AlexanderM said:
Yes, 100% agree. To me, the difference is that the rail guns require so much power that they aren't practical, that's why I was trying to come up with an alternate power source, and was dead wrong to think of fuel cells.
However, I suspect that the lasers won't require anywhere near the power of the rail guns. And, if the Germans can power that laser system on a mobile unit smaller than a tank, then the power requirements can't be that bad.
So, I'm thinking we need a number of gas turbine generators, and two or three banks of supercapacitors per laser, so that while one bank is firing the other banks can be charging. I expect these would charge much faster than the ones required for the rail guns. This equipment wouldn't necessarily be physically huge, but it would likely require one good sized power room, to power lasers fore and aft.
Have you actually run the numbers or are you taking generic quotes out of articles, and damning turbines (or other power sources) with generalities?
Let's dust off our electrical engineering textbooks and do that (address specifics) for a moment.
Current rail gun max energies: depending on the builder, 35-50 MJ (megajoules) or about one tenth the average power of a lightening strike (500 MJ).
Current power output of a GE LM2500+G4 gas turbine: 35 MW (megawatts) or about 48,000 shp.
Power is the rate at which work energy is created/expended: P = w / t (power = work / time). In SI Units the P=w/t formula is represented as: 1 Watt = 1 Joule per second ( 1 W = 1 J/s). Conversely, work energy equals power output times duration of the output - again in SI units, 1 Joule = 1 Watt•second)
Using rail gun energy and power source output, we will solve for time to determine how long the power source would have to work to charge up whatever storage system (capacitor/compensating alternator) the rail gun was using.
Power (of the LM2500) 35 MW = Energy required of the rail gun (let's use) 50 MJ ÷ time, t (sec)
35 x 106 W = 50 x 106 J ÷ t
t = 50 x 106 J (or W•s) ÷ 35 x 106 W
(Watts on the right side cancel out leaving just seconds)
t = 50/35 = 1.43s
So, specifics of the electrical storage medium notwithstanding, it would take just 1.43 seconds of an LM2500+G4's output to produce enough electrical energy to supply a 50 MJ rail gun. All other parts of the system appropriately designed, that means an LM2500 could be used to power a rail gun system with a cyclic rate of fire of 42 rounds per minute (60 sec/min ÷ 1.43 sec/rd).
The article you quoted used a lot of generalizations. Yes, power generation is a notable part of the system, but more so is its storage and other aspects of the weapon system. The real challenge will be in developing the storage system that can then expend huge amounts of potential electromechanical power AND do so while withstanding the huge forces (megaNewtons) during firing and rails not being ablated after just a few firings.
The whole issue comes down to: operationalization of the energy production/expenditure system in a reliable form factor that is smaller/more manageable than the current chemical/physical system whilst producing enhanced operationally employable lethality.
I would make that case that physical energy production (from say an LM2500 gas turbine) into the such a system is not the long pole in the tent.
Regards
G2G