MAKING A SIDEREAL CLOCK SHOW MEAN TIME. TG 
Let an additional correction, equal to that applied in each hour, be applied x times in 
a year. J 
. w æ = the yearly additional correction. 
Hence we must have 
vw x= 2%4y (1 — w — 7) 
24 y (1 — w — 7) 
w 

Whence % — 
Now, if x denotes the number of days in which the pins p p make one complete 
revolution, 
OT) 
Ww 

Upon substituting and working out we obtain 
n = 266 days very nearly. 
= 24 X 14 X 18 hours. 
Hence the pins p p must make one revolution while the arbor s makes 24 X 14 X 19 
revolutions. 
For the purpose of carrying the pins around they are fixed to the sleeve x (fig. 2) which 
is made to revolve backwards by the train of wheels shown. Wheel 12 is fixed to arbor 
s and meshes into wheel 36 which is supported by pivots in a fixed support. Wheel 36 
has on its arbor a tangent screw which works into wheel 38, which is also supported by a 
fixed support. Wheel 38 has a tangent screw working into wheel 56 which is fixed upon 
the sleeve (2) carrying the pins 2, and the direction of the tangent screw is such as to carry 
wheel 56, and therefore the pins backwards. 
" À À CIRE AUTRES 10 
Making now the final calculation, the arbor m is retarded (1 + 3x 38 x 86 )x 37x99 

reyolutions in one revolution of s. But this quantity 
88 X 168 + 1 EMTEC 
38 X 168 ~ 3663 6384 ” 3663 

X 
= .00273043 
“. While s makes one revolution, » makes 
1 — .00273043 = .99726957 revolutions, 
which is accurately the quantity r. 
The peculiarity of the solution here presented, consists in the movement of the pins 
Pp, 80 as to apply a secondary correction. 
As here illustrated, if s is the sidereal-hour arbor m will move along with s for about 
80m., when it will be in advance of its true place about 4% seconds ; a correction, quite 
