Quote:
| I'm looking at only the methematics of the science - not a particular task. |
There is a quick - and surprisingly accurate - way of computing the extra fuel needed :
The ratio TOW / LW is, for all intents and purposes constant for a low percentage increase on the TOW ( in the form of extra fuel, for instance). Let's call it
K K= TOW / LW = (LW+Bo) / LW where Bo is the burn-off
K = 1 +( Bo / LW)
so, for a small delta LW, ðBo = (K-1) x ðLW and that's what we're interested in.
For example, (rounded figures to protect the innocents, again) :
TOW = 260,000 kg
Bo = 74,700 kg
LW = 185,300 kg
So, our K is : 260 /185.3 =1.40
so ðBo = .40 x ð LW
You'll be burning 40 % of your extra fuel.
For info, it's for an air distance of 5350 Nm and an 11.10 hr trip
A 777.
Hopes it helps.
Edit : just discovered Kijangnim let the cat out of the bag : K is the transport factor