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Tough One! Lift/Drag Vs. Distance/Altitude  
User currently offlineJzucker From United States of America, joined Nov 1999, 100 posts, RR: 0
Posted (12 years 2 months 1 week 4 days 21 hours ago) and read 3437 times:

Ok guys...here is a tough one that I need to find an answer to before my final.

Why is the ratio of Lift/Drag proportional to Distance /Altitude??

I have already attempted to find a proportional relationship between Lift, Drag, and Potential and Kinetic airspeed. I failed. If anyone knows this you are amazing!

Thanks in advance. Josh

4 replies: All unread, jump to last
 
User currently offlineMD11Nut From , joined Dec 1969, posts, RR:
Reply 1, posted (12 years 2 months 1 week 4 days 19 hours ago) and read 3407 times:

Not too sure if this is what you're looking for... For cruise, the answer perhaps lies the Breguet range equation? Range is maximized at flight conditions which maximize M*L/D. With a constant Mach M, that means max L/D which means constant CL which means flying at a constant W/delta which means you continue to climb in cruise as fuel is burned off (follow optimum W/delta). I believe this is your L/D and Range/altitude relationship. If this is not what you have in mind, perhaps take a look at the equations of motion for climb/descent...One can deduce the relationship...maybe. I haven't touched the aero books in years.

Regards,
Nut


User currently offlineFredT From United Kingdom, joined Feb 2002, 2185 posts, RR: 26
Reply 2, posted (12 years 2 months 1 week 4 days 17 hours ago) and read 3388 times:

Ain't got the books with me but I'll have a shot at it anyway. Very interesting question and I have to admit I didn't really know the answer. I've always taken it as gospel - not a good thing!

In a glide, you're getting rid of altitude to compensate for the energy loss created by drag. Set up an equation for the work done by drag and by gravity.

dW = m*g*dh - dt*V*D

D = m*g*cos(ga)/(L/D), as L will then equal m*g*cos(ga)

where ga is the glide angle.

The ground distance travelled, ds is dt*V*cos(ga). As the work done on the aircraft will sum up to zero, we end up with

m*g*dh - dt*V*m*g*cos(ga)/(L/D) = 0

m*g * (dh - ds/(L/D)) = 0

divide by m*g as it can't be zero

dh = ds/(L/D)

ds/dh = L/D

Cheers,
Fred



I thought I was doing good trying to avoid those airport hotels... and look at me now.
User currently offlineFredT From United Kingdom, joined Feb 2002, 2185 posts, RR: 26
Reply 3, posted (12 years 2 months 1 week 4 days 17 hours ago) and read 3387 times:

I thought I'd clarify one step there to make it easier to follow.

[...]

m*g*dh - dt*V*m*g*cos(ga)/(L/D) = 0

m*g * (dh - dt*V*cos(ga) / (L/D) = 0

Substitute ds = dt*V*cos(ga)

m*g * (dh - ds/(L/D)) = 0

[...]

Cheers,
Fred



I thought I was doing good trying to avoid those airport hotels... and look at me now.
User currently offlineErasmus From Italy, joined Jun 2007, 0 posts, RR: 0
Reply 4, posted (12 years 2 months 1 week 4 days 16 hours ago) and read 3380 times:

L= Lift
T= Thrust
D= Drag
W= Weight
a=b= Glide Angle
Dis= Distance
A= Altitude

Imagine a glider in a constant speed, constant descent angle (a=b) glide.

Then : T= D and L=L1

tan (a) = A/Dis
tan (b) = T/L1 = D/L

since a=b : therefore A/Dis =D/L
or: ALTITUDE / DISTANCE = DRAG / LIFT

Is this clear enough?


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