By what way and with which variables could you determine a plane's maximum rate of climb per time? If I'm not mistaken, I'm looking for VY.

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    $\begingroup$ Are you asking "How is Vy calculated?", or "How many feet per minute can I get out of a given aircraft in a sustained best-rate (Vy) climb?" (The former is easy to answer. The latter, well, it depends...) -- Or are you asking something else entirely and I'm just being addle-minded? $\endgroup$
    – voretaq7
    Commented Dec 27, 2013 at 1:47
  • $\begingroup$ Start your plane, taxi to the runway, adjust for V_y and look at your vertical speed indicator… ;) probably the best way to actually know the real answer. $\endgroup$
    – Peter
    Commented Mar 21, 2015 at 0:10

3 Answers 3


To calculate your possible climb speed $v_z$, you will need

  1. Your engine's thrust $T$
  2. Your airplane's drag $D$
  3. Your airplane's mass $m$

Calculate how much power is needed to overcome drag, and any excess can be used for climbing: $$v_z = v\cdot sin\gamma = v\cdot\frac{T-D}{m\cdot g}$$

Note that this equation makes use of several simplifications, but works well for propeller and slow turbofan aircraft with moderate flight path angles $\gamma$.

To do this with more precision, you need to account for the fact that the aircraft should accelerate during the climb to stay at the same polar point. Now you further need:

  1. The gradient of air temperature over altitude (lapse rate $\Gamma$)
  2. The local speed of sound $a$, and
  3. The gas constant $R$ of air.

You need to add a correction factor $C$ which has several components: $$C = 1 + \frac{1}{2}\cdot\kappa\cdot R_w\cdot\Gamma_w\cdot Ma^2 + \frac{(1+0.2\cdot Ma^2)^{\frac{\kappa}{\kappa-1}}-1}{(1+0.2\cdot Ma^2)^{\frac{1}{\kappa-1}}}$$

where $\kappa$ is the ratio of the specific heats of air and is 1.405, the index w denotes the wet adiabatic gas constant and lapse rate of air, and $Ma$ is your flight Mach number. $\Gamma$ can vary between -0.004°/m and -0.0097°/m, but if you use the average of -0.0065°/m, this equation can be simplified to: $$C = 1 - 0.13335\cdot Ma^2 + \frac{(1+0.2\cdot Ma^2)^{3.5}-1}{(1+0.2\cdot Ma^2)^{2.5}}$$

which is probably the form you will find in most books which care to cover this topic. The second summand takes care of the reduction of atmospheric temperature with altitude and disappears in the stratosphere, and the third summand covers the additional energy needed for acceleration in terms of flight Mach number.

Graph of the acceleration factor over Mach number

Acceleration factor over the flight Mach number in the troposphere for the Standard Atmosphere

Now your climb speed becomes $$v_z = \frac{v}{C}\cdot sin\gamma = \frac{v}{C}\cdot\frac{T-D}{m\cdot g}$$

As you can see from the graph above, the correction factor is only important at higher speeds, but will cut the climb speed in half at Mach 2. Some jet aircraft need to climb at a high constant Mach number, and then the aircraft needs to decelerate while climbing. Now the correction factor becomes smaller than unity and the climb speed gets a boost because kinetic energy is converted into potential energy while climbing.

Optimum speeds

To pick the flight speed where the climb speed or the fight path angle reaches a maximum, you now need to describe how thrust will change with flight speed. To simplify things, we can say that thrust changes over speed in proportion to the expression $v^{n_v}$ where $n_v$ is a constant which depends on engine type. Piston aircraft have constant power output, and thrust is inverse with speed over the speed range of acceptable propeller efficiencies, hence $n_v$ becomes -1 for piston aircraft. Turboprops make some use of ram pressure, so they profit a little from flying faster, but not much. Their $n_v$ is -0.8 to -0.6. Turbofans are better in utilizing ram pressure, and their $n_v$ is -0.5 to -0.2. The higher the bypass ratio, the more negative their $n_v$ becomes. Jets (think J-79 or even the old Jumo-004) have approximately constant thrust over speed, at least in subsonic flow. Their $n_v$ is around 0. Positive values of $n_v$ can be found with ramjets - they develop more thrust the faster they move through the air.

The flight speed for maximum rate of climb ($v_y$) is reached at a lift coefficient $c_L$ of $$c_L = -\frac{n_v+1}{2}\cdot\frac{T\cdot\pi\cdot AR\cdot\epsilon}{m\cdot g}\cdot \sqrt{\frac{(n_v+1)^2}{4}\cdot\left(\frac{T\cdot\pi\cdot AR\cdot\epsilon}{m\cdot g}\right)^2 + 3\cdot c_{D0}\cdot\pi\cdot AR\cdot\epsilon}$$ which can be simplified for piston-powered airplanes with constant speed propellers as $$c_L = \sqrt{3\cdot c_{D0}\cdot\pi\cdot AR\cdot\epsilon}$$ The optimum flight speed for this specific case becomes $$v_y = \sqrt{\frac{2\cdot m\cdot g}{\rho\cdot S\cdot\sqrt{3\cdot c_{D0}\cdot\pi\cdot AR\cdot\epsilon}}}$$

whereas the steepest climb is possible with a $c_L$ of $$c_L = -\frac{n_v}{4}\cdot\frac{T\cdot\pi\cdot AR\cdot\epsilon}{m\cdot g}\cdot \sqrt{\frac{n_v^2}{16}\cdot\left(\frac{T\cdot\pi\cdot AR\cdot\epsilon}{m\cdot g}\right)^2 + c_{D0}\cdot\pi\cdot AR\cdot\epsilon}$$

$c_L \:\:\:$ lift coefficient
$T \:\:\:\:$ thrust
$m \:\:\:\:$ aircraft mass
$g \:\:\:\:\:$ gravitational acceleration
$\pi \:\:\:\:\:$ 3.14159$\dots$
$AR \:\:$ aspect ratio of the wing
$\epsilon \:\:\:\:\:$ the wing's Oswald factor
$c_{D0} \:$ zero-lift drag coefficient
$\rho \:\:\:\:$ air density
$S \:\:\:\:$ reference area

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    $\begingroup$ I have no idea if these formulae are correct, but an excellent answer :p $\endgroup$
    – Jon Story
    Commented Feb 17, 2015 at 12:15
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    $\begingroup$ @JonStory: I think your skepticism is quite appropriate. Don't trust any equation if you do not know the simplifications and assumptions that were used for deriving them! Maybe you are too generous with upvotes ;) $\endgroup$ Commented Feb 17, 2015 at 12:59
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    $\begingroup$ @PhilCrowther You are absolutely correct, the n$_v$ terms will become zero. But there is a sum under the square root, so a bit remains. I added another line to show what remains in case of n$_v$ = -1. $\endgroup$ Commented Jun 5, 2021 at 23:21
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    $\begingroup$ @PhilCrowther Yes, a bit too simplistic. Pounds is mass, not force (unless you mean pounds times 32.174 ft/s² which gives pound-force). This will become obvious when using SI units: Best climb speed v$_y$ needs mass times gravitational acceleration in the numerator. See my new edit for details. $\endgroup$ Commented Jun 6, 2021 at 18:27
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    $\begingroup$ @PhilCrowther Yes, to be precise, c$_L$ should be reduced. But these are approximations for small climb angles. Using a quadratic polar is already an approximation and assuming lift to be parallel to weight is another one. Still, the result is useful even though a precise calculation would look a lot more complicated and have a slightly different result. $\endgroup$ Commented Jun 8, 2021 at 4:16

Variables will have to include air conditions (temperature, humidity, altitude starting from), engine performance data at said altitude, and the vy speed for the airplane in question.


You are correct that Vy will give you the max RoC. However Vy is actually only a speed, not a climb rate, and will correspond to slightly different rates of climb depending on a few factors. The ones that spring to mind are A/C weight, temperature, and density altitude. Along with power setting but that one is somewhat self explanatory. As for determining the speed, normally you would just consult your PPOH or AOM to find a performance chart that will give you the base speed and you can correct it for each of the variables listed in order to have a precise climb rate.

  • $\begingroup$ It's important to note that reference speeds like Vx and Vy are not constants for a specific aircraft. They vary with air density and airplane weight. Many student pilots wrongly assume it's a fixed airspeed number. In some checklists, the pre-takeoff checklist mentions setting airspeed bugs for Vx, Vy and best glide based on current conditions and gross weight. $\endgroup$ Commented Dec 26, 2013 at 21:26
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    $\begingroup$ Also note, that Vy is only a sustained climb, if you have enough speed to bleed off, I'm betting you can get a better RoC by pointing the nose skyward, albeit not for very long. $\endgroup$
    – falstro
    Commented Dec 26, 2013 at 21:35
  • $\begingroup$ Yes, that's how we start our zero-G fun :-) $\endgroup$ Commented Dec 26, 2013 at 21:37
  • $\begingroup$ @roe Indeed, and if you've got excess airspeed and a rapidly approaching line of trees carefully executing that tradeoff of airspeed for altitude in the climb may be the difference between clearing the tops and having to climb the tree later to retrieve your tires. (Not that you should get yourself into such a situation to begin with, but we've all seen youtube videos...) $\endgroup$
    – voretaq7
    Commented Dec 27, 2013 at 1:46

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