New answers tagged aircraft-physics
11
Horsepower = weight(lbs) * sink_rate(fpm)/(550*60)
It really is that simple.
Background
At a constant airspeed glide, kinetic energy isn't changing, only potential energy. So we know that all the energy going into drag is coming from the descent.
A basic law of physics is:
$P = F*v$
where
$P$: the power
$F$: the force, in our case the force due to gravity
$...
9
That power is sink rate x glider weight. If you use m/s for sink rate and newton for weight, you'll get the power in watt. Advantages of SI...
It's easy to understand why it is so. Within a uniform gravitational field with acceleration g, a mass m is pulled down with a force m·g. The potential gravitational energy E at a given height h is E = m·g·h
Now, for ...
0
Excluding safety factor of a little excess airspeed before rotation and excess thrust/angle of climb considerations:
Weight = Lift = 1/2×rho×v$^2$×Area×Clift and Ws = Weight/Area
Multiplying W by g would convert your weight force into Newtons, otherwise your two formulas are interchangeable.
For these calculations, special attention to the AoA part of the ...
2
My understanding is that the lift fan (two sets of contra-rotating blades) is driven from the main engine via a permanently rotating drive shaft, but that the drive from the shaft to the lift fan is connected/disconnected from the lift fan via carbon clutch plates. In conventional flight it makes little sense to waste engine power driving the lift fan when ...
4
TLDR: Pretty much impossible with current technology as I understand the question.
I think you are focusing on the wrong part of the problem. Turboprop vs turbofan is pocket change compared to the amount of rocket fuel required to get to orbit. Let's put some numbers to it.
I'm assuming the following mission 1) Deorbit from low earth orbit, 2) landing (...
1
The F-35 Lightning has an approach to VTOL that is similar to what you are requesting. There is a propeller (strictly speaking, a ducted fan) pointed straight up in the middle of the fuselage and has doors that open above and below it. It is powered by the jet engine and used during vertical takeoff and landing.
1
Any VTOL arrangement will put a lot of blast on the landing surface, so a practical, off the shelf solution would be the Pratt & Whitney J58 hybrid turbojet/ram jet to get you off the ground and into near space at around Mach 3, before switching to rockets.
Although it is possible to mount a prop on this type of engine and fold it, drag and turbulence (...
Top 50 recent answers are included