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#1




The eccentricity constant of solar objects
The eccentricity constant of solar objects
The eccentricity constant X of solar objects can be calculated by the formula .5*sqrt(43(ab)^2/(a+b)^2) where a = the semi major axis and b = the semi minor axis. The eccentricity constant X of nine planets is equal to 1.0 as follows: semi major axis a semi minor axis b MER â€¦..57,909,175,000 .â€¦56,671,636,475 VEN â€¦108,208,930,000 ...108,206,447,840 EAR â€¦149,597,890,000 ...149,577,002,324 MAR â€¦227,936,640,000 ...226,939,989,085 JUP â€¦778,412,010,000 ...777,500,013,843 SAT 1,426,725,400,000 1,424,632,079,805 URA 2,870,972,200,000 2,867,776,762,478 NEP 4,498,252,900,000 4,498,087,097,722 PLU 5,906,380,000,000 5,720,641,563,568 â€¦â€¦â€¦.ab â€¦â€¦â€¦â€¦..a+b X= MER â€¦1237538525 .....114,580,811,475 1.0 VEN â€¦â€¦..2482160 .....216,415,377,840 1.0 EAR â€¦â€¦20887676 .....299,174,892,324 1.0 MAR â€¦..996650915 .....454,876,629,085 1.0 JUP â€¦..911996157 ..1,555,912,023,843 1.0 SAT â€¦2093320195 ..2,851,357,479,805 1.0 URA â€¦3195437522 ..5,738,748,962,478 1.0 NEP â€¦..165802278 ..8,996,339,997,722 1.0 PLU 185738436432 11,627,021,563,568 1.0 
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#2




The eccentricity constant of solar objects
Dne 03/01/2018 v 04:21 Peter Riedt napsal(a):
The eccentricity constant of solar objects The eccentricity constant X of solar objects can be calculated by the formula .5*sqrt(43(ab)^2/(a+b)^2) where a = the semi major axis and b = the semi minor axis. The eccentricity constant X of nine planets is equal to 1.0 as follows: So you say a circle has the eccentricity constant 1.0. Interesting. ..5*sqrt(43(rr)^2/(r+r)^2) = .5*sqrt(4) = 1  Poutnik ( The Pilgrim, Der Wanderer ) A wise man guards words he says, as they say about him more, than he says about the subject. 
#3




The eccentricity constant of solar objects
Dne 03/01/2018 v 09:12 Libor 'Poutnik' StÅ™Ã*Å¾ napsal(a):
Dne 03/01/2018 v 04:21 Peter Riedt napsal(a): The eccentricity constant of solar objects The eccentricity constant X of solar objects can be calculated by the formula .5*sqrt(43(ab)^2/(a+b)^2) where a = the semi major axis and b = the semi minor axis. The eccentricity constant X of nine planets is equal to 1.0 as follows: So you say a circle has the eccentricity constant 1.0. Interesting. .5*sqrt(43(rr)^2/(r+r)^2) = .5*sqrt(4) = 1 Similarly, any ellipse similar enough to a circle like those of planets has this constant 1.0, if rounded to 1 decimal place.  Poutnik ( The Pilgrim, Der Wanderer ) A wise man guards words he says, as they say about him more, than he says about the subject. 
#4




The eccentricity constant of solar objects
On Wednesday, January 3, 2018 at 4:23:40 PM UTC+8, Libor 'Poutnik' StÅ™Ã*Å¾ wrote:
Dne 03/01/2018 v 09:12 Libor 'Poutnik' StÅ™Ã*Å¾ napsal(a): Dne 03/01/2018 v 04:21 Peter Riedt napsal(a): The eccentricity constant of solar objects The eccentricity constant X of solar objects can be calculated by the formula .5*sqrt(43(ab)^2/(a+b)^2) where a = the semi major axis and b = the semi minor axis. The eccentricity constant X of nine planets is equal to 1.0 as follows: So you say a circle has the eccentricity constant 1.0. Interesting. .5*sqrt(43(rr)^2/(r+r)^2) = .5*sqrt(4) = 1 Similarly, any ellipse similar enough to a circle like those of planets has this constant 1.0, if rounded to 1 decimal place.  Poutnik ( The Pilgrim, Der Wanderer ) A wise man guards words he says, as they say about him more, than he says about the subject. I have improved the formula to read 13(ab)^2/(a+b)^2). All objects in closed orbits around the sun have an X constant between .8 and 1. Halley's comet has an X of about .85. A circular orbit has an X constant of 1. Orbits are subject to the Law of X. 
#5




The eccentricity constant of solar objects
Dne 03/01/2018 v 16:18 Peter Riedt napsal(a):
On Wednesday, January 3, 2018 at 4:23:40 PM UTC+8, Libor 'Poutnik' StÅ™Ã*Å¾ wrote: Dne 03/01/2018 v 09:12 Libor 'Poutnik' StÅ™Ã*Å¾ napsal(a): Dne 03/01/2018 v 04:21 Peter Riedt napsal(a): The eccentricity constant of solar objects The eccentricity constant X of solar objects can be calculated by the formula .5*sqrt(43(ab)^2/(a+b)^2) where a = the semi major axis and b = the semi minor axis. The eccentricity constant X of nine planets is equal to 1.0 as follows: So you say a circle has the eccentricity constant 1.0. Interesting. .5*sqrt(43(rr)^2/(r+r)^2) = .5*sqrt(4) = 1 Similarly, any ellipse similar enough to a circle like those of planets has this constant 1.0, if rounded to 1 decimal place. I have improved the formula to read 13(ab)^2/(a+b)^2). All objects in closed orbits around the sun have an X constant between .8 and 1. Halley's comet has an X of about .85. A circular orbit has an X constant of 1. Orbits are subject to the Law of X. Where is the improvement wrt the eccentricity e = sqrt(1(b/a)^2 ) ?  Poutnik ( The Pilgrim, Der Wanderer ) A wise man guards words he says, as they say about him more, than he says about the subject. 
#6




The eccentricity constant of solar objects
Dne 03/01/2018 v 21:00 Libor 'Poutnik' StÅ™Ã*Å¾ napsal(a):
Dne 03/01/2018 v 16:18 Peter Riedt napsal(a): On Wednesday, January 3, 2018 at 4:23:40 PM UTC+8, Libor 'Poutnik' StÅ™Ã*Å¾ wrote: Dne 03/01/2018 v 09:12 Libor 'Poutnik' StÅ™Ã*Å¾ napsal(a): Dne 03/01/2018 v 04:21 Peter Riedt napsal(a): The eccentricity constant of solar objects The eccentricity constant X of solar objects can be calculated by the formula .5*sqrt(43(ab)^2/(a+b)^2) where a = the semi major axis and b = the semi minor axis. The eccentricity constant X of nine planets is equal to 1.0 as follows: So you say a circle has the eccentricity constant 1.0. Interesting. .5*sqrt(43(rr)^2/(r+r)^2) = .5*sqrt(4) = 1 Similarly, any ellipse similar enough to a circle like those of planets has this constant 1.0, if rounded to 1 decimal place. I have improved the formula to read 13(ab)^2/(a+b)^2). All objects in closed orbits around the sun have an X constant between .8 and 1. Halley's comet has an X of about .85. A circular orbit has an X constant of 1. Orbits are subject to the Law of X. Where is the improvement wrt the eccentricity e = sqrt(1(b/a)^2 ) ? P.S.: As that has the very particular geometrical meaning of relative position of an ellipse focus on the major semiaxis.  Poutnik ( The Pilgrim, Der Wanderer ) A wise man guards words he says, as they say about him more, than he says about the subject. 
#7




The eccentricity constant of solar objects
On Thursday, January 4, 2018 at 4:19:46 AM UTC+8, Libor 'Poutnik' StÅ™Ã*Å¾ wrote:
Dne 03/01/2018 v 21:00 Libor 'Poutnik' StÅ™Ã*Å¾ napsal(a): Dne 03/01/2018 v 16:18 Peter Riedt napsal(a): On Wednesday, January 3, 2018 at 4:23:40 PM UTC+8, Libor 'Poutnik' StÅ™Ã*Å¾ wrote: Dne 03/01/2018 v 09:12 Libor 'Poutnik' StÅ™Ã*Å¾ napsal(a): Dne 03/01/2018 v 04:21 Peter Riedt napsal(a): The eccentricity constant of solar objects The eccentricity constant X of solar objects can be calculated by the formula .5*sqrt(43(ab)^2/(a+b)^2) where a = the semi major axis and b = the semi minor axis. The eccentricity constant X of nine planets is equal to 1.0 as follows: So you say a circle has the eccentricity constant 1.0. Interesting. .5*sqrt(43(rr)^2/(r+r)^2) = .5*sqrt(4) = 1 Similarly, any ellipse similar enough to a circle like those of planets has this constant 1.0, if rounded to 1 decimal place. I have improved the formula to read 13(ab)^2/(a+b)^2). All objects in closed orbits around the sun have an X constant between .8 and 1. Halley's comet has an X of about .85. A circular orbit has an X constant of 1. Orbits are subject to the Law of X. Where is the improvement wrt the eccentricity e = sqrt(1(b/a)^2 ) ? P.S.: As that has the very particular geometrical meaning of relative position of an ellipse focus on the major semiaxis.  Poutnik ( The Pilgrim, Der Wanderer ) A wise man guards words he says, as they say about him more, than he says about the subject. I have simplified the formula by removing the sqrt part. 
#8




The eccentricity constant of solar objects
Dne 04/01/2018 v 00:35 Peter Riedt napsal(a):
On Thursday, January 4, 2018 at 4:19:46 AM UTC+8, Libor 'Poutnik' StÅ™Ã*Å¾ wrote: I have improved the formula to read 13(ab)^2/(a+b)^2). All objects in closed orbits around the sun have an X constant between .8 and 1. Halley's comet has an X of about .85. A circular orbit has an X constant of 1. Orbits are subject to the Law of X. Where is the improvement wrt the eccentricity e = sqrt(1(b/a)^2 ) ? P.S.: As that has the very particular geometrical meaning of relative position of an ellipse focus on the major semiaxis. I have simplified the formula by removing the sqrt part. It is oversimplified in a not useful way. Similarly as if you would like to get rid of sqrt in the formula of Pythagoras.  Poutnik ( The Pilgrim, Der Wanderer ) A wise man guards words he says, as they say about him more, than he says about the subject. 
#9




The eccentricity constant of solar objects
On Thursday, January 4, 2018 at 3:39:00 PM UTC+8, Libor 'Poutnik' StÅ™Ã*Å¾ wrote:
Dne 04/01/2018 v 00:35 Peter Riedt napsal(a): On Thursday, January 4, 2018 at 4:19:46 AM UTC+8, Libor 'Poutnik' StÅ™Ã*Å¾ wrote: I have improved the formula to read 13(ab)^2/(a+b)^2). All objects in closed orbits around the sun have an X constant between .8 and 1. Halley's comet has an X of about .85. A circular orbit has an X constant of 1.. Orbits are subject to the Law of X. Where is the improvement wrt the eccentricity e = sqrt(1(b/a)^2 ) ? P.S.: As that has the very particular geometrical meaning of relative position of an ellipse focus on the major semiaxis. I have simplified the formula by removing the sqrt part. It is oversimplified in a not useful way. Similarly as if you would like to get rid of sqrt in the formula of Pythagoras.  Poutnik ( The Pilgrim, Der Wanderer ) A wise man guards words he says, as they say about him more, than he says about the subject. sqrt is not necessary for my formula but irreplaceable in Pythagoras 
#10




The eccentricity constant of solar objects
Peter Riedt wrote:
On Wednesday, January 3, 2018 at 4:23:40 PM UTC+8, Libor 'Poutnik' St?í? wrote: Dne 03/01/2018 v 09:12 Libor 'Poutnik' St?í? napsal(a): Dne 03/01/2018 v 04:21 Peter Riedt napsal(a): The eccentricity constant of solar objects The eccentricity constant X of solar objects can be calculated by the formula .5*sqrt(43(ab)^2/(a+b)^2) where a = the semi major axis and b = the semi minor axis. The eccentricity constant X of nine planets is equal to 1.0 as follows: I'm curious: Where does that formula come from? What *is* the eccentricity constant in the first place) How does it relate to e = sqrt(1(b/a)^2) ? You listed the semi minor axes (b) of the planets with 11 to 13 significant digits. Impressive!. Where did you get those numbers? (I have a hunch...) Even 8 digits for the semi major axes is quite a feat. Why do you round off to one decimal? Just to prove a point? Isn't it more interesting to explore the differences? 3 decimals? So you say a circle has the eccentricity constant 1.0. Interesting. .5*sqrt(43(rr)^2/(r+r)^2) = .5*sqrt(4) = 1 Similarly, any ellipse similar enough to a circle like those of planets has this constant 1.0, if rounded to 1 decimal place. Actually, the eccentricity of a circle is 0 (zero). Poutnik should maybe have pointed that out to you... I have improved the formula to read 13(ab)^2/(a+b)^2). How does .5*sqrt(43(ab)^2/(a+b)^2) simplfy to 13(ab)^2/(a+b)^2) ? Let's do the algebra. Oh bummer! ..5*sqrt(4) = 1 !!! Trust me, you'd fail any high school math test with that one (pun not intended). Orbits are subject to the Law of X. What's the Law of X.?  I recommend Macs to my friends, and Windows machines to those whom I don't mind billing by the hour 
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