Re: [math-fun] Kepler[t]. (Was: Two ellipse circumferences)
This solution is pretty well recognized: Let “t” be the angle around a unit circle, then A=1/2*t gives circular area, and A_t=1/2*e*sin(t) gives the area of a triangle such that the sectorial area can be written: A_s=Sqrt(1-e^2)*(A+A_t). Invert this equation to get the correct time parameter. I doubt that your pals Julian and Neil are the first to have noticed this. Your preferred calculation is very similar to the certificate method I was discussing recently. One area integral is easy to calculate and equals the sectorial area by the addition of a polygon volume. I have already shown *on this list* that the same tactic is capable of defining sectorial area relative to Cartesian area, which is calculated by a relatively simple trigonometric integral. I don’t have my computer with me at the moment, but later I will try to copy your Mathematica code with my alternative approach. Merry Ellipse-Mas, and a Happy New-Circle, —Brad
On Dec 25, 2019, at 6:03 PM, Bill Gosper <billgosper@gmail.com> wrote:
Yes! How could we possibly have gone this long without a universally recognized Kepler(a,b,t) function? —Bill
The following Mma code uses integral certificates to derive that "A_s=Sqrt(1-e^2)*(A-A_t)". This is not the same as the previous result, but equivalent when measuring sectorial area from the opposed focus. I have not seen the functions rzCert, xyCert elsewhere, but believe that L. EULER could already have known them. --Brad ========== Mma. Code ========== rad = (1 - e^2)/(1 + e Cos[z]); rzCert = (Sin[z] + (e/2)*Cos[z] Sin[z]) ((1 - e^2)/(1 + e Cos[z]))^2; xyCert = (1 - e^2) (1 - e^2)/(1 + e Cos[z]) Sin[z]; ADiff = Cancel[ Times[Factor[xyCert - rzCert], x y /(rad^2 Cos[z] Sin[z])]/e] xyAInt = Cancel[ ReplaceRepeated[ Sqrt[(1 - e^2)] Integrate[Sqrt[(1 - x^2)], {x, 1, t}][[1]], {t -> x + e, Sqrt[1 - (e + x)^2] -> y/Sqrt[1 - e^2]}]] Factor[xyAInt - ADiff] Factor[% /. {x -> Cos[t] - e, y -> Sqrt[1 - e^2] Sin[t]}] Out[]= (x*y)/2 (1/2)*(e*y + x*y - Sqrt[1 - e^2]*ArcCos[e + x]) (1/2)*(e*y - Sqrt[1 - e^2]*ArcCos[e + x]) (1/2)*Sqrt[1 - e^2]*(-ArcCos[Cos[t]] + e*Sin[t]) On Thu, Dec 26, 2019 at 8:53 AM Brad Klee <bradklee@gmail.com> wrote:
This solution is pretty well recognized: Let “t” be the angle around a unit circle, then A=1/2*t gives circular area, and A_t=1/2*e*sin(t) gives the area of a triangle such that the sectorial area can be written: A_s=Sqrt(1-e^2)*(A+A_t). Invert this equation to get the correct time parameter. I doubt that your pals Julian and Neil are the first to have noticed this.
Your preferred calculation is very similar to the certificate method I was discussing recently. One area integral is easy to calculate and equals the sectorial area by the addition of a polygon volume. I have already shown *on this list* that the same tactic is capable of defining sectorial area relative to Cartesian area, which is calculated by a relatively simple trigonometric integral.
I don’t have my computer with me at the moment, but later I will try to copy your Mathematica code with my alternative approach.
Merry Ellipse-Mas, and a Happy New-Circle,
—Brad
On Dec 25, 2019, at 6:03 PM, Bill Gosper <billgosper@gmail.com> wrote:
Yes! How could we possibly have gone this long without a universally recognized Kepler(a,b,t) function? —Bill
After Mike Stay's recent post, see p. 49-50 of: https://www.maa.org/sites/default/files/pdf/cms_upload/Frantz28946.pdf Mean anomaly is essentially the same quantity as sectorial area, so the above reference gives another explanation of the (+/-) symmetry inherent to "Kepler's equation". As mentioned previously, this symmetry has to do with the fact that an ellipse has two equivalent foci. The article, however, does not go into a detailed derivation or analysis of Kepler's equation. "It is well known that" angular momentum conserves, but how well known is it that Kepler's equation derives from the certificate functions given below? --Brad On Fri, Dec 27, 2019 at 12:58 PM Brad Klee <bradklee@gmail.com> wrote:
The following Mma code uses integral certificates to derive that "A_s=Sqrt(1-e^2)*(A-A_t)". This is not the same as the previous result, but equivalent when measuring sectorial area from the opposed focus. I have not seen the functions rzCert, xyCert elsewhere, but believe that L. EULER could already have known them. --Brad
========== Mma. Code ==========
rad = (1 - e^2)/(1 + e Cos[z]); rzCert = (Sin[z] + (e/2)*Cos[z] Sin[z]) ((1 - e^2)/(1 + e Cos[z]))^2; xyCert = (1 - e^2) (1 - e^2)/(1 + e Cos[z]) Sin[z]; ADiff = Cancel[ Times[Factor[xyCert - rzCert], x y /(rad^2 Cos[z] Sin[z])]/e] xyAInt = Cancel[ ReplaceRepeated[ Sqrt[(1 - e^2)] Integrate[Sqrt[(1 - x^2)], {x, 1, t}][[1]], {t -> x + e, Sqrt[1 - (e + x)^2] -> y/Sqrt[1 - e^2]}]] Factor[xyAInt - ADiff] Factor[% /. {x -> Cos[t] - e, y -> Sqrt[1 - e^2] Sin[t]}]
Out[]= (x*y)/2 (1/2)*(e*y + x*y - Sqrt[1 - e^2]*ArcCos[e + x]) (1/2)*(e*y - Sqrt[1 - e^2]*ArcCos[e + x]) (1/2)*Sqrt[1 - e^2]*(-ArcCos[Cos[t]] + e*Sin[t])
On Thu, Dec 26, 2019 at 8:53 AM Brad Klee <bradklee@gmail.com> wrote:
This solution is pretty well recognized: Let “t” be the angle around a unit circle, then A=1/2*t gives circular area, and A_t=1/2*e*sin(t) gives the area of a triangle such that the sectorial area can be written: A_s=Sqrt(1-e^2)*(A+A_t). Invert this equation to get the correct time parameter. I doubt that your pals Julian and Neil are the first to have noticed this.
Your preferred calculation is very similar to the certificate method I was discussing recently. One area integral is easy to calculate and equals the sectorial area by the addition of a polygon volume. I have already shown *on this list* that the same tactic is capable of defining sectorial area relative to Cartesian area, which is calculated by a relatively simple trigonometric integral.
I don’t have my computer with me at the moment, but later I will try to copy your Mathematica code with my alternative approach.
Merry Ellipse-Mas, and a Happy New-Circle,
—Brad
On Dec 25, 2019, at 6:03 PM, Bill Gosper <billgosper@gmail.com> wrote:
Yes! How could we possibly have gone this long without a universally recognized Kepler(a,b,t) function? —Bill
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Brad Klee