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Re: Explorer Power Gain Problem
- To: bmansur <bmansur@oc.edu>, David <David@InterWorld.com>, DotarSojat <DotarSojat@aol.com>, hous0042 <hous0042@maroon.tc.umn.edu>, jim <jim@bogie2.bio.purdue.edu>, KellySt <KellySt@aol.com>, kgstar <kgstar@most.fw.hac.com>, lparker <lparker@destin.gulfnet.com>, rddesign <rddesign@wolfenet.com>, stevev <stevev@efn.org>
- To: "T.L.G.vanderLinden" <T.L.G.vanderLinden@student.utwente.nl>, zkulpa <zkulpa@zmit1.ippt.gov.pl>
- Subject: Re: Explorer Power Gain Problem
- From: Brian Mansur <bmansur@oc.edu>
- Date: Mon, 25 Mar 96 09:42:00 PST
- Encoding: 49 TEXT
>From Brian,
>>Kelly
>> The formula is right, but the number is only valid when the velocity of
the
>> starship is 5 m/s. The problem with the document is that the meaning of
the
>> number is not told.
>
>Now I remember. I calculated the power the drive system would need for a
one
>second boost, and assumed it would be constant for the flight. I.E. if
>it takes X watts to push the ship at a ship G for a secound. Multiply that
>by the number of seconds of boost to get to desired speed, and ....
>
>Given that the engine should need to output the same power, to accelerate
the
>same mass, at the same rate, at diferent speeds. It seems like it should
>work.
>Tim
>Ah that makes some sence, you indeed can use this trick to calculate the
>power needed for a self-fueled ship. But keep in mind that the mass of the
>ship may decrease significantly due to the "burning" of the fuel.
Brian.
Don't bother sending me a reply to this e-mail (not because I don't like
you, but because I have to be going off-line for the rest of the school
year).
Over break I looked very closely at Kelly's Explorer paper power
requirements. I noted one very confusing thing that he did when he
calculated his power requirements. He said that Watts could be divided by
time (example: W/s).
After studying my physics textbook for a few nights, I now believe that
there is no such thing as a W/s because a Watt is already a Joule per
second. How can you have Joules per second per second? I don't believe you
can and I couldn't find anything in two different physics textbooks to
suggest otherwise.
I thought that I was missing something because I seemed to have remembered
hearing that the Earth received 1400 W/m^2 per second. That is wrong. The
texts available to me say that we get 1400 W/m^2 period.
Anyway, if you follow my reasoning, you can see why this error explains the
low numbers on the Explorer paper. Otherwise the math seemed good to me (I
think).