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*To*: KellySt@aol.com, kgstar@most.fw.hac.com, stevev@efn.org, jim@bogie2.bio.purdue.edu, zkulpa@zmit1.ippt.gov.pl, hous0042@maroon.tc.umn.edu, rddesign@wolfenet.com, David@InterWorld.com, lparker@destin.gulfnet.com, DotarSojat@aol.com*Subject*: OK, here the right calculus (I hope)*From*: T.L.G.vanderLinden@student.utwente.nl (Timothy van der Linden)*Date*: Sat, 30 Mar 1996 02:13:54 +0100

Optimal solution for a maser deceleration starship. The starship carries all the necessary repulsion mass and gets the energy needed to accelerate the repulsion mass from the maser. To do this we want to find the solution that consumes the least energy. Energy needed to decelerate a ship depends on it's empty mass, its start velocity, the exhaust velocity. So we need the formula E(Mo,Vstart,Vexh) Definition of g or gamma, also known as the energy factor: 1 ------------------ 2 (1) g = Vexh Sqrt[1 - ------] 2 c Momentum mass needed at time t to decelerate with a m/s/s M[t] (a+b) (2) j[t] = ------------ g Vexh b is the acceleration caused by the maser beam. The value of b depends on the power needed, this is yet unknown. Power needed at time t: P[t]=j[t] c^2 (g-1) Momentum caused by the maser beam: p=E/c or F=P/c Since F = M[t] b and P/c = j[t] c (g-1) we get M[t] b = j[t] c (g-1) Solving b gives: b = j[t] c (g-1)/M[t] Substituting b into (2) gives: M[t] a j[t] c (g-1) (3a) j[t] = --------- + -------------- g Vexh g Vexh Resolving j[t] gives: M[t] a (3b) j[t] = -------------------- g Vexh - c (g - 1) Definition of the time T that it takes to decelerate from Vstart to 0: c Vstart (4) T = --- ArcTanh[--------] a c Mass of the ship(Mo) and the repulsion mass(j[t]) at time t: T / (5) M[t] = Mo + | j[t] dt / t Substitute (3b) in (5): T a / (6) M[t] = Mo + -------------------- | M[t] dt g Vexh - c (g - 1) / t Solving this differential equation gives: a (T - t) (7) M[t] = Mo Exp[---------------------] g Vexh - c (g - 1) The total amount of energy needed to accelerate the repulsion mass to Vexh is defined by: T 2 / (9) Ek = c (g-1) | j[t] dt / 0 (Oops, forgot the "minus 1" in my previous letter) 2 a T (10) Ek = c (g-1) Mo (Exp [--------------------] - 1) g Vexh + c (1 - g) 2 c Vstart (11) Ek = c (g-1) Mo (Exp [-------------------- ArcTanh[--------]] - 1) g Vexh + c (1 - g) c Finally... E(Mo,Vstart,Vexh) Now we only need to find its minimum, that could be done by solving dE/dVexh=0 The minima are: Vstart Vexh optimal Fuel:ship-ratio Energy per kg of ship (in Joules) 0.1 0.062 5.36 7.45E14 0.2 0.121 5.84 3.25E15 0.3 0.180 6.40 8.11E15 0.4 0.240 7.06 1.64E16 0.5 0.300 7.87 2.97E16 0.6 0.364 8.91 5.23E16 0.7 0.433 10.38 9.21E16 0.8 0.512 12.72 1.73E17 0.9 0.615 17.75 4.04E17 0.99 0.803 52.00 3.12E18 0.9996 0.906 238.81 2.91E19 Note that the power of the maser-beam is NOT constant, it is supposed that it decreases while the ship gets lighter (because it repulses mass). Timothy P.S. Goodnight to all

**Follow-Ups**:**Re: OK, here the right calculus (I hope)***From:*Kevin C Houston <hous0042@maroon.tc.umn.edu>

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