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Re: Angular Distance Between Stars By Camera and Sextant
From: Marcel Tschudin
Date: 2012 Sep 20, 13:13 +0300
From: Marcel Tschudin
Date: 2012 Sep 20, 13:13 +0300
Paul, regarding your following explanation on how you obtained the refracted distance from Andrés calculation:
On Wed, Sep 19, 2012 at 11:50 PM, Paul Hirose <cfuhb-acdgw@earthlink.net> wrote:
Because, Andrés program starts from observations relative to the apparent Horizon and his following explanations given earlier:
On Tue, Sep 18, 2012 at 7:08 PM, Andres Ruiz <navigationalalgorithms@gmail.com> wrote:
I avoided to use his Ha values shown in the output for calculating the refracted distance. Indeed, if I would have tried it, I also would have received your value.
Instead of using Ha I used his unrefracted Hc which I expect to be related to the astronomical horizon (ZD=90°). Calculating first the unrefracted distance with Hc and Z (yes, only the difference in Z is relevant) results in an unrefracted distance of
Dtrue = 10.460896°
resulting in
Dref = 10.457595° (using Bill's Excel sheet with GHA, Dec and Hc from Andrés)
For verification one can also apply Frank's approximation that refraction lets the distance shrink by 0.1 moa per 5° distance for objects with Alt greater than about 45°:
Dref ≈ 10.457409°
differing in this case by less than 0.7 sec of arc.
I guess your smaller Dref = 10.455432° result from relating the altitude to the apparent horizon instead of the astronomical horizon, thus shifting the altitude further up relative to the astronomical horizon which results in an additional shrinking of the distance.
I hope that my above conclusions do not contain an other misunderstanding.
Marcel
Marcel Tschudin wrote:Marcel, here are the unrefracted and refracted coordinates from the program by Andrés.
But how did you, Paul, obtain for the
Alioth-Alkaid refracted distance the 10.455432°? I was not able to
reproduce this value. Transferring e.g. the Dec, GHA and Hc values from
Andrés' program into Bill's Excel sheet I obtain Alioth-Alkaid refracted
distance of 10.457595°, or when using Frank's refraction approximation
10.457409°.
>>>Calculated altitudes:
>>>Hc1 = 28.317414
>>>Z1 = 320.442970
>>>Hc2 = 34.115333
>>>Z2 = 310.248963
>> Refraction:Convert the refracted spherical coordinates to vectors. Although azimuth doesn't follow the normal convention for the "theta" angle in spherical coordinates - its zero point and direction of increase are different - both bodies are affected the same way. Thus, if we use azimuth as theta, the relative positions are still correct.
>>
>>>R1 = 0.028946
>>>R2 = 0.023032
>>>Apparent altitudes:
>>>Ha1 = 28.347441
>>>Ha2 = 34.139225
Alioth = (.678537, -.560478, .474817)
Alkaid = (.534770, -.631719, .561206)
Let x = dot product of the vectors = .983396
Let y = magnitude of the vector cross product = .181471
Convert (x, y) from rectanglar to polar. Angle = 10.455432°.
Because, Andrés program starts from observations relative to the apparent Horizon and his following explanations given earlier:
On Tue, Sep 18, 2012 at 7:08 PM, Andres Ruiz <navigationalalgorithms@gmail.com> wrote:
Hc = Hc( B, L, Dec, GHA )Z = Z( B, Dec, Hc, LHA( L, GHA ) );iterate to find Hs ( Hc = Ho = Hs + IE - dip - R )Ha = Hs + IE - dip;R = Refraccion( Ha, T, P );
I avoided to use his Ha values shown in the output for calculating the refracted distance. Indeed, if I would have tried it, I also would have received your value.
Instead of using Ha I used his unrefracted Hc which I expect to be related to the astronomical horizon (ZD=90°). Calculating first the unrefracted distance with Hc and Z (yes, only the difference in Z is relevant) results in an unrefracted distance of
Dtrue = 10.460896°
resulting in
Dref = 10.457595° (using Bill's Excel sheet with GHA, Dec and Hc from Andrés)
For verification one can also apply Frank's approximation that refraction lets the distance shrink by 0.1 moa per 5° distance for objects with Alt greater than about 45°:
Dref ≈ 10.457409°
differing in this case by less than 0.7 sec of arc.
I guess your smaller Dref = 10.455432° result from relating the altitude to the apparent horizon instead of the astronomical horizon, thus shifting the altitude further up relative to the astronomical horizon which results in an additional shrinking of the distance.
I hope that my above conclusions do not contain an other misunderstanding.
Marcel
