Re: superhero space navigation

On Tue, 08 Apr 2008 16:30:58 -0700, Erik Max Francis <max@[EMAIL PROTECTED]
>
wrote:

>Leonard Erickson wrote:
>
>> Well, I did some quick calcs and if we assume constant jerk sufficient
>> to reach the moon in an hour, then for a 10AU trip the turnover
>> velocity is 0.24c. 
>> 
>> If the time to the moon is only 30 minutes then turnover velocity for
>> the 10 AU trip is around .48 c.
>> 
>> Yow!
>
>I get different numbers.  The jerk for the first trip is 4.94 x 10^-2 
>m/s^3.  Applying that jerk to a 5 au trip (I presume by turnover you're 
>accelerating to midpoint and then decelerating the rest of the way), the 
>trip time is 4.50 x 10^4 s = 12.5 hr.  So the peak speed is 5.00 x 10^7 
>s = 16.7% c.  Not quite as much, but still pretty impressive.
>
>The second relation holds true, though:  Halve the trip time and hold 
>everything else constant will double the accumulated speed.
>
>> The peak accelerations are kinda scary too.
>
>Eh, a few hundred gees are peanuts for a superhero :-).

Well, I got a peak accel in the neighborhood of 20,000 m/s^2 for the
"fast" version of the 10AU trip.