DIY History | Transcribe | Scholarship at Iowa | Eccentricity of the Sextant by Frederic Furbish, 1893 | Eccentricity of the Sextant by Frederic Furbish, 1893, Page 82

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Eccentricity of the Sextant by Frederic Furbish, 1893

Eccentricity of the Sextant by Frederic Furbish, 1893, Page 82

[page][illegible][/page]

h = 90° - z , sin h [unclear][=?][/unclear] sin (90° - z) = cos z.
Cos z - sin h = sin δ sin φ sin[superscript]2[/superscript] (1/2) t = cos (φ - δ) - 2 cos φ cos δ sin[superscript]2[/superscript] (1/2) t   (2). Let φ - δ = z[subscript]0[/subscript] and 2 cos φ cos δ sin[superscript]2[/superscript] (1/2) t = y and (2) becomes cos z = cos z[subscript]0[/subscript] - 7   (3).

     [unclear][McIannis?][/unclear] formula is.-
z = z[subscript]0[/subscript] + ((dz)/(dy))y + ((d[superscript]2[/superscript]z)/(dy[superscript]2[/superscript]))((y[superscript]2[/superscript])/(L2)) + ((d[superscript]3[/superscript]z)/(dy[superscript]3[/superscript]))((y[superscript]3[/superscript])/(L3)) - - -
Differentiating this function and taking the derivative with respect to y remembering z = z[subscript]0[/subscript] when y = 0 we get,-     
cos z = cos z[subscript]0[/subscript] - y ; d cos z = - dy .      -sin z d z = -dy ; ((dz)/(dy)) = ((1)/(sin z)) .
Differentiating again and taking the derivative with respect to z we get;-
((d[superscript]2[/superscript]z)/(dy.dz)) = -((cos z)/(sin[superscript]2[/superscript]z)) . Multiplying both terms by ((dz)/(dy)) = ((1)/(sin z)) we get ((d[superscript]2[/superscript]z)/(dy.dz)) * ((dz)/(dy)) = -((cos z)/(sin[superscript]2[/superscript] z)) * ((1)/(sin z)) = -((cot z)/(sin[superscript]2[/superscript]z)) or ((dz[superscript]2[/superscript])/(dy[superscript]2[/superscript])) = -((cot z)/(sin[superscript]2[/superscript]z)) .
Differentiating again and taking the derivative with respect to z.-
((d[superscript]3[/superscript]z)/(dy[superscript]2[/superscript]dz)) = ((sin[superscript]2[/superscript] z cosec[superscript]2[/superscript]z + 2 cot z sin z cos z)/(sin 4 z))

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[page][illegible][/page] h = 90° - z , sin h [unclear][=?][/unclear] sin (90° - z) = cos z. Cos z - sin h = sin δ sin φ sin[superscript]2[/superscript] (1/2) t = cos (φ - δ) - 2 cos φ cos δ sin[superscript]2[/superscript] (1/2) t (2). Let φ - δ = z[subscript]0[/subscript] and 2 cos φ cos δ sin[superscript]2[/superscript] (1/2) t = y and (2) becomes cos z = cos z[subscript]0[/subscript] - 7 (3). [unclear][McIannis?][/unclear] formula is.- z = z[subscript]0[/subscript] + ((dz)/(dy))y + ((d[superscript]2[/superscript]z)/(dy[superscript]2[/superscript]))((y[superscript]2[/superscript])/(L2)) + ((d[superscript]3[/superscript]z)/(dy[superscript]3[/superscript]))((y[superscript]3[/superscript])/(L3)) - - - Differentiating this function and taking the derivative with respect to y remembering z = z[subscript]0[/subscript] when y = 0 we get,- cos z = cos z[subscript]0[/subscript] - y ; d cos z = - dy . -sin z d z = -dy ; ((dz)/(dy)) = ((1)/(sin z)) . Differentiating again and taking the derivative with respect to z we get;- ((d[superscript]2[/superscript]z)/(dy.dz)) = -((cos z)/(sin[superscript]2[/superscript]z)) . Multiplying both terms by ((dz)/(dy)) = ((1)/(sin z)) we get ((d[superscript]2[/superscript]z)/(dy.dz)) * ((dz)/(dy)) = -((cos z)/(sin[superscript]2[/superscript] z)) * ((1)/(sin z)) = -((cot z)/(sin[superscript]2[/superscript]z)) or ((dz[superscript]2[/superscript])/(dy[superscript]2[/superscript])) = -((cot z)/(sin[superscript]2[/superscript]z)) . Differentiating again and taking the derivative with respect to z.- ((d[superscript]3[/superscript]z)/(dy[superscript]2[/superscript]dz)) = ((sin[superscript]2[/superscript] z cosec[superscript]2[/superscript]z + 2 cot z sin z cos z)/(sin 4 z))