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Guest, the last post of this topic is over 60 days old and at this point you may not reply in this topic. If you wish to continue this conversation start a new topic.
Posted 09 May 2011 - 06:01 PM
Posted 09 May 2011 - 06:11 PM
Posted 09 May 2011 - 06:26 PM
10 / ( Root( 1-( (1/2)^2 / 1) ) ) = 11,5470
Posted 09 May 2011 - 06:33 PM
To keep the calculations simple, say that a traveler had a spaceship that could instantly accelerate to 1/2 of light speed (without killing him) and he started a journey to a point in space 10 light years away to finally instantly decelerate at the moment of arrival.
How many time would it pass from the point point of view of the traveler inside the spaceship?
Intuitive thinking may make one think it'd be 20 years but I know relativistic speeds don't work that way. What would be the equations to calculate that "t"?
Posted 09 May 2011 - 08:05 PM
Posted 10 May 2011 - 12:33 AM
Posted 10 May 2011 - 01:04 AM
Posted 10 May 2011 - 01:11 AM
Posted 10 May 2011 - 01:22 AM
20 * ( Root( 1-( (1/2)^2 / 1) ) ) = 17.32
or
time_for_traveler = time_for_observer * sqroot(1-(v/c)^2)
just pretending I'm not having sense of humor (which I'm really not having..)
10*(Root(1-(293800^2/299792^2))) = 1,9893 years (he states 1 week)
Posted 10 May 2011 - 01:24 AM
Posted 10 May 2011 - 01:25 AM
20 * ( Root( 1-( (1/2)^2 / 1) ) ) = 17.32
or
time_for_traveler = time_for_observer * sqroot(1-(v/c)^2)
just pretending I'm not having sense of humor (which I'm really not having..)
But, but, that doesn't very for the numbers provided by Asimov:10*(Root(1-(293800^2/299792^2))) = 1,9893 years (he states 1 week)
Posted 10 May 2011 - 01:32 AM
20 * ( Root( 1-( (1/2)^2 / 1) ) ) = 17.32
or
time_for_traveler = time_for_observer * sqroot(1-(v/c)^2)
just pretending I'm not having sense of humor (which I'm really not having..)
But, but, that doesn't very for the numbers provided by Asimov:10*(Root(1-(293800^2/299792^2))) = 1,9893 years (he states 1 week)
Well, you said 1/5 time at first.... What is the context in Asimov's?
Posted 10 May 2011 - 01:42 AM
299791/299792*10*(Root(1-( 299791^2/299792^2))) = 0,02583 years
which is 0.02583 1.3 weeks...
In a lapse that would appear of 60 years to them (the travelers) they'd reach the Galaxy of Andromeda, which is located at 2,300,000 light years from us.
Posted 10 May 2011 - 12:47 PM
A more interesting question:
How would we relativistically simulate a space-fighting game with all these time dilatations in a multiplayer game? All controls/the behavior of the spaceship would slow down compared to the outside scene? That is an interesting game design question....
Posted 11 May 2011 - 01:15 AM
Posted 11 May 2011 - 03:31 AM
Posted 11 May 2011 - 03:43 AM
in other words your first calculation was correct.
Posted 11 May 2011 - 03:49 AM
in other words your first calculation was correct.
Well, no, because lightyears as distance is defined in the Earth's coordinate system (not exactly, but you know what I mean). So 10 years means 10 years in cE not cT.
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