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*To*: stevev@efn.org*Subject*: Calculations*From*: T.L.G.vanderLinden@student.utwente.nl (Timothy van der Linden)*Date*: Fri, 06 Jan 1995 01:17:00 +0100

You advised to substitute sqrt(ax^2 + ay^2 + az^2) for a, I did it the other way around thus substitute (ax^2 + ay^2 + az^2) by a^2 and ... Wow, that really simplifies (after some other tricks): S|S + 2 a^2*(1+A|S)*(Cosh[a*T]-1) - 2*a^3*t*Sinh[a*T] = 0 A=[0 ax ay az] S=[t x y z] Solving gives: t' = 1/a * Ln[(d+Sqrt[4 f g a^4 + d^2])/(2g a^2)] ^ or a minus-sign With: d=2 a^2 (1+A|S) - S|S f=-1 - A|S - a t g= 1 + A|S - a t It looks much like your solution, but isn't the same: t' = 1/a * ln((-q - sqrt(q^2 - 4 * p * r)) / (2 * p)) > k = 1 - A|S > p = k - a * t = 1 - A|S - a t > q = a^2 * S^2 - 2 * k = a^2 S|S - 2 - 2 A|S > r = k + a * t = 1 - A|S + a t Timothy

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