morty_shworts
SOC-6
How many EPs does it take to jump?
Page 354 of the T20 Book gives the following step by step of a jump.
The Procedure for entering Jump is as follows:
1. Jump Plot is generated.
2. Powerplant brought to Jump Readiness.
3. Jump Plot is fed to the navigational computer.
4. Jump Drive is brought to Readiness.
5. Jump Drive begins to fast-burn its Jump Fuel, developing high levels of power.
(how many EP's would that be?)
6. Jump Grid is charged. (how many EP does that take?)
7. Astrogator or Captain gives final Jump command.
8. Jump Field is formed.
9. Vessel enters Jumpspace.
Its steps 5 and 6 that I'm wondering about. How many EP's are produced when the jump drive consumes its fuel? A fusion reactor produces 1 EP per round for 241,920 round (4 weeks) on 1 ton of fuel. Going with this as a base that would mean that a standard free trader's J-drive would produce 5,322,240 EP in a mater of seconds from its 22 tons of dedicated fuel. That can't be right, as page 268 of the T20 book states “A 1-ton capacitor (in a Jump drive or not) can store up to 36 Energy Points”, and it also says a jump drive has capacitors equal in size to (0.5% x Jump USP Rating) x Hull Tonnage. So our free trader's jump drive capacitors could only store 36 EP. In fact it would take 147,840 tons of capacitors to hold all 5,322,240 EP.
The T20 book also says on page 354 “This fuel is used up at an immense rate in a powerful, but inefficient process.” Exactly how inefficient could that be to eat up 5,322,240 EP in just a few seconds. These numbers just don't add up.
I think that it would be more reasonable to say that a jump drive needs EP equal to the maximum EP rating of its jump capacitors to enter jump space, and EP after that equal to what the jump drive needs per round to maintain the jump field.
With this as the bases for what I was pondering over, would it be fare to say that a ship could be designed to use capacitors charged by its reactor to make jump. Going back to my example of the standard free trader, it would require 13 rounds to charge the ships capacitors and then the ship could jump. Or would it be more fare to say a ship so rigged would need x10 times the capacitors to jump using this system. Thus it would take 130 rounds or 13 minutes to charge its jump drive before it could jump.
Page 354 of the T20 Book gives the following step by step of a jump.
The Procedure for entering Jump is as follows:
1. Jump Plot is generated.
2. Powerplant brought to Jump Readiness.
3. Jump Plot is fed to the navigational computer.
4. Jump Drive is brought to Readiness.
5. Jump Drive begins to fast-burn its Jump Fuel, developing high levels of power.
(how many EP's would that be?)
6. Jump Grid is charged. (how many EP does that take?)
7. Astrogator or Captain gives final Jump command.
8. Jump Field is formed.
9. Vessel enters Jumpspace.
Its steps 5 and 6 that I'm wondering about. How many EP's are produced when the jump drive consumes its fuel? A fusion reactor produces 1 EP per round for 241,920 round (4 weeks) on 1 ton of fuel. Going with this as a base that would mean that a standard free trader's J-drive would produce 5,322,240 EP in a mater of seconds from its 22 tons of dedicated fuel. That can't be right, as page 268 of the T20 book states “A 1-ton capacitor (in a Jump drive or not) can store up to 36 Energy Points”, and it also says a jump drive has capacitors equal in size to (0.5% x Jump USP Rating) x Hull Tonnage. So our free trader's jump drive capacitors could only store 36 EP. In fact it would take 147,840 tons of capacitors to hold all 5,322,240 EP.
The T20 book also says on page 354 “This fuel is used up at an immense rate in a powerful, but inefficient process.” Exactly how inefficient could that be to eat up 5,322,240 EP in just a few seconds. These numbers just don't add up.
I think that it would be more reasonable to say that a jump drive needs EP equal to the maximum EP rating of its jump capacitors to enter jump space, and EP after that equal to what the jump drive needs per round to maintain the jump field.
With this as the bases for what I was pondering over, would it be fare to say that a ship could be designed to use capacitors charged by its reactor to make jump. Going back to my example of the standard free trader, it would require 13 rounds to charge the ships capacitors and then the ship could jump. Or would it be more fare to say a ship so rigged would need x10 times the capacitors to jump using this system. Thus it would take 130 rounds or 13 minutes to charge its jump drive before it could jump.