I’ve heard of people ferrying General Aviation aircraft to Hawaii. I seem to remember reading about a single-engine Cessna (I don’t recall which model) having to ditch just offshore when it ran out of fuel. IIRC, that airplane had temporary fuel tanks installed.
I am by no means a math wizard. I’m prone to make simple mistakes if I’m not careful. So could a pilot Doper check me on this? I’ve used data from a 1985 Cessna 172 POH for the calculations.
Fuel capacity: 68 gallons with integral tanks, 62 gallons usable. (Standard fuel, BTW, is 43 gallons.)
Weight of fuel: 408 pounds (68 X 6)
Useful load: 974 pounds
Payload w/full tanks: 566 pounds (974 - 408)
Weight of pilot: 190 pounds
Payload w/full 68 gallons of fuel and 190 pound pilot: 376 (62.67 gallons of extra fuel. 556 - 190)
Maximum Range: 875 nm
Economy: 14 nmpg (875 / 62)
62.27 gallons X 14 nmpg = 877.33 nm
Total range: 1,752.33 (875 + 877.33)
Distance to Hawaii: 2,080 nm.
So by my calculations a Cessna 172 will run out of fuel 328 miles from Hawaii. At 14 nmpg the aircraft would require an additional 23.43 gallons, or 140.57 pounds of fuel.
So how can a Cessna fly from California to Hawaii? Can 140 pounds of “stuff” be stripped from the aircraft, allowing the necessary fuel to be carried? Do they try to get heroin addicts (just kidding – they’re typically skinny) as pilots? Or do the pilots try for the most favourable winds aloft, and hope the forcasts are right?
(Note: I’ve done a little rounding here and there, and have not taken into account the weight of the extra temporary fuel tanks, nor where they would fit.)