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*To*: Zenon Kulpa <zkulpa@zmit1.ippt.gov.pl>*Subject*: Re: starship-design: Help with formula*From*: Steve VanDevender <stevev@efn.org>*Date*: Wed, 22 Apr 1998 09:10:14 -0700 (PDT)*Cc*: starship-design@lists.uoregon.edu, stevev@efn.org*In-Reply-To*: <199804221305.PAA21025@zmit1.ippt.gov.pl>*References*: <199804221305.PAA21025@zmit1.ippt.gov.pl>*Reply-To*: Steve VanDevender <stevev@efn.org>*Sender*: owner-starship-design@lists.uoregon.edu

Zenon Kulpa writes: > > From: Steve VanDevender <stevev@efn.org> > > > > Zenon Kulpa writes: > > > Please, I have lost it somewhere, no time to search or derive, > > > what is the formula for Earth time & crew time > > > in the (one-way) flight with constant acceleration = 1g > > > (half way +, half way - ) depending on the distance (in ly)? > > > > > > Thanks, > > > > > > -- Zenon > > > Thank you, but: > > > For one-way acceleration: > > > Does that mean half-way acceleration, and then half-way deceleration > to stop at target (at distance d)? It covers one-half of the kind of trip you're talking about. > > t = (c / a) * acosh(1 + (a * d / c^2)) > > > > t = elapsed time > > > Earth time or crew (ship) time? Ooops, now that I look at it the formula gives ship time, not Earth time. Sorry. > > c = speed of light > > a = acceleration > > d = distance traveled > > acosh = inverse hyberbolic cosine function > > -- Zenon

**References**:**Re: starship-design: Help with formula***From:*Zenon Kulpa <zkulpa@zmit1.ippt.gov.pl>

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