Aero eq.s (was Re: Galileo Day)
Michael M. Butler (email@example.com)
Tue, 16 Feb 1999 18:00:12 -0800
> IAN: Mike, what I said is what I quoted from
> a physics book, how could that be my discovery?
> Every basic physics book covers the same fact.
Permit me to inject a few points before departing this conversation:
- Every basic physics book oversimplifies. That's why they call it "basic"
- The independence of horizontal and vertical motion is only true in
Furthermore, the notion of "horizontal" is a local one, as is "vertical".
- Aircraft pilots trade speed for altitude all the time. They are usually
taught to do it some time in the first 20 hours of dual flight training.
They couldn't do it if the "basic physics" notion about h and v speeds
were true for aerodynamic cases.
- The term "stall" does not refer to an aircraft having zero airspeed, nor
to an aircraft having zero speed over ground.
"Stall" is a technical term which refers to a significant, usually sharp
change in L/D (the lift-to-drag ratio), typically as the result of
excessive angle of attack--the angle between the midline of the lifting
surfaces and the airflow approaching the craft.
- An aircraft the size and shape of a 747 probably gets significant lift
from its fuselage ("body lift") at high angles of attack.
- The commonly-employed equation for subsonic aerodynamic drag is usually
rendered as (apologies for the poor ascii-isms):
D = 1/2 ( p * Cd * A * v**2 ).
p is "rho", the density of air, and it varies with weather and altitude; Cd
is the drag coefficient (and a bit of a fudge factor),
A is area, either cross-sectional or wetted, depending on what you are
using for Cd, and v is 'velocity' (more properly, airspeed).
'Terminal velocity' (so-called) is where D - F = 0, F being the force of
gravity on the object in question, and D being drag-in-the-upward-direction
(this terminal velocity calculation assumes "drag" is the same thing as
"lift", and that there are no remaining side forces on the object).
Figure out F, plug in the desired v and p, and you can calculate what Cd *
A would need to be.
If you figure that the engines were still on, add their thrust into the
force calculation: D - (F + T) = 0; this lets the Cd * A be higher for
the same velocity.
7) There's a popular number called the "ballistic coefficient", computed as
w / ( Cd * A ), where w is weight. A big ballistic coefficient means the
object in question acts more like a cannonball; a small one means it acts
more like a feather.
8) Handwaving about the plane being a parachute is not the same thing as
crunching the numbers and estimating ballistic coefficient.