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Re: Moon Parallax, Math Trivia (was Re: venus)
From: Frank Reed CT
Date: 2004 Oct 20, 16:19 EDT
From: Frank Reed CT
Date: 2004 Oct 20, 16:19 EDT
George H wrote:
"It seems to me that the expression I gave is completely rigorous, so I am not sure what Frank is telling us. Not that it makes the slightest difference, in practice."
It is a "rigorous" solution to SOME problem, yes. That much is certain!
And:
"This is purely geometry. There are no approximations that I can see. It all seems to be exact. So what's Frank on about?"
A quick speculation: the second equation and the first equation are exact solutions to slightly different geometric problems. One or the other is an approximation to its partner.
The original direct parallax equation is
OA = TA - HP cos OA .
Can you invert this and solve for OA in some "closed-form" equation? ....No. Can't be done. Proof? See any proof regarding the same question for Kepler's Equation. Kepler's Equation is the basic problem of elliptical motion in orbital dynamics (as I am sure anyone patient enough to reach this point in this discussion already knows). It reads
EA = MA + e sin EA ,
where EA is the eccentric anomaly, MA is the mean anomaly and e is the orbital eccentricity. Notice that this is essentially identical to the direct parallax equation. There were hundreds of methods used historically to solve Kepler's Equation for EA but the most common was simple iteration. You "guess" values for EA in some systematic fashion until the original equation is satisfied to some desired degree of accuracy. A good initial guess for low to moderate values of eccentricity could be provided by
EA0 = Arc-tan(cos MA / (sin MA - e)) .
For very small values of e, this first guess approximation may be accurate enough that no further iterations are required. Whether calculating EA0 in this fashion is easier or harder than two passes of brute force iteration is immaterial today.
The corresponding Arc-tan equation for the parallax problem is a valid approximation to the inversion of the direct parallax equation because HP is always a small angle. As I noted at the top, from a geometric stand-point (and by this I mean drawing triangles on paper and writing "rigorous" equations based on those triangles), both equations are exact solutions to some geometric puzzle. But the equations themselves are not exact inverses. They can't be! The direct equation has no "closed-form" inverse.
To reiterate (ouch), the difference I'm talking about is very small. If you iterate twice, the difference is less than a second of arc.
Frank R
[ ] Mystic, Connecticut
[X] Chicago, Illinois
"It seems to me that the expression I gave is completely rigorous, so I am not sure what Frank is telling us. Not that it makes the slightest difference, in practice."
It is a "rigorous" solution to SOME problem, yes. That much is certain!
And:
"This is purely geometry. There are no approximations that I can see. It all seems to be exact. So what's Frank on about?"
A quick speculation: the second equation and the first equation are exact solutions to slightly different geometric problems. One or the other is an approximation to its partner.
The original direct parallax equation is
OA = TA - HP cos OA .
Can you invert this and solve for OA in some "closed-form" equation? ....No. Can't be done. Proof? See any proof regarding the same question for Kepler's Equation. Kepler's Equation is the basic problem of elliptical motion in orbital dynamics (as I am sure anyone patient enough to reach this point in this discussion already knows). It reads
EA = MA + e sin EA ,
where EA is the eccentric anomaly, MA is the mean anomaly and e is the orbital eccentricity. Notice that this is essentially identical to the direct parallax equation. There were hundreds of methods used historically to solve Kepler's Equation for EA but the most common was simple iteration. You "guess" values for EA in some systematic fashion until the original equation is satisfied to some desired degree of accuracy. A good initial guess for low to moderate values of eccentricity could be provided by
EA0 = Arc-tan(cos MA / (sin MA - e)) .
For very small values of e, this first guess approximation may be accurate enough that no further iterations are required. Whether calculating EA0 in this fashion is easier or harder than two passes of brute force iteration is immaterial today.
The corresponding Arc-tan equation for the parallax problem is a valid approximation to the inversion of the direct parallax equation because HP is always a small angle. As I noted at the top, from a geometric stand-point (and by this I mean drawing triangles on paper and writing "rigorous" equations based on those triangles), both equations are exact solutions to some geometric puzzle. But the equations themselves are not exact inverses. They can't be! The direct equation has no "closed-form" inverse.
To reiterate (ouch), the difference I'm talking about is very small. If you iterate twice, the difference is less than a second of arc.
Frank R
[ ] Mystic, Connecticut
[X] Chicago, Illinois
