On Fri, 19 Nov 2004 at 22:53:26 in message
, alexy
wrote:
Jose wrote:

My first assumption is that for the same air density, the friction is
directly proportional to the speed of the aircraft.

Nope. To oversimplify, it goes as the cube at subsonic speeds. Once
supersonic other terms enter the equation. So at Mach 10 the scramjet
would
have to exert more than 1000 times the thrust as for Mach 1 at the
same altitude. And a scramjet can't run from a standing stop.

Jose
Jose, hopefully someone will correct me if I'm wrong, but the drag
(and the thrust needed to overcome it) increase with the square of
velocity. It is the power needed that increases with v^3.

Perhaps it is time to insert the simple fundamental equation of
aerodynamic drag.

Drag = 0.5 * (air density)* ( a representative area - commonly wing
area )*(velocity squared)*( a 'constant that depends mostly on shape)

If using pounds divide the result by 'g': 32.2 ft/(second squared) for
an answer in pounds.

The speed of sound in air is proportional to the square root of the
absolute temperature of the air. Thus in the standard atmosphere the
speed of sound falls from 1,117 ft/sec at sea level to 968.5 ft sec at
36,000 ft. Above that it is constant up to around 80,000 ft when it
starts to rise again up to about 175,000 ft when it starts to fall
again!.

So the original simple calculation is wrong. It is more complex than I
have said as drag also depends on Mach number and a number called
Reynolds Number.

Altitude ft Speed of sound density
S.L. 1,117 ft/second 0.076475 lb/(cubic foot)
50,000 ft 968.5 ft/second 0.011642
100,000 ft 1003.2 ft/second 0.0010332

E&OE (' Errors And Omissions Excepted. That means I hope it's right but
I don't guarantee it! Check the tables yourself.)

I apologise that this is not metric units.
--
David CL Francis