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 90

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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 90

Example I.

     Suppose that the time of transit over C[subscript]3[/subscript] of a star whose declination is -28° 42' is 23h 45m 58s. Find the time of transit over D[subscript]3[/subscript].

     Let I[subscript]0[/subscript] = 31.04s.  cos -28° 42' = 0.8771.
Therefore, I[subscript]δ[/subscript] = I[subscript][0?][/subscript]/(cos δ) = 31.04s/0.8771 = 35.39s.

Therefore the time of transit over D[subscript]3[/subscript] is 23h 45m 55s + 35.39s = 23h 46m 33.39s.

Example II.

     Suppose that in a transit observation, A, B, C[subscript]3[/subscript], and E[subscript]3[/subscript] are missed and the times are D[subscript]3[/subscript], E[subscript]3[/subscript], F, and G are noted. Let I[subscript]0[/subscript](W) denote the equatorial interval of [inserted]a wire[/inserted] [Ward or Wand?] D[subscript]3[/subscript]. Then the observations give for t, using the equatorial intervals, t = D[subscript]3[/subscript], t = E[subscript]3[/subscript] - I[subscript]0[/subscript](E[subscript]3[/subscript]) sec δ, t = F - I[subscript]0[/subscript](F) sec δ, and t = G - I[subscript]0[/subscript](G) sec δ. Averaging the times we have t = { (D[subscript]3[/subscript] + E[subscript]3[/subscript] + F + G) / 4 } - {(I[subscript]0[/subscript](E[subscript]3[/subscript]) + I[subscript]0[/subscript](F) + I[subscript]0[/subscript](G))/4}sec δ.

     If the time is noted on A in addition to the foregoing wires, t = { (A + D[subscript]3[/subscript] + E[subscript]3[/subscript] + F + G) / 5 } + {(I[subscript]0[/subscript](A) - I[subscript]0[/subscript](G) - I[subscript]0[/subscript](E[subscript]3[/subscript]) - I[subscript]0[/subscript](F))/5}sec δ 

or t = { (A + D[subscript]3[/subscript] + E[subscript]3[/subscript] + F + G) / 5 } - {(I[subscript]0[/subscript](E[subscript]3[/subscript]) + I[subscript]0[/subscript](F))/5}sec δ,

assuming (I[subscript]0[/subscript](A) = I[subscript]0[/subscript](G).

     Following are some observations from which we may determine I[subscript]0[/subscript] for the different wires.

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Example I. Suppose that the time of transit over C[subscript]3[/subscript] of a star whose declination is -28° 42' is 23h 45m 58s. Find the time of transit over D[subscript]3[/subscript]. Let I[subscript]0[/subscript] = 31.04s. cos -28° 42' = 0.8771. Therefore, I[subscript]δ[/subscript] = I[subscript][0?][/subscript]/(cos δ) = 31.04s/0.8771 = 35.39s. Therefore the time of transit over D[subscript]3[/subscript] is 23h 45m 55s + 35.39s = 23h 46m 33.39s. Example II. Suppose that in a transit observation, A, B, C[subscript]3[/subscript], and E[subscript]3[/subscript] are missed and the times are D[subscript]3[/subscript], E[subscript]3[/subscript], F, and G are noted. Let I[subscript]0[/subscript](W) denote the equatorial interval of [inserted]a wire[/inserted] [Ward or Wand?] D[subscript]3[/subscript]. Then the observations give for t, using the equatorial intervals, t = D[subscript]3[/subscript], t = E[subscript]3[/subscript] - I[subscript]0[/subscript](E[subscript]3[/subscript]) sec δ, t = F - I[subscript]0[/subscript](F) sec δ, and t = G - I[subscript]0[/subscript](G) sec δ. Averaging the times we have t = { (D[subscript]3[/subscript] + E[subscript]3[/subscript] + F + G) / 4 } - {(I[subscript]0[/subscript](E[subscript]3[/subscript]) + I[subscript]0[/subscript](F) + I[subscript]0[/subscript](G))/4}sec δ. If the time is noted on A in addition to the foregoing wires, t = { (A + D[subscript]3[/subscript] + E[subscript]3[/subscript] + F + G) / 5 } + {(I[subscript]0[/subscript](A) - I[subscript]0[/subscript](G) - I[subscript]0[/subscript](E[subscript]3[/subscript]) - I[subscript]0[/subscript](F))/5}sec δ or t = { (A + D[subscript]3[/subscript] + E[subscript]3[/subscript] + F + G) / 5 } - {(I[subscript]0[/subscript](E[subscript]3[/subscript]) + I[subscript]0[/subscript](F))/5}sec δ, assuming (I[subscript]0[/subscript](A) = I[subscript]0[/subscript](G). Following are some observations from which we may determine I[subscript]0[/subscript] for the different wires.