On Mon, 16 Oct 2006 23:22:40 GMT, Don Tuite
wrote:

On Mon, 16 Oct 2006 22:41:58 GMT, Ron Hardin
wrote:

Don Tuite wrote:

On 16 Oct 2006 13:28:37 -0700, "Snidely"
wrote:


Blanche Cohen wrote:
Aircraft Turn Calculator

www.csgnetwork.com/aircraftturninfocalc.html

Interesting -- but doesn't speak to asltitude issues (loss of, or
adjustments to avoid loss of).

Losing altitude at a constant rate (rather than accelerating downward)
would be the same as maintaining a constant altitude, wouldn't it?
(Vectors on a free-body diagram and all that)

Don

You get additional horsepower from going downhill, letting you maintain
your speed.

Telegraphic as ever, Ron.

So, because I'm using the same throttle position as I was using in
level flight, I'm descending while in a turn and banked at A degrees.
The airplane seat is pushing on my butt at an angle A degrees from
vertical. That can be resolved into a vertical vector and a horizontal
vector aimed at the center of the circular path I'm describing.
Meanwhile the back of the seat is exerting another force on my butt
tangential to the circular path I'm describing in the horizontal, er,
"plane." That force is a reaction to the thrust of the prop.

If I "maintain my speed," I get the same lift I would have got from
adding enough throttle so as not to lose altitude. That means the
magnitude of the vertical component of the lift vector is still equal
to mg and my rate of descent is constant.

You were agreeing with me? (Never happened before.) Maintaining
vertical equilbrium depends on maintaining the same speed as in
straight and level flight, doesn't it?

Don


I'll try this time:

altitude = potential energy.
Losing altitude means giving up potential energy.
Where does that energy go?
Into providing a greater (effective) angle of attack than
you could at constant altitude and airspeed and thrust
OR
providing more airspeed than you could at constant altitude
and AoA and thrust.

Untelegraphic enough?

Brian Whatcott Altus OK