Add or removeing mass problem - Help!

[rpetersson]

Ok - struggeling to understand what i have done wrong here, hopefully any kind souls can enlighten me...
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Q: An aeroplane has a zero fuel mass of 47800 and a performance limited take-off mass of 62600kg. The distances of the leading edge and trailing edge of the MAC from the daturm are 16m and 19.5m respectivly. What quantity of fuel in Imperial gallons , must be taken up to move the CG from 30% to 23% MAC if the tank arm is 16m aft of the datum and the fuel SG is 0.72?
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I caluclated the CG at 30% MAC to 17.05m and at 23% to 16.805m.
So the formula should be:

New Total Mass = Old total Mass + Change in mass
(47 800+m)*16.805 = 47800*17.05 + (16*m)

Answer= 14 547.8kg

And converting this to Imperial gallons gives 4444

But the right answer should be 4455..

What on earth am i doing wrong??

[keith williams]

I think that you have just made an minor error in converting the units.

Your equation.

New Total Mass = Old total Mass + Change in mass

Should read New moment = Old moment + Moment change

But your figures below are correct


(47 800+m)*16.805 = 47800*17.05 + (16*m)

Answer= 14 547.8kg


If you now use the conversion factor 1 Imp Gal = 4.546092 litres as given in the CAP 696 you will get the correct answer.

14547.8 kg divided by 0.72 SG = 20205.27778 litres

20205.2777b litres divided by 4.546092 litres/Impgal = 4444.54 Impgals

[rpetersson]

Thank you Keith for conforming im doing it right :)

[Da-20 monkey]

Yeah.
If you use oxford books, learn the weight/ volume schematic on page 37 of the perfo book. Very handy.

The logic behind a Imp gal is that it should be 10 pounds of water, hence x10xSG to go from Imp. Gal. to lbs.

To be honest I never found it to be necessary to use more than 2 decimals..

[ryan_sharara]

I need to start revising ! :S