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*To*: Steve VanDevender <stevev@efn.org>*Subject*: My derivation*From*: T.L.G.vanderLinden@student.utwente.nl (Timothy L.G. van der Linden)*Date*: Sun, 01 Oct 1995 13:44:46 +0100

>The primary text I use is _Spacetime Physics_ by Taylor and >Wheeler. They almost always use geometrized units, introducing >them almost immediately in the first chapter. I find they are >much easier to work with for most purposes. Probably easier, when you have worked with them a while. Maybe I will start using them too. >I really don't know how you could make antimatter in any quantity >using conventional technology. I doubt it would be very >efficient at all. Very very small quantities are produced when particles collide in these supercolliders. Most of the time these products are just left to decay. The problem is that one gets a myriad of particles while we are probably most interested in positrons anti-neutrons and anti-protons. >At best, you could make something like 2 million tons per second >if you could convert the total output of the Sun completely to >antimatter (and an equal amount of matter). Yes, but then it would become quite dark on Earth wouldn't it. :) If we could bundle a 1E-9 part of the Suns photons we could accelerate a 1 million ton ship with 1g by photon impulse. >I am about to leave on a trip to Seattle, so I can't promise I'll >have time to look it over. You are welcome to send it to me and >if I can find time to look it over I will; I'll still have email >access from there. Ok, thanks in advance. I will add it here. Greetings Timothy We want to build a spacevessel called Asimov which acclerates with a constant acceleration a relative to the Asimovs frame to a velocity Vend relative to Earth. For this it exhausts mass with a velocity Vexh backwards. The energy source can be of any kind and is stored on board. To know how much energy is needed one needs to know the total mass of the Asimov including the initial mass of the fuel and energy. But since the energy needed depends on the mass of the fuel and the fuel needed depends on the amount of energy the calculations may not be straight forward when one looks at the problem for the first time. Such kind of problems arise frequently in physics and the method to solve it is called "differential calculation". This method I will use here. Here are the variables and constants that may need explaination: M[t] The total mass of the Asimov at time t Mo The mass of the Asimov without fuel and energy (empty weight) j[t] The mass-exhaust at time t k[t] The mass of the energy used at time t to accelerate the exhaust-mass l[t] Mass that is leftover after the energy that it "contained" is used. (eg. the mass of the products of a fusion reaction) L[t] Mass that is unused and is dumped as soon as its "contained" energy is used. Vexh The exhaust velocity (assumed constant) g Relativistic mass increase due to Vexh (thus also constant) Vend End velocity relative to Earth a Constant acceleration perceived by the people in the Asimov T Total amount of time that the people in the Asimov experience during constant acceleration in their frame to Vend in Earths frame Extra explanation about l[t] and L[t]: In the case of matter & anti-matter reactions mass is completely transformed in energy so l[t]=0. In the case of fusion-reactions only a small fraction of the initial fusion matter is transformed in energy, so the largest part of the matter is left over after the reaction. Some of that matter can be used as exhaust mass, but in most cases not all. The part that cannot be used is called L[t]. This part is assumed to be dumped as soon as the energy it initially "contained" is "extracted". (Dumped means that the mass is released from the Asimov without giving it any extra velocity) Known formulas: (some explainations at the bottom of this document) 1 ------------------ 2 (1) g = Vexh Sqrt[1 - ------] 2 c M[t] a (2) j[t] = -------- g Vexh (3) k[t] = j[t] (g - 1) (3) l[t] = (f - 1) k[t] = (f - 1)(g - 1) j[t] (3.1) IF l[t]<j[t] THEN L[t] = 0 (3.2) ELSE L[t] = l[t]-j[t] = ((f-1)(g-1)-1) j[t] f=(Total mass)/(mass that can be converted to energy) f is always bigger than or equal to 1 For matter & anti-matter mixture f=1, L[t]=0, J[t]=j[t] For "De + 3He --> 4He + p" fusion (3.5E14 J/kg) f=257, L[t]=256k[t]-j[t], J[t]=0 c Vend (4) T = --- ArcTanh[------] a c T / (5) M[t] = Mo + | (j[t] + k[t] + L[t]) dt / t - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - Calculations: Let's substiute (3.1) in (5): T / (6.1) M[t] = Mo + | (1+(g-1)) j[t] dt / t Rewriting gives: T / (6.1a) M[t] = Mo + g | j[t] dt / t Now substitute j(t) in (6.1a): T a / (7.1) M[t] = Mo + ------ | M[t] dt Vexh / t Now there is our differential equation. Its solution is: a (T - t) c Vend (8.1) M[t] = Mo Exp[-----------] with T = --- ArcTanh[------] Vexh a c With formula (8.1) we can calculate the total initial mass M[0]. One also can calculate the total energy needed, By integrating k[t] from 0 to T we find the total mass Mk of the energy. T / (9) Mk = | k[t] dt / 0 We substitute (3), (2) and (8.1) in this order and get: T / Mo a (g - 1) a (T - t) (10.1) Mk = | -------- -------- Exp[-----------] dt / Vexh g Vexh 0 Solving this gives: Mo (g - 1) a T (11.1) Mk = ------------ (Exp[------] - 1) g Vexh It may seem that this formula depends on a but remember the substitution of T where 1/a occurs. 2 This can be converted to energy by E = Mk c - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - Same derivation as before only now we use (3.2) Let's substiute (3.2) in (5): T / (6.2) M[t] = Mo + | (1+(g-1)+(f-1)(g-1)-1) j[t] dt / t Rewriting gives: T / (6.2a) M[t] = Mo + (f g - f) | j[t] dt / t Now substitute j(t) in (6.2a): T f g - f a / (7.2) M[t] = Mo + --------- ------ | M[t] dt g Vexh / t Now there is our differential equation. Its solution is: (f g - f) a (T - t) c Vend (8.2) M[t] = Mo Exp[---------------------] with T = --- ArcTanh[------] g Vexh a c With formula (8.2) we can calculate the total initial mass M[0]. One also can calculate the total energy needed, By integrating k[t] from 0 to T we find the total mass Mk of the energy. T / (9) Mk = | k[t] dt / 0 We substitute (3), (2) and (8.2) in this order and get: T / Mo a (g - 1) a (T - t) f g - f (10.2) Mk = | -------- --------- Exp[----------- ---------] dt / Vexh g Vexh g 0 Solving this gives: Mo a T f g - f (11.2) Mk = ---- (Exp[------ ---------] - 1) f Vexh g It may seem that this formula depends on a but remember the substitution of T where 1/a occurs. 2 This can be converted to energy by E = Mk c =============================================================================== Here are some numbers with a=10 m/s and f=1: +-----------------+ +--------+--------+---------+-----------| Power/Mo | | Vexh/c | Vend/c | M[0]/Mo | Energy/Mo | at t=0 | at t=T | +--------+--------+---------+-----------+--------+--------+ | 0.4 | 0.76 | 12.18 | 8.40E16 | 7.63E9 | 6.26E8 | | 0.54 | 0.76 | 6.37 | 7.65E16 | 5.60E9 | 8.79E8 | Minimum energy | 0.9 | 0.76 | 3.75 | 1.03E17 | 5.71E9 | 1.88E9 | | 0.99 | 0.76 | 2.75 | 1.35E17 | 7.15E9 | 2.60E9 | | 0.99 | 0.99 | 14.49 | 1.04E18 | 3.8E10 | 2.60E9 | +--------+--------+---------+-----------+--------+--------+ Here a number with a=10 m/s and f=257: +------------------+ +--------+--------+---------+-----------| Power/Mo | | Vexh/c | Vend/c | M[0]/Mo | Energy/Mo | at t=0 | at t=T | +--------+--------+---------+-----------+---------+--------+ | 0.4 | 0.76 | 1.9E23 | 6.91E37 | 1.23E32 | 6.26E8 | +--------+--------+---------+-----------+---------+--------+ =============================================================================== Derivation of (2): dp relativistic M a ---- = F --> j Vexh = M a ---------------> g j Vexh = M a --> j = -------- dt mass increase g Vexh ------------------------------------------------------------------------------- Derivation of (3): The energy needed to acclerate mass m to Vexh is: 2 Ekin = m c (g - 1) The mass n needed for this energy can be calculated with 2 2 E = n c E = n 3.5E14 = n 0.0039 c (fusion) Since E = Ekin we find: n = m (g - 1) n = 257 m (g - 1) (fusion) ------------------------------------------------------------------------------- Derivation of (4): Rewriting of: a t' v = c Tanh[-----] where T=t' and Vexh=v c (This formula can be found in some relativity books. It gives the velocity for a object which accelerates constant with a in its own frame during time t' in its own frame) ------------------------------------------------------------------------------- Derivation of (5): M(t) = (Total initial mass) - (Mass used for propulsion during time t) T / Total initial mass = Mo + | [j(t) + k(t)] dt / 0 t / Mass used for propulsion = | [j(t) + k(t)] dt / 0 Thus the substraction of both gives: T / M(t) = Mo + | [j(t) + k(t)] dt / t ------------------------------------------------------------------------------- Pascal program to calculate certain values CONST c:Double=3E8; a:Double=10; f:Double=257; Mo:Double=1; VAR Vend,Vexh,P1,P2,Ek,M,TT,t,g:Double; FUNCTION ArcTanh(x:Real):Double; BEGIN {Arctanh(x) x element of [0,1]} ArcTanh:=Ln((1+x)/Sqrt(1-Sqr(x))); END; BEGIN Vend:=0.76159*c; Vexh:=0.54*c; g:=1/Sqrt(1-Sqr(Vexh/c)); TT:=c/a*ArcTanh(Vend/c); t:=0; IF (f-1)*(g-1)<1 THEN BEGIN M:=Mo*Exp(a*(TT-t)/(Vexh)); Ek:=Mo*(g-1)/g*(Exp(a*TT/Vexh)-1)*Sqr(c); t:=0;P1:=Sqr(c)*Mo*a/Vexh*(g-1)/g*Exp(a*(TT-t)/Vexh); t:=TT;P2:=Sqr(c)*Mo*a/Vexh*(g-1)/g*Exp(a*(TT-t)/Vexh); END ELSE BEGIN M:=Mo*Exp((f*g-f)*a*(TT-t)/(g*Vexh)); Ek:=Mo/f*(Exp(a*TT/Vexh*(f*g-f)/g)-1)*Sqr(c); t:=0;P1:=Sqr(c)*Mo*a/Vexh*(g-1)/g*Exp(a*(TT-t)/Vexh*(f*g-f)/g); t:=TT;P2:=Sqr(c)*Mo*a/Vexh*(g-1)/g*Exp(a*(TT-t)/Vexh*(f*g-f)/g); END; Write('M[0]=',M:12,' Ek=',Ek:12,' P[0]=',P1:12); WriteLn(' P[T]=',P2:12,' T=',T/86400:0:0); END.

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