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Re: Explorer Power Gain Problem
To: Brian Mansur
Cc: bmansur; David; hous0042; jim; Kelly Starks x7066 MS 10-39; L. Parker;
rddesign; Steve VanDevender; T.L.G.vanderLinden
Subject: Re: Explorer Power Gain Problem
Date: Monday, March 11, 1996 8:45AM
>Damn, I thought I caught up with all of these then I found some people sent
>to my office but not my home!!!
No more need to worry about any mail avalanches coming from me. I need to
do school work and I'm limiting myself to a max of 1 hr. a day for e-mail
response. Probably less actually and perhaps not every day even.
>Your numbers plugged in like this for your Explorer power requirements for
a >10m/s^2 acceleration.
>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.
I'm starting to see a pattern to all of our starship design problems.
However bad we think something is. Multiply that problem by a factor of E9
and that should keep us in the gremlin's ballpark (and in the solar system
>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. Can you
>I can't confirm this, but I do know the energy numbers I worked out in the
>Explorer page were sloppy.
Does ANYONE know the proper equations? I'll try to find time to ask my