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 Tough One! Lift/Drag Vs. Distance/Altitude
 Jzucker From United States of America, joined Nov 1999, 100 posts, RR: 0Posted Tue May 14 2002 06:18:04 UTC (13 years 9 months 4 days 3 hours ago) and read 4848 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
 MD11Nut From , joined Dec 1969, posts, RR: Reply 1, posted Tue May 14 2002 08:07:11 UTC (13 years 9 months 4 days 1 hour ago) and read 4818 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
 FredT From United Kingdom, joined Feb 2002, 2185 posts, RR: 25 Reply 2, posted Tue May 14 2002 10:29:35 UTC (13 years 9 months 3 days 23 hours ago) and read 4799 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.
 FredT From United Kingdom, joined Feb 2002, 2185 posts, RR: 25 Reply 3, posted Tue May 14 2002 10:35:22 UTC (13 years 9 months 3 days 23 hours ago) and read 4798 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.
 Erasmus From Italy, joined Jun 2007, 0 posts, RR: 0 Reply 4, posted Tue May 14 2002 11:21:41 UTC (13 years 9 months 3 days 22 hours ago) and read 4791 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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