|
|
Post by randommechanicumguy on Jun 10, 2017 17:42:45 GMT
 An interplanetary liner, which is also my favorite one. With that pathetic delta-v of just 2260 km/s, t's not an interstellar spacecraft. It can carry 2 kt of cargo and its crews to any planet in the Solar System.  code or it did'nt happen |
|
|
|
Post by Kerr on Jun 10, 2017 17:55:11 GMT
The radiators are so thin that it is literally impossible to hit them. I even tried hitting them at 10km with radiators targetted. lol If they were oriented perpendicular to the fire, though, you're dead. And how could that occur? If a enemy is roughly 50Mm away I will just orient my ship relative to the fire. And even If it would hit, how much damage would it even do? The radiators are so thin that projectiles just pass through and leave a hole the size of their cross-section. Lasers? At the 50Mm mark the most lasers would make the radiator glow red, that is. And every ship would be dead if their radiators were destroyed. |
|
|
|
Post by bigbombr on Jun 10, 2017 19:32:28 GMT
lol If they were oriented perpendicular to the fire, though, you're dead. And how could that occur? If a enemy is roughly 50Mm away I will just orient my ship relative to the fire. And even If it would hit, how much damage would it even do? The radiators are so thin that projectiles just pass through and leave a hole the size of their cross-section. Lasers? At the 50Mm mark the most lasers would make the radiator glow red, that is. And every ship would be dead if their radiators were destroyed. I've also found lasers to be largely ineffective against radiators. IMO, lasers would be mainly effective against sensors, followed by other lasers and fuel tanks. |
|
|
|
Post by Kerr on Jun 10, 2017 19:39:34 GMT
And how could that occur? If a enemy is roughly 50Mm away I will just orient my ship relative to the fire. And even If it would hit, how much damage would it even do? The radiators are so thin that projectiles just pass through and leave a hole the size of their cross-section. Lasers? At the 50Mm mark the most lasers would make the radiator glow red, that is. And every ship would be dead if their radiators were destroyed. I've also found lasers to be largely ineffective against radiators. IMO, lasers would be mainly effective against sensors, followed by other lasers and fuel tanks. Especially when these radiators are supposed to survive the x-ray flux of a nuclear bomb for months. |
|
|
|
Post by The Astronomer on Jun 11, 2017 2:37:06 GMT
Wait a minute. I've already subtracted the neutron/gamma ray from my current rocket thrust power. Does that mean I have to divide it by 0.75 then multiply it by 0.25 to get the actual heat power?
|
|
|
|
Post by samchiu2000 on Jun 11, 2017 3:02:50 GMT
Wait a minute. I've already subtracted the neutron/gamma ray from my current rocket thrust power. Does that mean I have to divide it by 0.75 then multiply it by 0.25 to get the actual heat power? Yea actually how is the heat calculated Kerr ? |
|
|
|
Post by The Astronomer on Jun 11, 2017 3:03:30 GMT
 The rocket engines should be at the rear end of the main spacecraft body, dragging the radiator tail along, but then there's the problem about turning, so I added 50 t H2 tank with 4 gas core NTRs. Now I just have to hope that the tail (20 cm in diameter, overall length is 750 m) won't break due to centrifuge force. |
|
|
|
Post by Enderminion on Jun 11, 2017 4:33:12 GMT
does it count as a fusion ship if I use thermonukes in a orion drive?
|
|
|
|
Post by The Astronomer on Jun 11, 2017 5:02:42 GMT
does it count as a fusion ship if I use thermonukes in a orion drive? Probably no, if the main reaction is fission, it's fission drive. If you use fusion bombs, it's fusion. Either way, it's a nuclear pulse drive. Qswitch should add nuclear pulse drive as a module  |
|
|
|
Post by Kerr on Jun 11, 2017 6:02:29 GMT
The Astronomer samchiu2000 You take your Thrust power, for example 1TW, then you multiply this value by the percentage of non-charged particles. D-T: 80% Neutrons, 20% Thermal (80% heat) D-D: 12% Thermal, 38% Neutrons, 50% X-rays. (88%) D-He³: 75% Thermal, 5% Neutrons, 20% X-rays (25%) p-B11: 100% Thermal. (0%) Bremsstrahlung isn't calculated in already. Thermal: Helium-4 Ions and protons. They can be directed using a magnetic nozzle. So they don't contribute in heating.
|
|
|
|
Post by samchiu2000 on Jun 11, 2017 8:38:10 GMT
The Astronomer samchiu2000 You take your Thrust power, for example 1TW, then you multiply this value by the percentage of non-charged particles. D-T: 80% Neutrons, 20% Thermal (80% heat) D-D: 12% Thermal, 38% Neutrons, 50% X-rays. (88%) D-He³: 75% Thermal, 5% Neutrons, 20% X-rays (25%) p-B11: 100% Thermal. (0%) Bremsstrahlung isn't calculated in already. Thermal: Helium-4 Ions and protons. They can be directed using a magnetic nozzle. So they don't contribute in heating. I mean how you reduce heat like 65.5 TW to 1.5.. |
|
|
|
Post by Kerr on Jun 11, 2017 8:41:10 GMT
The Astronomer samchiu2000 You take your Thrust power, for example 1TW, then you multiply this value by the percentage of non-charged particles. D-T: 80% Neutrons, 20% Thermal (80% heat) D-D: 12% Thermal, 38% Neutrons, 50% X-rays. (88%) D-He³: 75% Thermal, 5% Neutrons, 20% X-rays (25%) p-B11: 100% Thermal. (0%) Bremsstrahlung isn't calculated in already. Thermal: Helium-4 Ions and protons. They can be directed using a magnetic nozzle. So they don't contribute in heating. I mean how you reduce heat like 65.5 TW to 1.5.. Inverse square law, it uses the firefly design, which means no engine bell which absorbs heat, 95% most radiates into space, the 1.5TW rest is just the bottom fuel tanks and the very small radiator area. |
|
|
|
Post by The Astronomer on Jun 11, 2017 9:20:08 GMT
I mean how you reduce heat like 65.5 TW to 1.5.. Inverse square law, it uses the firefly design, which means no engine bell which absorbs heat, 95% most radiates into space, the 1.5TW rest is just the bottom fuel tanks and the very small radiator area. I DONT GEDDIT Math and in-depth explanation please |
|
|
|
Post by Kerr on Jun 11, 2017 9:24:55 GMT
Inverse square law, it uses the firefly design, which means no engine bell which absorbs heat, 95% most radiates into space, the 1.5TW rest is just the bottom fuel tanks and the very small radiator area. I DONT GEDDIT It doesn't have a engine made out of matter, the fusion fuel is ignited 10m away from the ship using Z-pinch, neutrons and x-ray can just escape the core into space, the exhaust plume is being formed through very thin superconducting coils, it's open-cycle. |
|
|
|
Post by Kerr on Jun 11, 2017 9:38:48 GMT
Inverse square law, it uses the firefly design, which means no engine bell which absorbs heat, 95% most radiates into space, the 1.5TW rest is just the bottom fuel tanks and the very small radiator area. I DONT GEDDIT Math and in-depth explanation please Okay, How I get that heat value was. Getting the heat of the engine, 65,5TW. Then I used my deuterium fuel radius to find out the area, the fusion will be ignited 10m away from the fuel tanks. 4*pi*10^2 = 1256.6, 65.5TW/1256.6 = 52.1GW/m² pi*2.5^2 = 19.6m². 19.6x52.1 = 1.023TW. I added 50% as safety range. |
|
