Re: superhero space navigation

On Mon, 07 Apr 2008 14:46:13 -0700, Erik Max Francis <max@[EMAIL PROTECTED]
>
wrote:

>If you're asking about constant jerk (the rate of acceleration), then 
>you're just integrating three times instead of two:
>
>	j = j_0
>
>	a = int_0^t j dt = j_0 t + a_0
>
>	v = int_0^t a dt = (1/2) j_0 t^2 + a_0 t + v_0
>
>	s = int_0^t v dt = (1/6) j_0 t^3 + (1/2) a_0 t^2 + v_0 t + s_0.
>
>Where the naught (naughty?) subscripts are all the initial values of 
>jerk, acceleration, velocity, and displacement, respectively, at t = 0. 
>  If you're using constant jerk and acceleration, velocity, and 
>displacement all start at zero, then
>
>	s = (1/6) j_0 t^3,
>
>where j_0 is the constant jerk.  If jerk isn't constant, then it's a 
>more general integration problem, but it's still solved the same way: 
>integrating three times.
>
>> Of course I'd probably run into *nasty* relativistic effects on
>> flights beyond Mars or so if I go that route.
>
>Well, that depends on the jerk.

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!

The peak accelerations are kinda scary too.

And if the super can carry 100 kilos or so along, you've got planet
wrecker capability.