On Sun, 01 Jan 2012 22:42:56 -0500, STJensen wrote:

Let us see if I get what William Elliot, James Waldby, and David Bernier
are saying.

Time it would take to go the entire 62,000 miles: Elliot: 76 minutes
Waldby: 3548.9 If the time is in seconds, that would equal 59.1
minutes. Bernier: Sorry, I couldn't find a time.

Velocity human would be traveling at end of space elevator: Elliot: 28
miles per second Waldby: 61,686.8 meters per second. I'm assuming
you mean meters and not miles so that would be ~ 38 miles per second
Bernier: 19.6 miles per second

If the above is correct, why so much difference between the numbers?

First, "the above" is partly wrong: David Bernier said nothing about
19.6 miles per second; he wrote 19.6 m/sec/sec, where m is meters, and
said that that is the acceleration equal to 2*g.

Second, the numbers William Elliot supplied are for the case where
a uniform 1 g acceleration is applied for a little over 75 minutes,
causing one to arrive at the top of the elevator at a speed of about
27.5 miles per second. The numbers are from s = (1/2)*a*t^2 and
v=a*t with s = 62000 miles, a = g. If you apply a uniform 2 g
acceleration (as suggested in Barry Schwarz's post) and experience
about 3G's on takeoff and about 2G's near the top, you arrive at
the top in about 53 minutes going 39 miles per second. In my
solution, with about 2G's total all the way up it takes about 59
minutes and top velocity is about 38.3 miles per second.

--
jiw