the hour under the meridian subtracted from 12 will 
be the hour of his rising, and the point of the horizon 
coinciding with his. place will be his eastern amplitude. 
In like manner, ea the globe westward, till the 
‘sun’s place be again in the horizon, the hour under the 
aidia subtraeted from 12, will be the hour of his 
setting, and the point of the horizon, coinciding with 
_ his place, will be his western amplitude. 
In solving this problem by the celestial globe, the 
© ‘method now shewn, is the most obvious and natural; 
because in using the célestial globe, the heavens are 
supposed to 6; px! point rising above the hori- 
zon in the east, and sinking Detieath it in the west, 
‘while the earth itself remains fixed, and the hour cir- 
cle is graduated accordingly. But in solving the pro- 
‘blem by the terrestrial globe, when the sun is sup- 
Feo to be stationary at his place in the ecliptic for 
» and the time of his rising or setting to any 
place, is the time when that place, by the revolution of 
; the earth from west to east, is on the west or east side 
\ > of the circle of’ illumination, the following method, 
» when the hours’of rising and setting only are wanted, 
is perhaps more natural. 
. Find ‘the ‘circle of illumination 
the given place and the 12th hour 
meridian ; then ‘turning the globe westward till the 
place is'in the horizon, the hour under the meridian 
will shew the time of the sun’s risin » and turning it 
eastward till the place is again in the orizon, the hour 
. under the meridian will shew the time of his setting. 
Ttis an advantage attending’ this method of’ solution, 
that the horary shews the time at once, and by one 
rectification of the fiobe, may be found the Jensth 
of the day, in any latitude between the polar circles, 
as well as those places within the polar circles where 
the sun never rises or never sets on that day. 
ind the Pros. XXIX.—To find the length of any day or 
thot night, at “4 given place. < 
days &c Find the hour of sun rising and sun setting by Prob, 
XXVIII. double the latter will be the length of the day, 
‘and double the former the length of the night. 
- Pros. XXX.—To find when, and how long, the sun 
is ant to, or absent from, a given place within the 
zone. 
~Rectify the globe for the 
while it revolves on ‘its axis, mark the points of the 
po og which coincide with the north or south point 
of the horizon, according as the place is in the north or 
south frigid zone. Find the days'in the calendar cor- 
__ ¥esponding to these points, and they will be the limits 
of the time, during which the sun never sets at the 
given place. If the points of the ecliptic intersected 
by the opposite point of the horizon be marked, the 
days corresponding to these points, will be the limits of 
the time, during which the sun never rises at ‘the place. 
The f hcg may be solved without rectifying the 
ig a ‘ 
joa 
a Find the sun’s places in the ecliptic when his decli- 
nation is equal to the co-latitude of the place, and on 
equator ; the days, corresponding 
‘ 5 be the limits of the time during 
: which the sun is hepa to the place. In like manner, 
w c vhen his declination is equal to the co- 
__ latitude of the given Pe but on the opposite side of 
the days, ing to these 
for the day ; bring 
of the horary to the 
nt to 
‘day, not one of thé inoxes, 
%, &. Jess than 24 hours, _ 
GEOGRAPHY. 
‘posing the globe to the sun’s rays. 
latitude of the place, and. 
159 
Rectify the globe for the sun’s place in the ecliptic, Mathemati- 
and bring the first, or any other meridian, with 12 of cal Geogra- 
the hour circle, to the brazen meridian ; then turning — Phy. 
the globe eastward, till the hour denoting half the gi- 
ven Fength of the day, be under the brazen meridian, 
the point of the first meridian, intersected by the eastern 
side of the horizon, will be the latitude required. 
The problems, for finding in what latitudes the longest 
day is of any given length, and for determining the 
boundaries of the different climates, between the polar 
circles, are only particular cases of the preceding ge. 
neral problem. 
Pros. XXXII.—To find the hour of the day by ex- To find the 
hour of the. 
Place the globe so that the wooden horizon may be day, &c. 
level, and the brazen meridian may coincide with the 
meridian of the place. Rectify the globe for the lati. 
tude, and bring the sun’s place in the ecliptic, and 12 
of the hour circle, to the brazen meridian. In the sun’s 
place fix a small pin or needle perpendicular to that 
place, and turn the globe till the pin has no shadow, 
that is, till it point directly towards the sun ; then the 
hour under the brazen meridian, subtracted from 12, 
will be the hour of the day. 
For various other methods of solving the preceding 
ae see Adam’s Astronomical and Geographical 
issays. 
Pros. XXXIII.—To find when the twilight begins To find the 
in the morning, and ends at night on any day, - beginning, 
Rectify the globe for the latitude of the place: bring &«. of twi- 
the 12th hour of the horary and the sun’s place in the ish 
ecliptic to the brazen meridian, and fix the quadrant of 
altitude on the zenith. Turn the globe westward, till 
the point of the ecliptic diametrically opposite to the 
sun’s place cut the quadrant of altitude 18° above the 
eastern side of the horizon, and the hour denoting 
the beginning of twilight in the morning, will be. un- 
der the brazen meridian. By turning the globe in the 
opposite direction, till the point opposite the sun’s place, 
be 18° above the western side-of the horizon, the hour 
circle will shew the end of twilight inthe evening. In 
those latitudes where the sun-does not sink more than 
18° beneath the horizon, the twilight continues all night. 
See Astronomy, vol. ii. p. 660. : 
Pros. XXXIV.—To construct a horizontal dial bY To con 
the globe, 
Rectify the globe for the latitude of the place, and dial. 
bring the first meridian to the brazen meridian ; the 
arches of the horizon, intercepted between the first me- 
ridian, and the meridians passing through every fifteenth 
degree of the equator, will be the measures of the 
angles, which the hour lines must make, with the meri- 
dian line. 
Example. It is required to construct a horizontal dial 
for Edinburgh, supposing its latitude to be 56°. 
Elevate tf. north pole 56° above the horizon, and 
bring the meridian of London to the brazen meridian. 
While the globe-remains fixed in this position, the me. 
ridians on each side of the first meridian, and 15° distant 
from it, will intersect the horizon 77 $°, or more accu- 
rately 77° 28! from the east and west points, that is 
12° 32’ from the south point, therefore the hour lines 
for 11 forenoon, and 1 a! ternoon, must make with the 
meridian. line, each an angle of 12° $2’. The meridi- 
ans on each side of the first meridian, and 30° distant 
from it, will intersect the horizon in 64° 25’, that is 25° 
35’ from the south point, which is therefore the angle 
that the hour lines, of 10 and 2, must make with the 
meridian, In the same manner may be found the angles 
struct a 
