Aero eq.s (was Re: Galileo Day)

Michael M. Butler (
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:

  1. Every basic physics book oversimplifies. That's why they call it "basic" physics.
  2. The independence of horizontal and vertical motion is only true in airless environments. Furthermore, the notion of "horizontal" is a local one, as is "vertical".
  3. 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.
  4. 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.
  5. An aircraft the size and shape of a 747 probably gets significant lift from its fuselage ("body lift") at high angles of attack.
  6. 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.