# What is the general mathematical relationship between increasing an airplane's weight and the needed thrust to fly?

I understand that increasing an aircraft's weight without changing any other properties means that its takeoff, stall and cruising speeds all increase since you need to generate additional lift. So of course that means a plane needs to generate more thrust to achieve these higher speeds. But I'm trying to figure out what exactly is the mathematical relationship? If I make the plane 20% heavier will I typically need about 20% more thrust?

• I will only answer part of your question, that may be enough to help form the rest of your intuition. In the left formula, lift increases with the square of the speed, all other factors equal. But put the other way that means speed increases with the square root of lift required. So a close-enough approximation for speeds would be the speed increase with the square root of weight. If you increase the weight of the airplane by 120%, your stall speed increases by 109.5%. This doesn’t account for the increase in drag at the higher speed, so thrust would also be a non linear relationship. Aug 31, 2022 at 0:59
• @Max Drag (and therefore thrust required) goes up with the square of speed, so does that cancel out your square root? Aug 31, 2022 at 3:35
• @StephenS It's not quite that simple. (a) I may be at a slightly different angle of attack and C(L) and C(D) are specific to an AOA. (b) I think it is "intuitively correct" to think of D = T. But 99.9% of the time I will be flying with some positive angle of attack and thus my thrust vector and drag vector are not colinear. In fact in a climb, my propellor is contributing a meaningful vertical component to counteract gravity. At any positive angle of attack I am generating slightly more thrust than what is contributing to horizontal motion. Aug 31, 2022 at 4:34

While the takeoff and stall speeds increase, the cruising speed doesn't need to change. Assuming that the cruising speed is kept constant, the thrust increase can be derived as follows:

At cruise, the aerodynamic lift $$L$$ (which depends on the air density $$\rho$$, the cruise speed $$V_C$$, the projected wing surface area $$S$$ and the lift coefficient $$C_L$$) is equal to the weight $$W$$ (which depends on the mass $$m$$ and the gravitational acceleration $$g$$).

$$\begin{equation} L= \frac{\rho}{2}\cdot V_C^2 \cdot S \cdot C_L = m \cdot g = W \end{equation}$$

At cruise, we can also assume that the thrust $$T$$ is equal to the drag $$D$$ ($$C_{D0}$$ and k are two constants).

$$\begin{equation} D = (C_{D0} + k \cdot C_L^2) \cdot \frac{\rho}{2}\cdot V_C^2 \cdot S = T \end{equation}$$ Joining both equations to get rid of $$C_L$$, we get the required thrust to fly with a certain mass:

$$\begin{equation} T = \left(C_{D0} + k \cdot \left( \frac{m \cdot g}{\frac{\rho}{2}\cdot V_C^2 \cdot S} \right)^2 \right) \cdot \frac{\rho}{2}\cdot V_C^2 \cdot S \end{equation}$$

$$\begin{equation} \frac{\mathrm{d}T}{\mathrm{d}m} = \frac{4 \cdot k \cdot g^2}{\rho \cdot V_C^2 \cdot S} \cdot m \end{equation}$$

The right side of the equation depends on the mass $$m$$, which means that the relationship between the mass increase and the thrust increase required to maintain the same cruising speed is not constant.

Instead of increasing the thrust, you could fly with the same thrust and a lower cruising speed. Another option would be to fly at the best endurance speed. In all cases, you can derive the desired relationship similarly.

• Nice derivation! Yet another option is to fly at a different altitude as weight changes, which is what airliners typically do. This would result in changes to $\rho$ in the equation. Aug 31, 2022 at 6:03
• This is showing thrust required scaling with m^2, but other answers are suggesting that typically it is about linear. Are you maybe holding something constant that isn't required, like angle of attack? Or is it purely down to holding ρ constant, flying in thicker air when heavier to get more lift from the same wing area and (true?) air speed? Aug 31, 2022 at 21:20
• @PeterCordes It depends on the constraints. If you assume $C_D/C_L$ to be constant, it is indeed linear. For this to be true, you would need to cruise at a different speed and angle of attack after increasing the mass. As I wrote at the end, we can make many assumptions, and I've only derived it for the case where the cruising speed is kept constant. Aug 31, 2022 at 22:54
• I see. So same speed in same air density, requiring a higher angle of attack and thus much more drag, scaling with $C_L^2$; that's where the quadratic scaling comes in. Interesting that it's so much costly to get more lift at the same speed. Aug 31, 2022 at 23:46
• Interesting. I guess I should have asked my question with more specifics. The only thing I was clear wouldn't change is the other properties of the plane. In my head I was thinking about a change in the minimum thrust needed to cruise. Sep 2, 2022 at 19:57

For the majority of planes and flights in practice, it's untrue that cruise speed increase with weight. Heavier flights would fly a higher AoA instead, maintaining the same cruise speed.

Every plane has a (mostly) constant lift/drag ratio (approximately the same as glide ratio) at cruise speed and altitude. In a straight and level flight where lift=weight and drag=thrust, the statement that "20% heavier flight needs 20% more thrust" isn't far from the fact at all.

This is a very interesting question! And the simple answer is...

$$T = W \frac{C_D}{C_L}$$

where $$T$$ is thrust in lbs, $$W$$ is weight in lbs, and $$C_D$$, $$C_L$$, are coefficient of drag and coefficient of lift, respectively. Of course, this simply explains that the thrust is in direct relation to weight. We are, of course, discussing conditions for an aircraft in flight. So if the weight is increased by 20 percent, the thrust must also increase accordingly by 20 percent.

Keep in mind that the coefficient of lift and coefficient of drag are affected by atmospheric density. My supposition is that you have a handle on those already. And the conversions to metric can be done anytime.

But, you say, maybe you would like to know thrust horse power... also given here in English units.

$$T_{HP} = \frac{T\cdot\ V}{550} = \frac{W V}{550}\frac{C_D}{C_L}$$

where $$V$$ is velocity in feet per second. Now we can see, for instance, that increasing the weight and/or the velocity will increase the thrust horsepower required to maintain flight.

Just some brief insights on this question, a very good question, indeed.