Imagine that we have a wing with constant thrust in level flight. Then imagine that we add mass to that wing. In order to maintain level flight, I would expect that the angle of attack would need to be increased, and that speed would decrease.

As the mass increases, I would also anticipate that speed would decrease and at some point the wing would stall.

What is the mathematical relationship between mass (x), speed (y) and stall speed where all other variables are constants?

Please assume that the thrust vector is always horizontal, and does not change with the angle of attack.

  • $\begingroup$ @sophit Thanks, but no, that doesn't help. I need to be able to set constants and graph mass vs speed and when a stall would occur. $\endgroup$
    – Monty Wild
    Commented Dec 8, 2023 at 9:26
  • $\begingroup$ The answer shows a formula that provides a relationship between speed and lift. In steady flight, you know that L = W, which is equal to M*g, so that helps you set up a relationship between speed. You can then plot mass versus speed. If you manage to do it, you can provide the solution here as an answer to your own question to share the knowledge :) $\endgroup$
    – ROIMaison
    Commented Dec 8, 2023 at 10:48
  • $\begingroup$ @sophit The trouble is that I don't know how to calculate the relationship between AoA and lift... or is there no way of calculating it and it's worked out empirically? $\endgroup$
    – Monty Wild
    Commented Dec 8, 2023 at 11:12
  • $\begingroup$ For that, you need to know about the airfoil type. If you know the airfoil, as a first guess you can look up the relationship between AoA and lift for the airfoil. Or you can assume that it has an airfoil of similar aircraft, see The Incomplete Guide to Airfoil Usage $\endgroup$
    – ROIMaison
    Commented Dec 8, 2023 at 11:21
  • 4
    $\begingroup$ You can't answer this question with a simple equation. For one thing, it would most certainly depend on which side of the power-required curve or thrust-required curve the aircraft was initially flying on. $\endgroup$ Commented Dec 8, 2023 at 13:53

3 Answers 3


You can't do this. (see comments below) Expanding on quietflyer's comment, the result (consequences), of adding mass in a constant thrust scenario depends on where the aircraft (wing) is on the power curve. This is a graph representing the relationsip between airspeed (horizontal scale), and the power or thrust required (vertical), to maintain that airspeed. The curve is U-shaped. The two points where the red and blue lines intersect are the only two stable points on the graph. Power curve

For all aircraft there is a speed (the bottom or lowest point on the blue U-shaped curve), where the power(thrust) required to maintain that speed is a minimum. I.e., to maintain any faster, (or slower) speed, requires more thrust or power. This is because the drag due to lift increases as AOA increases, and at slow speeds, (high AOA), this increase is greater than the reduction in drag from just going slower.

So, if above, (faster), than that airspeed corresponding to L/Dmax, reducing thrust, (moving the red line down), and slowing down brings you closer to that minimum thrust required airspeed, (assuming you continue to trim to maintain level flight), and then, unless you reduced thrust below the thrust required for that slower speed, when you have slowed to the speed that your reduced thrust can maintain, the aircraft will stabilize.

If, otoh, you start this exercise from below your L/Dmax airspeed, then the scenario you describe would happen.

  • $\begingroup$ Yes, you can do this. Nevertheless, quiet flyer makes a valid statement. OP's question presents an important quandary: How do we assess flight performance under specific conditions? What are the constraints in the mathematics of flight? What are the requirements on paper to answer OP's question? $\endgroup$ Commented Dec 12, 2023 at 4:07
  • $\begingroup$ @ThomasPerry in a practical sense we vary thrust with weight (by way of Velocity) to maintain optimal AoA. What the OP wants (constant thrust) is not often practiced in aviation any more than it is in trains and automobiles. $\endgroup$ Commented Dec 12, 2023 at 10:29
  • $\begingroup$ I admit that the first sentence in my answer (You can't do this.), might be a bit over the top. Perhaps it should be amended to read "You can't make these assumptions...", or "you can't have a mathematical relationship ... for constant thrust in level flight" - but "You can't do this." was shorter, and a bit more attention grabbing. $\endgroup$ Commented Dec 12, 2023 at 13:33
  • $\begingroup$ @CharlesBretana your graph more than makes up for it. It boils down to how much speed can one get for that amount of thrust. How much weight can one lift at that speed. The OP hinted at variable AoA. $\endgroup$ Commented Dec 12, 2023 at 14:33
  • $\begingroup$ @sophit and Charles Bretana -- My comment was posted before sophit's answer-- I suspect he is probably right, but I haven't pored through either his answer or this answer carefully enough to be certain either way -- $\endgroup$ Commented Dec 12, 2023 at 18:02

What is the mathematical relationship between mass and speed for constant thrust in level fight?

What you're looking for can be easily obtained from the standard equations for lift and drag.

Thrust $T$ wins drag $D$, which can be as usual expressed as:

$$D=½\rho V²SC_d=T$$

where $\rho$ is air's density, $S$ wing surface and $C_d=C_{d_0}+kC²_l$; $k$ is a constant which depends on the actual wing geometry while $C_{d_0}$ represent the parasite drag of the whole airplane; $C_l$ can be simply derived from the usual equation for lift $L$ that equals weight $W$:

$$L=½\rho V²SC_l=W \Rightarrow C_l=\frac{W}{½ \rho V²S}$$

Substituting we get:


which is the equation you're looking for ($q$ is short for $½\rho V²$).

Since you've made the hypothesis of constant $T$, this equation can be rewritten as:

$$q²S²C_{d_0} - qST + kW²=0$$

which is the equation of a quartic in $q$ (i.e. a biquadratic in $V$) whose solution is well known. This solution gives you the requested

relationship between mass and speed for constant thrust


You have an interesting question but the wrong ending.

the maximum weight carrying capacity will be at V min power, not at stall

There is a practical application to the constant thrust/less weight application as the aircraft consumes its fuel.

Now the plane can reduce its angle of attack (and fly faster) because more thrust can be used for forward speed and less for lifting.

Thrust required for lifting = induced drag

Thrust required for forward speed = form drag

As one goes faster, the proportion of thrust for induced drag decreases, and the proportion of thrust for form drag increases.

The sum of these 2 drags forms a minimum at Vbg.

Propeller aircraft have V min power at a V less than Vbg because of greater prop thrust output per RPM at lower airspeeds. Jets (and gliders) have their maximum weight carrying capacity at Vbg.

Relevant reading here. Adding weight shifts stall speed, Vmin power, Vbg, and Vy to higher airspeeds.


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