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For a very basic airplane, say like Cessna 150 is built with Wing Area (S)=15 meter squared (say it 15m*1m). In another part said that the Lift (L)=0.5*rho*(V)^2*Cl*S. Mean, Lift is affected by the wing area. Yes, it is clear. My question, should I consider the wing area as fix area during the take off, cruise, and during landing? If angle of attack of the wing is 15 degree during take off, what is the wing area?

From the Lift formula above that is written here, S (wing area) is planform (projected). Where it to be projected? If like above, if during the take off the Angle Of Attack is 15 degree, what is the wing area? Very appreciate if explanation provided with reference.

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    $\begingroup$ Doesn't wing area change as the flaps are extended (takeoff and landing) and retracted? My airplane for example, I take off with 10 degrees flaps, and land with 30 degrees, but cruise with 0 degrees. The flaps extend back and droop. Cessna 150 has similar, and I think in some years had 40 degrees of flaps for landing. $\endgroup$ – CrossRoads Oct 26 '18 at 11:30
  • $\begingroup$ Cessna 150L has wing area of Trainer version: 157 square feet, and Commuter version: 159.5 square feet (different wing tips I think, so wings are 6 inces wider) $\endgroup$ – CrossRoads Oct 26 '18 at 11:33
  • $\begingroup$ Angle of attack of 15 degrees for a Cessna (and most fixed wing aircraft) is either stalling, or right at the edge of a stall. Are you sure this is what you mean (AoA rather than pitch angle)? Typical take-off Aoa is usually less than 10 degrees. $\endgroup$ – Ron Beyer Oct 26 '18 at 14:17
  • $\begingroup$ Could you please define the other variables in your equation? V velocity, rho?, Cl? In the real world, unless the wing is reconfigured, area will not change. Let's see if the math will fit. $\endgroup$ – Robert DiGiovanni Oct 26 '18 at 16:58
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Wing area does not change with angle of attack. The only way wing area can change in flight is if there are devices on the wing that can extend and retract, making the wing longer, shorter or wider.

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  • $\begingroup$ Sir, after I did long study, wings area is changed with AoA. You must not see from the relative wind side, but from the Lift, from top of the wing. Lift (L)=0.5*rho*(V)^2*Cl*S. Now, we talk about the S. As surface area (S) is opposes the gravitational force while Lift (L) is orthogonal to the wing, then the surface area must be multiplied with cos(AoA). Just need to open Google to see the force component here, which F * cos (AoA) is Lift (oppose the m * g), and F * sin (AoA) is induced drag. $\endgroup$ – AirCraft Lover Dec 10 '18 at 14:32
  • $\begingroup$ Of course lift changes with AoA. But Lift <> Wing Area. $\endgroup$ – Juan Jimenez Dec 11 '18 at 15:29
  • $\begingroup$ @AirCraftLover, your mistake stems from the assumption that "Lift (L) is orthogonal to the wing". This is wrong. Lift, by definition, is orthogonal to the wind (and drag is along the wind). Gravity has nothing to do here whatsoever. ALL the aerodynamic forces/coefficients are calculated in the wind axes. Lift is linear wrt angle of attack only in the wind axes. Wing area is the "working space" of the flow, so to speak, and so its area is not changing with $\alpha$. Only with this condition lift is linear with $\alpha$, which is the most useful and gratifying property in aerodynamics. $\endgroup$ – Zeus Jul 5 at 7:49
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Simple answer: the surface area is the area of the wing when you look at it straight down. The question then of course is, why?

The answer is quite trivial: because it's easier that way. In a way, your idea that the lift is dependent on the wing area projected in the direction of the incoming wind seems quite reasonable (surely, this is what the incoming air "sees"). This projected area equals $$S_p=S\sin(\alpha)$$ A good reason not to use $S_p$ is that we like the equation to contain real geometric parameters, so that we can easily scale our equation up and down. So, why not define the lift equation as $$L=\frac{1}{2}\rho c_l S\sin(\alpha) v^2$$ This looks nice right? Everything is in there, even the angle of attack!

Sadly, curve of the lift versus the angle of attack does not follow a nice $\sin(\alpha)$ curve. This equation is thus sadly not only misleading, but you would have to divide all existing lift curves by $\sin(\alpha)$, which would make them erratic (especially around $\alpha=0$) and much less insightful. So, lumping the angle of attack into the lift coefficient is in the end the best way to go.

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  • $\begingroup$ Either that or only consider one angle of attack at a time and take Cl off the graph. "taking a bigger bite of air" would go towards drag, which is on the L/D graph. No need for sadness, you engineers do a great job getting us in the ball park before a single model is built and tested. $\endgroup$ – Robert DiGiovanni Oct 26 '18 at 19:10
  • $\begingroup$ @sanchises, wings area is changed due to AoA change. See my comment to Robert DiGiovanni below and Juan Jiminez above. See the area changed related to the lift, which lift is oppose the gravity, while Lift is orthogonal to the wing. This is make sense. The surface area is projected to the earth (parallel to the gravity). $\endgroup$ – AirCraft Lover Dec 10 '18 at 14:48
  • $\begingroup$ @AirCraftLover I disagree. It's not a question of geometry, but a question of definitions. The wing area is defined as the chord length times the span length. The apparent shortening due to AoA is factored into your typical $C_L$ graphs. $\endgroup$ – Sanchises Dec 10 '18 at 15:34
  • $\begingroup$ Yes, I maybe lack in defining my question as lack of my understanding. But after I read again my question, the last paragraph is clearly I asked about the projection, "where it to be projected". $\endgroup$ – AirCraft Lover Dec 10 '18 at 20:28
  • $\begingroup$ @AirCraftLover I understand, but I maintain that S should be projected at 0° AoA at all times, or equivalently perpendicular to the wing chord line (at least if you're planning on using any available data for $C_L$). The gravity and lift vector do not matter, because $S$ is defined as a geometric constant and should never change during flight. $\endgroup$ – Sanchises Dec 10 '18 at 21:17
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Here is the 150M planform, you can see the flaps from the fuselage out, a foot or so wider than the elevator. enter image description here

The 150 uses Fowler flaps, described in Wikipedia:

Fowler flap

A split flap that slides backwards, before hinging downward, thereby increasing first chord, then camber.[13] The flap may form part of the upper surface of the wing, like a plain flap, or it may not, like a split flap, but it must slide rearward before lowering. As a defining feature - distinguishing it from the Gouge Flap - it always provides a slot effect. [14] Invented by Harlan D. Fowler in 1924, and tested by Fred Weick at NACA in 1932. They were first used on the Martin 146 prototype in 1935, and in production on the 1937 Lockheed Super Electra,[15] and are still in widespread use on modern aircraft, often with multiple slots.

https://en.wikipedia.org/wiki/Flap_(aeronautics)#/media/File:Airfoil_lift_improvement_devices_(flaps).png

I can't find numbers on how much the chord/camber is changed on a 150.

Here's a great discussion of the Fowler flap from 1942

https://www.flightglobal.com/pdfarchive/view/1942/1942%20-%200783.html

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An analysis of the variables shows Coefficent of Lift, CL, varies with change in angle of attack. ρ is air density. So, check your AoA/CL graph. Area should be constant.

L = 0.5 * ρ * V2 * CL * S

Post Script: this equation may be of value if one wanted to swap airfoils in a design. For AOA comparison of lift, one might make a giant constant out of the entire equation (L) and the run various AOA off the AOA/Cl graph. Although laborious by hand, this would be tailor made for a spreadsheet and would make a nice presentation with graphics.

Also, if anyone knew if coefficient of lift for an airfoil was done at a specific AOA by convention, this would be extremely good information. I would imagine it may be done at AoA for best lift/drag ratio.

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  • $\begingroup$ Yes, it is. After I did long study, wings area is changed with AoA. We must not see from the relative wind side, but from the Lift, from top of the wing. Lift (L)=0.5*rho*(V)^2*Cl*S. Now, we talk about the S. As surface area (S) is opposes the gravitational force while Lift (L) is orthogonal to the wing, then the surface area must be multiplied with cos(AoA). Just need to open Google to see the force component here, which F * cos (AoA) is Lift (oppose the m * g), and F * sin (AoA) is induced drag. $\endgroup$ – AirCraft Lover Dec 10 '18 at 14:35
  • $\begingroup$ Force components change with angle of attack (and also with bank 😊), but this does not change wing area. However, on terms of lift, I see how you might call it EFFECTIVE wing area based on the VERTICAL lift component, opposing m×g. $\endgroup$ – Robert DiGiovanni Dec 10 '18 at 19:17
  • $\begingroup$ That probably the word should be, the effective wing area. In my question I asked, where the wing area to be projected. That was confusing me as I read in wikipedia. $\endgroup$ – AirCraft Lover Dec 10 '18 at 20:32
  • $\begingroup$ Dear Robert DiGiovanni, in calculation, how to use the projected wing area and the coefficient of lift (Cl)? As we know, the Cl is related to the AoA. (Here is explained that the surface area (S) is defined as the planform (projected) wing area)). But at the same time, the Lift Coefficient is different for every AoA, as also explained in that link. $\endgroup$ – AirCraft Lover Dec 10 '18 at 22:11
  • $\begingroup$ It is true that F x cos(AoA) will have a higher percentage of Total lift vector at lower AoA, but remember Total lift vector gets much bigger as AoA increases, right up to stall. Now what we need to find out is what the Lift value on lift vs AOA and lift vs drag charts is! Is it total lift or cos(total lift for that angle). I bet @Sanchises or Jan Hudec knows! $\endgroup$ – Robert DiGiovanni Dec 11 '18 at 0:03

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