On Wednesday, May 7, 2014 11:49:46 PM UTC+12, James Metcalfe wrote:
At 06:55 07 May 2014, Bruce Hoult wrote:
[1] handy formula: X knots of kinetic energy is worth (X/5)^2 feet of
gravitational potential energy. e.g. 70 knots = (70/5)^2 = 14^2 = 196 ft.
50 knots = (50/5)^2 = 10^2 = 100 ft. Less drag loses of course.
You'll never turn speed into quite that much height, and you'll need more
height than that to get speed. But the differences are large in a high
performance glider at moderate speeds.

..or you could use my rule of thumb: a change of speed of 10 knots IAS
gives
you (or costs you) the number of feet in height of the speed you arrive at.

For example:

70 knots to 60 knots : plus 60 feet
60 knots to 70 knots: minus 60 feet
70 knots to 50 knots (i.e. 70 to 60, then 60 to 50) : (60+50) = plus 110
feet

It's nearly the same formula. If you made it "every 12.5 knots IAS" instead of 10 then it'd be nearly exact.

The derivative of my formula (X/5)^2 i.e. X^2/25 is X/12.5. So gaining or losing X feet at X knots takes roughly 12.5 knots of speed change.

Doing it using the speed you're changing *to* is clever, as it makes some allowance for drag losses. However your height estimates are systematically biased 20% too large.

If you said 70 to 50 is (60+50)-20% = 110-22 = a gain of 88 ft then that would be good.
And 50 to 70 is (60+70)-20% = 130-26 = a loss of 104 ft, which is pretty good too.

If you're doing something like "how high can I zoom to after a 140 knot low pass and still have 60 knots for the circuit?" then I'm not convinced that adding up eight numbers is easier than calculating two squares and subtracting them :-)

By me: 28^2 - 12^2 = ~800 - ~150 = 650 ft (784-144 = 640 if you do it exact)
By you: 130+120+110+100+90+80+70+60 = 760 ft
By you with my -20% correction: 760 - 152 = 608

(remember folks, that's 140 knots at the END of the low pass, not the start!)