# How to determine climb performance for various flight speeds?

I have the following climb performance chart:

and I need to evaluate the maximum rate of climb in different flight conditions. The problem is, this chart is valid only for 85KIAS. Is there a way to evaluate the maximum rate of climb for other flight speeds, given the RoC and climb gradient @ 85KIAS?

edit: The zero-lift drag coefficient and aspect ratio of the aircraft are given. And it is not important if it is an optimum climb or not, I simply need to predict the maximum achievable rate of climb, given flight speed, altitude and aircraft mass.

Aircraft performance charts all work the same way. Unfortunately, this one only gives the climb speed at 85 knots indicated airspeed (KIAS).

To find the climb speed at different speeds needs more knowledge about the airframe. Then you can apply a simple approximation following this procedure.

The chart does not tell where on the polar the aircraft is. Will it climb better when flying faster or not? This is impossible to tell. Since you have only one speed and one power setting, but a range of masses, the aircraft will not be at its optimum climb setting for most points. We can make an assumption and declare the reference mass of 1700 kg as the point where the cited conditions are at the optimum. But then I would need at least the zero-lift drag and the aspect ratio to make further assumptions.

From the previous answer we take the climb speed equation $$v_z = \frac{v}{C}\cdot sin\gamma = \frac{v}{C}\cdot\frac{T-D}{m\cdot g}$$ and set the correction factor C = 1 for now. The resulting error is small at low speeds. Now we need drag and thrust.

First drag: The drag coefficient $c_D$ is approximately $$c_D = c_{D0} + \frac{c_L^2}{\pi\cdot AR\cdot\epsilon}$$ with the known aspect ratio $AR$ and an assumed Oswald factor $\epsilon$ of 0.85. To arrive from here at the drag we need to multiply this with the dynamic pressure $q = ½\rho\cdot v^2$ and the reference area $S$: $$D = ½\rho\cdot v^2\cdot S\cdot c_{D0} + \frac{(1700\cdot g)^2}{½\rho\cdot v^2\cdot S\cdot\pi\cdot AR\cdot\epsilon} = ½\rho\cdot v^2\cdot S\cdot c_{D0} + \frac{(1700\cdot g)^2}{½\rho\cdot v^2\cdot\pi\cdot b^2\cdot\epsilon}$$ when we substitute $AR = b^2/S$ with $b$ the span of your airplane. As I said before, induced drag depends on span, not on aspect ratio.

Now for the thrust. In a propeller aircraft power is constant and thrust is inverse with airspeed. Not indicated, but true airspeed, so we need to be careful. At sea level both are equal, and then thrust is straightforward: $$T = \left(½\rho\cdot v_{ref}^2\cdot S\cdot c_{D0} + \frac{(1700\cdot g)^2}{½\rho\cdot v_{ref}^2\cdot\pi\cdot b^2\cdot\epsilon} + \frac{5.5}{v_{ref}}\cdot1700\cdot g\right)\cdot\frac{v_{ref}}{v}$$

with $v_{ref}$ = 43.7278 m/s, which is 85 kts in sane units. The first two terms in the bracket look familiar: They are the drag contribution. The third term accounts for the climb speed of 5.5 m/s at the reference point in sea level altitude, so this accounts for the change in potential energy. If you need the thrust at higher altitudes, please correct $v_{ref}$ with the square root of the density ratio.

If those equations look daunting, continue solving for the climb speed: $$v_z = \frac{v}{C}\cdot\frac{\left(D + \frac{v_{z_{ref}}}{v_{ref}}\cdot m\cdot g\right)\cdot\frac{v_{ref}}{v} - D}{m\cdot g}$$ $$v_z = \frac{1}{C}\cdot\left(\frac{D\cdot\left(v_{ref}-v\right)}{m\cdot g} + v_{z_{ref}}\right)$$

• I edited my question with further details. Thanks. – mezzanaccio Aug 21 '15 at 11:04
• since that chart comes from a pilot handbook, as a pilot, how would you solve this set of equations in your head while flying/planning the flight? – Federico Aug 21 '15 at 13:48
• @Federico: When you are sitting in the plane already, use this analog computer - it will give you the most precise results! Set the conditions and watch the instruments. When planning the flight, finish solving for the climb speed equation. – Peter Kämpf Aug 21 '15 at 14:33
• @Federico As a pilot, using the pilot handbook, you use the information that they give you. If you want to achieve the listed performance, you climb at the airspeed that they provide. Assuming that there are no other charts, using anything else essentially makes you a test pilot and you use trial and error to determine the results. :-) – Lnafziger Aug 21 '15 at 17:48
• I believe that the two drag terms in the second last equation are not the same. The first is the "reference drag", that is the drag at V_ref and the second one will be the one at the actual speed. So I am not sure you can simplify the formula like you did. Anyway, how reliable is the assumption of constant power with airspeed and therefore that thrust is inverse with airspeed? Is it just a theoretical trick or can it be used for actual calculations? – mezzanaccio Aug 27 '15 at 8:26