Originally posted by HoHun:
Hi Dwarf,
Your low speed example is misleading because you implicitely assume that initially, the plane is in level flight and has to pull up to enter a climb. You're right it can't pull up while flying at 1 G stall speed without stalling. However, if the initial situation is a climb, this is not necessary, and aircraft can climb quite well at 1 G stall speed.
I agree entirely. Once a climb has been established it can be maintained, even at speeds down to Stall.
And once either climb or accel has been established, Ps will do a useful job of providing rate data.
All I've been trying to do is get across the idea that Ps will NOT tell you whether you can get from your current flight condition to that established state in all cases.
Your high speed example is slightly flawed, I'm afraid. At top speed, an aircraft uses all of its power to overcome drag, and there's no power left to climb. Only by going slower, the aircraft can begin to climb - just like it could accelerate by descending.
The math isn't that complicated:
Ps=Pe/(m*g)=(Pt-Pd)/(m*g)
=> acceleration: a=(Ps*g)/v, climb rate: Vv=Pe/W=Pe/(m*g)=Ps
The conclusion:
a=Vv*g/v
Acceleration and climb are directly and linearly interdependend.
The derivation of the above formula requires no knowledge of aerodynamics at all, it's a fairly simple application of Newton's axioms.
Regards,
Henning (HoHun)[/qb]
Please reread my posts. In the first instance, I specified that some part(s) of the structure had encountered critical mach and its attendant drag rise. Whether the equation(s) we have for deducing Ps accurately would take that drag into account isn't apparent. At that point, the aircraft could no longer accelerate, but it could still climb.
In the second instance, I specified terminal velocity, not merely top speed. The aircraft is in a dive and can go no faster no matter what you do, but it still has enough altitude to effect recovery.
For the sake of discussion, let's take a hypothetical aircraft. It has a top level speed of 400 mph. Due to it's configuration, not even WEP and gravity can get it to exceed 600 mph.
It enters a dive at near top speed. Once it passes above that 400 mph top speed, Ps becomes negative. Soon it reaches its 600 mph terminal velocity. Now nothing can make it go faster, but it can still recover (climb). Not because there is some Ps hiding somewhere, but because of the inertia it carries. So long as the stresses don't cause disintegration, inertia will carry it through the transition into climbing flight, after which it will slow until Ps again becomes positive and it can sustain some rate of climb.
At the point at which recovery is begun, it can climb but it can no longer accel. Ps would say it can do neither.
At all points between (slightly above) Stall Speed and Terminal Velocity, the value of Ps doesn't really matter so far as entering a maneuver goes. Both climb and accel can be initiated at will. Both can be sustained at some rate and for some period of time.
Even at those speeds between top level speed and terminal velocity, the pilot can still accel or climb at will. At such speeds Ps can no longer provide any accel rate information and can only tell you that climbing will reduce your speed. It may assist somewhat as to how much accel is being retarded and decel would occur in the climb, but probably not all that much.
Dwarf
[ 01-08-2002: Message edited by: Dwarf ]