Maximum theoretical lift of a plane as a function of engine power and surface of the wings?

I know that the maximum static thrust of a propeller turned by an engine of power $P$ can not be greater than: $$Thrust_{max}=\sqrt[3]{2\pi(\frac{D}{2})^2 \rho P^2}$$ where $\rho$ is the density of the air and $D$ the diameter of the propeller.

I am looking for a formula that gives the maximum theoretical lift of a plane which flies horizontally. After half a page of demonstrations I have found a relation:

$$Lift_{max}=\sqrt[3]{2S \rho P^2}$$ but I am not sure if it is correct ($S$ is the total surface of the wings).

Update 1

Two examples that show the formula for max lift gives lifts smaller than the loaded weight of the plane. I do not know why.

1) Cessna 172: Gross weight - $1,111kg$, Powerplant - $160 hp$, Wing area - $16.2 m^2$

$Lift_{max}=(2 * 16.2 m^2 * 1.2 kg/m^3 * (160 hp)^2)^{1/3}=837 kgf<1,111kgf$

2) de Havilland DHC-6 100 Twin Otter: Gross weight - $5,246kg$, Powerplant - $2*579 eshp$, Wing area - $39 m^2$

$Lift_{max}=(2 * 39 m^2 * 1.2 kg/m^3 * (2 * 579 hp)^2)^{1/3} = 4,198 kgf<5,246kgf$

Update 2

I have tried a different demonstration that stars from the known relations: $$Lift={0.5 \rho S C_L V^2}$$ $$Drag={0.5 \rho S C_D V^2}$$ $$Power=Drag * V$$ By eliminating the speed of the plane, $V$, the lift result as being: $$Lift = \sqrt[3]{2S \rho P^2 \frac{C_L^3}{4 C_D^2}}$$ It appears that this new lift can be greater than $\sqrt[3]{2S \rho P^2}$ as long as $\frac{C_L^3}{4 C_D^2}$ does not seem to be restricted to always remain less than 1.

Update 3

Using the formulas for $C_D$ an $L_{max}$, suggested to me by Peter Kampf and mnunos, I have made some evaluations.

$$C_D = C_{D0} + \frac{C_L^2}{\pi\cdot AR\cdot\epsilon}$$

$$L_{max}=\frac{1}{2}\cdot \sqrt{3\pi AR\epsilon}\cdot \sqrt[3]{\frac{P^2\eta^2\rho S}{4 \sqrt{C_{D0}}}}$$

For the assumed values: $\eta=0.7$, $\epsilon=0.8$, $C_{D0}=0.03$, $\rho=1.2\: kg/m^3$ and the data taken from the Cessna 172 general characteristics: $P = 160\:hp$, $S=16.2\:m^2$ and $AR=7.32$, I get: $$L_{max,\:\eta=0.7}=2199\:kgf$$ However, only $57.451\:hp$ are necessary for lifting $1111\:kg$ (horizontal flight), the rest up to $160\:hp$ can be used for climbing.

From the relation:

$$m_{plane}gV_{climb} = \eta(160\:hp - 57.451\:hp)$$

I get $V_{climb}= 4.913\:m/s$ which is significantly greater than the rate of climb 3.66 m/s the Cessna 172 has.

It seems that working with $C_{D0}$ and $\epsilon$ gives only a quite rough estimate of the max lift. Without knowing the actual $C_L=C_L(C_D)$ little can be said about the real maximum possible lift of a certain plane.

Update 4

The maximum theoretical thrust of an ideal helicopter propeller having the diameter identical to the wingspan, $11\:m$, of a Cessna 172:

$$T_{max,Diameter\:prop=11\:m}=(2 * pi * (11 m / 2)^2 * 1.2 kg/m^3 * (160 hp)^2)^{1/3} = 1 509.95\:kgf$$

is considerably smaller than the max lift of the plane, $L_{max,\:\eta=0.7}=2199\:kgf$, I calculated in Update 3. Is such a result logical? Does it have an explanation?

  • $\begingroup$ Shouldn't the theoretical L/D max ratio of the wing be taken into consideration in the second equation? In turn L/D max depends on the Reynolds number, hence on the assumed speed. $\endgroup$
    – mins
    Commented Jan 6, 2016 at 8:08
  • $\begingroup$ The problem is that I only know the total surface of the wings and the power of the engine. The ratio L/D and the speed are unknown in my case. The max thrust for a propeller, which is a wing that rotates so it generates lift and drag, is calculated using the momentul theory (see: en.wikipedia.org/wiki/Momentum_theory), a procedure that does not make use of CD and CL. I applied the same method for evaluating the max lift of a plane. $\endgroup$ Commented Jan 6, 2016 at 8:21
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    $\begingroup$ I'm not an aerodynamics expert, but one of the pre-requisites of the MT is that "Viscous effects are not considered (no drag, no momentum diffusion)". That doesn't seem to match the wing conditions. There are a bunch of formulas in this summary, in case you can do something with them. $\endgroup$
    – mins
    Commented Jan 6, 2016 at 8:49
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    $\begingroup$ I think the main problem is that you're using horsepower, while all the other units are in SI. If you substitute 746 W = 1hp, your answers might make more sense $\endgroup$
    – costrom
    Commented Jan 6, 2016 at 18:06
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    $\begingroup$ ....and now @Peter Kampf is going to fly in like and angel and solve all your troubles magically. :) $\endgroup$ Commented Jan 6, 2016 at 19:49

2 Answers 2


Your thrust equation is only valid for the static case (initial speed at infinite distance from the propeller is zero), and neglects the propeller's efficiency. For the flight case, I recommend to make a bold assumption for propeller efficiency and to assume that most power will be converted to thrust:

$$T = \frac{P}{v}\cdot\eta_{Prop}$$

A large, slowly turning propeller will have an efficiency approaching 90%, while a small, fast-spinning one will only reach 70%. If operated outside of its design speed, efficiency might even become negative. If you look for an upper boundary of thrust, the equation above sans efficiency will do nicely.

For the maximum lift answer you should expand your drag coefficient a little, as recommended already in this answer. $$c_D = c_{D0} + \frac{c_L^2}{\pi\cdot AR\cdot\epsilon}$$

Now you can equate drag and thrust to get: $$P\cdot\eta_{Prop} = \left(c_{D0} + \frac{c_L^2}{\pi\cdot AR\cdot\epsilon}\right)\cdot S\cdot\frac{\rho}{2}\cdot v^3$$

and solve for lift: $$L = c_L\cdot S\cdot\frac{\rho}{2}\cdot v^2 = \sqrt{\pi\cdot AR\cdot\epsilon\cdot\left(\frac{P}{v}\cdot\eta_{Prop} - c_{D0}\cdot S\cdot\frac{\rho}{2}\cdot v^2\right)}\cdot\sqrt{S\cdot\frac{\rho}{2}\cdot v^2}$$

The minimum power is consumed when the ratio $\frac{c_L^3}{c_D^2}$ reaches its maximum. Expressed differently, this means that $$c_L = \sqrt{3\cdot\pi\cdot AR\cdot\epsilon\cdot c_{D0}}$$

Plugging that back into the lift equation I leave as an exercise to the reader.

As you can see, the answer really depends on the aerodynamic quality of your airfoil. The best you can reach with regular airfoils is a $c_L$ between 1.5 and 1.8, depending on Reynolds number. With slotted flaps this can be doubled, but $c_{D0}$ will go up substantially, so the best ratio $\frac{c_L^3}{c_D^2}$ can be gained with a clean airfoil.

$c_L \:\:\:$ lift coefficient (normally between 0 and 1.5)
$S \:\:\:\:\:$ Wing surface area
$\pi \:\:\:\:\:$ 3.14159$\dots$
$AR \:\:$ aspect ratio of the wing (ratio of span to mean chord)
$\epsilon \:\:\:\:\:$ the wing's Oswald factor (use 0.8 when in doubt)
$c_{D0} \:$ zero-lift drag coefficient (use 0.02 when in doubt and 0.03 with a fixed gear)

  • $\begingroup$ In the case of Cessna 172, using 0.8 for the Oswald factor and 0.3 for the zero-lift drag coefficient and considering the efficiency of the propeller as being 70%, I get a max lift that is more than double the gross weight of the plane which is 1111 kg. Is such a result realistic? $\endgroup$ Commented Jan 8, 2016 at 7:21
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    $\begingroup$ @RobertWerner: First of all, it should be 0. 0 3 for the zero-lift drag and yes, the maximum lift must be more than the weight in order to have reserves for turns. Real lift almost never is close to maximum lift, but is what is needed to stay aloft. $\endgroup$ Commented Jan 8, 2016 at 8:05
  • $\begingroup$ I made the calculations with 0.03. By mistake I wrote 0.3 in my previous comment. I get $L_{max100}$ = 2790 kgf for the theoretical case when the efficiency is 100% and $L_{max60}$ = 1984 kgf for $\eta=$ 60%. It appears that even for a quite low efficiency like 0.6 there is a huge surplus of lift 1984 - 1111 = 873 kgf that seems unrealistic. I will update the initial question to further explain way. $\endgroup$ Commented Jan 8, 2016 at 19:01
  • $\begingroup$ @RobertWerner: Which power setting did you use? Piston engines are rarely used at their full rated power; something around 70% is more likely for continuous use. $\endgroup$ Commented Jan 9, 2016 at 3:01
  • $\begingroup$ I used the power specified in the general characteristics of Cessna 172: Powerplant: 1 × Lycoming IO-360-L2A four cylinder, horizontally opposed aircraft engine, 160 hp (120 kW). I calculated that Cessna 172 can climb at its specified 3.66 m/s if the engine delivers only $0.745*160\:hp$ and a propeller 70% efficient is utilized. I have to verify if that 3.66 m/s really corresponds to using only around 75% of the engine power. This should be written somewhere. $\endgroup$ Commented Jan 9, 2016 at 9:08

Try to write down step by step how did you get your first formula for $Lift_{max}$. Check not only the math but also the physical meaning of each formula and you'll probably find the error.

If you want to calculate the maximum possible lift of an airplane for a given geometry you need to know its lift and drag polars as you wrote in the second update. Starting with the lift formula in the second update,

$$L=\sqrt[3]{\frac{S\rho P^2}{2}\cdot \frac{C_L^3}{C_D^2}}$$

and assuming a quadratic function for the drag,

$$C_D=C_{D0}+k \cdot C_L^2$$

we can maximize L:

$$\frac{\partial L}{\partial C_L}=0 \implies \frac{\partial}{\partial C_L}(\frac{C_L^3}{(C_{D0}+k \cdot C_L^2)^2})=0 \implies C_{L,Lmax}=\sqrt{\frac{3C_{D0}}{k}}$$

Substituting $C_L$ with your calculated $C_{L,Lmax}$ we get an expression for the maximum lift:

$$L_{max}=\frac{1}{2}\cdot \sqrt{\frac{3}{k}}\cdot \sqrt[3]{\frac{P_{t}^2\eta^2\rho S}{4 \sqrt{C_{D0}}}}$$

I also substituted $P=P_{t}\cdot \eta$. $\eta$ is the thrust efficiency due to losses in the gear, propeller,...

Now lets see if it makes qualitatively sense:

  • $\downarrow k$ and $\downarrow C_{D0} \implies \uparrow L_{max}$
  • $\uparrow P_t$ and $\eta \implies \uparrow L_{max}$

This seems to be OK. Now let us calculate some values.

Using $T\cdot V=\eta P_t$ and $T=D=C_D\frac{\rho}{2}V^2$ we get the velocity we are flying at: $$V=\sqrt[3]{\frac{2P_t\eta}{C_D S \rho}}\\ \implies V_{Lmax}=\frac{1}{\sqrt[3]{2}}\cdot\sqrt[3]{\frac{P_t\eta}{C_{D0} S \rho}}$$

For comparision we can also calculate this values for $L/D_{max}$, the best glide point:

$$\frac{\partial L/D}{\partial C_L}=0 \implies \frac{\partial}{\partial C_L}(\frac{C_L}{C_{D0}+k \cdot C_L^2})=0 \implies C_{L/Dmax}=\sqrt{\frac{C_{D0}}{k}} \\ \implies V_{L/Dmax}=\sqrt[3]{\frac{P_t\eta}{C_{D0} S \rho}}$$

The relationship between the velocities at the best glide and maximum lift point is: $$\frac{V_{Lmax}}{V_{L/Dmax}}=\frac{1}{\sqrt[3]{2}}\approx 0.79$$

If we look at a Cessna 172 we get: $$\frac{V_x}{V_{bg}}\approx \frac{60 KCAS}{73 KCAS}=0.82$$ This is close to the theoretical value we had calculated. (I have assumed that flying with best climb angle speed is also flying with maximum lift.)

(source: fiu.edu)
Drag, and thrust of a Cessna 172. Picture source

While I do not have exact values for the Cessna, for $\eta=0.4$, $k=0.04$ and $C_{D0}=0.03$ at sea level I get $L_{max}\approx 17600 N$, which is also a plausible result. Therefore I suppose that the formula for $L_{max}$ is correct.

  • $\begingroup$ The formula of the max lift, you obtained, appears to be correct and quite useful for me. A question, why did you take the efficiency so low, 40%? $\endgroup$ Commented Jan 8, 2016 at 6:22
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    $\begingroup$ @RobertWerner Because the pilot forgot to clean the dead flies on the propeller and after thirty years the motor isn't so powerful anymore... It's the only value I didn't think much about. Take a higher efficiency if you want. To calculate $k$ and $C_{D0}$ I used the first polar with measured values I found in Google-images. $\endgroup$
    – Gypaets
    Commented Jan 8, 2016 at 7:53
  • $\begingroup$ @RobertWerner I don't have enough reputation to answer on the other comment, but assuming a you get a maximum lift twice the weight you would have half of your power available for climb. With $G=10900N$, $P=120kW$ and $\eta=0.7$ you get $G\cdot V_{climb}=P\cdot \eta\cdot 0.5 \implies V_{climb}= 3.85m/s$. This is a quite accurate estimation of real climb speeds. $\endgroup$
    – Gypaets
    Commented Jan 8, 2016 at 8:19
  • $\begingroup$ For $\eta=$ 70% I get $V_{climb}$ = 4.9 m/s. I will update the question to explain why. $\endgroup$ Commented Jan 8, 2016 at 19:05

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