lift / drag ratio question
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Join Date: Dec 2006
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lift / drag ratio question
My brain is tired. Here is a question that I am not getting correct:
A glider is flying from A to C. With a normal L/D ratio of 20:1 and a constant airspeed of 40 MPH, what minimum altitude AGL is needed at B to arrive over C at 800 feet AGL with no sinking air?
The distance from A to B is 15 miles and from B to C is 20 miles. We have a tailwind of 20 mph.
The answer is listed as 4320 feet.
Thx. I'm sure it is simple, but I keep getting a different answer!
A glider is flying from A to C. With a normal L/D ratio of 20:1 and a constant airspeed of 40 MPH, what minimum altitude AGL is needed at B to arrive over C at 800 feet AGL with no sinking air?
The distance from A to B is 15 miles and from B to C is 20 miles. We have a tailwind of 20 mph.
The answer is listed as 4320 feet.
Thx. I'm sure it is simple, but I keep getting a different answer!
Join Date: May 2007
Location: Manchester, UK
Age: 38
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Hi lpokijuhyt,
The solution comes in knowing the air distance the aeroplane has travelled. You might find it helpful to draw a diagram. Ignore the A to B section, seems a bit pointless.
We need to know how much height the aircraft will lose in the 20 miles from B to C. With a 20mph tailwind, it will take 20 minutes to cover this distance. In order to calculate the air distance travelled (so we can get the altitude loss), we multiply the wind component by the journey time, and add it to the distance.
20 miles + (-20 * 1/3)
= 13 1/3 miles.
Convert this to feet:
= 70,380 feet.
Divide by 20 (L/D ratio):
= 3,520 feet.
We need to be at 800 ft at C, so add 800 to 3520:
= 4,320 feet.
Hope this helps.
Dave.
The solution comes in knowing the air distance the aeroplane has travelled. You might find it helpful to draw a diagram. Ignore the A to B section, seems a bit pointless.
We need to know how much height the aircraft will lose in the 20 miles from B to C. With a 20mph tailwind, it will take 20 minutes to cover this distance. In order to calculate the air distance travelled (so we can get the altitude loss), we multiply the wind component by the journey time, and add it to the distance.
20 miles + (-20 * 1/3)
= 13 1/3 miles.
Convert this to feet:
= 70,380 feet.
Divide by 20 (L/D ratio):
= 3,520 feet.
We need to be at 800 ft at C, so add 800 to 3520:
= 4,320 feet.
Hope this helps.
Dave.