From: firstname.lastname@example.org (Badwater Bill)
Subject: Re: Wing profile
Date: Tue, 10 Oct 2000 18:42:45 GMT
If I recall correctly the bending theory for lift is much more
simplistic than you fellows are making it out to be. If you take
Cameron's flat plate with no differential in top or bottom surface,
you do still get lift when you tilt the AOA up in a wind. What you
are doing is bending the flow of the oncoming air downward. This
simply means that you are deflecting mass downward. If you deflect
mass in a certain unit of time, you get force from Newton's equation.
F = ma = dp/dt where p is "momentum.mass times velocity" and
remember it's a vector.
So, you don't have to generate more or less mass to get lift, what you
have to do is change it's direction only. Then you get force (lift).
Of course in rocket engines the dp/dt translates to
d(mv)/dt = mdv/dt + v dm/dt but since "v" is nozzle velocity and is
constant, dv/dt = 0 you use dm/dt or mass flow rate times the nozzle
velocity for your acceleration figure. Different problem all
Let's just do a simple calculation on my RV-6. Let's assume I have
100 square feet of wing area or about 10 square meters. I'm going to
have to do this in the metric system because it's ten times easier
(pun intended here).
The mass of my airplane is 600 kilograms at normal flying weight.
Since gravitational acceleration on Earth is 9.8 meters/sec-sec I
weigh about 5900 Newtons. That's the force I have to generate to
keep me at a constant altitude.
At 200 miles per hour I'm going about 90 meters per second and my
wingspan is 8 meters.
Let's assume I'm at sea level and the density of air is 1.2 kg/m^3.
Another assumption is that I affect a cylinder of air as I fly through
it that has a cross sectional diameter equal to my wing span which is
8 meters (radius is 4 meters). We know this is not exactly true but
it's a good first order approximation. So, the total mass I effect
each second is:
Cross section = pi x r^2 = 3.14 x 4^2 = 3.14 x 16 = 50 square meters.
Since I'm sweeping through 90 meters a second I'm effecting
90 meters/s x 50m^2 or about 4500 cubic meters of air each second. At
1.2 kg mass per cubic meter that's a mass equal to 5400 kilograms of
From Newton's equation
F=dp/dt we can write F= d(mv)/dt = 5400kg x (Velocity change) in the
air we are deflecting, divided by one second.
So, we know we need 5900 Newtons of force and we can write the
5900 Nt = 5400 kg x dv/dt
Dividing each side of the equation by 5400 and solving for dv/dt we
get about 1.1 m/s^2. Integrating over our one second interval gives
us the number 1.1 m/s final air velocity of this horizontal column of
air we've disturbed.
What all this means is: If I am at sea level in my RV-6, the effect
of me cruising along the beach at 10 feet and 200 mph is to
effectively deflect the air down as I go by. If I were a helicopter
I'd have to pump air down from above in the same fashion. An airplane
is no different. It's an air pump. It pumps air down as it passes.
I take what was relatively still air and throw 5400 kilograms (nearly
12,000 pounds of air) of it down each second at 1.1 meters per second
which is about two and a half (2.5) miles per hour defection velocity.
I think this is what is meant when people talk about bending the air
down. It's a matter of what reference system you are in. If you are
in the airplane then you are in a stationary inertial reference frame
(no accelerations) and to you it looks like the wind is blowing by
you. In this case, you take a 200 mph wind and defect it down about
2.5 miles per hour and you get the lift you need to stay afloat.
Isn't that interesting that it's that small of a number. On the order
of 1% of your horizontal speed. I find that interesting that it's
such a small number. If you look at pressure differential on a wing
to produce this pumping effect, it's a tiny number too.
It's short and simple to calculate so I'll throw it in here for fun
too. One hundred square feet of wing area supports 1500 pounds on my
RV-6. That's 15 pounds/sq.-ft. Since each sq-ft is 144 sq inches,
that's 15/144=0.104 pounds/sq.-inch (1.6 ounces).
Since atmospheric pressure at STP is about 15 pounds/sq-inch, this
represents a change in atmospheric pressure of 0.104/15= 0.0069
atmospheres (atm) or in percentage, 0.69%.
So, the wing doesn't really have to do that much. All it has to do on
average is change the pressure of the atmosphere less than 1% as it
goes by to produce all the lift necessary for flight. I find that
sort of amazing.