Physics/Canopy Question

[pilotdave 0]

It works but that's not what is normally used. Most airplanes are said to generate zero lift at a negative angle of attack. SOME reference line must be used. Normally the chord line is chosen. For a non-symmetric airfoil, the zero lift angle of attack will be negative. It's easy to find lift slope curves (lift vs angle of attack) for most standard airfoils, and I think they all base AOA on the chord line. That way the lift at zero angle of attack is visible on the graph. If AOA was defined based on the zero lift angle of attack, it would be impossible to compare two airfoils on the same graph.

Dave

[quade 3]

Quote

. . . Most airplanes are said to generate zero lift at a negative angle of attack. . . .

Well, that just don't make no sense at all.

What I -think- you mean is . . .

. . . because most aircraft have a slightly positive angle of incidence . . .



Does this sound familiar? Dive bombers in WWII having to be pitched over slightly more than 90 degrees in order to maintain a straight down approach to target. This was due to the angle of incidence of the aircraft. In order to achieve zero lift (and therefore not drift off target due to lift) they had to pitch down more than 90 degrees. 90 + the angle of incidence.

Quote

If AOA was defined based on the zero lift angle of attack, it would be impossible to compare two airfoils on the same graph.

And that makes no sense whatsoever. SURE you could. Absolutely. You can pick any arbitrary datum point you want and still make a graph.

The reason the zero lift line makes the most sense is that it works for every shape between a barn door and a baseball.

quade -
The World's Most Boring Skydiver

[teason 0]

Quote

The amount of lift created by a wing has no direct relationship to its attitude relative to the horizon.

Was that what I said? I didn't think that's what I said

I think I said that lift is 90 degrees to glide angle or angle of attack (no matter what that angle is).

Then I postulated that a flatter more horizontal glide would produce lift in a direction that was more verticle.

I then lept to the conclusion (in my basic rudementry way) that the lift was more usable for keeping the canopy up as it would be more verticle.

I wanted to know if that conclusion was off or not.
and I think pilotdave answered that one for me
While I appreciate the link, I guess I wasn't clear in my post. Or maybe I didn't understand your response.

I would rather be a superb meteor, every atom of me in magnificent glow, than a sleepy and permanent planet.

[quade 3]

Yeah, I think the basic issue is a a lack of common knowledge between the two of us.

All of my understanding of aerodynamics comes from teaching students to fly airplanes. Pretty basic stuff and not exactly rocket science.

So, anyway, when somebody says lift is created in so-and-so direction, I have a tendancy to revert back to basics because, well, that's what I think they'll understand. The problem is, most skydivers don't have even the most basic knowledge of aerodynamics that we teach student pilots, so, they usually can't relate in proper terms to begin with.

Hey, not your fault either.

Look at the good Professor, Pilot Dave and myself bantering about the exact definition of AoA. And I can guarantee we all passed as least the most basic levels of aerodynamics years ago. Pretty esoteric stuff and for 99.9% of all conversations the angle between the relative wind and the chord line works just fine.

As to the direction lift is created . . .

Lift from the wing is created perpendicular to the flight path.

IF I read your last posting correctly, then yes, actually you sort of do have an understanding of what is going on but it should be made clear that lift and the concept of "up" or "down" have nothing to really do with one another, which is what I was trying to do in my last post to you.

quade -
The World's Most Boring Skydiver

[kallend 1,623]

Quote

It works but that's not what is normally used. Most airplanes are said to generate zero lift at a negative angle of attack. SOME reference line must be used. Normally the chord line is chosen. For a non-symmetric airfoil, the zero lift angle of attack will be negative. It's easy to find lift slope curves (lift vs angle of attack) for most standard airfoils, and I think they all base AOA on the chord line. That way the lift at zero angle of attack is visible on the graph. If AOA was defined based on the zero lift angle of attack, it would be impossible to compare two airfoils on the same graph.

Dave

What if the chord line is adjustable, as it is on a flexible (fabric ram-air) airfoil. Pull the brakes and what happens to the chord line?

...

The only sure way to survive a canopy collision is not to have one.

[pilotdave 0]

No clue what parachute manufacturers use (if this type of analysis is even done). But my guess would be they do something similar to how flaps on planes are analyzed. Coefficient of lift is a made up measure of how much lift a wing can produce. When flaps are dropped, Coefficient of lift isn't really "recalculated" based on the new wing shape. It is compared to the coefficient of lift of the original wing shape. In other words... fowler flaps increase wing area, which is part of the calculation of C_L. But when determining the effect of flaps, wing area is held constant. If you used two different wing areas, their C_L's wouldn't be directly comparable. So my guess for parachutes is that angle of attack is held constant as the brakes are pulled, and the theoretical lift, based on that "arbitrary" angle of attack, can be measured.

Could be wrong though!

Edit before someone bashes me... Yes I know wing area is needed to calculate LIFT, not coefficient of lift. But usually we'd be back tracking using wind tunnel test results which start with lift, and you need to figure out coefficient of lift.

And Quade... an airfoil CAN produce zero lift at a negative angle of attack. I'm NOT talking about angle of incidence. A symmetrical airfoil produces no lift when AOA=0. But any airfoil with positive camber produces SOME lift at AOA=0. Therefore in order to produce zero lift, AOA has to be negative. Seems strange, but it's true.

Dave

[Icon134 0]

I say we get a scale model of a sabre2 (we have anyone from PD here?) get it all rigged up with a/b/c/d and brake lines (adjustable of course) hang a small figurine (ensuring a proper wingloading for his skill level... say 1:1 or so... once he's proven his abilities of course we can increase that some.)

Then generate a series of experiments under different brake sets (in an aneochic chamber... or some other controlable environment.) and find out once and for all which one will land first.

personally, I think A will land first because as I understand it speed equals lift under the same wingloading and since the braked canpoy will have a lower speed it will subsiquently have less lift...

but of course my fluid dynamics is extremely rusty and my engineering expertice is submicroscopic not macroscopic.

Scott

please don't tear me up to badly.

Livin' on the Edge... sleeping with my rigger's wife...

[pilotdave 0]

You don't want me to start with scaling effect/reynolds numbers...

Dave

[Chrisky 0]

Quote

What happens for the few seconds after the brakes are applied has little bearing on the question at hand. Co

If these few seconds are enough to create a difference in altitude between jumper A and B, it has substantial bearing on the question at hand, as once the canopies are descending at the same rate again, the braked one is a little higher and thus will land a little later.
Answers the question, but not the physics inquiry.

The mind is like a parachute - it only works once it's open.
From the edge you just see more.
... Not every Swooper hooks & not every Hooker swoops ...

[Icon134 0]

Quote

You don't want me to start with scaling effect/reynolds numbers...

Well then... looks as if we need a really big hanger...

You don't scare me... scaling effect and reynolds numbers... I scoff in you're general direction.

ok, then explain to me why do airplane manufacturers construct scale models... that is if they don't effectively mimic the full size products...

Scott

[teason 0]

OK, I think I'm piecing this together now
(wow! I'm really enjoying this thread!)

All jumpers with the same exit weight will produce the same amount of lift even if one jumps a Manta and the other jumpes a 58 Xaos 27.

That is because lift = weight.

That means lift has nothing to do with which jumper lands first.
Could it be the speed at which the lift is obtained?
Does the braked canopy produce lift at a slower speed than the unbraked canopy?
If so, then the braked canopy will land last having acheived lift = weight at a slower decent.

Am I getting closer or did I just leave the rails?

I would rather be a superb meteor, every atom of me in magnificent glow, than a sleepy and permanent planet.

[kallend 1,623]

Quote

Quote

You don't want me to start with scaling effect/reynolds numbers...

Well then... looks as if we need a really big hanger...

You don't scare me... scaling effect and reynolds numbers... I scoff in you're general direction.

ok, then explain to me why do airplane manufacturers construct scale models... that is if they don't effectively mimic the full size products...

Scott

Because they know how to correct for the scale effects if they know the Re of the model and of the full size product.

...

The only sure way to survive a canopy collision is not to have one.

[quade 3]

Quote

personally, I think A will land first because as I understand it speed equals lift under the same wingloading and since the braked canpoy will have a lower speed it will subsiquently have less lift...

Please note any use of the phrase "speed equals lift" generally makes me want to pound my head against the wall.

Speed does NOT equal lift!

If it did, we'd never need parachutes to begin with since we have way more speed in freefall than under an open canopy.

No.

Lift = ((rho*V^2)/2 )*area)*Cl

The phrase "airspeed equals lift" is just about as accurate as the statement "peanut butter equals sandwich". Yes, if you add more peanut butter you'll have a bigger sandwich, but without the bread, all ya got a LOT of goo on your hands.

That is to say, yes, airspeed is -a- factor in the lift formula but is NOT the entire story.

quade -
The World's Most Boring Skydiver

[quade 3]

Quote

That is because lift = weight.

No, that's incorrect as well.

The interaction of the shape of the wing and the air, which past it is flying, creates lift perpendicular to the direction of travel.

If you'd like to fool around with a program by NASA that will give you an idea of how a wing works, you might try visiting this web site. http://www.grc.nasa.gov/WWW/K-12/airplane/foilsimu.html

quade -
The World's Most Boring Skydiver

[quade 3]

Quote

And Quade... an airfoil CAN produce zero lift at a negative angle of attack. I'm NOT talking about angle of incidence. A symmetrical airfoil produces no lift when AOA=0. But any airfoil with positive camber produces SOME lift at AOA=0. Therefore in order to produce zero lift, AOA has to be negative. Seems strange, but it's true.

I'll conceed the point.

quade -
The World's Most Boring Skydiver

[teason 0]

... crap ...
I was going on what pilotdave said

Quote

Both canopies are producing the same amount of lift... L=weight for any wing in unnaccelerated flight.

Thanks for the link.

If I could figure out a rubics cube when I was 10, then I should figure this out! 54 secs, best time! (he says embarressed yet scraping up his last shreds of dignity and defending his intelligence)

I would rather be a superb meteor, every atom of me in magnificent glow, than a sleepy and permanent planet.

[pilotdave 0]

Hehe... I was kidding... but since you asked...

Wind tunnel models can give descent results, but if you need exact results, the model needs to have the same geometry (obviously), and the mach number and reynolds number need to be the same as the flow you are simulating. This can be done using pressurized or chilled wind tunnels.

Dave

[quade 3]

Well, here's the key point about the speed equals lift thing.

When we take a student pilot up on one of his very first flights we show him how an airplane flies at various speeds. If the airplane is in straight, level and unaccelerated flight, then lift opposes weight and thrust opposes drag. The four forces balance each other out.

So, I can take an airplane up to say 2000 feet AGL, apply full power and trim it for hands off level flight. At this point the lift created by the wing -does- equal the weight of the airplane, but it's not the weight of the plane that's causing the lift. It's the speed of the airplane, the density of the air, the shape of the wing and it's angle of attack against the relative wind that is causing the lift.

Any of the factors can be changed with various results, but the most common one (in a student's life anyway) is to slow the airplane down while increasing the angle of attack and therefore create the -same- amount of lift. Above a certain AoA this will no longer work and the amount of lift it creates will -rapidly- decrease. That's called a stall.

If the -same- amount of lift can be created at different airspeeds by varying the AoA then -clearly- airspeed, in and of itself, does not equal lift.

Lest you think this only applies to airplanes with power . . . watch a swooper as he makes a long swoop. As he slows down, he much constantly increase the amount of rear risers or brakes to increase its AoA.

quade -
The World's Most Boring Skydiver

[pilotdave 0]

Quote

The interaction of the shape of the wing and the air, which past it is flying, creates lift perpendicular to the direction of travel.

Net lift may be defined as perpendicular to direction of travel, but I was talking about the vertical component of lift. Again, disregarding the vertical component of drag, which I think is reasonable for shallow descent angles and speeds, the vertical component of lift equals weight for any wing in unnaccelerated flight. Even a wing in a constant rate climb or descent has to be producing a vertical component of lift equal to the weight. Otherwise vertical forces would be out of balance and the wing would be accelerating.

Dave

[quade 3]

Actually, if the wing is climbing, it's creating slightly more lift than its weight . . . that's why its climbing.

quade -
The World's Most Boring Skydiver

[pilotdave 0]

Awww geez... c'mon physics boy... We've got 4 forces, right? Thrust, drag, lift, and weight. What happens when thrust is greater than drag? We accelerate. What happens as drag increases to finally match thrust? We stop accelerating. Meaning we maintain a constant speed. Well what happens when lift is greater than weight? We accelerate upward. Well now it gets comlicated since we've got an upward component of thrust and a downward component of drag. But for simplicity lets just assume the climb is shallow so the upward thrust and the downward component of drag are minimal. As long as L>W, the aircraft will accelerate upward. But what's gonna happen? Eventually an equilibrium will be reached, right? Can't accelerate upward forever (unless you're flying a space shuttle). Once you've entered a steady climb rate, your upward lift is equal to your weight. Your total lift could be higher, but the lift that actually pulls you upward is not.

Dave

[rehmwa 2]

PD - Just as well quit.

Teason - you're close enough for comprehension. good job

other - straight and level flight does not mean the lift component of the airfoil is straight up. (PD's fit of 'net lift' is closer to it). and, just because you're climbing, doesn't mean lift > weight --- think 'constant' climb rate and the F=ma approximation

...
Driving is a one dimensional activity - a monkey can do it - being proud of your driving abilities is like being proud of being able to put on pants

[billvon 2,396]

> Actually, if the wing is climbing, it's creating slightly more lift than its
>weight . . . that's why its climbing.

If that were true, you'd feel a constant slight acceleration upwards since the forces didn't balance. If you were feeling 1.1G, after a minute you'd be doing 1200mph upwards. That usually doesn't happen (outside of aircraft like the X-15.)

The one time this IS true (for GA aircraft) is in a turn. If you are in a stabilized, gentle turn, you will be generating more lift than the aircraft's static weight, since you need the additional lift to counteract the "centrifugal force" (really the airplane's inertia) as you change its direction.