Page:Madras Journal of Literature and Science, series 1, volume 6 (1837).djvu/162

140 a function the error is given of the difference of longitude, by changing the formula.

In the diagram (Pl. 4 fig. 3) let A D be the meridian of the first station, B C the perpendicular of any other station, then A being the pole, D C will be the parallel of latitude of C, and A D will be the co-latitude of E, and not A B, and B D will be the error or the difference of latitude between the foot of perpendicular and the parallel.

$$ \begin{align} \text{Take } & \mathrm{A\ B} = 90 - \lambda \\ & \mathrm{B\ C} = \pi \\ & \mathrm{A\ C} = 90 - (\lambda - x) \\ & \mathrm{D\ B} = x. \text{ then by right angled spherics we have} \\ & \sin\ (\lambda - x) = \sin\ \lambda\ \cos\ \pi \\ \sin\ \lambda.\ \cos\ x\ -\ & \sin\ x.\ \cos\ \lambda\ = \\ & \text{dividing by}\ \sin\ \lambda\ \text{we have} \\ & \cos\ x - \sin\ x.\ \cot\ \lambda = \cos\ \pi \\ \end{align} $$

and the $$\cos\ x$$ being nearly $$= 1$$, and substituting the value of $$\cos\ \pi$$

$$ \begin{align} & 1 - \sin\ x.\ \cot\ \lambda = 1 - 2\ \overline{\sin \tfrac{1}{2}}^2\ \pi \\ & \sin\ x. = 2\ \overline{\sin \tfrac{1}{2}}^2\ \pi \ \operatorname{tang}\ \lambda \\ \end{align} $$

and taking $$x''$$ and $$\pi'$$ for the sines

$$x = \tfrac{1}{2} \pi ^2. \operatorname{tang} \lambda. \sin l''$$ which is the formula before given,—but the angle B A C is the difference of longitude between B and C, and in the right angled triangle A B C

$$ \begin{align} \text{Tang } \mathrm{A} & = \operatorname{tang}\ \pi.\ \sec\ \lambda \\ \mathrm{A}'' & = \pi.\ \sec\ \lambda\ \text{and transposing} \\ \pi & = \cos\ \lambda.\ \mathrm{A} \\ & = \cos\ \lambda.\ \text{diff. longitude} \end{align} $$

substituting this in the last formula

$$ \begin{align} x & = \tfrac{1}{2}\ \delta\ ^2. \overline{\cos \lambda.}^2 \operatorname{Tang}\ \lambda.\ \sin\ l'' \\ & = \tfrac{1}{2}\ \delta''\ ^2. \cos \lambda.\ \sin\ \lambda.\ \sin\ l'' \end{align} $$

or become $$\cos\ \lambda.\ \sin\ \lambda = \tfrac{1}{2}\ \sin\ 2 \lambda$$

$$x = \tfrac{1}{2}\ \delta^2 \sin.\ 2\ \lambda.\ \sin\ l$$ from this formula the following table is computed x'' always subtractive:—