4 ms·

It is all about costs. Those engines are producing energy to keep the plane flying at 500mph. If you pull some power to spin an alternator then you need to ru

by evgen 16y ago

It is all about costs. Those engines are producing energy to keep the plane flying at 500mph. If you pull some power to spin an alternator then you need to run the engines faster and burn more fuel (which means you need even more fuel on board at takeoff to lug all of this extra fuel to the destination.) Add in the weight of the various inverters and equipment necessary to make this system safe on an airplane and I am surprised they even bother trying to run provide power to the seat at all.

nostrademons 16y ago
Some math: 120V * 2A = 240W * 300 passengers = 72 kW. That's 1/50th of a percent of the existing power output of a 747. If you figure a $600 plane ticket, I'd gladly pay 1/50th of a percent ($0.30) of that for electrical power on the flight. The numbers get a bit more complicated when you figure in the weight for transformers, alternators, inverters, etc. But fuel is not a major component of this. The math really doesn't support the assertion that it's not cost-effective.
- mrb 16y ago
  72 kW is 1/19th of a percent of the output of a 747 (140MW). More to the point, 240W is irrealistic for a laptop. Most laptops consume under 40W. With this better estimation, this is 1/117th of a percent.
- neutronicus 16y ago
  "fuel is not a major component of this ... The math really doesn't support the assertion that it's not cost-effective." How do you figure? Say some extra weight delta_w in additional equipment is required, and say that the fuel required to complete a given flight is a function f of the plane's weight, and that the cost of the fuel is C. Further, let the lifetime maintenance cost of the new equipment be E, and the lifetime maintenance of the plane's structural components be g, also a function of its weight. As a rough approximation, the additional cost should be E + C df/dw|_W * delta_w + dg/dw|_W * delta_w (Note that, really, f is a function of both passenger power consumption and of weight, both of which we're varying here, but I'll buy your argument that df/dp|_P * delta_p is neglible in comparison to f(P,W)) The article asserts that one or both of the derivates in this equation are large enough that its product with delta_w is non-negligible (in fact, prohibitive). I'm guessing that df/dp is not the problem, which is the only term you really addressed.
- pavel_lishin 16y ago
  Sure. But don't forget that adding a new feature to an airplane is going to take work. People have to figure out how all of this works, then other people have to test it, other people have to make sure it passes safety inspections, etc., etc., etc. Plus, then you have to retrofit old planes - or build new ones. It's more than just fuel costs.
chii 16y ago
a (good!?) method would be, to install small power generating bikes on the plane, and use that to suppliment the power usage of passengers. You market the bikes as exsercise machines for the long journey to prevent blood clots.

by evgen 16y ago

nostrademons 16y ago

Some math: 120V * 2A = 240W * 300 passengers = 72 kW. That's 1/50th of a percent of the existing power output of a 747. If you figure a $600 plane ticket, I'd gladly pay 1/50th of a percent ($0.30) of that for electrical power on the flight. The numbers get a bit more complicated when you figure in the weight for transformers, alternators, inverters, etc. But fuel is not a major component of this. The math really doesn't support the assertion that it's not cost-effective.

mrb 16y ago
72 kW is 1/19th of a percent of the output of a 747 (140MW). More to the point, 240W is irrealistic for a laptop. Most laptops consume under 40W. With this better estimation, this is 1/117th of a percent.
neutronicus 16y ago
"fuel is not a major component of this ... The math really doesn't support the assertion that it's not cost-effective." How do you figure? Say some extra weight delta_w in additional equipment is required, and say that the fuel required to complete a given flight is a function f of the plane's weight, and that the cost of the fuel is C. Further, let the lifetime maintenance cost of the new equipment be E, and the lifetime maintenance of the plane's structural components be g, also a function of its weight. As a rough approximation, the additional cost should be E + C df/dw|_W * delta_w + dg/dw|_W * delta_w (Note that, really, f is a function of both passenger power consumption and of weight, both of which we're varying here, but I'll buy your argument that df/dp|_P * delta_p is neglible in comparison to f(P,W)) The article asserts that one or both of the derivates in this equation are large enough that its product with delta_w is non-negligible (in fact, prohibitive). I'm guessing that df/dp is not the problem, which is the only term you really addressed.
pavel_lishin 16y ago
Sure. But don't forget that adding a new feature to an airplane is going to take work. People have to figure out how all of this works, then other people have to test it, other people have to make sure it passes safety inspections, etc., etc., etc. Plus, then you have to retrofit old planes - or build new ones. It's more than just fuel costs.

chii 16y ago

a (good!?) method would be, to install small power generating bikes on the plane, and use that to suppliment the power usage of passengers. You market the bikes as exsercise machines for the long journey to prevent blood clots.