# relativity length contraction for twin clock experiment

The question im going to ask is based on scenario given here https://www.scientificamerican.com/article/how-does-relativity-theor/ Only thing that's different is that spacecraft has a length of 1 light year ($1ly$) and traveler is sitting on extreme rear. When the traveler reach the star after 8 years (in traveler's clock), the extreme front end has a explosion. The coordinate of this event for traveler is $(1 ly, 0, 0, 8)$. The transformed coordinate for homebody is $((1 ly +8*0.6c)*1.25, 0, 0, (8+1*0.6c/c^2)*1.25)$. Where 1.25 is $\gamma$ at $v=0.6c$. It means the position of explosion as seen homebody (i.e. the x-coordinate of explosion) is $(1 ly +8*0.6c)*1.25=7.25 ly$.

So here is my question: if the spacecraft is contracted in length due to travelling at $v=0.6c$, the coordinate of explosion should come out to be between 6ly and 7ly for the homebody because the star is already 6ly distance away (proper distance) and the $1ly$ long spaceship has contracted due to length contraction formula. But it is even more than 7ly. Where am i going wrong with it?

It should be self-explanatory but... $(t_1,x_1)=(0,0)$ is the traveler's ($T$) position in his own reference frame at the beginning of the journey, $(t_2,x_2)=(0,1)$ is the position of the other end of his spaceship; $(t_3,x_3)=(8,0)$ and $(t_4,x_4)=(8,1)$ are the same at the end of the journey, and the prime denotes the same events in the "homebody's" ($H$) frame.