Its exactly the same thing except I have used very basic theory where as you have resorted to using the greek alphabet.
Well you wrote the same as I did as far as the energy conservation is concerned, but we drew different inferences out of this, pardon my greek letters, they were just for the symbology.
At Vmo what is the drag going to be at various weight say 50% 75% and 100% of MTOW. In relation to each other?
I will get the data for this from a real aircraft, B737-700
Code:
L/D Data for 737-700 @ 0.78/280/250 KIAS
UNITS
Weight 40.0 50.0 60.0 70.0 (1000 kgf)
L/D 15.4 17.8 19.6 20.6
Drag 2.6 2.8 3.1 3.4 (1000 kgf)
ERR 2.6 2.8 3.1 3.4 (250kts)*(1000kgf)
ERR' 0.65 0.56 0.51 0.49 (25kts)
ERR(fpm)1621 1405 1277 1213 (fpm)
ERR = Energy Removal Rate
ERR' = Energy Removal Rate Per Unit Weight
ERR'(fpm) = Energy Removal Rate Per Unit Weight (expressed in feet per minute)
Or put another way when are you going to have maximum energy removal from the system?
Maximum energy removal rate is for the heaviest aircraft.
But maximum per unit weight energy removal rate is highest for the lightest aircraft. Interestingly enough it has the same units as speed and can be conveniently expressed in fpm.
Your graph is fine and my graph is fine but there are differences.
1. Your graph has speed on its horizontal axis, my graph has angle of attack on horizontal axis
2. Your graph doesn't depict L/D directly, my graph does.
The way to infer L/D curve from your graph is by assuming that lift is constand and speed is varying, that way L/D curve from your graph will be reciprocal of total drag curve.
Now a few arguments why my graph includes the parasite drag as well as induced drag just like your graph. My graph sure doesn't break them down in two components.
At very high speeds where angle of attack approaches zero, lift induced drag must be zero, your graph says that. My graph at zero angle of attacks shows a positibe coefficient of drag, this is parsite drag coefficient.
Please note that my graph doesn't depict lift or drag, instead it depicts the lift and drag coefficients, thats why the shape of the curve is different from yours.
As you can see the max drag is at the highest airspeed which is limit by Vmo. Therefore to get max rate of decent you need to go as fast as possible. Which is where my very simple equation comes into play.
Its not the max drag that matters, its the least L/D which will give fastest and steepest descent. Your graph hence cannot be used to describe the phenomenon under consideration.