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Post by amimai on Dec 9, 2016 17:57:52 GMT
the main advantage of kinetic guns is the ability to flyby a target with the gunboats adding their velocity onto bullets creating silly velocities in the slugs, so barrel velocity does not mater as much.
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Post by randomletters on Dec 9, 2016 22:17:34 GMT
Has anyone managed a ~12 km/s non MPD thruster? The closest I've gotten is with HD resistojets.  |
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Post by newageofpower on Dec 10, 2016 0:58:39 GMT
I was bored.Got any high Thrust Decane/Methane resistojets? Somewhere between 10 to 200 GW is nice.
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Post by The Astronomer on Dec 10, 2016 2:02:45 GMT
The Astronomer : Yup, the top gun is about 300% efficient. (roughly 38.4MW of kinetic energy leaving the barrel when firing) I've stopped worrying about it because power is so cheap in CoaDE. Basically big EM gun ships will look more like doom laser ships once this gets fixed or they will fire slower. Bump the reload time to 24ms and you have a realistic and still very nasty gun. How do I calculate this? My first attempt yields just 2.8 W. |
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Post by lawson on Dec 10, 2016 3:30:14 GMT
The Astronomer : Yup, the top gun is about 300% efficient. (roughly 38.4MW of kinetic energy leaving the barrel when firing) I've stopped worrying about it because power is so cheap in CoaDE. Basically big EM gun ships will look more like doom laser ships once this gets fixed or they will fire slower. Bump the reload time to 24ms and you have a realistic and still very nasty gun. How do I calculate this? My first attempt yields just 2.8 W. Kinetic energy per shell is (1/2 m * V^2). when "m" is in kilograms and "V" is in m/s the energy is in Joules (J). Power or watts can also be expressed as Joules/second. So to get the kinetic energy leaving the barrel per second (aka power), divide the energy per shell by the reload time and you'll get the power output of the gun. Efficiency is then (power_output/power_input*100%) Note: it's highly unlikely the gun will be 90-100% efficient, 5-50% efficiency is a much more realistic target. |
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Post by The Astronomer on Dec 10, 2016 4:19:56 GMT
How do I calculate this? My first attempt yields just 2.8 W. Kinetic energy per shell is (1/2 m * V^2). when "m" is in kilograms and "V" is in m/s the energy is in Joules (J). Power or watts can also be expressed as Joules/second. So to get the kinetic energy leaving the barrel per second (aka power), divide the energy per shell by the reload time and you'll get the power output of the gun. Efficiency is then (power_output/power_input*100%) Note: it's highly unlikely the gun will be 90-100% efficient, 5-50% efficiency is a much more realistic target. Doing so, I got 2.8 W out of 36 MW... |
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Post by newageofpower on Dec 10, 2016 4:39:37 GMT
Kinetic energy per shell is (1/2 m * V^2). when "m" is in kilograms and "V" is in m/s the energy is in Joules (J). Power or watts can also be expressed as Joules/second. So to get the kinetic energy leaving the barrel per second (aka power), divide the energy per shell by the reload time and you'll get the power output of the gun. Efficiency is then (power_output/power_input*100%) Note: it's highly unlikely the gun will be 90-100% efficient, 5-50% efficiency is a much more realistic target. Theoretically, linear motors (which electromagnetic guns are) can exceed 99.9% efficiency. Coilguns in particular are basically straightened out brushless electric motors in weaponized format, and real life electric motors regularly exceed 95% efficiency. Of course, the stress of performing roles in industrial machinery or transportation tend to be much lower than in weapons systems, so it's almost certain we'll see a loss in efficiency, but I wouldn't be surprised if we still had 80-90% efficient weapons. Due to switching speed limitations in conventional coilguns, I believe their maximal fire rates will be far lower than currently possible in game. |
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Post by dragonkid11 on Dec 10, 2016 4:42:33 GMT
Kinetic energy per shell is (1/2 m * V^2). when "m" is in kilograms and "V" is in m/s the energy is in Joules (J). Power or watts can also be expressed as Joules/second. So to get the kinetic energy leaving the barrel per second (aka power), divide the energy per shell by the reload time and you'll get the power output of the gun. Efficiency is then (power_output/power_input*100%) Note: it's highly unlikely the gun will be 90-100% efficient, 5-50% efficiency is a much more realistic target. Doing so, I got 2.8 W out of 36 MW... Just use google and get your a webpage to calculate the joule for you. www.csgnetwork.com/kineticenergycalc.html |
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Post by The Astronomer on Dec 10, 2016 4:48:21 GMT
Surprisingly, I power'd mass instead of velocity... Thanks, anyways. |
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Post by lawson on Dec 10, 2016 5:51:51 GMT
Theoretically, linear motors (which electromagnetic guns are) can exceed 99.9% efficiency. Coilguns in particular are basically straightened out brushless electric motors in weaponized format, and real life electric motors regularly exceed 95% efficiency. Of course, the stress of performing roles in industrial machinery or transportation tend to be much lower than in weapons systems, so it's almost certain we'll see a loss in efficiency, but I wouldn't be surprised if we still had 80-90% efficient weapons. Due to switching speed limitations in conventional coilguns, I believe their maximal fire rates will be far lower than currently possible in game. Theoretically, electric motors can be over 99% efficient. Unfortunately the real world isn't that nice. Wires and iron poles have eddy current losses, magnetic steels have hysteresis loops, power electronics and supply wires have resistance, stray rotating fields cause eddy currents in surrounding metals, adjacent coilgun coils couple like transformers, etc. Wikipedia and other sources mention 5-50% efficiency for current coilgun prototypes, especially when they launch to >1km/s. |
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Post by newageofpower on Dec 10, 2016 6:30:59 GMT
Theoretically, linear motors (which electromagnetic guns are) can exceed 99.9% efficiency. Coilguns in particular are basically straightened out brushless electric motors in weaponized format, and real life electric motors regularly exceed 95% efficiency. Of course, the stress of performing roles in industrial machinery or transportation tend to be much lower than in weapons systems, so it's almost certain we'll see a loss in efficiency, but I wouldn't be surprised if we still had 80-90% efficient weapons. Due to switching speed limitations in conventional coilguns, I believe their maximal fire rates will be far lower than currently possible in game. Theoretically, electric motors can be over 99% efficient. Unfortunately the real world isn't that nice. Wires and iron poles have eddy current losses, magnetic steels have hysteresis loops, power electronics and supply wires have resistance, stray rotating fields cause eddy currents in surrounding metals, adjacent coilgun coils couple like transformers, etc. Wikipedia and other sources mention 5-50% efficiency for current coilgun prototypes, especially when they launch to >1km/s. Did you even read my post? In the real world, many electric motors operate at 95% (or more!) efficicency. Boosting our 5-50% linear motor weapon systems to 80+ is an optimization problem, not an insurmountable challenge. The first electric motors too, had lower efficiencies. |
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Post by The Astronomer on Dec 10, 2016 15:54:55 GMT
Theoretically, electric motors can be over 99% efficient. Unfortunately the real world isn't that nice. Wires and iron poles have eddy current losses, magnetic steels have hysteresis loops, power electronics and supply wires have resistance, stray rotating fields cause eddy currents in surrounding metals, adjacent coilgun coils couple like transformers, etc. Wikipedia and other sources mention 5-50% efficiency for current coilgun prototypes, especially when they launch to >1km/s. Did you even read my post? In the real world, many electric motors operate at 95% (or more!) efficicency. Boosting our 5-50% linear motor weapon systems to 80+ is an optimization problem, not an insurmountable challenge. The first electric motors too, had lower efficiencies. Speaking of efficiency, I have adjusted my railguns so they have reasonable firing rate and save some power. |
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Post by thorneel on Dec 11, 2016 0:43:33 GMT
I was bored.Got any high Thrust Decane/Methane resistojets? Somewhere between 10 to 200 GW is nice. Voici quelques 10 GW resistojets for decane, methane and neon that I just threw together. They may be slightly more optimised, and come without turrets because reaction wheels always consumed more power than the coils themselves to get any decent turn rate. Note: I have not found any positive effect to changing the length of the chamber or the radius of the coil, so there may be optimisation paths there that I missed.    (For neon, regenerative cooling had no effect either way, so I got rid of it to keep the thing smoother. Still, I am surprised there isn't at least a mass cost to the circuit.) |
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Post by newageofpower on Dec 11, 2016 4:02:15 GMT
Thank you very much.
Query: Is it possible to build a good (non-MPD) Mercury thruster?
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Post by jasonvance on Dec 11, 2016 4:57:10 GMT
Thank you very much. Query: Is it possible to build a good (non-MPD) Mercury thruster? If you are just looking for something to complement an MPD for quick maneuvering thrust something like this might work:  |
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