What is the maximum theoretical lift of a plane as a function of engine power and surface of the wings?

Question

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?

Answer 1

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.

Nomenclature:
$c_L \:\:\:$ 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)

Answer 2

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.