Sender: <marv@lancaironline.net> (Marvin Kaye)
To: lml
Date: Mon, 13 Jan 2003 16:00:46 -0500
Message-ID: <redirect-1982426@logan.com>
From: "Larry Graves" <rlg34@attbi.com>
References: <list-1982094@logan.com>
Subject: Re: [LML] wing loading and stall speed
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Brian wrote:

>> but if I could just understand the relationship of wing loading to stall
speeds, I'd be happy.

~~~~~~~~~~~~

OK, here goes. Lot's of background, then a conclusion. Let's use an example
of three full-size model Lancair IVs, each identical in size and shape, but
made of different materials. Each mockup has a 100 sq ft wing (just to make
it easy) with the same airfoil shape. Each has a weight in the nose to
assure the same CG. The first model is made entirely of balsa wood and
weighs, let's say, 100 pounds. The second is hollow and made of carbon fiber
lay-ups, and weighs 1000 lbs. The third is made of sheet steel and weighs
10,000 lbs.

All three of our identical wings have identical capabilities to create lift.
They will each create the same amount of lift at the same airspeed and angle
of attack. This is another way to say they have identical lift coefficients.
This is because they have the same wing shape and size (area). What a wing
is made out of or how strong it is has nothing to do with how much lift it
can create (unless it breaks, of course). Lift equals (coefficient of lift)
times (wing area) times (velocity squared).

Each of these three wings also has the same "critical angle of attack;" that
is, the point at which the angle of attack, if exceeded, will cause airflow
separation from the wing, and a stall. Let's say for example that the
critical angle of attack for our three models is +15 degrees. If the
relative wind is flowing past our wing at +15 degrees, the wing will be
developing the most lift it can develop, for all airspeeds below the speed
of sound. If we hold our angle of attack (alpha) at exactly +15 degrees, the
lift will vary directly with the square of the velocity.

At +16 degrees, each wing will be stalled and developing no lift, regardless
of airspeed. For alpha angles less than the critical angle of attack of +15
degrees, each wing will be developing lift regardless of airspeed, just less
lift than at +15 degrees alpha.

So let's pick some numbers and say that this particular wing size/shape has
the capability of developing 3200 pounds of lift at +15 degrees angle of
attack at, say, 100 knots TAS. We pilots would call 100 kts TAS the stall
speed at 3200 pounds, and it is the same airspeed for all three wings **at
1G wing loading,** regardless of their construction.

Now, let's take our 100-pound balsa wood model airplane out to the runway on
a no-wind day and tip it up on its main gear until it has a wing angle of
+15 degrees, and start accelerating it down the runway. (We tip it up by
pushing down on the tail, just as the elevators do.) At the airspeed where
the wing develops slightly more than 100 pounds of lift (airplane weight
plus the couple of pounds of down-force on the tail), the plane will take
off.

Let's say that airspeed is 15 kts for the balsa plane. At 14 kts, it is
developing less than 100 lbs of lift, and our model will stay on the ground.
At 16 kts, it will begin climbing. At 20 kts it will be climbing faster, and
at 100 kts (if it holds together), it will be going practically straight up.
The wing loading for our balsa model is just a tad over 1 lb/sq ft. The
stall speed for our 100 lb balsa model is 15 kts at 1G -- very slow, and
that is because the plane weighs very little. It has a very low wing
loading. If it is a gusty day, we will have a very rough ride because our
low wing loading will allow every gust to push us around. The gust
velocities will be a high percentage of our overall velocity.

OK, now take the 1000 lb carbon fiber model and push down on the tail so
that the wing is tipped up to +15 degrees and start accelerating it down the
runway. Somewhere around 50 kts, that plane will take off, because (I'm
guessing) that's about the airspeed at which the lift formula will give us
slightly more than 1000 pounds of lift from that wing size and shape. The
stall speed for our 1000 lb model is around 50 kts at 1G. As you continue
accelerating past 50 kts, the plane's climb rate will continue to increase.
It will have a better ride on a gusty day because it has a higher wing
loading and will not get tossed around as much by the gusts as the balsa
model.

Now take the sheet steel airplane and begin accelerating it (with wing angle
at +15 degrees) down the runway. You're going to be on that runway for a
long time. Keep on accelerating it past 100 kts, 150 kts, 200 kts. Keep
going. Let's guess that somewhere around 250 kts that sheet steel airplane
will rise off the ground -- at that point the velocity of the airflow past
the wing is generating more than 10,000 lbs of lift. We would say the stall
speed for the 10,000 lb steel airplane is 250 kts at 1G. You will have a
very smooth ride on a gusty day in this "highly loaded" airplane, because
the gusts are such a small fraction of your total airspeed.

We could also calculate what the stall speed of any of these airplanes would
be at 2G's, 3G's, 4G's, etc., etc. The more the plane weighs, the higher the
wing loading, and the higher the stall speed, in all flight regimes.

The balsa model's wing loading at 1G is 1 lb/sq ft and it stalls at 15 kts.
The carbon model's wing loading at 1G is 10 lb/sq ft and it stalls at 50
kts.
The steel model's wing loading at 1G is 100 lb/sq ft and it stalls at 250
kts.
All the same wing size and shape. Just the weights (wing loadings) are
different.

Stall speed increases with increasing wing loading. That's why the first
reaction to most stalls should be to "lower the nose", thus decreasing the
weight or force on the wing and at the same time decreasing the angle of
attack to less than critical. The wing "starts flying again" because airflow
re-attaches to the wing and it once again is developing lift.

I did a Google search on "formula for lift" and found this nifty website:
http://www.planemath.com/activities/pmenterprises/airfoils/airfoils_teacher.
html

(If your e-mail program wraps this website address around on two lines, be
sure to copy/paste the whole thing un-broken into your browser's address
line. There is an underscore between the words airfoils and teacher.)

Sorry for the ramble, but I enjoy the challenge of trying to explain things!
Hope it helped more than it hurt!

Larry Graves