GRAPHIC REDUCTION OP STAR PLACES. 
177 
First term .—To find a A = (3 S .073 + I s . 337 sin a tan 3) X A (see 
plate 9). 
The tangent of 53° 7' is the line cd — 1.33. In order to 
verify this value reference must be had to plate 7; but in 
the regular work the determinations are made on the same 
sheet on which the curves are drawn. The sine of 16 h 33 m .5 
is a b = — 0.930 X 20. Project b to f and draw / o. The 
vertical line g h, at a distance from the axis of ordinates 
equal to c d and included between the line / o and the axis 
of abscissas, is the product of sin a tan 3 , or 1.24. 
This follows from the proposition 
Hence 
or 
gh = 
e f: e o :: g h : g o 
ab : eo ::gh :cd 
ab x cd — 0.930 x 20 x 1.33 
= 20 
= — 1.24. 
Draw k o so that j k = 1.337 times / o. The horizontal line 
l m, passing through the point h and included between the 
axis of ordinates and the line k o, is therefore equal to the 
quantity 1.337 sin a tan 3 , or —1.66. This follows from the 
fact that in the triangle / o k each abscissa is 1.337 times the 
corresponding ordinate. The total value of the quantity 
within the parenthesis, or a, is therefore 3.07 — 1.66, or 
d- 1.41. 
This quantity is to be multiplied by the value of A on 
June 9, or 0.507. The necessary lines for the multiplication 
of this quantity by any factor have already been drawn in 
the case of the declinations, and in actual work their appli¬ 
cation to the right ascensions is directly made without new 
construction. The method is as follows: The value of A 
projected to the line p n gives the point q, and a vertical line, 
r s, included between q o and the axis of abscissas, and at a 
distance from the origin equal to 1.44, gives the value of a A, 
or + 0.71. 
We therefore have 
a A = (3 S .073 + l s .337 sin a tan 3) x A = + 0 S .71. 
