Cessnaflyer, I just went back and re-read your post again and it came through my mind in a whole different way this time. (I should stop reading and replying at 2:30am.:cwm27
It makes sense, complete sense. But the way you replied it sounded like you said that increased weight
does not increase the stall speed. That's what got me confused. You went on to write, "The reason why the aircraft stalls faster is because it needs more of an AoA to maintain that weight at equal speeds." Does the statement "...
why the aircraft stalls faster..." not equal "added weight increase stall speed?" Doesn't an aircraft, (lets say a 152) loaded to max gross of 1670 not stall sooner than the same aircraft loaded at 1350? The stalling sooner is a result of a greater angle of attack needed to keep heavier loaded aircraft in the air as opposed to a lighter load on the same aircraft. Is this not correct. or am I just thinking about this all wrong?
Now it sounds like you've got it. I always think about the lift equation when I think about stall speeds. So for the example of 1 g flight, lift is equal to the weight of the airplane. The wing must produce the same lift as the weight.
Okay, well, how do we produce lift? Let's look at the lift equation: L=1/2*CL*rho*V^2*S. Okay, the 1/2 is a constant, we can't affect that. The rho is the density of the air, so on a given day that the density is whatever it is. We can affect it by changing altitudes, but let's ignore it for right now and assume that we have to fly at a particular altitude. "S" is the planform area of the wing (which is like the "bird's eye view area"). You can change that on some airplanes with high lift devices (moving flaps, variable geometry wings like the F-14, etc.) but for the sake of arguement let's assume we don't have any of those options. What does that leave?
Well, it leaves 2 quantities, both of which the pilot can affect. The first is "V squared", which is just velocity. We can certainly affect that by speeding up or slowing down. So let's look at the other term. That's CL. CL is the "coefficient of lift" and it depends upon several factors, but the biggest two are the angle of attack, and the shape of the wing. Let's for sake of arguement throw out the shape of the wing, and assume that we aren't going to play around with flap settings or variable sweep, or slats, etc. and that leaves one major factor for CL. That major factor is angle of attack (sometimes written with the greek letter alpha or written as AOA.)
Ok, so now I have a desired amount of lift that I am trying to produce (it's equal to whatever my weight is), and I have two quantities that I can change to produce it: velocity and CL (which depends mostly on AOA). Unfortunately, for a given wing, there is a maximum value that my CL can obtain. The CL increases as I increase AOA, up to some "critical AOA" (also called the "stall AOA") and if I increase AOA any more past that, CL DECREASES (fairly rapidly). That maximum CL establishes a minimum speed at which I can generate a particular amount of lift.
So let's see how this works with some trial numbers. Let's imagine a wing in which the CL is equal to .08 per degree of angle of attack, and imagine it has a stall or critical angle of attack of about 15 degrees. That means the MAXIMUM possible CL is 1.2 (which is .08*15).
Ok, let's imagine this wing is on a 2000lbs. airplane, with a 20 square foot wing planform area (that's S), and imagine that the density is sea level standard day (which happens to be .002377 slugs/cubic foot). What is the minimum speed at which I could produce that 1000lbs. of lift?
Well, that would occur at the maximum CL of 1.2. Here's how: rearranging the lift equation to solve for velocity I can divide both sides of the equation through by 1/2*CL*rho*S, and I get V^2=L/(1/2*CL*rho*S). Then for the numbers I gave that means v^2=2000/(.5*1.2*.002377*20)=70116 ft squared/second squared. But I want velocity, not velocity squared, so if I take the square root of that I get 265 ft/second, which is equal to 157 knots. So for this particular airplane, loaded this way, the stall speed is 157 knots.
Now, if I made it a 1000lbs airplane but kept everything else the same, I could STILL fly it at 157 knots if I wanted to: I'll just rearrange the lift equation again to solve for CL: CL=L/(.5*rho*S*V^2). So, if I had the same plane flying at 157 knots (265 ft/second) the CL required would be: 1000/(.5*.002377*20*265*265)=.599. Since I get .08/degree of AoA for my CL, that equates to an angle of attack of .559/.08 or approximately 7.5 degrees. So to keep the SPEED the same, but reduce the lift produced (which is what I would need to do to fly 1 g at a lighter weight), I need to reduce the angle of attack.
One last exercise. Since we know that the stall speed is 157 KTAS at 2000lbs, let's calculate what it would be at 1000lbs of lift. Rearranging the same way I did two paragraphs ago V^2=L/(.5*Cl*rho*S). That's V^2=1000/(.5*1.2*.002377*20). Did you notice I used the MAXIMUM CL of 1.2 since I'm talking about stall? So V^2=35058, which means V=187 ft/sec. 187 ft/second is the same as 111 knots true. The weight goes up, so the stall speed goes up, the weight goes down, the stall speed goes down.
But actually it's not really the WEIGHT, it's the amount of LIFT that you want to produce. It's just that for most flight conditions we want to produe an amount of lift that is equal to our weight. Sometimes, though, we want to produce more lift than our weight, like in a turn. For example in a two g turn, we want to produce twice our weight in lift. So that 1000lbs. airplane with a stall speed of 111 knots at 1 g, would have the same stall speed as the 2000lbs airplane (157 knots) if it were in a 2 g turn.
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