350 Mr. Ivory on the Properties of a Line of shortest 
And hence, 
= \jpaein  - \ 
fi = Sia = 00039150 
f* = 00000153. 
It will be convenient to multiply the coefficients m, p, g by the 
seconds in an arc equal to the radius in order to have all the 
terms of the formula expressed in seconds of a degree: then, 
log m = 5*3139988, 
log p = 2°78254, 
log g = 0°170, 
The arc zis the hypothenuse of a right-angled triangle in 
which the latitude subtends the angle 2’: therefore, 
sin 2(0) — in? ; 
sin z 
or, when 7’ and / are very nearly equal, we may use this for- 
mula, 
cot. 2° = ¥ sin ( Ce) : 
sin / 
2° = 86° 28! 19""0. 
This quantity must now be substituted in the formula (2) to 
find m s°; the amount of the terms.to be subtracted is only 
37"-31; therefore, 
m® s = 86° 27! 41"-69. 
In order to get m (s?+ s) we must add the degrees in the arc 
between Seeberg and Dunkirk, viz. 5° 16! 48':48, which is 
found by adding the log. of m to the log. of the distance be- 
tween the two places; then, 
m (s° + s) = 91° 44! 30""17. 
This value being found, we must next compute the correspond- 
ing quantity z° + z by the formula (2): the correction to be 
applied is only — 18'*4; wherefore, 
20 4+ 2 = 91° 44! 11-77, 
Finally, we have, 
sin u = sin?’ x sin (z° + 2); 
the lat. of Dunkirk, « = 51° 2! 12!-7, 
In order to find the difference of longitude, I shall resume 
the expression of d¢ before found, writing 7’ for 7, and sin z 
for cos z, and likewise, for the sake of brevity, putting A* = 
cos 72’ + sin ?z' cos?z = 1 — sin ?2’ sin?z: then, 
cosi' dz 1 
do=———- vie 
By expanding the radical we get, 
ae - cosidz$ = —3 4 AR + Ze At— &e.}. 
If now we substitute what A? stands for, and in place of the 
powers of sin z write the equivalent expressions in the oo 
of the 
d¢= 
