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Re: Explorer Power Gain Problem

At 12:59 PM 3/15/96, Timothy van der Linden wrote:
>>> 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
>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.

In my case dramatically!  Thats why I droped that and went to using fusion
rockets and standard rocket and specific impulse fuel mass consumption



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