Now we have to find the semi diameters of Sun and Moon which last

varies as the Altitude.

Sun's Semi diameter.

O's true Semi diam. 16′ 2′′.6 962".6

Moon's Semi diameter.

By a rough trial we find that the beginning, the middle and the end happen at about 3h; 4h 30m and Sunset respectively, for which last therefore the 's semi diameter will receive no augmentation.

D's true Semi diameter 14′ 41′′.4

At 4h 30m, D's Alt. 18° 15′

881".4

D's Semi diam. (Augmentation 4′′.3) = 885 .7
D's Alt. 39° 16'

At 3h

▷'s Semi diam. (Augmentation 8′′.7) = 890′′.1

CONSTRUCTION.

With radius R = 3224".6 describe a circle draw the diameter AB which represents the ecliptic. Draw OG a perpendicular at the centre and take Oc=2958".1 and drawing cP parallel to AB take cP = 1237".1, P will be the pole of the earth by the principle of projection. Join OP and produce and in it take Or 1223".04; r will be the centre of the ellipse described by Calcutta. At r draw rb perpendicular to OP and take rb = 2980".8 the semi axis major, and in rP take rn 316".2 the semi axis minor. With centre r and radii ru, rb, describe two circles or two quadrants, as is here required and divide them into equal parts of 5° each or 20m of time. From the division of the larger quadrant draw parallels to the axis minor, and from those of the smaller, parallels to the axis major. The points of intersection will determine the path of the ellipse, and the times at which the Sun is in each of them.

In OG take Ov 1877".96 the Moon's latitude at 6 and draw IuM making an angle OuM to the right (for the latitude of the is increasing) equal to the complement of 5° 45′ 28".92. LM will be the relation orbit of the D. At u mark 3h 13m 26s.68 the Calcutta apparent time of true ecliptic conjunction and with 271".5 the relative motion in 10m take off and mark the hours and parts of hours.

Now see by trial the distance of what two points of the orbit and ellipse corresponding in time is the least, and it will soon be found that mS is that distance. With centre m and radius = 885".7 describe a circle which will represent the D, and with centre S and radius 962".6 describe the circle representing the Sun. This will represent the

greatest obscuration and eu the digits eclipsed and as by our scale we find Su to be 670′′ and 962: 670 :: 6 : 4.18 + 6 = 10.18 the digits eclipsed are 10.18.

Again with 962".6 + 890".1 the sum of the semi diameters see the distance of what two corresponding points will be equal to it and it will be found that the position in question is m' S' the sum.

In the same manner with 962′′.6 +881.4 find that m" S" is equal to it and this will be position of lost contact and as the ellipse touches the circle to radius of projection at S"; S" will denote the time of the setting of the Sun whence the eclipse ends setting. As for the times of beginning, middle and end we have them from the positions of S, S', S".