snrdg082102
SOC-14 1K
Howdy Mike Wightman,
First my apologies if I have aready made a comment on this, the fun time I had yesterday put me behind looking at the my email for new posts. The big kicker is that clicking on the associated link in the email isn't taking me to the post. I'm going to the newest one, which means I have to scroll back to the correct one.
Okay, back to the post.
My interpretation, as probably mentioned once or twice, is that when a ship's captain in combat decides to break off the enegagement by jumping needs to provide twice the required EP needed by the jump drive a routine insertion from normal space to jump space.
In order to determine the EP required for a normal insertion the formula is 0.01 x M x Jn when one has the luxury of doing the calculations while transiting to the 100 D limit and the jump point.
Calculating a jump to break off combat does not permit the luxury of prolonged calculations. The navigator has to do a down dirty calculation which the powers that be say needs to be 2x the required EP of a normal insertion into jump space which alters the formula to 2 x 0.01 EP = 0.02 EP x M x Jn.
In the Standard Ship's Operating Procedures (SSOP) the normal jump space insertion formula is Required Jump EP = 0.01 x M x Jn and the emergency, which probably lists what conditions consitutes an emergency, jump space insertion formula is EP = 0.02 x M x Jn.
So they are both correct depending on the conditions.
The EP required to jump out of combat is different since the time frame to refine the calculation for a routine jump is not there. The navigator has to run the calculation with whatever data, not as precise, in 20 to 40 minutes under fire.
I'll agree capacitors charge and discharge rates are under 20 minutes, however a majority of the 20 minutes appears to making the preparations to make the jump.
From page 42: "If a black globe absorb's energy and the ship's capacitors are already full, the ship is destroyed."
Here is how I see the rule: The capacitors installed in the jump drive can store 3,600 EP and requires 300 EP for a jump. The ship has a black globe and is preparing to break of by jumping. The capacitors have absorbed 3,100 EP prior to the jump. Dumping 2 x required EP pushes the capacitors from 3,100 + (2 x 300) = 3,100 + 600 = 3,700 EP which exceeds the storage capacity destroying the ship by gutting engineering at a minimum.
IIRC there are pictures showing the wreckage of the USS Thresher strewn on the sea floor after going past crush depth. This is similar to how I see when the jump capacitors go off when exceeding the maximum EP rating.
First my apologies if I have aready made a comment on this, the fun time I had yesterday put me behind looking at the my email for new posts. The big kicker is that clicking on the associated link in the email isn't taking me to the post. I'm going to the newest one, which means I have to scroll back to the correct one.
Okay, back to the post.
HG is pretty clear - 2 turns output from a power plant rated equal to the jump drive - or at least the jump being performed
It then goes on to put a formula in parenthesis that contradicts this
2% of the hull tonnage per jump number worth of EP to initiate the jump according to the text, 1% of the hull tonnage per jump number worth of EP according to the formula.
They both can't be right.
My interpretation, as probably mentioned once or twice, is that when a ship's captain in combat decides to break off the enegagement by jumping needs to provide twice the required EP needed by the jump drive a routine insertion from normal space to jump space.
In order to determine the EP required for a normal insertion the formula is 0.01 x M x Jn when one has the luxury of doing the calculations while transiting to the 100 D limit and the jump point.
Calculating a jump to break off combat does not permit the luxury of prolonged calculations. The navigator has to do a down dirty calculation which the powers that be say needs to be 2x the required EP of a normal insertion into jump space which alters the formula to 2 x 0.01 EP = 0.02 EP x M x Jn.
In the Standard Ship's Operating Procedures (SSOP) the normal jump space insertion formula is Required Jump EP = 0.01 x M x Jn and the emergency, which probably lists what conditions consitutes an emergency, jump space insertion formula is EP = 0.02 x M x Jn.
So they are both correct depending on the conditions.
The example that follows clarifies things it is the text that is correct and the formula that is wrong since a power plant 8 jump 5 takes 2 turns.
The formula should read EP required = 0.02MJn
Plus you need to then burn the fuel for the jump.
The EP required to jump out of combat is different since the time frame to refine the calculation for a routine jump is not there. The navigator has to run the calculation with whatever data, not as precise, in 20 to 40 minutes under fire.
Marc's jumpspace article and the rules for drop tanks also show that the capacitors can discharge in less than a 20 minute combat turn.
I'll agree capacitors charge and discharge rates are under 20 minutes, however a majority of the 20 minutes appears to making the preparations to make the jump.
And filling your capacitors does not destroy the ship - the rules state that you have to take more damage on top of having full capacitors![]()
From page 42: "If a black globe absorb's energy and the ship's capacitors are already full, the ship is destroyed."
Here is how I see the rule: The capacitors installed in the jump drive can store 3,600 EP and requires 300 EP for a jump. The ship has a black globe and is preparing to break of by jumping. The capacitors have absorbed 3,100 EP prior to the jump. Dumping 2 x required EP pushes the capacitors from 3,100 + (2 x 300) = 3,100 + 600 = 3,700 EP which exceeds the storage capacity destroying the ship by gutting engineering at a minimum.
IIRC there are pictures showing the wreckage of the USS Thresher strewn on the sea floor after going past crush depth. This is similar to how I see when the jump capacitors go off when exceeding the maximum EP rating.