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Saturday, April 30th, 2005
9:57a - Just when you give up... you give up.
I was doing an update showing how far I'd gotten in solving the problems from last post, specifically "given a trip of known distance at constant acceleration a halfway and -a the other half, how long does it take?" (I'd gotten pretty much nowhere). So I realized I'd forgotten to write down part of it, and did a google search to find the Newtonian formula for position as a function of time and acceleration, and found THIS:

t=2c*sinh(aT)/G, where t is time for rest observers, T is time for accelerating observers, and G is SOMETHING, I have *no* idea what.

The same site gives distance

d=2c²(cosh(gT/2c)-1)/a

so,

2c(arccosh(.5ad/c²)+1)/g=T

I *think* little g is the acceleration due to the ambient gravitational field, effectively 0, while big G is almost definitely the gravitational constant. But that definition for g gives a divide by zero error, so that can't be right; plus it makes the time experienced by the crew more dependent on the gravity than the distance, which makes no sense at all. I am completely stymied.

So... I'm more or less done trying to work it out. I'll keep an eye out for a spreadsheet or program that will accept d and a as inputs, and spit out t and T, but I'm not going to actively search for a while, and, until I find it, this project is on indefinite hold.

This likely brings an end to the world-building kick, for now... back to the political and philosophical wankage.


current mood: frustrated and confused

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