Calculating the approximate top speed of a vehicle is straightforward.

$v = \sqrt[3]{\frac{2P}{c \cdot D \cdot A}}$


  • v = velocity of the vehicle
  • P = power of the engine
  • c = coefficient of friction
  • D = density of the air
  • A = the area of the front of the vehicle

Using this equation you can get a rough idea of the top speed of a car if you know its engine power, the body’s coefficient of friction and the area of its front profile.

Is it possible to use this same equation to determine an approximate top speed for subsonic aircraft? No doubt engineers at Boeing have to account for 100 additional variables, but is it good enough to produce a decent approximation for props that fly less than 500mph?

Also does the picture I attached below show the correct way to account for variables $A$ and $c$?

A = front profile of the aircraft body, wings and tail; c = coefficient of friction of entire aircraft

If $c$ represents the friction along the entire surface of the aircraft does this suggest that all things being equal an aircraft with twice the surface area would have twice the coefficient of friction? Or am I looking at $c$ incorrectly?

  • $\begingroup$ You have no terms for the efficiency of the propeller or induced drag. Your results are likely to be optimistic, at best. $\endgroup$
    – user30284
    Apr 12 '18 at 2:31
  • $\begingroup$ @Hobo Sapiens I believe decent propeller efficiency is around .7 so I could use that for an estimate. But doesn't induced drag drop the faster a plane travels? So induced wouldn't be a significant factor in top speed? $\endgroup$
    – DR01D
    Apr 12 '18 at 2:37

The short answer is no, that will not give you a very accurate estimate. As other mentioned, your equation does not include lift. Given some basic engine specs and aerodynamic properties, the top speed for aircraft can be estimated by determining the intersection of required power and available power. Available power is based on selected engines and propellers, and required power is based on the wing loading, lift coefficient, and other parameters that describe the amount of power required to sustain minimum lift.

Power curves


Will this simple equation produce a rough estimate of the top speed of a prop driven aircraft?

Yes, but only with the correct reference area.

You are on the right track, but need to consider what a friction coefficient is. A coefficient is a dimensionless value which, in the case of friction, needs to be multiplied with an area and a dynamic pressure to yield a force. Equating this force to the available thrust does indeed produce a good estimate of maximum speed.

However, just applying some friction coefficient will not suffice. You need to know how much of the surface has a laminar boundary layer, what the local roughness and the Reynolds number of the flow are and you need to add correction factors for Mach and relative thickness, as demonstrated in this answer. And the method only works for sleek shapes - a blunt base like on the Shuttle will add a lot of pressure drag which is neglected by this method.

If you have done all that, use the full wetted surface of the whole aircraft (not just its wing surface!) for the drag calculation. The front profile will not help you - now you need the top and side view to arrive at a useable result. For good measure, you should also add a correction factor to account for surface imperfections (gaps, rivet heads, antennas …).

But lift is not considered, you might say. Yes, but drag due to lift is insignificant at top speed for well motorized aircraft. If you desire a more precise result, of course you should include induced drag.

If c represent the friction along the entire surface of the aircraft does this suggest that all things being equal an aircraft with twice the surface area would have twice the coefficient of friction?

No. The coefficient will stay the same, but the reference area will double, so the force will double, too.

  • $\begingroup$ All fascinating! Now the coefficient makes sense. It needs a reference area. But I have a followup question in regards to wetted area. In your answer to this question aviation.stackexchange.com/questions/50462/… you mentioned that the surface area would be reduced to 70.7% of the original. But in your equation I don't see reference to 70.7%, only 1/2. Is that being accounted for in some other way? Or maybe your equation was just a rough estimate? Thanks so much this is very helpful! $\endgroup$
    – DR01D
    Apr 22 '18 at 23:02
  • $\begingroup$ if the Space Shuttle's blunt shape produced additional pressure drag is that why aircraft like the HE-219 were so narrow? Of was the 219 narrow simply to reduce the wetted area? You can see an image of it here. airandspace.si.edu/collection-objects/… $\endgroup$
    – DR01D
    Apr 22 '18 at 23:06
  • 1
    $\begingroup$ @DR01D: The size comment referred to the length change. An aircraft with 70.7% size will have half the area. And yes, the reduction in frontal area was believed to be key to drag reduction 70 or 80 years ago, but does not result in the smallest wetted area. Some two-engine designs even moved the engine instruments into the nacelles in order to reduce the cockpit size. $\endgroup$ Apr 23 '18 at 5:53

The key aspect of flight is lift and your automobile equations don't have it so no, you can't estimate any speed for the subject aircraft. There is no weight, lift coefficient, wing span, or wing area. A coefficient of friction applies to surface area not to frontal area, where you need a drag coefficient. Coefficients do not change when applied to a larger reference area.

Here is a NASA page dealing with lift:


enter image description here


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