232 
Proceedings of Royal Society of Edinburgh. [sess. 
consecutive points from which, the curve has been traced. This 
implies taking dt = 1 throughout the three orbits corresponding to 
undisturbed distances from the central line equal respectively to 
0, *6, ’8, and throughout the other semi-orbits, except for the 
portions next the corner, which correspond essentially to intervals 
each <1. The plan followed is sufficiently illustrated by the 
accompanying Table III., which shows the whole process of 
calculating and summing the parts for the orbit corresponding to 
undisturbed distance 7. 
Table IV. shows the sums for the ten orbits and the products of 
each sum multiplied by the proper value of r', to prepare for the 
final integration, which has been performed by finding the area of 
a representative curve drawn on conveniently squared paper as 
described in § 6 above. The result thus found is ‘02115. It is 
very satisfactory to see that, within ‘1 per cent., this agrees with 
the simple sum of the widely different numbers shown in col. 3 of 
Table IV. 
Table III. 
Orbit r = ‘7. 
ds. 
<fo2. 
dt. 
dsfdt. 
‘006 
•000036 
0-14 
•000257 
•137 
•018769 
1-00 
•018769 
•112 
•012544 
1-00 
•012544 
•077 
•005929 
1-00 
•005929 
•050 
•002500 
1-00 
•002500 
*048 
*002304 
1-00 
•002304 
•050 
•(02500 
1'00 
•002500 
‘052 
•002704 
1-00 
•002704 
Sum 
•047507 
Table IV. 
r'. 
Jds3/d«. 
■lxr’ .J ds 2 /dt . 
•o 
•0818 
•ooooo 
•1 
•0804 
•00080 
•2 
•0781 
*00156 
•3 
•0769 
•00231 
•4 
•0722 
•00289 
*5 
•0670 
•00335 
•6 
•0567 
•00340 
•7 
•0475 
•00332 
•8 
•0310 
•00248 
*9 
•0114 
•00102 
Sum 
•02113 
§ 14. Using in (13) the conclusion of § 13, and taking ^=1, 
we find 
k = 2tt.‘002115 (14). 
A convenient way of explaining this Result is to remark that 
it is -634 of tbe kinetic energy 
of an ideal globe 
t 
