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*To*: stevev@efn.org*Subject*: Re: The dreaded derivation...*From*: DotarSojat@aol.com*Date*: Sun, 1 Sep 1996 16:30:17 -0400*cc*: T.L.G.vanderLinden@student.utwente.nl

Hi Steve Thanks for the derivation in your email of 8/28. I'll save it as a reference to help understand Lorentz dot-product manipulations if I ever run across them again. Actually, the description of the space-time diagram in your 8/26 email was more effective in con- veying the relation to my intuition. Having gone through that thought process, I can now rewrite the derivation more succinctly: The distance, x, of the sail after accelerating at a level, a, for a proper time, t', is x = [cosh(a * t') - 1]/a . The apparent (Earth) time, t, at which the sail reaches x is t = [sinh(a * t')]/a . The time, delta-t, that it takes light from Earth to reach the sail at x is delta-t = t - te = x/c or = x for c = 1 lt-yr/yr, where te is the Earth time at which the light was emitted. Solving for te gives te = t - x = [1 + sinh(a * t') - cosh(a * t')]/a . Replacing sinh and cosh with their exponential forms gives te = [1 - exp(-a * t')]/a . And solving for t' gives t' = -[ln(1 - a * te)]/a . Q.E.D. I'll be looking forward to the way you'll treat the "dreaded derivation" for the SD Group. I'd feel honored if you used any part of the above approach. You started a very interesting learning process with your email of 8/20 to the Group. "If mankind is ever to achieve the ultimate goal of traveling to the stars, we will need an engineering understanding of relativ- ity" (RGF, IDA Paper P-2361, Oct 1990). Regards, Rex

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