DIY History | Transcribe | Scholarship at Iowa | Theory of the astronomical transit instrument applied to the portable transit instrument Wuerdemann no.26: a compilation from various authorities, with original observations by Harry Edward Burton, 1903 | Theory of the astronomical transit instrument applied to the portable transit instrument Wuerdemann no. 26: a compilation from various authorities, with original observations by Harry Edward Burton, 1903, Page 59

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Theory of the astronomical transit instrument applied to the portable transit instrument Wuerdemann no.26: a compilation from various authorities, with original observations by Harry Edward Burton, 1903

Theory of the astronomical transit instrument applied to the portable transit instrument Wuerdemann no. 26: a compilation from various authorities, with original observations by Harry Edward Burton, 1903, Page 59

Draw S[subscript]1[/subscript]S[subscript]2[/subscript] (arc of a great circle perpendicular to BP.
Then S[subscript]1[/subscript]S[subscript]2[/subscript] = C[superscript]s[/superscript]. Let P[subscript]1[/subscript][superscript]s[/superscript] = angle S[subscript]0[/subscript]PS[subscript]1[/subscript] , the correction due to C[superscript]s[/superscript] for sub-polar transits.

     In rt. [image: symbol similar to part of circle] PS[subscript][2?][/subscript]S[subscript]1[/subscript] , we have 
((sin S[subscript]1[/subscript]PS[subscript]2[/subscript])/(sin S[subscript]1[/subscript]S[subscript]2[/subscript])) = ((sin PS[subscript]2[/subscript]S[subscript]1[/subscript])/(sin PS[subscript]1[/subscript]))
whence -((sin P[subscript]1[/subscript][superscript]s[/superscript])/(sin C[superscript]s[/superscript])) = ((2)/(cos δ)) . 

P[subscript]1[/subscript][superscript]s[/superscript] and C[subscript]s[/subscript] being very small, we may write 
-((sin P[subscript]1[/subscript][superscript]s[/superscript])/(sin C[superscript]s[/superscript])) = -((P[subscript]1[/subscript][superscript]s[/superscript])/(C[superscript]s[/superscript])) = ((1)/(cos δ)) , or P[subscript]1[/subscript][superscript]s[/superscript] = -((C[subscript]s[/subscript])/(cos δ)) = ((C[subscript]s[/subscript])/(cos (180° - δ)))        (2).

     By comparing (1) and (2) we see that (1) will hold true for sub-polar transits if we take δ at its supplementary value and consider P[superscript]s[/superscript] the desired correction.

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Draw S[subscript]1[/subscript]S[subscript]2[/subscript] (arc of a great circle perpendicular to BP. Then S[subscript]1[/subscript]S[subscript]2[/subscript] = C[superscript]s[/superscript]. Let P[subscript]1[/subscript][superscript]s[/superscript] = angle S[subscript]0[/subscript]PS[subscript]1[/subscript] , the correction due to C[superscript]s[/superscript] for sub-polar transits. In rt. [image: symbol similar to part of circle] PS[subscript][2?][/subscript]S[subscript]1[/subscript] , we have ((sin S[subscript]1[/subscript]PS[subscript]2[/subscript])/(sin S[subscript]1[/subscript]S[subscript]2[/subscript])) = ((sin PS[subscript]2[/subscript]S[subscript]1[/subscript])/(sin PS[subscript]1[/subscript])) whence -((sin P[subscript]1[/subscript][superscript]s[/superscript])/(sin C[superscript]s[/superscript])) = ((2)/(cos δ)) . P[subscript]1[/subscript][superscript]s[/superscript] and C[subscript]s[/subscript] being very small, we may write -((sin P[subscript]1[/subscript][superscript]s[/superscript])/(sin C[superscript]s[/superscript])) = -((P[subscript]1[/subscript][superscript]s[/superscript])/(C[superscript]s[/superscript])) = ((1)/(cos δ)) , or P[subscript]1[/subscript][superscript]s[/superscript] = -((C[subscript]s[/subscript])/(cos δ)) = ((C[subscript]s[/subscript])/(cos (180° - δ))) (2). By comparing (1) and (2) we see that (1) will hold true for sub-polar transits if we take δ at its supplementary value and consider P[superscript]s[/superscript] the desired correction.