• Welcome to the new COTI server. We've moved the Citizens to a new server. Please let us know in the COTI Website issue forum if you find any problems.

Calculation needed...

F

Fox's Master

SOC-9
I know there are some big-brain math folks out here...

So, you have a spaceship which burned at max. 2G acceleration for a period of 30 minutes, then maintaining that velocity...If we take-off in a ship capable of 5G acceleration, how long do we need to burn at 5G acceleration, before decelerating (or as my old Physics professor preferred, "applied negative acceleration")? And moreover, when would we catch-up or more precisely maintain an exact velocity with the other ship for the purpose of boarding?

I appreciate any and all responses!
 
Fox's Master said:
I know there are some big-brain math folks out here...

So, you have a spaceship which burned at max. 2G acceleration for a period of 30 minutes, then maintaining that velocity...If we take-off in a ship capable of 5G acceleration, how long do we need to burn at 5G acceleration, before decelerating (or as my old Physics professor preferred, "applied negative acceleration")? And moreover, when would we catch-up or more precisely maintain an exact velocity with the other ship for the purpose of boarding?

I appreciate any and all responses!
Click to expand...

I may be a little rusty on this but it's late and I can't sleep.

frist assuming a starting velocity of 0 meters per second for both vehicals

after 30 minutes the 2G vehical (A) has a final velocity of 36,000 meters per second or 36 kilometers per second.

in those 30 minutes A has travelled 32,400 kilometers.

just to match velocities ship B with 5 g acceleration will take 720 seconds or 12 minutes.

in that time ship B will have travelled 12,960 Kilometers

so now both ships are travelling the same speed in the same direction but there still quite a distance betwen the two ships.

(32400-12960 = 19440) + (720*36 = 25,920) = 45360 kilometers seperation the two.

now fo our purposes to calculate time we just have to calculate how long it will take a ship accelerating at 5Gs to travelle 45360 KM with turn around at mid point, or at the 22,680KM mark.


22680000 M = 25T squared 0r 952.5 seconds to turn around point. or about 16 minutes to turn around.

so for ship A leaving with a 30 minute head start a 5g ship will take approximately 44 minutes from launch to catch and match velocites.

But I am willing to be corrected by some one else a little more current with thier physics.

R
 
Last edited:
Hmm, I'm a bit rusty on this, but I suspect there's a flaw in your calculation Rover in that you seem to assume both vehicles left at the same time, in which case there would be no chase...
...and you're doing it in two stages...
I dunno quite what's wrong, but I'm sure something is - maybe several somethings.

Looking at the OP, he hasn't specified how long the ship has been travelling at constant speed before the second ship gives chase. without that info, you can't solve the problem.

Even with that info, I think it's a simultaneous equation problem - but I could be wrong. Hell, just how rusty have I got?? Maybe I should work on this just for the exercise - I obviously need it! :eek:
Gimme the launch interval.
 
The way understood the problem is:

ship A leaves the starting point with an initial velocity of zero. Accelerates for 30 minutes at 2g.(20 meters per second squared) then stops accelerating but continues to travel at a constant speed.

ship B leaves the same starting point with the same initial velocity of zero 30 minutes after A.

Ship B wants to match velocities and position with A so it can board.

I broke the problem down like this:

what is the velocity, and distance travelled by A in the first 30 minutes? 36 kilometres per second and 32,400 kilometres.

Vfinal = Vinitial + a * t

V = 0 + 20 * 1800

Vfinal = 36000 meters per second.

Vaverage + (Vinitial + Vfinal) over 2

V average = 18000 kilometres per second.

Distance travelled = Vaverage*time

distance = 1800 * 180

Distance = 3,240,000 meters or 32400 kilometres


how long does it take ship B, leaving 30 minutes after ship A to first match the velocity of ship A? 12 minutes and it has travelled 12,960 kilometres. Ship a has travelled an addition 720 seconds at 36 kilometres per second or 25,920 kilometres

Vfinal = Vinitial> + a * t

36,000 = 0 + 50 * t

solve for t = 720 seconds.


Both A and B are now travelling along the same line and at the same speed only 45,360 kilometres apart.

At this point Ship A has been travelling for 42 minutes and ship B 12. Ship A and B both stationary relative to each other at this instant, and the velocity of ship A does not change. For calculation purposes we can now treat ship A as if it were stationary. We now have to calculate how long it will take ship B to get to same relative position as A. It accelerates at full thrust 5g for half the relative distance then turns around and decelerates (negative acceleration) so it will be at the same

Distance = 0.5*Acceleration * (time squared)

22,680,000 meters = (0.5*50 meters per second squared) * (time squared)

22,680,000 = 25 * time squared

907,200 = time squared

time = square root of 907,200

time = 952.5 seconds. (rounded to 1 place)

so it will take and additional (2 * 952.5 seconds) for ship B to match position and velocity of A.

That is 1905 seconds or 31.75 minutes (I rounded to 32)

So from the point ship B leaves it will take approximately 44 minutes (12+32)

Ship A will have been running for 74 minutes
 
Last edited:
Yep. Now I've sat down and thought, instead of skim-reading it, you are correct - IF your initial assumption holds that ship B starts out at the moment ship A stops accelerating.

That split not only seems valid, it's easier. You are obviously less rusty than me. ;)

Over to Fox's Master to confirm (or otherwise) the assumption.
 
Rover and Icosahedron,

Thank you ~ I was not very clear with some of the points. Yes, the second ship gave chase approximately 30 minutes after the first ship passed the space station. Your elaboration of the problem is exquisite...thank you very much!
 
It has been a while since I calculated anything like that. It was kind of fun. I don't think i could hav calculated an interception point if Ship A ha been under constant Acceleration.

Thanks and you're welcome.

R
 
Thank you very much for the original post and also for the very entertaining replies. I've not had to, as many of us I suspect, use that part of my brain in a very long time.

Of course instead of giving chase you could have simply press the GM remote destruct button and then let the fools roll up new characters. :devil:
 
One change to the initial assumption...the ship that we're trying to intercept is 1 million km from the station which we start our chase.
 
Fox's Master said:
One change to the initial assumption...the ship that we're trying to intercept is 1 million km from the station which we start our chase.
Click to expand...

So A and B are not starting from the same place?
Talking without engaging my brain properly again, I'd say that adds to the distance B has to cover during the 'A is relatively stationary' phase. Have a go at figuring that yourself using Rover's example.

Edit: Whoa!... I just looked upthread - you say Ahip A is passing a space station. Do you mean that Ship A's trajectory passes Ship B's launch point at a distance of 1Mkm, so that B has to travel at an angle to A's trajectory in order to catch up? (A is headed 'North' but is 1Mkm 'East' of B's launch point?) That's a whole different ball-game. I'm not sure I even want to imagine that one. If I scrape that much rust off my brain there may be nothing left! Any professional Astrogators here? Anyone work for NASA (and not as a janitor)?
 
Last edited:
Hey - what have you got against Janitors? :D (re: http://en.wikipedia.org/wiki/Good_Will_Hunting)

Well, never worked for NASA (but am related to those who have) nor have I ever been a Janitor (but am related to those who have), however...

Think he just means B has a million km head start... just add that distance to the second part of Rover's solution.

Handling direction (more than +/-) could be done readily enough by handling each x,y,z coordinate but would require vectors to be stated in the original problem. However, given the assumptions (pursuer's initial velocity of 0 and instantaneous acceleration changes in an inertial coordinate system sans gravity and other forces) this is simply a 2D problem. ;)
 
OK I don't think we need a Z coordinate as ship a and b are only two points and two points define a a line or ray.

if all other things are the same what i need is initial velocities for both ships A and B and the displacement between them. With that I can calculate time to intercept.

Jim
 
OK, if the only thing that is different is a million KM head start, then ship B will take 225 minutes from the time ship B launches to catch Ship A.
 
Rover said:
OK I don't think we need a Z coordinate as ship a and b are only two points and two points define a a line or ray.

if all other things are the same what i need is initial velocities for both ships A and B and the displacement between them. With that I can calculate time to intercept.

Jim
Click to expand...
Sure. The current problem only requires one coordinate, if I've read the posts correctly.

If the pursuer had an initial velocity one might need to deal with other coordinates, in order to compensate for initial direction. I.e. opposing acceleration in relevant components.

But, as is, its a simple 2D problem, er, correction :eek:... 1D problem ;)
 
Much thanks!

Rover, BytePro, and Icosahedron,

I greatly appreciate your patiece and your mathematical capabilities ~ I'm simply a lowly historian. People who possess a Masters Degree in International Relations short circuit calculators when they get to close to them.
Again, I can't thank you enough for your rapid assistance.
 
Fox's Master said:
Rover, BytePro, and Icosahedron,

I greatly appreciate your patiece and your mathematical capabilities ~ I'm simply a lowly historian. People who possess a Masters Degree in International Relations short circuit calculators when they get to close to them.
Again, I can't thank you enough for your rapid assistance.
Click to expand...

You are very welcome. It was no problem.

R
 
I've done nothing but ask a few questions and bitch about the effects of Ageing on my Int and Edu Characteristics. It's all Rover's work. :)

Anyway, if your Master's helps to short circuit wars, who cares about a few calculators? ;)
 
You must log in or register to reply here.

Similar threads

S
CT Only: Talking M-Drive Acceleration
2
Replies
30
Views
3K
Supplement Four
S
H
Update on my project and what it means
2
Replies
21
Views
2K
Hal
H
Back
Top