the power, but redesigning the airframe (and adding retractable gear).

  • $\begingroup$ OP stated: same aircraft. $\endgroup$
    – Koyovis
    Nov 23, 2017 at 23:44
  • $\begingroup$ The question was not about practical considerations, but about theoretical implications. $\endgroup$
    – Sanchises
    Nov 25, 2017 at 9:12
  • $\begingroup$ It isn't a valid theory if it can be easily disproven. You can't just consider power when doubling the speed of the aircraft. You have to beef up the airframe, or redesign it unless you want to risk catastrophic failure, which adds weight, which requires even more power. Plus, the wings will probably produce more drag than the calculations provide, because they were designed for lower speeds. $\endgroup$
    – tj1000
    Nov 25, 2017 at 15:19

Your thinking is correct if thrust won’t change with speed. But it does.

Since you prescribe power, I have to assume a piston or turboprop engine. In that case, thrust is power divided by speed and your thrust will decrease with the inverse of speed. Not only will drag grow, but thrust will decrease at the same time, which will give you a speed increase of only 1.26 when you double engine power.

Mathematically speaking: $$P = \frac{T\cdot v}{\eta_{Prop}} = \frac{\rho\cdot S_{ref}\cdot c_D}{2\cdot\eta_{Prop}}\cdot v^3 \Rightarrow v = \sqrt[\LARGE{3}]{\frac{P\cdot\eta_{Prop}}{\frac{\rho}{2}\cdot S_{ref}\cdot c_D}}$$

I'm afraid that just swapping engines will result in a poorly matched propeller, so your $\eta_{Prop}$ will also decrease. Now much depends on the lift coefficient at which the 100 hp version flew. Flying faster at the same density will lower your lift coefficient and with it the induced drag, so the drag coefficient at the higher speed will be lower. By how much depends on the initial lift coefficient. If an underpowered aircraft gains a bit more power, the possible speed increase can indeed be proportional to the square root of the power increase. Normally, however, induced drag is almost negligible at the top speed of powerful propeller airplanes and the cube root law reflects reality best.

You will need approximately an 800 HP engine and matching propeller (ideally of the same mass as the 100 HP engine and propeller) to fly twice as fast.

$P\;\;\;\;\;\;\:$engine power
$v\;\;\;\;\;\;\;$flight speed
$\rho\;\;\;\;\;\;\;$air density
$c_D\;\;\;\;\;$drag coefficient
$S_{ref}\;\;\;$reference area

  • $\begingroup$ But the thrust of a rocket, for example, stays the same at any speed, because the speed of the ejected gases is relative to the rocket itself, not to the 'stationary' observer... The same should be the case of the air accelerated by a propeller. $\endgroup$
    – xxavier
    Nov 23, 2017 at 22:10
  • $\begingroup$ I've been watching a debate on the physics stack exchange about whether the horsepower requirement scales with the square, the cube, or the 4th power of airspeed. I thought it was with the square... $\endgroup$ Nov 23, 2017 at 22:24
  • $\begingroup$ @xxavier Yes, but do you know a rocket where thrust is specified in horsepower? $\endgroup$ Nov 23, 2017 at 22:49
  • 1
    $\begingroup$ I went out to another website after googling airspeed versus horsepower and got the v^3 dependency quoted by Peter Kaempf, the derivation was a little easier to follow: drag scales as v squared, power is force x velocity, equate force with drag, there's your v cubed dependency. thanks to all, I enjoy reading your comments. -NN $\endgroup$ Nov 23, 2017 at 23:45
  • 1
    $\begingroup$ Let us continue this discussion in chat. $\endgroup$
    – Koyovis
    Nov 24, 2017 at 8:21

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .