trk miles consumed in a S/E scenario

[downwind]

g'day guys,

Wanting to know: here is the scenario;

for a multi engine aircraft lets say has a 300 FPM rate of climb for a single engine ops scenario and a S/E climb speed of 90 kts, the missed approach started at 1000' AMSL and the missed approach ends at 4000' AMSL which is MSA.

My question is how many nm's over the ground will the aircraft use to get to 4000' AMSL MSA from the 1000' AMSL missed approach point from the instrument approach?

Is there a handy formula to use that will be inter changeable for different aircraft performance catergorys and rate of climb and g/s, but give the same result all the time that is nm's travelled over the ground?

Want to know the rough figures so I have a bit a of knowledge of how many nm's you burn up when you lose a engine!

thank DW.

[BEagle]

4000ft - 1000ft = 3000 ft

3000ft at 300 ft/min takes 10 min (Good grief!)

90 KIAS = 1.5 nm/min

10 min at 1.5 nm/min = 15 nm.

(Assuming still air and ignoring the TAS difference at 4000 ft - which at only 90 KIAS won't be significant)