Posted by Wreckage on 2016-04-14 18:26:55

I always wanted to recalculate this stuff based on how long it takes for the traveller. Can you tell me how it is done?

Posted by Verminardo on 2016-04-14 18:29:55

Personally I prefer travelling by farcaster.

Posted by Reisender on 2016-04-14 18:58:40

36.9 years on fumbbl!? congratz

Posted by xnoelx on 2016-04-14 19:13:40

Wreckage: it's all on the Wikipedia page for Time Dilation. (Other websites are available)

Posted by KhorneliusPraxx on 2016-04-14 19:14:59

36.9 years is not the oldest FUMBBL coach.

Posted by Wreckage on 2016-04-14 19:30:53

A shut up xnoelx, I'm still hoping for an answer, awambawamb.

Posted by pythrr on 2016-04-14 19:47:38

is this a quitting blog?

Posted by OTS on 2016-04-14 21:01:21

High blog.

Posted by awambawamb on 2016-04-14 22:15:33

well, I'll be quick and absolutely not educational, but the practical passages are:
- you need γ[gamma] to solve Lorentz transform, so you can start from there. we know that γ=1/ sqrt[1-(v/c)^2] . gamma is what makes Galileo's and Maxwell's going together, since it's a value that is veeeeery little different for low speeds but goes into the flurbzillions the sooner you get close to c (lightspeed)
- next, we want to solve it for the time, and back to Lorentz Transform the stuff to find t1 (the time on the spaceship) is t1=γ*[t+(vx/c^2)].
remember we are going at 0.99999c, and since the spaceship is our frame of reference, doesn't change. indeed if we took the Earth's frame of reference, x is the distance from Earth to the Galactic Centre and the time is the distance to our target divided by our speed.
Plug in the numbers and play with relativity

Posted by Uedder on 2016-04-14 22:51:56

So when will the space traveller quit FUMBBL and why?

How long will it take for him to play a game against Dominik? Will he reach the Galactic Centre before Second Half begins?

Posted by awambawamb on 2016-04-15 09:52:12

errata:
*remember we are going at 0.99999c, and since the spaceship is our frame of reference, x doesn't change.

edit:
x is the "distance" in our frame of reference (xyz coordinates, for simplicity we orient them so that our travel will take place on the x axis), v is the speed we're going.

Posted by Wreckage on 2016-04-15 11:14:36

just a couple of thoughts:
what i have to calculate now is:
t1 = 1/[1-(v/c)^2]^2 * [t+(vx/c^2)] ?

And x is the distance?

And t is... the time passing on earth? Which I then also have to figure out?

And there is some confusion about the speed I think...

0.99999c would equate to 99.999% not 99.99999% which i imagine would be a significantly slower for the traveller.

Posted by awambawamb on 2016-04-15 11:38:58

yes, but since we're calculating it for the spaceship, x is 0, as we don't change (we don't mind going to the bathroom on the spaceship). t is the time necessary for the spaceship to cover the distance from a spatial point of view, so [distance to destination]/[speed we are going to]

about the 99 thing, won't be really a big issue. Faster speed will give you surely a very high gamma but I would suggest to keep it within three digits to retain mental health during calculations :D for my calculations I've just multiplied c*0.999999

Posted by awambawamb on 2016-04-15 11:41:38

oh I never mixed up gamma calculations with Lorentz transorm tho

Posted by awambawamb on 2016-04-15 12:08:57

whops, and there's a plus where it should ahve been a minus. my pinky slipped on it, I've noticed it now that I'm doing calculations again:

t1=gamma[t-(vx/c^2)]
use also 0.999999 to get to my result.
bloody 5 or 6 nines...

paper, paper is great.

Posted by awambawamb on 2016-04-15 12:12:44

http://i.imgur.com/YTpSRPw.png

Posted by Nextflux on 2016-04-15 16:31:23

Its 26500 light years to the center of the galaxy, not 36.8 light years.

Posted by awambawamb on 2016-04-15 17:30:14

Nextflux you are not taking into consideration the time stretching of relativity. 36.8 (or somewhat there) it's time you would perceive it would pass if you could travel at 0.999999c.

you wouldn't see anything outside but a tiny flashlight in front of your eyes, tho...

Posted by Wreckage on 2016-04-15 21:32:56

I'm trying to make sense of this but my numbers don't add up to something that makes sense.

You say x is 0 so doesn't that mean that this is essentially:
t1=gamma[t-(vx/c^2)] for x=0 -> t1=gamma[t] ? and for t we have d/v with d=distance so -> t1= gamma[d/v] and gamma being ->
t1 = 1/[1-(v/c)^2]^2 * d/v ?