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 56

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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 56

     When δ > φ it is evident that the correction must be negative, as the star will cross the instrumental meridian to the west of the true meridian and be seen too late; but in this case sin (φ - δ) is negative and the sign of P[superscript]s[/superscript] takes care of itself.

     Let S[subscript]1[/subscript] be the position of a sub-polar star on the instrumental meridian. In the [image: symbol of part of circle] PZS[subscript]1[/subscript] , we have 
((sin ZPS[subscript]1[/subscript])/(sin PZS[subscript]1[/subscript])) = ((sin ZS[subscript]1[/subscript])/(sin PS[subscript]1[/subscript])) .
Let P[subscript]1[/subscript] = ∠ S[subscript]1[/subscript]PB = 180° - ∠ ZPS[subscript]1[/subscript] , and let the declination of the star when on the meridian be measured thru the zenith and pole.

     Then ((sin S[subscript]1[/subscript]PB)/(sin PZS[subscript]1[/subscript])) = ((sin P[subscript]1[/subscript][superscript]s[/superscript])/(sin a[superscript]s[/superscript])) = ((sin (δ - φ))/(sin (δ - 90°))) or ((P[subscript]1[/subscript][superscript]s[/superscript])/(a[superscript]s[/superscript])) [strikethrough] = (-(sin (φ - δ))/-(cos δ))[/strikethrough]  = ((sin (φ - δ))/(cos δ)) .

Therefore P[subscript]1[/subscript][superscript]s[/superscript] = ((sin (φ - δ))/(cos δ)) a[superscript]s[/superscript] .   (2). Comparing (2) with (1) we see that [strikethrough]we may use (1)[/strikethrough] [inserted](1) may be used[/inserted] for sub-polar transits if we take δ and P[superscript]s[/superscript] at their supplementary values.

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When δ > φ it is evident that the correction must be negative, as the star will cross the instrumental meridian to the west of the true meridian and be seen too late; but in this case sin (φ - δ) is negative and the sign of P[superscript]s[/superscript] takes care of itself. Let S[subscript]1[/subscript] be the position of a sub-polar star on the instrumental meridian. In the [image: symbol of part of circle] PZS[subscript]1[/subscript] , we have ((sin ZPS[subscript]1[/subscript])/(sin PZS[subscript]1[/subscript])) = ((sin ZS[subscript]1[/subscript])/(sin PS[subscript]1[/subscript])) . Let P[subscript]1[/subscript] = ∠ S[subscript]1[/subscript]PB = 180° - ∠ ZPS[subscript]1[/subscript] , and let the declination of the star when on the meridian be measured thru the zenith and pole. Then ((sin S[subscript]1[/subscript]PB)/(sin PZS[subscript]1[/subscript])) = ((sin P[subscript]1[/subscript][superscript]s[/superscript])/(sin a[superscript]s[/superscript])) = ((sin (δ - φ))/(sin (δ - 90°))) or ((P[subscript]1[/subscript][superscript]s[/superscript])/(a[superscript]s[/superscript])) [strikethrough] = (-(sin (φ - δ))/-(cos δ))[/strikethrough] = ((sin (φ - δ))/(cos δ)) . Therefore P[subscript]1[/subscript][superscript]s[/superscript] = ((sin (φ - δ))/(cos δ)) a[superscript]s[/superscript] . (2). Comparing (2) with (1) we see that [strikethrough]we may use (1)[/strikethrough] [inserted](1) may be used[/inserted] for sub-polar transits if we take δ and P[superscript]s[/superscript] at their supplementary values.