How do I create an Excel form for computing Einsteinian Time
Compression for a ship using constant acceleration to travel between
the stars. I need Excel specific formulas for computing the amount of
time a trip will take both from the frame of reference of the people
inside the ship and back in the 'normal' frame of reference (Earth
normal). I also need to be able to change the speed of acceleration
(in gravities) to recompute real and perceived time. Note that this
system presumes half the trip will be acceleration and half the trip
will be deceleration at that constant rate of speed.

In other words, I need to be able to define distance (in light years)
and speed (in Gravities, each gravity defined as 9.8 m/s/s) as
variables, and get back length of time the actual trip will take and
the length of time that will seem to pass for the crew.

Thanks!

Ok. I found some formulas in
http://www2.corepower.com:8080/~relfaq/rocket.html
and converted them to excel by hand

The only unit you specified was acelleration in Gravities, so for
simplicity i used lightyears and years as the other units (if you need
something else, post a 'request for clarification' and i can convert
it for you)

i have my excel worksheet online at
http://www.boomspeed.com/jes5199/relativity_rocket.xls

the formulas are
----------------
initial values are set like so:
B2 = 1 (lightspeed is 1 lightyear/year)
B3 = ?? g (your acceleration, usually one)
D3 =1.03*B3  (convert Gravities to lightyears per year per year)
B4 = ?? lightyears (distance to travel)

and one-way calculations are like this:
B6 =SQRT((D4/B2)^2+(2*(D4/D3)))   (time elapsed, as seen from Earth)
B7 =(B2/D3)*ACOSH((D3*D4)/(B2^2)+1)   (time elapsed, as seen from the
ship)

and round trip times are just doubles of the above
B10 =2*B6
B11 =2*B7


if you need anything else, feel free to ask.
thanks.

---

Request for Answer Clarification by scholarman-ga on 30 Jul 2002 14:30 PDT

The units are all correct (time in years, distance in light years,
acceleration in gravities). However, when testing your sheet against
some of the prefigured examples at
http://www2.corepower.com:8080/~relfaq/rocket.html, I discovered it
wasn't yielding the correct answers. For example, a one way trip to
Alpha Centauri -- 4.3 LY will take (relative time) 3.6 years according
to the web site, but your tool gives a result of 2.3 years. A trip to
Vega (27 LY) will take (ship's time) 6.6 years according to the web
site, but your tool claims 3.9 years.

So, there's something wrong in the formula, so far. Is it fixable?

---

Request for Answer Clarification by scholarman-ga on 30 Jul 2002 14:41 PDT

One thing that might not have been clear -- this has to assume slow
speed start to slow speed finish. As was said in the initial question,
this means that the ship will be accelerating for half the trip, then
<i>decelerating</i> for the second half of the trip. Therefore, the
ship will arrive at its destination moving relatively slowly and at
parity in terms of local time.

---

Clarification of Answer by jes5199-ga on 30 Jul 2002 18:22 PDT

aha! that's the problem. for some reason i was arriving at full
speed... which could be a bad idea... let me see how i can fix that

---

Clarification of Answer by jes5199-ga on 30 Jul 2002 18:34 PDT

my modified formulas -- which i made by instinct, i'm assuming
de-celleration is basically the same as acelleration when you look at
it over a length of time -- simply divide the distance by 2, and then
double the result. seems to match the numbers on the page.

here's my updated spreadsheet:
http://www.boomspeed.com/jes5199/relativity_rocket_two.xls

hope that helps. if you need anything else, feel free to ask.

Excellent work, and exactly what I needed. When I needed some changes
to fit my parameters, they were swift in coming. This is what makes
this service so great.

I don't see the correction in here for the frame in which acceleration
is measured.  If the people on the ship believe they're accelerating
at 1 g, the people on the Earth will disagree with that quickly.

I think this problem is probably harder than it appears.

In the parameters of the question, I intentionally chose not to
address the factors of energy expenditure and requirement, mass
change, and other such factors. In cross comparing known trip times
with the generated formula, keeping those factors as read, the revised
formulae and spreadsheets do indeed provide accurate answers.

It may seem like acceleration is different, but quite honestly, the
formula does factor what seems like 9.8 m/s^2 acceleration from the
ship's frame of reference versus the Earth's frame of reference, even
though a 'second' becomes a very different thing in the two frames of
reference. Relativity is a strange, strange beast.

In the parameters of the question, I intentionally chose not to
address the factors of energy expenditure and requirement, mass
change, and other such factors. In cross comparing known trip times
with the generated formula, keeping those factors as read, the revised
formulae and spreadsheets do indeed provide accurate answers.

It may seem like acceleration is different, but quite honestly, the
formula does factor what seems like 9.8 m/s^2 acceleration from the
ship's frame of reference versus the Earth's frame of reference, even
though a 'second' becomes a very different thing in the two frames of
reference. Relativity is a strange, strange beast.

Let me retract that & apologize.  I've since looked at
http://math.ucr.edu/home/baez/physics/Relativity/SR/rocket.html  and
the spreadsheet does appear to be correct.

The formulas already include the fact that Earthbound observers won't
see the same acceleration as those on the rocket.  The acceleration
used here is the one felt by the rocket's occupants.