# Re: Mistake

```Hello Steve,

> > Sorry for the confusion
>
>errors.  And not only are these fine points of relativistic physics
>often counterintuitive, but they are also quite hard to talk about
>without resorting to elaborate mathematics.

Thanks

>I am interested to know the equation you got that produces the same
>results as mine without looking at all the same.

The result of my best calculations was:

a t0 (-2 c + a t)
c ArcSinh[------------------]
t1 =             2 c (-c + a t)
-----------------------------
a

where t is the emit-time in Earth's frame
and t1 is the recieve-time in the ship's frame

Note that in this formula I did not use your definition of c=1. This morning
I tried to do the same with c=1 but it gave me even more elaborate formulas.

te=Emit time in Earth's frame  (your t)
te'=Emit time in the ship's frame
f[x]=Function f determined by variable x

The main idea is that I solve:

x[tr]=c(tr-te)

I first try to get a formula for x that does have a t in it instead of a t':

x[t]=1/a Cosh[ArcSinh[a t]]

Then I try to solve te, which in this case gives me a formula that I don't
want to write down.

Having found te, I can easely find te', just by using:

t'=1/a ArcSinh[a t]

>Once good point you had in your last post was that you really don't have
>only two years to send power for the boost and deceleration phases of a
>1 g constant-acceleration mission.  You do, however, have only one year
>to send the power for the boost phase.  The situation is actually the
>worst for the so-called "Forward sail" design, involving the use of a
>detachable reflecting sail that reflects the beamed power back to the
>ship for deceleration -- in that case the power continues to accelerate
>the reflecting sail and therefore more drastically limits how long you
>have to send power to be used through the entire mission.

I think that when one uses the "Forward sail" the time would even be shorter
than 2 years. The reason for this is that the sail will (after it is
detached) probably accelerate faster than the ship, since it has less mass.
I cannot directly write down what the maximum power-emit time would be since
it involves 2 different accelerations.

Timothy

```