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Jenab

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  1. We now find the velocity vector, VJ, in the transfer orbit at the non-apsidal endpoint of the intended trajectory. Canonical velocity at non-apsidal endpoint for hyperbolic orbits VxJ’’’ = -(a/rJ) { GMsun / a }^0.5 sinh uJ VyJ’’’ = +(a/rJ) { GMsun / a }^0.5 (e^2 - 1)^0.5 cosh uJ VzJ’’’ = 0 Canonical velocity at non-apsidal endpoint for elliptical orbits VxJ’’’ = -sin QJ { GMsun / [ a (1-e^2) ] }^0.5 VyJ’’’ = (e + cos QJ) { GMsun / [ a (1-e^2) ] }^0.5 VzJ’’’ = 0 Rotation to heliocentric ecliptic coordinates VxJ’’ = VxJ’’’ cos w - VyJ’’’ sin w VyJ’’ = VxJ’’’ sin w + VyJ’’’ cos w VzJ’’ = VzJ’’’ VxJ’ = VxJ’’ VyJ’ = VyJ’’ cos i VzJ’ = VyJ’’ sin i VxJ = VxJ’ cos L - VyJ’ sin L VyJ = VxJ’ sin L + VyJ’ cos L VzJ = VzJ’ By replacing J with K in all the above, you can find the sun-relative velocity in the transfer orbit at the apsidal endpoint, also. Finding the delta-vees on each end of the intended trajectory goes like this: dVx1 = Vx1 - Vx(preburn) dVy1 = Vy1 - Vy(preburn) dVz1 = Vz1 - Vz(preburn) dVx2 = Vx(rendezvous) - Vx2 dVy2 = Vy(rendezvous) - Vy2 dVz2 = Vz(rendezvous) - Vz2 In my example problem, V(preburn) is the orbital velocity of Vesta at the moment of departure, and V(rendezvous) is the orbital velocity of Earth at the moment of arrival. You can work it out explicitly, if you want to. Their magnitudes will be, roughly: dV1 = 9.2 kilometers per second dV2 = 20.4 kilometers per second THE END Jerry Abbott
  2. We have reached a point where we must begin treating hyperbolic and elliptical orbits somewhat differently. The proposed hyperbolic transfer orbit B = cosh uJ = (1/e) (1 + rJ /a) uJ’ = ln [ B + (B^2 - 1)^0.5 ] If sin QJ < 0 then uJ = -uJ’ else uJ = uJ’ MJ = e sinh uJ - uJ Here, the uJ is the hyperbolic eccentric anomaly in the transfer orbit of the non-apsidal endpoint of the intended trajectory, and MJ is the corresponding hyperbolic mean anomaly. When calculating MJ it is necessary that uJ be in radians. You will want to convert [a] from astronomical units to meters. GMsun = 1.32712440018E+20 m^3 sec^-2 1 astronomical unit = 1.49597870691E+11 meters m = (86400 sec/day) ( GMsun / a^3 )^0.5 The hyperbolic mean motion will result to units of radians per day. Calculation of the transit time, dt. Once we have the transit time, we can compare it to the time difference required by the known times of departure and of arrival. If they match, or very nearly match, then we have found our transfer orbit. If they do not match, we are disgusted at having done so much work for nothing. Case #1: Perihelion at Departure: K=1, J=2. dt = M2/m, if Q1=0 (thus M1=0) Case #3: Perihelion at Arrival: K=2, J=1. dt = -M1/m, if Q2=0 (thus M2=0) Note that hyperbolic orbits do not have aphelia. The proposed elliptical transfer orbit sin uJ = (rJ /a) sin QJ / (1-e^2)^0.5 cos uJ = e + (rJ /a) cos QJ uJ = arctan2(sin uJ , cos uJ) MJ = uJ - e sin uJ The uJ is the eccentric anomaly in the transfer orbit of the non-apsidal endpoint of the intended trajectory, and MJ is the corresponding mean anomaly. When calculating MJ it is necessary that uJ be in radians. Although you generally don’t intend to follow an elliptical transfer orbit around its complete circuit, its orbital period can be found from P = (365.256898326 days) a^1.5 The mean motion (radians per day) of a spaceship in the transfer orbit would be m = 2 pi radians / P Calculation of the transit time, dt. Once we have the transit time, we can compare it to the time difference required by the known times of departure and of arrival. If they match, or very nearly match, then we have found our transfer orbit. If they do not match, we are disgusted at having done so much work for nothing. Apside at Departure: K=1, J=2. Case 1: dt = M2/m, if Q1=0 (thus M1=0) Case 4: dt = (M2-pi)/m, if Q1=pi (thus M1=pi) Apside at Arrival: K=2, J=1. Case 3: dt = (2 pi-M1)/m, if Q2=0 (thus M2=0) Case 2: dt = (pi-M1)/m, if Q2=pi (thus M2=pi) If necessary, correct dt to the interval [0,P) by adding or subtracting the appropriate multiple of P. Plugging in the numbers... Proposed hyperbolic transfer orbit r1 = 2.17821044 e = 6.287121148 a = 0.190141423 AU Q1 = 5.09773397 radians u1 = -1.30600661 radians M1 = -9.44655271 radians m = 0.20747527 radians per day dt = 45.5 days Whoops! This transit time does not equal the required time difference of 225.1 days! So, although the hyperbolic orbit does connect r1 with r2, a spaceship travelling along it would go from r1 to r2 too quickly, and the destination planet would not have arrived yet. So this orbit won't work. Proposed elliptical transfer orbit r2 = 1.00530074 e = 0.651493744 a = 1.318933507 AU Q2 = 4.32697753 radians sin u2 = -0.53583329 cos u2 = +0.36494920 u2 = 5.31030870 radians M2 = 5.84877378 radians P = 553.26446728 days m = 0.0113565676 radians per day dt = 238.4 days Hm, that's what I get for using orbital elements different from those I used in my last episode of working out this problem. Ok, the spaceship pilot sees that his trip will take a little too long, and so he applies a course correction. Jerry Abbott
  3. I define a subscript variable J to designate the non-apsidal endpoint of the intended trajectory. If the apside is at departure (K=1), then J=2. If the apside is at arrival (K=2), then J=1. We desire a canonical position vector for the non-apsidal endpoint of the intended trajectory. That is, we want to find the heliocentric vector to the non-apsidal endpoint in the coordinate system in which the XY plane contains the transfer orbit, with the +X axis extended through the transfer orbit's perihelion. xJ’ = yJ sin L + xJ cos L yJ’ = yJ cos L - xJ sin L zJ’ = zJ xJ’’ = xJ’ yJ’’ = zJ’ sin i + yJ’ cos i zJ’’ = zJ’ cos i - yJ’ sin i xJ’’’ = yJ’’ sin w + xJ’’ cos w yJ’’’ = yJ’’ cos w - xJ’’ sin w zJ’’’ = zJ’’ This triple-primed position vector is the one we were looking for. Important check: Within a reasonable allowance for roundoff error, the value of zJ’’’ should be zero. QJ = arctan2 (yJ’’’ , xJ’’’) Finding the true anomaly of the non-apsidal endpoint of the intended trajectory is a milestone in solving the transfer orbit problem. Plugging in the numbers... The proposed hyperbolic transfer orbit x1 = +0.5867243 y1 = -2.097687 z1 = -0.008034048 L = 6.17056860 radians i = 0.00398025345 radians w = 0 x1' = +0.81874324 y1' = -2.01846369 z1' = -0.008034048 x1'' = +0.81874324 y1'' = -2.01847968 z1'' = -8.50786288E-9 Notice how small z1'' is. Actually, it would be zero, but for roundoff error. x1''' = +0.81874324 y1''' = -2.01847968 z1''' = -8.50786288E-9 Since w=0 for this orbit, the triple-primed vector is equal to the double-primed vector. Q1 = 5.09773397 radians The proposed elliptical transfer orbit x2 = +0.9989326 y2 = -0.1129745 z2 = 0 L = 6.17104239 radians i = 0.00398025345 radians w = 1.95573400 radians x2' = +1.00530063 y2' = -4.76296468E-4 z2' = 0 x2'' = +1.00530063 y2'' = -4.76292695E-4 z2'' = +1.89577565E-6 x2''' = -0.37793320 y2''' = -0.93155573 z2''' = +1.89577565E-6 Q2 = 4.32697753 radians Jerry Abbott
  4. I write this post in regard to the essay in support of Intelligent Design having the title, "Was There Ever Nothing?" and found at http://everystudent.com/journeys/nothing.html The essay is nonsense. In particular,
  5. The inclination [symbolized as i] of an orbit is the angle between the plane of the orbit and the plane of the ecliptic. Herein the ecliptic plane is assumed to be the plane of Earth's orbit. These two planes intersect in a line called "the line of nodes." The orbit itself crosses the ecliptic at two points, called nodes. At one node, called the ascending node, the planet (or whatever the orbiting object happens to be) crosses the ecliptic with the Z component of its velocity in the direction of the North Ecliptic Pole. At the other node (the descending node), the planet crosses the ecliptic with the Z component of its velocity toward the South Ecliptic Pole. The heliocentric ecliptic coordinate system is defined such that the sun is at the origin, with +X axis extending toward the Vernal Equinox, and +Z axis normal to the Earth's orbit. It is a right-handed system, in which the +Y axis is 90 degrees counterclockwise from the +X axis to an observer positioned somewhere up the +Z axis looking back toward the origin. The longitude of the ascending node [L] is the angle, subtended at the sun (measured in the ecliptic counterclockwise by the aforementioned observer) from the Vernal Equinox to the ascending node. The argument of the perihelion [w] is the angle, subtended at the sun, measured in the plane of the orbit in the direction of motion, from the ascending node to the orbit's perihelion. The true anomaly [Q] is the angle, subtended at the sun, measured in the plane of the orbit in the direction of motion, from the orbit's perihelion to the current position of the planet. cos(w+QK) = (xK cos L + yK sin L) / rK if (i = 0) or (i = pi radians), then sin(w+QK) = (yK cos L - xK sin L) / rK if (i<>0) and (i<>pi radians), then sin(w+QK) = zK / (rK sin i) w = arctan2[ sin(w+QK) , cos(w+QK) ] - QK If necessary, adjust w to the interval [0, 2 pi). Plugging in the numbers... The proposed hyperbolic transfer orbit x2 = +0.9989326 y2 = -0.1129745 z2 = 0 r2 = 1.00530074 i = 0.00398025345 radians L = 6.17056860 radians cos(w+Q2) = 1.00000000 (to the limit of justified precision) sin(w+Q2) = 0 w + Q2 = 0 This is the Case #3, having perihelion at arrival, so... Q2 = 0 w = 0 The zero result for the argument of the perihelion should not surprise anyone. The arrival occurs at the perihelion of the transfer orbit and also in the plane of Earth's orbit - i.e., in the ecliptic. The ascending node and the perihelion are in the same spot, and thus the argument of the perihelion is zero. The proposed elliptical transfer orbit x1 = +0.5867243 y1 = -2.097687 z1 = -0.008034048 r1 = 2.17821044 i = 0.00398025345 radians L = 6.17104239 radians cos(w+Q1) = +0.37543977 sin(w+Q1) = -0.92666979 w + Q1 = 5.09732665 radians This is the Case #4, having aphelion at departure, so... Q1 = pi radians w = 1.95573400 radians Jerry Abbott
  6. The angular momentum per unit mass, h, is equal to the cross product of the heliocentric radius vector and the velocity vector at the apsidal endpoint: rK x VK. Angular momentum is a conserved quantity, and the components of h are constant for the entire transfer orbit. You will want to convert xK, yK, zK from astronomical units to meters. hX = yK VzK - zK VyK hY = zK VxK - xK VzK hZ = xK VyK - yK VxK h = [ (hX)^2 + (hY)^2 + (hZ)^2 ]^0.5 The longitude of the ascending node, L, of the transfer orbit can be calculated as follows: L = arctan2(hX , -hY) Plugging in the numbers... Proposed hyperbolic transfer orbit hX = -5.39471769E+12 m^2 sec^-2 hY = -4.77006702E+13 m^2 sec^-2 hZ = +1.19075189E+16 m^2 sec^-2 h = 1.19076157E+16 m^2 sec^-2 L = 6.17056860 radians Proposed elliptical transfer orbit hX = -1.72886746E+12 m^2 sec^-2 hY = -1.53519637E+13 m^2 sec^-2 hZ = +3.88213341E+15 m^2 sec^-2 h = 3.88216415E+15 m^2 sec^-2 L = 6.17104239 radians Comment. Strictly, the longitude of the ascending node for both of these proposed transfer orbits should be equal, as they both contain the same two heliocentric radius vectors and therefore occur in the same plane. The most likely reason that they are not quite equal is roundoff error. Their closeness provides a check on my work so far. Jerry Abbott
  7. I define a subscript variable K to designate the apsidal endpoint of the trajectory. When K=1, the apside is at departure. When K=2, the apside is at arrival. We must determine the sun-relative velocity of the spaceship at the apsidal end of its intended trajectory. Although it would be possible to use the general method shown in Post #12, there is another way to proceed in this special circumstance. We found a unit vector normal to the transfer orbit Rn = [Xn , Yn, Zn] in Post #13. We also have the heliocentric position vector to the apsidal endpoint, rK = [xK , yK, zK], which was calculated in Post #11. We will obtain the cross product Rn x rK. VxK'' = Yn zK - Zn yK VyK'' = Zn xK - Xn zK VzK'' = Xn yK - Yn xK We find the magnitude of this double-primed vector. VK'' = [ (VxK'')^2 + (VyK'')^2 + (VzK'')^2 ]^0.5 We divide the components of the double-primed vector by the magnitude of the double-primed vector, obtaining a unit vector in the direction of the velocity in the transfer orbit at the apsidal endpoint. VxK' = VxK'' / VK'' VyK' = VyK'' / VK'' VzK' = VzK'' / VK'' We refer to the Vis Viva equation to get the sun-relative speed in the transfer orbit at the apsidal endpoint. Elliptical orbits: VK = [ GMsun ( 2/rK - 1/a ) ]^0.5 Hyperbolic orbits: VK = [ GMsun ( 2/rK + 1/a ) ]^0.5 Again, remember to keep the units consistent. You may want to convert [a] and [rK] from astronomical units to meters, for example. VxK = VK VxK' VyK = VK VyK' VzK = VK VzK' The vector VK = [VxK , VyK , VzK] is the velocity in the transfer orbit at the apsidal endpoint of the intended trajectory, referred to heliocentric ecliptic coordinates. This method only works at apsides, not in general. It works at apsides because the heliocentric position vector and the sun-relative velocity are perpendicular to each other there. Plugging in the numbers... Possible hyperbolic transfer orbit Perihelion at arrival at Earth on JD 2453265.4 K=2 Xn = -0.0004472955 Yn = -0.0039550340 Zn = +0.9999920788 xK = +0.9989326 yK = -0.1129745 zK = 0 VxK'' = +0.11297361 VyK'' = +0.99892469 VzK'' = +0.00400135 VK'' = 1.00530074 VxK' = +0.11237792 VyK' = +0.99365757 VzK' = +0.00398025 GMsun = 1.32712440018E+20 m^3 sec^-2 1 astronomical unit = 1.49597870691E+11 meters a = 2.8444752E+10 meters rK = 1.50390850E+11 meters VK = 80190.5 m/s VxK = +9011.6 m/s VyK = +79681.9 m/s VzK = +319.2 m/s If this hyperbolic orbit turns out to be an actual transfer orbit - which we don't yet know for certain - we would determine the necessary delta-vee for matching velocity with Earth. Possible elliptical transfer orbit Aphelion at Departure at Vesta on JD 2453040.3 K=1 Xn = -0.0004472955 Yn = -0.0039550340 Zn = +0.9999920788 xK = +0.5867243 yK = -2.097687 zK = -0.008034048 VxK'' = +2.09770216 VyK'' = +0.58671606 VzK'' = +0.00325880 VK'' = 2.17821044 VxK' = +0.96303926 VyK' = +0.26935692 VzK' = +0.00149609 GMsun = 1.32712440018E+20 m^3 sec^-2 1 astronomical unit = 1.49597870691E+11 meters a = 1.97309644E+11 meters rK = 3.25855643E+11 meters VK = 11913.7 m/s VxK = +11473.4 m/s VyK = +3209.1 m/s VzK = +17.8 m/s If this elliptical orbit turns out to be an actual transfer orbit - which we don't yet know for certain - we would determine the necessary delta-vee for entering the transfer orbit from initially being at rest with respect to Vesta. Significance of our progress so far. We now have complete state vectors for two proposed transfer orbits at whichever endpoint of their intended trajectories occurs at an apside: [xK, yK, zK, VxK, VyK, VzK] The proposed hyperbolic transfer orbit has this state vector: x2 = +0.9989326 AU y2 = -0.1129745 AU z2 = 0 Vx2 = +9011.6 m/s Vy2 = +79681.9 m/s Vz2 = +319.2 m/s The proposed elliptical transfer orbit has this state vector: x1 = +0.5867243 AU y1 = -2.097687 AU z1 = -0.008034048 AU Vx1 = +11473.4 m/s Vy1 = +3209.1 m/s Vz1 = +17.8 m/s Jerry Abbott
  8. That's what the eMode test said after declaring me a Visionary Philosopher with a 142 IQ. We find the lengths of the three sides of a triangle having the sun, the point of departure, and the point of arrival as their vertices. r1 = [ (x1)^2 + (y1)^2 + (z1)^2 ]^0.5 r1 is the distance from the sun to the point of departure. r2 = [ (x2)^2 + (y2)^2 + (z2)^2 ]^0.5 r2 is the distance from the sun to the point of arrival. d = [ (x2-x1)^2 + (y2-y1)^2 + (z2-z1)^2 ]^0.5 d is the line-of-light separation between the points of departure and arrival. In order to get the eccentricity and semimajor axis of the transfer orbit from these three values, we must restrict our allowed transfer orbits to those having an apside at either endpoint of the intended trajectory. That is, perihelion or aphelion must occur at departure or at arrival. (Both instances of "or" in the previous sentence are exclusive. If one endpoint of the intended trajectory has an apside, the other endpoint may not have the other one, or this procedure will fail.) We proceed by considering, first, the Law of Cosines, understanding that the angle between r1 and r2 is the difference in true anomaly Q2-Q1. cos(Q2-Q1) = ( r1^2 + r2^2 - d^2 ) / (2 r1 r2) Secondly, we solve the polar equation of the orbit, with the pole at the solar focus, for the cosine of the true anomaly. cos Q = [ a (1 - e^2) - r ] / (e r) There are four general cases for a subsequent solution: 1. Departure occurs from the perihelion of the transfer orbit. Outward bound trajectory, Q1 = 0. 2. Arrival occurs at the aphelion of the transfer orbit. Outward bound trajectory, Q2 = pi radians. 3. Arrival occurs at the perihelion of the transfer orbit. Inward bound trajectory, Q2 = 0. 4. Departure occurs from the aphelion of the transfer orbit. Inward bound trajectory, Q1 = pi radians. Now, I'm not about to solve all four cases. I'm just going to solve case #1 explicitly, to provide proof of principle. You can solve the other three cases yourself, if you want to check my work. I'm only going to provide the results for cases #2, #3, and #4. INSIST Q1=0. Therefore, r1 = a(1-e). Why? Because Q1=0 means that departure occurs at the transfer orbit's perihelion, and the heliocentric distance at perihelion equals the product of the semimajor axis and of one minus the eccentricity. We thus have two equations for cos Q2, namely, cos Q2 = ( r1^2 + r2^2 - d^2 ) / (2 r1 r2) cos Q2 = [ r1 (1-e^2) / (1-e) - r2 ] / (e r2) Meaning that... e r2 ( r1^2 + r2^2 - d^2 ) = 2 r1^2 r2 (1+e) - 2 r1 r2^2 ...solve, solve, solve... e = 2 r1 (r1 - r2) / (r2^2 - r1^2 - d^2) a = r1 / (1-e) How easy that was. Case 1. Perihelion at Departure: Q1 = 0. e = 2 r1 (r1 - r2) / ( r2^2 - r1^2 - d^2 ) If e is in [0,1) then an elliptical transfer orbit exists, and a = r1 / (1-e) If e>1 then a hyperbolic transfer orbit exists, and a = r1 / (e-1) If e<0 then there is no transfer orbit having perihelion at departure. Case 2. Aphelion at Arrival: Q2 = pi radians. e = 2 r2 (r1 - r2) / ( r1^2 - r2^2 - d^2 ) If e is in [0,1) then an elliptical transfer orbit exists, and a = r2 / (1+e) If e is not in [0,1) then there is no transfer orbit having aphelion at arrival. Case 3. Perihelion at Arrival: Q2 = 0. e = 2 r2 (r2 - r1) / ( r1^2 - r2^2 - d^2 ) If e is in [0,1) then an elliptical transfer orbit exists, and a = r2 / (1-e) If e>1 then a hyperbolic transfer orbit exists, and a = r2 / (e-1) If e<0 then there is no transfer orbit having perihelion at arrival. Case 4. Aphelion at Departure: Q1 = pi radians. e = 2 r1 (r2 - r1) / ( r2^2 - r1^2 - d^2 ) If e is in [0,1) then an elliptical transfer orbit exists, and a = r1 / (1+e) If e is not in [0,1) then no transfer orbit exists having aphelion at departure. Since our example problem involves a transfer from Vesta to Earth, the trajectory will be inward bound and the transfer orbit we seek will be found in either Case #3 or Case #4. Plugging in the numbers... r1 = 2.178210435 r2 = 1.005300740 d = 2.027082617 Case #3. e = 6.287121148 a = 0.190141423 A hyperbolic orbit, with perihelion at arrival point. Case #4. e = 0.651493744 a = 1.318933507 An elliptical orbit, with aphelion at departure point. It is not at all likely that both of these orbits are valid transfer orbits. One of them might be (ok, is...I've already checked), but the other will turn out to have an inappropriate transit time. A spaceship using the wrong orbit will reach the "arrival point" either much sooner or much later than the Earth does, which would be kind of foolish. The spaceship pilot will have to choose the right orbit, if he wants to remain employed. Jerry Abbott
  9. Given two heliocentric position vectors to points on an orbit, excepting a pair differing in true anomaly by pi radians, it is possible to determine the inclination of the orbit to the ecliptic. Position vector 1, R1, has these components. x1 = +0.5867243 y1 = -2.097687 z1 = -0.008034048 Position vector 2, R2, has these components. x2 = +0.9989326 y2 = -0.1129745 z2 = 0 We find the cross product, R1xR2. Xn' = y1 z2 - z1 y2 Yn' = z1 x2 - x1 z2 Zn' = x1 y2 - y1 x2 We find the magnitude of the primed normal vector, Rn'. Rn' = [ (Xn')^2 + (Yn')^2 + (Zn')^2 ]^0.5 We divide the primed vector's components each by the magnitude of the primed vector. Xn = Xn' / Rn' Yn = Yn' / Rn' Zn = Zn' / Rn' The vector [Xn, Yn, Zn] is a unit normal vector with respect to the transfer orbit. i = pi/2 - arcsin(Zn) Where is the inclination of the transfer orbit to the ecliptic plane. Plugging in the numbers... Xn' = -0.0009076426 Yn' = -0.0080254725 Zn' = +2.0291630445 Rn' = 2.02917911803 Xn = -0.0004472955 Yn = -0.0039550340 Zn = +0.9999920788 i = 0.00398025345 radians Jerry Abbott
  10. There are probably environmental factors in the formation of character, but the genetic factors are probably dominant. Intelligence is about 80% inherited (Arthur Jensen, "How Much Can We Boost IQ and Scholastic Achievement?", Harvard Educational Review, 1969). Metabolic efficiency is probably inherited, too. Harmonal balances, likewise. If, say, one race had more homone-induced aggression and a lower intelligence than another, then it would not be surprising if members of the (relatively) stupid and aggressive race were arrested more frequently for violent crimes. Cloning wouldn't be used universally. It would be reserved for those persons having an exceptionally good genetic endowment, as measured by objective tests of physical and mental abilities. Jerry Abbott
  11. France has limited domestic energy resources, and dependence on imported fuels is a politically hot topic there. France generates 77% of its electricity from nuclear reactors. It's automobiles and tractors run on the same kind of gasoline we use. France gets its plastics and its agricultural fertilizers from petroleum, just as we do. Fossil fuels currently provide 3/4 of the world's energy. In order to supplant fossil fuels with nuclear reactors, it would be necessary to build 5,000 to 10,000 new reactors, and there is no longer time in which this can be done. I predict that you will soon live in a world in which working cars and trucks are much less abundant than they are today, due to the depletion of fossil fuels. Eventually, you'll also see the breakdown of industrial agriculture and the depopulation of the world by starvation. No, it isn't. A dysgenic trend in the White race from the beginning of the 20th century, is obvious. The principal causes of that trend are wars, feminism, and liberal social programs. When the fitter men were sent to war, inspired by largely fictitious propaganda to kill each other, other men, being mentally and physically unfit for military service, stayed home and enjoyed enhanced access to women. Feminism took the smarter and more motivated women out of the home and put them into the workplace. Instead of being mothers first, they became [put job title here] first, and motherhood often had to wait, and wait, and...oops, too late. The number of children per woman declined. The White birthrate for the past 50 years has been below replacement, with the shortfall in births being mainly in the genetically better women. Feminism culled good qualities out of our race worse than the wars did. Liberal social programs burdened the more capable people so that the less capable could have unearned money. Politically, the idea was sold as "safety nets," but what it really did was transfer childbirth from the better human stock to the lower grades. Workers had to carry non-workers on their backs, paying to raise their children as well as any they had themselves. Why do you suppose "undesirables" are undesired? Because they make trouble. They cause problems. At best' date=' they are a burden that no one wants to carry and should not be required to. At worst, they are an active evil that seeks the destruction of the host people who unwisely let them in, being to a nation what a virus is to an organism. It's a defect in one's judgment to imply that the Nazis always got everything wrong. Maybe once in a while they did the right thing. When you really think about results - "what are the longterm consequences of these trends?" - you will usually arrive at the conclusion that the Nazis were right about eugenics. Normal reproduction should be controlled on the basis of genetic quality, with seriously defective persons being permitted no children at all, average people being permitted one or two, somewhat gifted people being allowed three or four, and, for the very best, "no limit", incentives to have more babies, and cloning to insure that an especially propitious genotype does not perish from the world.
  12. Maybe each universe is made of all the others. We know that in space there's a continuous flux of zero energy in vacuum. It goes on everywhere and all the time. But suppose we're missing a connection, namely, that this flux of other universes is what our universe is composed of. I think that Stephen Hawking's self-contained "the universe would just BE" idea might have an element of truth. However, instead of this singular universe just being, it's the whole system of tangent universes just being. In some other universe, ours appears as a transient vacuum fluctuation: it's there, somewhere and at some moment, but for all practical purposes beneath notice. Jerry Abbott
  13. Refer to my previous post for nomenclature. We will eventually want to know the velocity of Vesta at departure and the velocity of Earth at arrival, in order to calculate the necessary delta-vee at each end of the intended trajectory. Let Q be the true anomaly of an object in its orbit around the sun. (I don't have access to the greek letters here.) To determine its value: Q = arctan2(y''' , x''') Where x''' and y''' are two of the components of the canonical position vector (see previous post). The arctan2 function is the two-argument arctangent function. It is related to the single-argument arctan function as follows: Function arctan2(y,x); begin if x=0 then begin if y>0 then Q:=+pi/2 if y=0 then Q:=0 if y<0 then Q:=-pi/2 end else begin Q:=arctan(y/x) if x<0 then Q:=Q+pi if x>0 and y<0 then Q:=Q+2 pi end arctan2:=Q end GMsun is the solar gravitational constant. GMsun = 1.32712440018E+20 m^3 sec^-2 When using the equations below, make sure to keep the units consistant. You will probably want to enter the semimajor axis in meters, rather than attempt to adjust GMsun. 1 astronomical unit = 1.49597870691E+11 meters. Vx''' = -sin Q { GMsun / [ a (1 - e^2) ] }^0.5 Vy''' = (e + cos Q) { GMsun / [ a (1 - e^2) ] }^0.5 Vz''' = 0 In order to adjust this canonical velocity vector to heliocentric ecliptic coordinates, you'd rotate it in the same way that a position vector would be rotated. Again, see my preceding post. Plugging in the numbers... Orbital velocity of Vesta at JD 2453040.3 x''' = +1.857826 y''' = +1.137138 Q = 0.5492546 radians a = 2.36160912 e = 0.08871313 Canonical velocity Vx''' = -10158.3 m/s Vy''' = +18322.5 m/s Vz''' = 0 ...intermediate steps skipped... HEC velocity Vx = +20241.4 m/s Vy = +4735.7 m/s Vz = -2601.2 m/s Orbital velocity of Earth at JD 2453265.4 x''' = -0.3339160 y''' = -0.9482244 Q = 4.37380123 radians a = 1.00000011 e = 0.01671022 Canonical velocity Vx''' = +28097.7 m/s Vy''' = -9396.8 m/s Vz''' = 0 ...intermediate steps skipped... HEC velocity Vx = +2862.5 m/s Vy = +29488.7 m/s Vz = 0 Jerry Abbott
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