Is anybody aware of or have a calculator that will determine the time to intercept with matched velocities? I.e. A derelict vessel drifting at xG constant velocity needs to be intercepted by a rescue craft. How long will it take a rescue craft to reach the derelict and match velocity if the starting distance between the objects is D and the rescue craft is capable of aG acceleration. Assume no outside gravity affects, i.e. planets, stars, etc. If it calculated turn over (when the chase ship starts to decelerate) that would be a nice touch.
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- Thread starter D. Foxx
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Ben W Bell
SOC-14 1K
Are you sure your question is what you mean? A derelict vessel that is still under power and accelerating? A derelict shouldn't be under acceleration other than by planetary and stellar bodies exerting an external influence upon it.
xG's is a measure of acceleration, that is, change in velocity over time, not velocity itself.
THe G's of a derelict are ALWAYS 0G...
5x6min turns = 3600x5x10x2 m/s: 360km/s.
Icosahedron
SOC-14 1K
Aside from nitpicking (your request was clear enough to me), You may need to figure angles, too.
If both ships are travelling in the same direction (the pursuer coming up directly behind the derelict) it should be fairly straightforward to calculate (though I don't have a pre-gen calculator and I don't have the time to figure it out just now).
However, if the derelict is crossing your bows at an angle and you need to not only catch up with its speed, but also change direction, that would be much harder to calculate. Off the top of my head, you could be looking at calculus or iterative procedures to figure that out. It's too early in the morning to even think about what would be necessary, sorry.
If both ships are travelling in the same direction (the pursuer coming up directly behind the derelict) it should be fairly straightforward to calculate (though I don't have a pre-gen calculator and I don't have the time to figure it out just now).
However, if the derelict is crossing your bows at an angle and you need to not only catch up with its speed, but also change direction, that would be much harder to calculate. Off the top of my head, you could be looking at calculus or iterative procedures to figure that out. It's too early in the morning to even think about what would be necessary, sorry.
Break up the problem,
First problem is you need to match vectors, second problem is you need to match position, third problem is how much time is available in the scenario, and forth, how much fuel the rescue vehicle has available (is there fuel aboard the derelict?) The third and fourth items become moot and uninteresting with a reactionless drive, you just use a least time course everwhere.
Needs 10 g hours to match vector in this scenario, that could be 2 hours for a 5g rescue craft, at which time the other ship gained another 10 units in distance to go with the (1+3+5+7+9)=25 that she already had.
Next you need to cover 35 units of distance in a time dictated by the avaliable fuel (if not using reactionless drives) the close the distance could take 26 hours if only 2 g hours are used 13 1/2 hours if 4 g hours are used and so on till you get to the least time/most fuel of constant acceleration with a turn over at the middle, or 5 hours at 5 g's.
So a 5 g craft could effect rescue in 7 hours starting from the location the derelict started from , needing 45 g hours of fuel to come to a stop relative to the craft's base some 35+ (10*the number of hours the rescue procedures take.) distance units away. Any fuel remaining at this point can be used to return to base, the more fuel left, the quicker the return may be made.
If the intent is to rescue the ship and tow it back, that's a different problem.
First problem is you need to match vectors, second problem is you need to match position, third problem is how much time is available in the scenario, and forth, how much fuel the rescue vehicle has available (is there fuel aboard the derelict?) The third and fourth items become moot and uninteresting with a reactionless drive, you just use a least time course everwhere.
Needs 10 g hours to match vector in this scenario, that could be 2 hours for a 5g rescue craft, at which time the other ship gained another 10 units in distance to go with the (1+3+5+7+9)=25 that she already had.
Next you need to cover 35 units of distance in a time dictated by the avaliable fuel (if not using reactionless drives) the close the distance could take 26 hours if only 2 g hours are used 13 1/2 hours if 4 g hours are used and so on till you get to the least time/most fuel of constant acceleration with a turn over at the middle, or 5 hours at 5 g's.
So a 5 g craft could effect rescue in 7 hours starting from the location the derelict started from , needing 45 g hours of fuel to come to a stop relative to the craft's base some 35+ (10*the number of hours the rescue procedures take.) distance units away. Any fuel remaining at this point can be used to return to base, the more fuel left, the quicker the return may be made.
If the intent is to rescue the ship and tow it back, that's a different problem.
