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 88

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 88

[image: drawing of angles involved in observation of star; labels include B, C, D, F, H, O, P, Q, S]
Let B be the position of a star on a side wire. Draw BD perpendicular to the observer's meridian plane CPQ meeting it at D. Let FDH be the middle wire of the reticle and suppose it to be in the meridian plane. Let BS be the path of the star from B to the meridian. Pass a plane then BD and OD cutting PQ at C and making BC the arc of a great circle.
Then in rt. [image: spherical angle symbol?] PBC, ∠Ps = I[subscript]δ[/subscript]s; arc PB = [co?]-δ, and arc BC = I[subscript]0[/subscript]s. We have then
((sin BC)/(sin P)) = ((sin PB)/(sin 90°)) or ((sin I[subscript]0[/subscript]s)/(sin I[subscript]δ[/subscript]s)) = cos δ ; (1)
whence sin I[subscript]0[/subscript]s = sin I[subscript]δ[/subscript]s cos δ.
As an approximation we may take I[subscript]0[/subscript] (radians) = sin I[subscript]δ[/subscript] cos δ ;
i.e. I[subscript]0[/subscript]s = sin I[subscript]δ[/subscript]s cos δ x 206,264.8 ,
or I[subscript]0[/subscript]s = (206,264.8/15) sin I[subscript]δ[/subscript]s cos δ ,
or I[subscript]0[/subscript]s = ((sin I[subscript]δ[/subscript]s cos δ)/(15 sin 1["?])) .

[image: drawing of angles involved in observation of star; labels include B, C, D, F, H, O, P, Q, S]
Let B be the position of a star on a side wire. Draw BD perpendicular to the observer's meridian plane CPQ meeting it at D. Let FDH be the middle wire of the reticle and suppose it to be in the meridian plane. Let BS be the path of the star from B to the meridian. Pass a plane then BD and OD cutting PQ at C and making BC the arc of a great circle.
Then in rt. [image: spherical angle symbol?] PBC, ∠Ps = I[subscript]δ[/subscript]s; arc PB = [co?]-δ, and arc BC = I[subscript]0[/subscript]s. We have then
((sin BC)/(sin P)) = ((sin PB)/(sin 90°)) or ((sin I[subscript]0[/subscript]s)/(sin I[subscript]δ[/subscript]s)) = cos δ ; (1)
whence sin I[subscript]0[/subscript]s = sin I[subscript]δ[/subscript]s cos δ.
As an approximation we may take I[subscript]0[/subscript] (radians) = sin I[subscript]δ[/subscript] cos δ ;
i.e. I[subscript]0[/subscript]s = sin I[subscript]δ[/subscript]s cos δ x 206,264.8 ,
or I[subscript]0[/subscript]s = (206,264.8/15) sin I[subscript]δ[/subscript]s cos δ ,
or I[subscript]0[/subscript]s = ((sin I[subscript]δ[/subscript]s cos δ)/(15 sin 1["?])) .