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Post by oprean on Oct 12, 2016 16:01:22 GMT
ok, I understand rail guns and coil guns can get outrageous efficiency's, but how do I check if they have sane efficiency? how do i do the math? is the kinetic damage mass of the projectile times velocity squared? where does time come in the formula? isn't it all supposed to be divided by time?
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Post by ross128 on Oct 12, 2016 16:22:42 GMT
It's based on power in vs power out.
Step 1: To calculate power out, first you find the kinetic energy of your bullet. This is (mv^2)/2, an easy way to remember it is that it's very similar to Einstein's E=mc^2 (c is the speed of light, v the speed of your bullet).
Step 2: Power is how fast you're using energy. So to get power, you divide your energy from step 1 (energy/bullet) by your reload time in seconds (seconds/bullet, so 10ms for example would be 0.01 seconds). The bullets cancel out, and you get energy/seconds (Watts).
Step 3: Laugh maniacally and enjoy your supergun. Or you can adjust it to make it realistic if you want by raising the power consumption, reducing the rate of fire, or reducing the energy of the projectile, whichever you prefer.
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Post by elouda on Oct 12, 2016 16:29:18 GMT
The kinetic energy per projectile is
KE = 0.5 x M x V x V
where
M is the projectile mass in kilograms (so 1g is 0.001kg)
V is the projectile velocity in m/s (so 5.4km/s is 5400m/s)
The power output of the gun is
P = KE / D
where D is the delay between shots in seconds (so 23ms is 0.023s)
As a worked example, take the stock 13MW 286mm;
KE = 0.5 x M x V x V
In this case;
M = 10kg
V = 5.14km/s = 5140m/s
Therefore;
KE = 0.5 x 10 x 5140 x 5140
KE = 132098000J = 132.098MJ
As for power;
P = KE / D
Here D = 125ms = 0.125s
Therefore;
P = 132.098 / 0.125
P = 1056.784MW = 1.056GW
As the gun has an input power of 13MW, this gives an efficiency of (1056.784/13) * 100 = 8129%
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Post by redparadize on Oct 12, 2016 17:15:02 GMT
8129%  Impossibru! |
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Post by RA2lover on Oct 12, 2016 18:21:49 GMT
note the linear motor force is only applied for as long as the projectile is accelerated. Efficiencies will still be off-the-charts regardless.
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Post by ross128 on Oct 12, 2016 20:10:50 GMT
My stance on superguns for now is to enjoy them while they last, then whenever they get fixed we'll see where that puts us and adjust our builds accordingly.
I do think post-fix we might want to ask for capacitor banks though. If it actually is trying to model reactor power being fed directly to the barrel, low-power EM guns are going to need REALLY long barrels to get any kind of impressive velocity.
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Post by morrigi on Oct 12, 2016 22:29:42 GMT
Yeah, I'd really like to see capacitors as well.
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Post by oprean on Oct 13, 2016 15:21:50 GMT
"As the gun has an input power of 13MW, this gives an efficiency of (1056.784/13) * 100 = 8129%"
Of course it has, why wouldn't it, thanks for explaining it like this to me, I was unsure what reactor output meant, if it was W/h of W/s, until the fix, a nice calculator to compute all this would be nice.
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Post by cuddlefish on Oct 13, 2016 16:07:35 GMT
"As the gun has an input power of 13MW, this gives an efficiency of (1056.784/13) * 100 = 8129%" Of course it has, why wouldn't it, thanks for explaining it like this to me, I was unsure what reactor output meant, if it was W/h of W/s, until the fix, a nice calculator to compute all this would be nice. Watts is actually a rate measure already (which then sometimes leads to terms like Watt-Hours to discuss specific quantities of energy - in that case, the amount you'd get at 1 watt over an hour), which takes a step out of it. But yeah, doing the googling to figure out what the different design stats meant has been surprisingly informative - and they said kids never learn things playing these newfangled vidyagames. |
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Post by ross128 on Oct 13, 2016 16:13:44 GMT
Ain't physics fun?  |
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Impossibru!