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Re: Explorer Power Gain Problem

Damn, I thought I caught up with all of these then I found some people sent
to my office but not my home!!!

At 5:27 PM 3/9/96, Brian Mansur wrote:
>>From Brian
>Kelly, I was looking over your explorer paper and the power section, trying
>to understand how you got a power gain of 1E9 W per second.  Your equation
>went something like:
>P = F dx/dt = m*a*dx/dt
>The mass of the ship was 100E6 kg.  The acceleration was 10m/s^2.  The
>duration of acceleration was 1E7 seconds.  You plugged in the numbers like
>100E6kg * 10m/s^2 * 5m/s = 5E9W
>Where did that 5m/s come from.  I thought dx/dt was a distance traveled in
>one second.  My physics textbook tells me power is F * v where v is
>velocity.  Since our average speed is 5E7 m/s (average between our terminal
>speed of 1E8 m/s and 0 m/s) the equation should look like this:
>100E6kg * 10m/s^2 * 5E7m/s = 5E16W.

5m/s is the amount of distence it would travel in the first second.  The
equation was computing only the power needed for one secound of boost.  I'm
not sure if the equation is set up right, but it would suggest your 5e16w
number for the full boost to .3c.

>If I'm right, that is a hefty difference in power gain which we couldn't
>hope to meet without loading tremendous amounts of fusion fuel.  Please tell
>me that I'm the one who is wrong.  Also, when I ran the numbers for K.E.
>using the relativity equation and the basic physics equation (these are on
>your explorer paper and I don't have time to write them here) I came up with
>a 100+ fold difference in power.  Can you confirm this?  Do you even have
>time to read this letter? :)

I can't confirm this, but I do know the energy numbers I worked out in the
Explorer page were sloppy.



Kelly Starks                       Internet: kgstar@most.fw.hac.com
Sr. Systems Engineer
Magnavox Electronic Systems Company
(Magnavox URL: http://www.fw.hac.com/external.html)