DIY History | Transcribe | Scholarship at Iowa | Eccentricity of the Sextant by Frederic Furbish, 1893 | Eccentricity of the Sextant by Frederic Furbish, 1893, Page 84

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Eccentricity of the Sextant by Frederic Furbish, 1893

Eccentricity of the Sextant by Frederic Furbish, 1893, Page 84

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In this equation the items 2 sin[superscript]2[/superscript] (1/2)t , 2 sin[superscript]4[/superscript] (1/2)t , 2 sin[superscript]6[/superscript] (1/2)t arc expressed in terms of the radius. To render the eqation [equation?] homogeneous we divide these terms by sin 1"

     Let 
((cos φ cos δ)/(sin Z[subscript]0[/subscript])) = A

A[superscript]2[/superscript] cot Z[subscript]0[/subscript] = B
A[superscript]3[/superscript] (2/3)(1 + 3 cot[superscript]2[/superscript]Z[subscript]0[/subscript]) = C
2 sin[superscript]2[/superscript](1/2)t = m
sin[superscript]4[/superscript](1/2)t = n
sin[superscript]6[/superscript](1/2)t = 0
Now let Z = 90° - [n or h?] and Z[subscript]0[/subscript] = 90° - h[subscript]0[/subscript] and substitute in (4).

     90° - h = 90° - h[subscript]0[/subscript] + Am - Bn + C[subscript]0[/subscript] - - - or -h = -h[subscript]0[/subscript] + Am - Bn + C[subscript]0[/subscript] ---- or (5) h[subscript]0[/subscript] = h + Am - Bn + C[subscript]0[/subscript] - - - - which is the formula we shall use.

     In this A, B and C are constants which are computed once for all, (h) is the observed altitude and (m) and (n) are found in the tables. The last term C[subscript]0[/subscript] is seldom used being extremely small when having any finite value. (0) is not given in the tables.

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[page][illegible][/page] In this equation the items 2 sin[superscript]2[/superscript] (1/2)t , 2 sin[superscript]4[/superscript] (1/2)t , 2 sin[superscript]6[/superscript] (1/2)t arc expressed in terms of the radius. To render the eqation [equation?] homogeneous we divide these terms by sin 1" Let ((cos φ cos δ)/(sin Z[subscript]0[/subscript])) = A A[superscript]2[/superscript] cot Z[subscript]0[/subscript] = B A[superscript]3[/superscript] (2/3)(1 + 3 cot[superscript]2[/superscript]Z[subscript]0[/subscript]) = C 2 sin[superscript]2[/superscript](1/2)t = m sin[superscript]4[/superscript](1/2)t = n sin[superscript]6[/superscript](1/2)t = 0 Now let Z = 90° - [n or h?] and Z[subscript]0[/subscript] = 90° - h[subscript]0[/subscript] and substitute in (4). 90° - h = 90° - h[subscript]0[/subscript] + Am - Bn + C[subscript]0[/subscript] - - - or -h = -h[subscript]0[/subscript] + Am - Bn + C[subscript]0[/subscript] ---- or (5) h[subscript]0[/subscript] = h + Am - Bn + C[subscript]0[/subscript] - - - - which is the formula we shall use. In this A, B and C are constants which are computed once for all, (h) is the observed altitude and (m) and (n) are found in the tables. The last term C[subscript]0[/subscript] is seldom used being extremely small when having any finite value. (0) is not given in the tables.