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#21
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The eccentricity constant of solar objects
Peter Riedt wrote:
On Friday, January 5, 2018 at 3:20:27 AM UTC+8, Anders Eklöf wrote: 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(4-3(a-b)^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(4-3(r-r)^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 1-3(a-b)^2/(a+b)^2). How does .5*sqrt(4-3(a-b)^2/(a+b)^2) simplfy to 1-3(a-b)^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.? Satellites ecc 1-3(a-b)^2/(a+b)^2) Moon 0.0549 0.99999978624 Io 0.0041 0.99999999999 Europa 0.0090 0.99999999985 Ganymed 0.0013 1.00000000000 Calli 0.0074 0.99999999993 Mimas 0.0202 0.99999999607 Encela 0.0047 0.99999999999 Tethys 0.0200 0.99999999624 Dione 0.0020 1.00000000000 Rhea 0.0010 1.00000000000 Comets Halley 0.9670 0.85758592313 Encke 0.8470 0.96428781552 Tempel1 0.5190 0.99769836444 Planets MER 0.2056 0.99995625439 VEN 0.0068 0.99999999995 EAR 0.0167 0.99999999817 MAR 0.0934 0.99999819976 JUP 0.0484 0.99999987116 SAT 0.0542 0.99999979788 URA 0.0472 0.99999988373 NEP 0.0086 0.99999999987 PLU 0.2488 0.99990429852 Asteroid Pallas 0.2313 0.99992918251 I repeat: Where does that formula (.5*sqrt(4-3(a-b)^2/(a+b)^2))come from? What *is* the eccentricity constant (X) in the first place? Where do your 11-13 digit values for the semi minor axis com from? How does .5*sqrt(4-3(a-b)^2/(a+b)^2) simplfy to 1-3(a-b)^2/(a+b)^2) ? -- I recommend Macs to my friends, and Windows machines to those whom I don't mind billing by the hour |
#22
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The eccentricity constant of solar objects
Peter Riedt wrote:
Your points are valid. However, the formula using .5*sqrt(4)... produces the same result as using 1-3.... No - they don't, except for a circle. For Mercury 5*sqrt(4-3(a-b)^2/(a+b)^2) gives 0,999956 and 1-3(a-b)^2/(a+b)^2) gives 0.999650. The difference doesn't look big, but the devation from 1 differs by an order of magnitude. The comet Halley produces .85 for X. Only with 1-3(a-b)^2/(a+b)^2) as you listed. Just try using .5*sqrt(4-3(a-b)^2/(a+b)^2) instead. Since I don't have your values for a and b I can't check. -- I recommend Macs to my friends, and Windows machines to those whom I don't mind billing by the hour |
#23
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The eccentricity constant of solar objects
On Sunday, January 7, 2018 at 7:08:23 AM UTC+8, Anders Eklöf wrote:
Peter Riedt wrote: Your points are valid. However, the formula using .5*sqrt(4)... produces the same result as using 1-3.... No - they don't, except for a circle. For Mercury 5*sqrt(4-3(a-b)^2/(a+b)^2) gives 0,999956 and 1-3(a-b)^2/(a+b)^2) gives 0.999650. The difference doesn't look big, but the devation from 1 differs by an order of magnitude. The comet Halley produces .85 for X. Only with 1-3(a-b)^2/(a+b)^2) as you listed. Just try using .5*sqrt(4-3(a-b)^2/(a+b)^2) instead. Since I don't have your values for a and b I can't check. -- I recommend Macs to my friends, and Windows machines to those whom I don't mind billing by the hour The values for the semi major axis were obtained from Princeton.edu and the values for the semi minor axis were calculated by me with the formula semi minor axis = semi major axis * sqrt(1-e^2): smajora smina e MER 57,909,231,029 56,672,064,712 0.2056 VEN 108,209,525,401 108,207,023,568 0.0068 EAR 149,598,319,494 149,577,457,301 0.0167 MAR 227,943,771,564 226,947,353,141 0.0934 JUP 778,342,761,465 777,430,569,626 0.0484 SAT 1,426,714,892,866 1,424,617,764,212 0.0542 URA 2,870,633,540,862 2,867,434,101,795 0.0472 NEP 4,498,393,012,162 4,498,226,658,512 0.0086 PLU 5,906,438,090,764 5,720,709,449,730 0.2488 The two formulas for X differ indeed: .05*sqrt(4-3(a-b)^2/(a+b)^2) 1-3(a-b)^2/(a+b)^2) MER 0.999956281 0.999650256 VEN 1.000000000 1.000000000 EAR 0.999999998 0.999999985 MAR 0.999998201 0.999985606 JUP 0.999999871 0.999998969 SAT 0.999999797 0.999998377 URA 0.999999883 0.999999067 NEP 1.000000000 0.999999999 PLU 0.999904311 0.999234522 |
#24
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The eccentricity constant of solar objects
On 03/01/2018 15:18, Peter Riedt wrote:
On Wednesday, January 3, 2018 at 4:23:40 PM UTC+8, Libor 'Poutnik' StÅ™Ã*ž wrote: Dne 03/01/2018 v 09:12 Libor 'Poutnik' StÅ™Ã*ž 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(4-3(a-b)^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(4-3(r-r)^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 1-3(a-b)^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. It isn't a law though it is a mutilated form of eccentricity which is already defined geometrically from the parameters a,b of an ellipse as: e^2 = (1-(b/a)^2) Hence e^2a^2 = a^2 - b^2 b = a.sqrt(1-e^2) This can be stuffed into your so called "law" to understand why it maps all low eccentricity orbits to very approximately 1. (and insanity check the limiting case of e=1 a parabola is X = -2). "X" = 1 -3a(1-sqrt(1-e^2))^2/a(1+sqrt(1-e^2))^2 = 1 - 3*(2-e^2 -2sqrt(1-e^2))/(2-e^2+2sqrt(1-e^2)) Taking sqrt(1-e^2) taylor series expansion for small e as sqrt(1-e^2) ~ 1 - e^2/2 - 1/8e^4 for small e = 1 - 3*( 2-e^2-2+e^2+1/8e^4)/(4-2e^2) ~ 1 - 3e^4/16/(2-e^2) So for small eccentricity e it looks like 1 - e^4/32 For the record at larger e it has no merit either. "X" = e at about e= 0.80481 and "X"=0 at about e= 0.963433 Neither of these having any physical significance. -- Regards, Martin Brown |
#25
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The eccentricity constant of solar objects
Dne 07/01/2018 v 05:01 Peter Riedt napsal(a):
The values for the semi major axis were obtained from Princeton.edu and the values for the semi minor axis were calculated by me with the formula semi minor axis = semi major axis * sqrt(1-e^2): If looking at https://www.princeton.edu/~willman/p...y_systems/Sol/ ( as you did not provided a particular page ), the semi major axis are not as accurately known as you provided, neither the excentricity. So you are just throwing numbers without much sense in it. smajora smina e MER 57,909,231,029 56,672,064,712 0.2056 VEN 108,209,525,401 108,207,023,568 0.0068 EAR 149,598,319,494 149,577,457,301 0.0167 MAR 227,943,771,564 226,947,353,141 0.0934 JUP 778,342,761,465 777,430,569,626 0.0484 SAT 1,426,714,892,866 1,424,617,764,212 0.0542 URA 2,870,633,540,862 2,867,434,101,795 0.0472 NEP 4,498,393,012,162 4,498,226,658,512 0.0086 PLU 5,906,438,090,764 5,720,709,449,730 0.2488 The two formulas for X differ indeed: .05*sqrt(4-3(a-b)^2/(a+b)^2) 1-3(a-b)^2/(a+b)^2) MER 0.999956281 0.999650256 VEN 1.000000000 1.000000000 EAR 0.999999998 0.999999985 MAR 0.999998201 0.999985606 JUP 0.999999871 0.999998969 SAT 0.999999797 0.999998377 URA 0.999999883 0.999999067 NEP 1.000000000 0.999999999 PLU 0.999904311 0.999234522 Cannot you see that a nearly constant X parameter value means the parameter is useless ? -- Poutnik ( The Pilgrim, Der Wanderer ) A wise man guards words he says, as they say about him more, than he says about the subject. |
#26
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The eccentricity constant of solar objects
On Sunday, January 7, 2018 at 10:16:22 PM UTC+8, Martin Brown wrote:
On 03/01/2018 15:18, Peter Riedt wrote: On Wednesday, January 3, 2018 at 4:23:40 PM UTC+8, Libor 'Poutnik' StÅ™Ã*ž wrote: Dne 03/01/2018 v 09:12 Libor 'Poutnik' StÅ™Ã*ž 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(4-3(a-b)^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(4-3(r-r)^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 1-3(a-b)^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. It isn't a law though it is a mutilated form of eccentricity which is already defined geometrically from the parameters a,b of an ellipse as: e^2 = (1-(b/a)^2) Hence e^2a^2 = a^2 - b^2 b = a.sqrt(1-e^2) This can be stuffed into your so called "law" to understand why it maps all low eccentricity orbits to very approximately 1. (and insanity check the limiting case of e=1 a parabola is X = -2). "X" = 1 -3a(1-sqrt(1-e^2))^2/a(1+sqrt(1-e^2))^2 = 1 - 3*(2-e^2 -2sqrt(1-e^2))/(2-e^2+2sqrt(1-e^2)) Taking sqrt(1-e^2) taylor series expansion for small e as sqrt(1-e^2) ~ 1 - e^2/2 - 1/8e^4 for small e = 1 - 3*( 2-e^2-2+e^2+1/8e^4)/(4-2e^2) ~ 1 - 3e^4/16/(2-e^2) So for small eccentricity e it looks like 1 - e^4/32 For the record at larger e it has no merit either. "X" = e at about e= 0.80481 and "X"=0 at about e= 0.963433 Neither of these having any physical significance. -- Regards, Martin Brown ok, thanks. |
#27
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The eccentricity constant of solar objects
Peter Riedt wrote:
On Sunday, January 7, 2018 at 7:08:23 AM UTC+8, Anders Eklöf wrote: Peter Riedt wrote: Your points are valid. However, the formula using .5*sqrt(4)... produces the same result as using 1-3.... No - they don't, except for a circle. For Mercury 5*sqrt(4-3(a-b)^2/(a+b)^2) gives 0,999956 and 1-3(a-b)^2/(a+b)^2) gives 0.999650. The difference doesn't look big, but the devation from 1 differs by an order of magnitude. The comet Halley produces .85 for X. Only with 1-3(a-b)^2/(a+b)^2) as you listed. Just try using .5*sqrt(4-3(a-b)^2/(a+b)^2) instead. Since I don't have your values for a and b I can't check. The values for the semi major axis were obtained from Princeton.edu and the values for the semi minor axis were calculated by me with the formula semi minor axis = semi major axis * sqrt(1-e^2): So, since you use the eccentricity e to calculate the semi major axis, what is the point of introducing a new "eccentricity constant" X? 1. What is the geometrical mening of X? 2. In what way is it useful? smajora smina e MER 57,909,231,029 56,672,064,712 0.2056 VEN 108,209,525,401 108,207,023,568 0.0068 EAR 149,598,319,494 149,577,457,301 0.0167 MAR 227,943,771,564 226,947,353,141 0.0934 JUP 778,342,761,465 777,430,569,626 0.0484 SAT 1,426,714,892,866 1,424,617,764,212 0.0542 URA 2,870,633,540,862 2,867,434,101,795 0.0472 NEP 4,498,393,012,162 4,498,226,658,512 0.0086 PLU 5,906,438,090,764 5,720,709,449,730 0.2488 The two formulas for X differ indeed: .05*sqrt(4-3(a-b)^2/(a+b)^2) 1-3(a-b)^2/(a+b)^2) MER 0.999956281 0.999650256 VEN 1.000000000 1.000000000 EAR 0.999999998 0.999999985 MAR 0.999998201 0.999985606 JUP 0.999999871 0.999998969 SAT 0.999999797 0.999998377 URA 0.999999883 0.999999067 NEP 1.000000000 0.999999999 PLU 0.999904311 0.999234522 So much for you simplification of the formula. Can you still not see where the error is? You say my points are valid, and choose to ignore them. -- I recommend Macs to my friends, and Windows machines to those whom I don't mind billing by the hour |
#28
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The eccentricity constant of solar objects
On Monday, January 8, 2018 at 4:44:46 AM UTC+8, Anders Eklöf wrote:
Peter Riedt wrote: On Sunday, January 7, 2018 at 7:08:23 AM UTC+8, Anders Eklöf wrote: Peter Riedt wrote: Your points are valid. However, the formula using .5*sqrt(4)... produces the same result as using 1-3.... No - they don't, except for a circle. For Mercury 5*sqrt(4-3(a-b)^2/(a+b)^2) gives 0,999956 and 1-3(a-b)^2/(a+b)^2) gives 0.999650. The difference doesn't look big, but the devation from 1 differs by an order of magnitude. The comet Halley produces .85 for X. Only with 1-3(a-b)^2/(a+b)^2) as you listed. Just try using .5*sqrt(4-3(a-b)^2/(a+b)^2) instead. Since I don't have your values for a and b I can't check. The values for the semi major axis were obtained from Princeton.edu and the values for the semi minor axis were calculated by me with the formula semi minor axis = semi major axis * sqrt(1-e^2): So, since you use the eccentricity e to calculate the semi major axis, what is the point of introducing a new "eccentricity constant" X? 1. What is the geometrical mening of X? 2. In what way is it useful? smajora smina e MER 57,909,231,029 56,672,064,712 0.2056 VEN 108,209,525,401 108,207,023,568 0.0068 EAR 149,598,319,494 149,577,457,301 0.0167 MAR 227,943,771,564 226,947,353,141 0.0934 JUP 778,342,761,465 777,430,569,626 0.0484 SAT 1,426,714,892,866 1,424,617,764,212 0.0542 URA 2,870,633,540,862 2,867,434,101,795 0.0472 NEP 4,498,393,012,162 4,498,226,658,512 0.0086 PLU 5,906,438,090,764 5,720,709,449,730 0.2488 The two formulas for X differ indeed: .05*sqrt(4-3(a-b)^2/(a+b)^2) 1-3(a-b)^2/(a+b)^2) MER 0.999956281 0.999650256 VEN 1.000000000 1.000000000 EAR 0.999999998 0.999999985 MAR 0.999998201 0.999985606 JUP 0.999999871 0.999998969 SAT 0.999999797 0.999998377 URA 0.999999883 0.999999067 NEP 1.000000000 0.999999999 PLU 0.999904311 0.999234522 So much for you simplification of the formula. Can you still not see where the error is? You say my points are valid, and choose to ignore them. -- I recommend Macs to my friends, and Windows machines to those whom I don't mind billing by the hour X is more useful than SR and GR which cannot calculate any real elements of solar orbits. |
#29
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The eccentricity constant of solar objects
Dne 08/01/2018 v 01:16 Peter Riedt napsal(a):
On Monday, January 8, 2018 at 4:44:46 AM UTC+8, Anders Eklöf wrote: The values for the semi major axis were obtained from Princeton.edu and the values for the semi minor axis were calculated by me with the formula semi minor axis = semi major axis * sqrt(1-e^2): So, since you use the eccentricity e to calculate the semi major axis, what is the point of introducing a new "eccentricity constant" X? 1. What is the geometrical mening of X? 2. In what way is it useful? So much for you simplification of the formula. Can you still not see where the error is? You say my points are valid, and choose to ignore them. X is more useful than SR and GR which cannot calculate any real elements of solar orbits. An evasive attempt. X is useless. BTW, If you cannot calculate it via SR/GR, it does not mean it cannot be done. -- Poutnik ( The Pilgrim, Der Wanderer ) A wise man guards words he says, as they say about him more, than he says about the subject. |
#30
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The eccentricity constant of solar objects
Den 08.01.2018 01.16, skrev Peter Riedt:
On Monday, January 8, 2018 at 4:44:46 AM UTC+8, Anders Eklöf wrote: Peter Riedt wrote: On Sunday, January 7, 2018 at 7:08:23 AM UTC+8, Anders Eklöf wrote: Peter Riedt wrote: Your points are valid. However, the formula using .5*sqrt(4)... produces the same result as using 1-3.... No - they don't, except for a circle. For Mercury 5*sqrt(4-3(a-b)^2/(a+b)^2) gives 0,999956 and 1-3(a-b)^2/(a+b)^2) gives 0.999650. The difference doesn't look big, but the devation from 1 differs by an order of magnitude. The comet Halley produces .85 for X. Only with 1-3(a-b)^2/(a+b)^2) as you listed. Just try using .5*sqrt(4-3(a-b)^2/(a+b)^2) instead. Since I don't have your values for a and b I can't check. The values for the semi major axis were obtained from Princeton.edu and the values for the semi minor axis were calculated by me with the formula semi minor axis = semi major axis * sqrt(1-e^2): So, since you use the eccentricity e to calculate the semi major axis, what is the point of introducing a new "eccentricity constant" X? 1. What is the geometrical mening of X? 2. In what way is it useful? smajora smina e MER 57,909,231,029 56,672,064,712 0.2056 VEN 108,209,525,401 108,207,023,568 0.0068 EAR 149,598,319,494 149,577,457,301 0.0167 MAR 227,943,771,564 226,947,353,141 0.0934 JUP 778,342,761,465 777,430,569,626 0.0484 SAT 1,426,714,892,866 1,424,617,764,212 0.0542 URA 2,870,633,540,862 2,867,434,101,795 0.0472 NEP 4,498,393,012,162 4,498,226,658,512 0.0086 PLU 5,906,438,090,764 5,720,709,449,730 0.2488 The two formulas for X differ indeed: .05*sqrt(4-3(a-b)^2/(a+b)^2) 1-3(a-b)^2/(a+b)^2) MER 0.999956281 0.999650256 VEN 1.000000000 1.000000000 EAR 0.999999998 0.999999985 MAR 0.999998201 0.999985606 JUP 0.999999871 0.999998969 SAT 0.999999797 0.999998377 URA 0.999999883 0.999999067 NEP 1.000000000 0.999999999 PLU 0.999904311 0.999234522 So much for you simplification of the formula. Can you still not see where the error is? You say my points are valid, and choose to ignore them. -- I recommend Macs to my friends, and Windows machines to those whom I don't mind billing by the hour X is more useful than SR and GR which cannot calculate any real elements of solar orbits. A very strange idea. X = X(e) = √(4-3⋅(1-√(1-e²))²/(1+√(1-e²))²)/2 e = 0.00 X = 1.0000000000 e = 0.04 X = 0.9999999399 e = 0.08 X = 0.9999990338 e = 0.12 X = 0.9999950691 e = 0.16 X = 0.9999842377 e = 0.20 X = 0.9999609449 e = 0.24 X = 0.9999175202 e = 0.28 X = 0.9998438029 e = 0.32 X = 0.9997265649 e = 0.36 X = 0.9995487110 e = 0.40 X = 0.9992881691 e = 0.44 X = 0.9989163362 e = 0.48 X = 0.9983958716 e = 0.52 X = 0.9976775062 e = 0.56 X = 0.9966953251 e = 0.60 X = 0.9953596037 e = 0.64 X = 0.9935455742 e = 0.68 X = 0.9910751195 e = 0.72 X = 0.9876855065 e = 0.76 X = 0.9829727662 e = 0.80 X = 0.9762812095 e = 0.84 X = 0.9664653233 e = 0.88 X = 0.9512997861 e = 0.92 X = 0.9256679486 e = 0.96 X = 0.8733242883 e = 1.00 X = 0.5000000000 So you have made a function which evaluates to something very close to 1 for most eccentricities. What's the point with that? Wouldn't the function X = 1.0 be equal useful? What does X tell us about the planetary orbits? Mercury e = 0.2056 X = 0.9999562811 Venus e = 0.0086 X = 0.9999999999 Earth e = 0.0167 X = 0.9999999982 Mars e = 0.0934 X = 0.9999982007 Jupiter e = 0.0484 X = 0.9999998711 Saturn e = 0.0541 X = 0.9999997986 Uranus e = 0.0472 X = 0.9999998834 Neptun e = 0.0086 X = 0.9999999999 Pluto e = 0.2488 X = 0.9999043107 You said: "X is more useful than SR and GR which cannot calculate any real elements of solar orbits." Can you please explain what elements of solar orbits you can calculate using X? -- Paul https://paulba.no/ |
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