In terms of IAS, Vx (best climb angle speed) increases with altitude, and Vy (best climb rate speed) decreases with altitude, until they meet at the plane's absolute ceiling. http://cospilot.com/documents/Why%20Vx%20and%20Vy%20Change%20with%20Altitude.pdf

In a jet powered aircraft, is this achieved at a constant angle of attack (AOA) regardless of altitude and other conditions (temperature, pressure, etc.)?

  • 2
    $\begingroup$ Question was edited to be more clear, narrow and to-the-point. If you see fit, please vote to reopen. $\endgroup$ Commented Jul 3, 2018 at 7:51
  • $\begingroup$ The link works. I get a 3-page pdf by Rod Machado, including four complicated plots. $\endgroup$ Commented Jun 21, 2022 at 22:13
  • $\begingroup$ I had better luck with this. Worth reading. $\endgroup$ Commented Jun 22, 2022 at 0:24

2 Answers 2


It might help to derive both speeds from first principles. We assume that the drag polar of the aircraft can be described by a parabola, like this: $$c_D = c_{D0}+\frac{c_L^2}{\pi\cdot AR\cdot\epsilon}$$ The symbols are:
$\kern{5mm} c_D \:\:\:$ drag coefficient
$\kern{5mm} c_{D0} \:$ zero-lift drag coefficient
$\kern{5mm} c_L \:\:\:$ lift coefficient
$\kern{5mm} \pi \:\:\:\:\:$ 3.14159$\dots$
$\kern{5mm} AR \:\:$ aspect ratio of the wing
$\kern{5mm} \epsilon \:\:\:\:\:\:$ the wing's Oswald factor

Next, we describe thrust $T$ over speed $v$ with $$T = T_0·v^{n_v}$$

Now first to the maximum climb angle. This is reached when the condition $$\frac{\delta \gamma}{\delta c_L} = 0$$ holds true. No change in lift coefficient $c_L$ will improve the climb angle $\gamma$, it is only downhill from here to both sides. In order to get a grip on the climb angle, we look at the force equilibrium in steady flight at full power, assuming small values for $\gamma$: $$sin\gamma = \gamma = \frac{v_z}{v} = \frac{T - c_D\cdot \frac{\rho}{2}\cdot v^2\cdot S_{ref}}{m\cdot g} = \frac{T_0·\left(\sqrt{\frac{2\cdot m\cdot g}{\rho\cdot c_L\cdot S_{ref}}}\right)^{n_v}}{m\cdot g} - \frac{c_{D0}+\frac{c_L^2}{\pi\cdot AR\cdot\epsilon}}{c_L}$$ The symbols are:
$\kern{5mm} m\:\:\:\:$ aircraft mass
$\kern{5mm} g\:\:\:\:\;$ gravitational acceleration
$\kern{5mm} \rho\:\:\:\:\:$ air density
$\kern{5mm} v\:\:\:\:\:$ velocity
$\kern{5mm} v_z\:\:\;$ climb speed
$\kern{5mm} S_{ref} \:$ wing area

Ideally, we would also multiply the climb angle with an acceleration factor, but I leave this out here for simplicity.

Now we can derive the expression for the climb angle with respect to the lift coefficient and get $$\frac{\delta \gamma}{\delta c_L} = -\frac{n_v}{2}·c_L^{-\frac{n_v}{2}-1}·\frac{T_0·(m·g)^{\frac{n_v}{2}-1}}{\left(\frac{\rho}{2}·S_{ref}\right)^{\frac{n_v}{2}}}+\frac{c_{D0}}{c_L^2}-\frac{1}{\pi·AR·\epsilon}$$ 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 ($n_v = 0$) 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 propeller aircraft, we have less luck and get a much longer formula. This is the one for propellers ($n_v = -1$): $$c_{L_{{\,\gamma_{\,max}}}} = \frac{T·\pi·AR·\epsilon}{4·m·g}+\sqrt{\left(\frac{T·\pi·AR·\epsilon}{4·m·g}\right)^2+c_{D0}·\pi·AR·\epsilon}$$ So yes, pure turbojets have an optimum lift coefficient for maximum climb angle that uses only terms which are constant over altitude. They do indeed climb steepest at a constant lift coefficient.

But the thrust-dependent optimum for other engine types hints at an altitude dependency which might affect the other optimum, that for best climb speed.

In order to find the conditions for maximum climb speed, repeat the process above with an expression where both sides are multiplied by speed: $$v_z = \frac{T\cdot v - c_D\cdot \frac{\rho}{2}\cdot v^3\cdot S_{ref}}{m\cdot g} = \frac{T_0·\left(m\cdot g\right)^{\frac{n_v-1}{2}}}{\left(c_L\cdot\frac{\rho}{2}\cdot v^2\cdot S_{ref}\right)^{\frac{n_v+1}{2}}} - \sqrt{\frac{2\cdot m\cdot g}{\rho\cdot c_L\cdot S_{ref}}}\cdot\frac{c_{D0}+\frac{c_L^2}{\pi\cdot AR\cdot\epsilon}}{c_L}$$

Now the solution for turbojets becomes the more complicated one, but that needs to be the case - how else would those optima converge at altitude? $$c_{L_{{\,n_{z_{\,max}}}}} = \sqrt{\left(\frac{T·\pi·AR·\epsilon}{2·m·g}\right)^2 + 3\cdot c_{D0}·\pi·AR·\epsilon} - \frac{T·\pi·AR·\epsilon}{2·m·g}$$

While the angle of attack for steepest climb is constant over altitude, the angle of attack for best climb speed increases as excess thrust disappears with rising altitude. Thus, the question can be answered: No.


...constant AoA at all altitudes

Well, let's throw in a third propulsion system, a rocket, and compare thrust output vs altitude.

Excluding Mach effects, constant thrust means constant IAS means constant AoA.

Vx and Vy do not converge until decreasing thrust forces the aircraft to seek its lowest drag airspeed, between Vx and Vy.

We can see by drawing thrust vs airspeed curves for all three what happens as altitude increases.

The piston/prop has two strikes against it as propeller RPM must increase and the turbocharger has to work harder to maintain mixture ratio for more fuel. At some point, available thrust decreases, then Vx and Vy must converge at the absolute ceiling.

The turbojet, operating on excess air, can dump in more fuel to power its compressor and maintain thrust a little longer, but it too reaches a point where thrust begins to drop off, and Vx and Vy converge.

The rocket never experiences a loss of thrust, therefor its Vx and Vy AoA remain constant.

The key is: constant thrust at constant IAS gives constant AoA. Both Vx and Vy are higher drag configurations than Vbg, one for higher climb angle, the other for higher climb speed. They can only be flown if there is adequate thrust at their respective IAS.

  • $\begingroup$ Speaking of thrust, the aircraft could hold Vx or Vy IAS and AoA as thrust reduces by reducing their pitch to the horizon, thereby reducing the gravity component acting against thrust. In all cases the plane has 0 excess thrust available at absolute ceiling. It can no longer climb. Flying this profile, the IAS would not converge until the very end. $\endgroup$ Commented Jun 22, 2022 at 21:04

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