Is it possible to determine the initial climb rate of a craft with a few key pieces of data such as top speed, engine power and weight?

For example the XB-42 was an experimental bomber design from the WW2 era.

XB-42 in flight

Its top speed was 660 km/h at 7,145 meters.

Its empty weight was 9,475 kg and maximum weight was 16,194 Kg.

Its twin engines produced a total of 2,498 kW at takeoff and 2,834 kW using war emergency power.

Is that enough information to determine the approximate climb rate? Thanks so much for any help!


4 Answers 4


No, the top speed performance is not enough. To calculate any climb speed one needs to know at which flight speed this climb takes place. Per definition, the climb speed is zero at top speed.

Next, it helps to have some more information about the airframe. The wing's aspect ratio is an important factor because climb happens at lower speed when induced drag is a higher proportion of total drag.

So I need to make two more assumptions in order to find the climb speed:

  1. You want to know the maximum climb speed at sea level.
  2. I am allowed to use information from this answer to complete my knowledge.

Now for the climb speed calculation, which can already be found in several older answers: The flight speed for maximum climb rate is when induced drag is three times as large as zero-lift drag. The first step is to determine the correct polar point. This is a bit more tricky than it sounds because the zero-lift drag coefficient depends on the flight speed and air temperature. Let's start with 0.029 and see where that brings us to: $$c_{L_{opt.\,climb}} = \sqrt{3\cdot c_{D0}\cdot\pi\cdot AR\cdot\epsilon} = \text{1.48}$$ That corresponds to a flight speed of $$v_{opt.\,climb} = \sqrt{\frac{2\cdot m\cdot g}{c_{L_{opt.\,climb}}\cdot\rho\cdot S}} = \text{39.7 m/s}$$ calculated with the MTOW of the XB-42 of 15,060 kg. Now we can determine the ratio of the Reynolds numbers between climb and maximum speed which is 0.2915. If I assume that the zero-lift drag changes in proportion to $\left(\frac{1}{Re}\right)^{0.2}$, the zero-lift drag at best climb speed rises to 0.02907. Close enough.

However, this speed is rather close to what I suspect is the stall speed of a clean XB-42, and extending flaps will create more drag. Therefore, it will be better to climb at a higher speed, say 50 m/s, and see how fast it climbs there. The zero-lift drag at that speed is scaled to 0.0276 and the lift coefficient to 0.933. Now to the climb speed: $$v_z = \frac{P_{mot}\cdot\eta_P}{m\cdot g} - \frac{\frac{\rho\cdot v^3}{2}\cdot S\cdot\left(c_{D0}+\frac{c_L^2}{\pi\cdot AR\cdot\epsilon}\right)}{m\cdot g} = \text{12.57 m/s}$$

The symbols used are:
$\kern{5mm} \rho\:\:\:\:\:$ air density = 1.225 kg/m³ at sea level
$\kern{5mm} v\:\:\:\:\:$ flight speed
$\kern{5mm} S\:\:\:\:\:$ wing surface area = 51.6 m²
$\kern{5mm} c_{D0} \:$ zero-lift drag coefficient at 183.3 m/s and 7140 m
$\kern{5mm} c_L \:\:\:$ lift coefficient
$\kern{5mm} \pi \:\:\:\:\:$ 3.14159$\dots$
$\kern{5mm} AR \:\:$ aspect ratio of the wing = 8.95
$\kern{5mm} \epsilon \:\:\:\:\:\:$ the wing's Oswald factor = 0.9
$\kern{5mm} m \:\:\:\;$ aircraft mass = 15,080 kg
$\kern{5mm} g \:\:\:\:\;$ gravitational acceleration
$\kern{5mm} P \:\:\:\:\:$ engine shaft power = 2498 kW
$\kern{5mm} \eta_P \:\:\;$ propeller efficiency = 0.85

  • $\begingroup$ Thank you so much Peter, I'm going to program this and your other equation with Javscript and produce dozens of different aircraft! Thanks so much for your knowledge!!! I'm going to try out a small XB-42 in 1940 with DB601 engines. $\endgroup$
    – DR01D
    Commented Apr 15, 2018 at 16:07
  • $\begingroup$ +1. Insightful answer! $\endgroup$
    – Hans
    Commented Feb 22, 2020 at 16:47

Yes, but the process will be indirect. For a rough estimate, the engine power- which is the rate at which it can perform work- can be thought of as lifting the mass of the aircraft upwards against gravity. Since the units of power are foot-pounds per second and velocity is in feet per second, the engine power and the aircraft's mass can be used to solve for the climb rate in feet per second (be sure to use the right units for mass). this ignores airframe drag and propeller efficiency losses and as such yields an absolute upper limit for the maximum possible rate of climb. Your mileage will be lower!

Note that engine power depends on density altitude, and hence this estimate holds only for the density altitude at which the engine's power rating was measured & specified.

  • $\begingroup$ I know that top speed is largely a function of (power - drag). So climb rate is largely a function of (power/weight)? $\endgroup$
    – DR01D
    Commented Apr 5, 2018 at 16:03
  • 1
    $\begingroup$ yes. more power + less weight = faster climb. $\endgroup$ Commented Apr 5, 2018 at 17:17

I would say you can make a pretty good approximation given the data you provided. What is missing is the engine thrust, but maybe you can find it somewhere in the internet, or make an approximation based on that you know that the speed was 660 km/h at 7,145 meters. Drag coefficients do also need to be obtained or guessed.

For sources on how to perform a performance calculation I would use the equations provided in Appendix B in ECAC doc 29, Volume 2.

As a professional you would use the BADA database from Eurocontrol for all performance calculations. Military aircraft is not included in the BADA-database.

  • $\begingroup$ Thanks for the equations at the ECAC those are really helpful. I've got some reading to do! $\endgroup$
    – DR01D
    Commented Apr 5, 2018 at 16:01

To answer your question directly - NO, speed, HP, and weight are not enough.

A possible solution is to find weight, hp, and ceiling for it and another aircraft that are very close to each other. The ROC should be similar.

The XB-42 had a Service ceiling of 29,400 ft. You can also use this with the formula given below with a power curve that looks normal by eyeballing.

If you choose to try to calculate ROC, you will have to guess at the power available.

enter image description here

rc = roc/ft/min
Pa = power available
Pr = power required
W = weight


enter image description here enter image description here

  • $\begingroup$ Thanks so much @jwzumwalt! So in a nutshell the climb rate of an aircraft is a function of POWER/WEIGHT. As weight increases climb rate decreases. For WW2 era piston engine aircraft would drag be a significant factor or is POWER/WEIGHT most of the effect? $\endgroup$
    – DR01D
    Commented Apr 5, 2018 at 15:57

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