using the table provided, and how it says the Velocity should be in mph, is it correct to say that the units of Hp/W from this table is mph also?

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    $\begingroup$ Some context for where this table comes from might help you get an answer. This may also be better suited to Physics.SE or Engineering.SE. $\endgroup$ Commented Mar 29, 2022 at 15:59
  • $\begingroup$ This is from Aircraft Design: A Conceptual Approach by Daniel Raymer, as @Raketenolli guessed. (page 81) $\endgroup$ Commented Mar 30, 2022 at 3:56

2 Answers 2


This is an empirical formula showing a simple relationship observed between power-to-weight ratio and maximum speed for different types of airplane. Its source is most likely an aircraft design book by Raymer (it does look familiar!) or Roskam.

For the formula to work, all physical quantities have to be made dimensionless, because you can't really raise a speed to a non-integer power (and even for integer powers, it only makes sense in a few cases). So you have to know exactly which units were assumed for power (presumably hp = 746 W), weight or mass (lbf = 4.45 N, lbm = 0.454 kg) and speed (mph = 0.447 m/s).

An example: Assume we are designing a metal homebuilt airplane that is supposed to fly 120 mph max. We check Row 2 of the diagram to find the proper values of $a$ and $C$, then plug those values into the given formula: $$ \begin{align} \frac PW &= a {V_{max}}^C \\ &= 0.005 \cdot 120^{0.57} \\ &= 0.0766 \end{align} $$

Assuming horsepower and pounds and an MTOW of 1000 lb, we find our engine needs to have

$$ \begin{align} P &= \frac PW \cdot W \\ &= 0.0766\, \frac{\mathrm{hp}}{\mathrm{lb}} \cdot 1000\,\mathrm{lb} \\ &= 77\,\mathrm{hp} \end{align} $$

which seems like a very reasonable value for a small 1000 lb airplane.

If you wanted to use different units, e.g. SI, then you would need to adapt the coefficients $a$ and $C$ from the table.


Yes. It would be correct to say that "power divided by weight" is a unit of speed. Given that the equivalent speed is given in miles per hour, the power-to-weight ratio will also be in miles per hour. Note that when I say "weight" here I really do mean "weight," that is, the force exerted due to gravity acting on the mass of an object; the colloquial definition of "power-to-weight ratio" really means "power-to-mass ratio," and that is not equivalent to a speed.

Using SI units, power divided by weight is given as Watts per Newton:

$$ \frac WN = \frac {\mathrm{kg} \cdot \mathrm{m}^2 \cdot \mathrm{s}^{-3}} {\mathrm{kg} \cdot \mathrm m \cdot \mathrm s ^{−2}} = \frac {\mathrm m}{\mathrm s} $$

which is sufficient to convince us that "power-to-weight ratio" is nothing more than a complicated way of expressing a speed, in particular the resultant speed of an object of some given weight when it is pushed by some given power.

In the English units here, "power" is in horsepower and "weight" can be assumed to be pounds-force:

$$ \begin{align} 1 \, \frac {\mathrm{hp}} {\mathrm{lbf}} &= \frac {550 \mathrm{ft} \cdot \mathrm{lbf} \cdot \mathrm s^{-1}} {\mathrm{lbf}} \\ &= 550 \, \frac{\mathrm{ft}}{\mathrm s} \times \left( \frac {1 \, \mathrm{mile}} {5280 \, \mathrm{ft}} \right) \left( \frac {3600 \, \mathrm{s}} {1 \, \mathrm{hr}} \right) \\ &= 375 \, \mathrm{mph} \end{align} $$

So we expect that an object which weighs one pound and is acted upon by a force of one horsepower will equalize at a speed of 375 miles per hour. But! the power given in the equation is not the true force acting on the object; rather it is the power generated by the power plant, which cannot be 100% efficient. Also, our 375 mph figure assumed no air resistance; in fact air resistance does exist, and it is a complicated topic, but generally (for high-speed flow) increases with the square of the object's speed. So when your textbook says

$$ \frac{\mathrm{hp}}{W_0} = a V^C $$

they really mean

$$ V_\mathrm I = a V_\mathrm R ^C $$

where $V_\mathrm I$ represents the "ideal" speed in a vacuum and using a power plant whose energy output is completely translated to speed with no losses along the way, and $V_\mathrm R$ represents the actual speed you will see in the real world.


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