Questions about Time Dilation

[Best of the Worst]

For example, one could travel to the "edge" of the observable Universe in about 25 years, ship-time, if the craft travels at a very high percentage of c. For an observer, the elapsed time is much, MUCH greater (too many zeroes to write down here). But hold on - time dilation is an observed effect from the outside - the actual traveler thinks that everything is going along just as normally as ever; her clocks ticks at the same rate, light bounces around the cabin off mirrors as she expects, yet we are told that to her, the total journey time is very much compressed because of her high speed.
Could someone explain this to me? Unless my math is wrong, the trip should take ~1.52 x 10^6 years, not 25 years (I'm assuming that the "trip to the edge of the observable universe" is referring to the radius of the universe, which I've read to be 1 500 000 light years), to the one in the spaceship. How is it that time is compressed for her?

Also, I've been having trouble finding a good explanation of the Twin Paradox. I know that because the twin in the spaceship has to change directions, he jumps from one frame of reference to another, but I haven't been able to find an explanation of why that means that their observations aren't contradictory. Anyone care to explain this to me, too?

Thanks in advance :).

[Ich]

It´s 1.52E6 universe time and 25 ship time, I assume. BTW: the observable universe is much larger.
Think of changing velocity similar to changing direction in spacetime. The one changing direction took a detour, ie not the straight line between two events. In Minkowski geometry, a detour is always shorter than the straight line. And path length in Minkowski space is proper time.

[Doc Al]

Could someone explain this to me? Unless my math is wrong, the trip should take ~1.52 x 10^6 years, not 25 years (I'm assuming that the "trip to the edge of the observable universe" is referring to the radius of the universe, which I've read to be 1 500 000 light years), to the one in the spaceship. How is it that time is compressed for her?
From the traveler's viewpoint, the distance has "contracted" to a small fraction of 1 500 000 light years, so her travel time is quite small according to her clocks. (To state a specific time, one would need to know her speed.) What you have calculated, I presume correctly, is the time of her trip according to Earth observers. (Per special relativity, Earth observers measure the space ship clocks to be running slowly, so everything is consistent.)

[Best of the Worst]

From the traveler's viewpoint, the distance has "contracted" to a small fraction of 1 500 000 light years, so her travel time is quite small according to her clocks.
Hmm... I didn't really read up much on Length Contraction, but what I read was that it's the object itself, and not the distance travelled, that contracts as its speed approaches c. In any case, what's the principle that states that travel distance contracts as speed approaches c?

[JesseM]

Hmm... I didn't really read up much on Length Contraction, but what I read was that it's the object itself, and not the distance travelled, that contracts as its speed approaches c. In any case, what's the principle that states that travel distance contracts as speed approaches c? If two objects are moving at the same constant velocity in my frame (two galaxies, say), then whatever the distance between them in their own mutual rest frame, that distance will be shrunk in my frame by the same factor as each object's length is shrunk. Either type of shrinkage is called "Lorentz contraction".

[Best of the Worst]

Cool. Thanks for the replies.