# Re: C-ship

```Steve VanDevender writes:
> Things are substantially more complicated when dealing with
> accelerated worldlines.  I've got a preliminary solution stated
> in similar terms as the above discussion; perhaps you'd like to
> check my math :-).
>
> You are at the point S = [ t x y z ] attempting to view an object
> whose coordinates are P(t') =
>
> 1/a^2 * [ a * sinh(a * t')
>           ax * (cosh(a * t') - 1)
> 	  ay * (cosh(a * t') - 1)
> 	  az * (cosh(a * t') - 1) ]
>
> where the acceleration is represented by A = [ 0 ax ay az ] (in
> the object local frame) and a = sqrt(ax^2 + ay^2 + az^2) (the
> magnitude of the acceleration).
>
> So again, we want to solve the equation (S - P(t'))^2 = 0.  Of
> course, the components of P(t') are much more complicated.  I
> won't bore you with the full derivation, other than to note that
> it becomes easier to isolate t' by writing the sinh and cosh
> terms in terms of their definitions using exp (e^x).
>
> Eventually, you get:
>
> exp(a * t')^2 * (1 - A|S - a * t') +
> exp(a * t') *   (a^2 * S^2 - 2 * (1 - A|S)) +
>                 (1 - A|S + a * t')
> = 0

Sigh.  That should actually be:

exp(a * t')^2 * (1 - A|S - a * t) +
exp(a * t') *   (a^2 * S^2 - 2 * (1 - A|S)) +
(1 - A|S + a * t)
= 0

I wrote t' rather than the t component of S in a couple wrong
places.

> It's convenient to make some substitutions for common
> subexpressions:
>
> k = 1 - A|S
> p = k - a * t'
> q = a^2 * S^2 - 2 * k
> r = k + a * t'

And these should be

p = k - a * t
q = a^2 * S^2 - 2 * k
r = k + a * t

> So then applying the quadratic formula and isolating t' gives:
>
> t' = 1/a * ln((-q - sqrt(q^2 - 4 * p * r)) / (2 * p))

Fortunately I still copied that correctly.

```

• References:
• Re: C-ship
• From: T.L.G.vanderLinden@student.utwente.nl (Timothy van der Linden)
• Re: C-ship
• From: Steve VanDevender <stevev@efn.org>