I know that the maximum angle of climb is reached at maximum specific excess thrust (S.E.T.) for jet airplanes, or minimum drag or maximum $\frac{L}{D}$.

Is this true, and if yes, is this valid also for propeller airplanes, or is $\gamma_{max}$ reached at another condition for props?


2 Answers 2


What you say is true only for turbojets and aircraft with fixed-pitch propellers. Generally, all optimum points for variable-pitch propeller driven aircraft are at lower speeds than those of jet aircraft. The reason is the variation of thrust with speed: For propellers, thrust is inverse to speed, while it is roughly constant over speed for turbojet aircraft in the subsonic speed range.

Specifically, the optimum angle of climb condition can be expressed as $$\frac{\delta \gamma}{\delta c_L} = 0$$ If we assume a quadratic polar, constant propeller efficiency over speed (which means a variable-pitch propeller) and an expression for thrust which lets us model an exponential variation of thrust over speed ($T = T_0·v^{n_v}$), we can write this condition as $$\frac{\delta \gamma}{\delta c_L} = -\frac{n_v}{2}·c_L^{-\frac{n_v}{2}-1}·\frac{T_0·(m·g)^{\frac{n_v}{2}-1}}{\left(\frac{\rho}{2}·S_{ref}\right)^{\frac{n_v}{2}}}+\frac{c_{D0}}{c_L^2}-\frac{1}{\pi·AR·\epsilon}$$ The general solution is $$c_{L_{{\gamma_{max}}}} = -\frac{n_v}{4}·\frac{T·\pi·AR·\epsilon}{m·g}+\sqrt{\frac{n_v^2}{16}·\left(\frac{T·\pi·AR·\epsilon}{m·g}\right)^2+c_{D0}·\pi·AR·\epsilon}$$ For jets and fixed-pitch propeller aircraft ($n_v = 0$) the solution is quite simple, because the thrust terms are proportional to the thrust coefficient $n_v$ and disappear: $$c_{L_{{\gamma_{max}}}} = \sqrt{c_{D0}·\pi·AR·\epsilon}$$ For turbofan and variable-pitch propeller aircraft, we have less luck and get a much longer formula. This is the one for propellers ($n_v = -1$): $$c_{L_{{\gamma_{max}}}} = \frac{T·\pi·AR·\epsilon}{4·m·g}+\sqrt{\left(\frac{T·\pi·AR·\epsilon}{4·m·g}\right)^2+c_{D0}·\pi·AR·\epsilon}$$ To arrive from here at a flight speed, I recommend to look up the speed in a polar. Solving this analytically will get messy. Below I have plotted a generic climb speed over air speed chart for different thrust loadings of a typical turbofan. The blue lines show thrust (right Y-axis) and the green lines the resulting climb speed. The two black lines show how the optimum flight speed for best climb speed and best climb angle (steepest climb) vary over thrust loadings. They are easy to find graphically: Pick the tops of the green curves for best climb and the steepest tangent from the origin of the coordinate system to the green lines for best climb angle. Note that they cross when moving from positive to negative climb speeds. With a propeller, the results will look similar, however, the best climb speed line would be vertical. This will collect all optima at the low speed range of the plot - using a turbofan makes the effects easier to see because the optima are spread out more.

Climb speeds for different thrust loadings

Climb optima for different thrust loadings (own work)

The optimum climb speed (which is proportional to $\frac{1}{c_L^2}$) varies inversely with the square of the thrust loading ($\frac{T_{ref}}{m·g}$)² of the aircraft. With lots of excess thrust, the optimum is limited by the stall speed (the black line bends into a vertical trend), while with no excess thrust both optimum speeds $v_x$ and $v_y$ converge. This makes sense: If thrust is just sufficient to keep the airplane from descending at one speed, this speed will both give the best flight path angle and the best vertical speed (unfortunately, both will be 0 at this point). It also helps to reduce induced drag, so aircraft with a high aspect ratio wing will climb steepest at a higher lift coefficient (= lower speed).

The steepest climb optimum looks a little more compact if we use the induced drag coefficient directly: $$c_{L_{\gamma_{max}}} = \frac{2·m·g·(c_{D0}-c_{Di})}{T}$$ but since the lift coefficient is hidden again in the induced drag term, and so sits on both sides of the equation, it is much harder to draw conclusions from this version.

$c_L \:\:\:$ lift coefficient
$n_v \:\:\:$ thrust exponent, as in $T \sim v^{n_v} $
$T \:\:\:\:$ thrust
$m \:\:\:\:$ 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
$c_{Di} \:\:$ induced drag coefficient


The maximum angle of climb for all aircraft is achieved when the specific excess thrust available is maximum.

$sin \ \gamma_{max} \ = \frac{(T-D)_{max}}{W}$

However, the thrust varies differently with speed in case of propeller and jet engines.

Climb Performance

Source: code7700.com

For turbojet aircraft, the thrust is approximately constant with speed. So, $(T-D)_{max}$ (and maximum angle of climb) occurs are $D_{min}$. This velocity is the $V_{min_{T_{R}}}$, the velocity of minimum thrust (drag) required and also the velocity for maximum angle of climb, $V_{\gamma_{max}}$. i.e. for jet aircraft, $V_{min_{T_{R}}}$ = $V_{\gamma_{max}}$.

In case of propeller aircraft, the thrust varies with speed. In general, thrust decreases with speed. As a result, the maximum excess thrust (i.e. the maximum SET) does not occur at the speed of minimum drag, but usually before it. As a result, for propeller aircraft, $V_{\gamma_{max}}$ < $V_{min_{T_{R}}}$.

The condition is the same (max. excess thrust), but the speeds are different.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .