Lift / drag ratio question.

[lpokijuhyt]

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!

[DFC]

Ground distance from B to C is 20 miles

The air distance flown between B and C is ground distance * TAS / groundspeed

= 20 * 40/60 = 13.3333333 miles

13.3333333 miles is 13.333333 * 5280 feet = 70400

Glide ratio 20:1

So for 70400 forward the aircraft descends 70400/20 = 3520 feet

Plus the 800ft required at C = 4320.

[lpokijuhyt]

Thanks for taking the time to explain it to me! Appreciate your help!:ok:

[palou89]

I solved this problem using "crossed multiplication" or "regla de 3 simple" (in spanish) and got the same result..

For every foot I fall down I go forward 20 feet (In no wind condition, this happens at 40mph). At 60mph how many feet will I go forward for every foot I drop?

(60mph x 20 ft) / 40 mph = 30 ft

So glide ratio with the tailwind is now 1:30

Distance from B to C is 20 miles or 105600ft.

If for every 30 feet I go forward I fall down 1 foot, how many feet (vertically) do I need to cover 105600ft horizontally?

105600ft x 1 ft / 30 ft = 3520ft

Add 800ft to cross C at 800ft AGL and you get 4320ft.