GLOBE. 
eclipse, and what the beginning or ending ; 
by using the moon’s place instead of the 
sun’s place in the problem. 
17 . “ To find the bearing of one place 
from another, and their angle of position.” 
Bring the one place to the zenith, by recti- 
fying the globe for its latitude, and turning 
-the globe till that place come to the meri- 
dian ; then screw the quadrant of altitude 
•upon the meridian at the zenith, and make 
it revolve till it come to the other place on 
the globe ; then look on the wooden hori- 
zon for the point of the compass, or num- 
ber of degrees from the south, where the 
quadrant of altitude cuts it, and that will 
be the bearing of the latter place from the 
former, or the angle of position sought. 
18. “ The day and hour of a solar or lu- 
nar eclipse being given, to find all those 
places in which the same will be visible.” 
Find the place to which the sun is vertical 
at the given instant, and elevate the globe 
to the latitude of the place ; then, in most 
of those places above the horizon will the 
sun be visible during his eclipse ; and all 
those places below the horizon will see the 
moon pass through the shadow of the earth 
in her eclipse. < 
19. “ The length of a degree being given, 
to find the number of miles in a great cir- 
cle of the earth, and thence the diameter of 
the earth.” Admit that one degree con- 
tains 69§ English statute miles ; then multi- 
ply 360 (the number of degrees in a great 
circle) by 69*, and the product will be 
25,020, the miles which measure the cir- 
cumference of the earth. If this number 
be divided by 3.1416, the quotient will be 
7,963-,!! miles, for the diameter of the earth. 
20. “ The diamefer of the earth being 
known, to find the surface in square miles, 
and its solidity in cubic miles.” Admit the 
diameter be 7,964 miles ; then multiply the 
square of the diameter by 3,1416, and the 
product will be 199,250,205 very near, 
which are the square miles in the surface of 
the earth. Again multiply the cube of the 
diameter by 0.5236, and the product 
264,466,789,170 will be the number of the 
cubic miles in the whole globe of the earth. 
21. “ To express the velocity of the diur- 
nal motion of the earth.” Since a place in 
the equator describes a circle of 25,020 
miles in twenty-four hours, it is evident, 
that the velocity with which it moves is at 
the rate of 1,042* in one hour, or 17$ miles 
per minute. The velocity in any parallel 
of latitude, decreases in the proportion of 
the co-sine of the latitude to the radius. 
Thus, for the latitude of London, 51° 30', 
say, 
As radius. . . 10.000000 
To the co-sine of lat. 51° 30'... 9.794149 
So Bthe velocity in the equa- J 2<238046 
To the velocity of the city of \~ c „ ~ 
London, 10$' .,.,.-5 
That is, the city of London moves about 
the axis of the earth at the rate of 10$ miles 
every minute of time : but this is far short 
of the velocity of the annual motion about 
the sun ; for that is at the rate of more than 
65,000 miles per hour, 
PROBLEMS ON THE CELESTIAL GLOBE. 
1. “ To rectify the globe.” Raise or ele- 
vate the pole to the latitude of the place ; 
screw the quadrant of altitude in the zenith ; 
set the index of the hour-circle to the upper 
xu ; and place the globe north and south by 
the compass and needle ; then is it a just re- 
presentation of the heavens from the given 
day at noon. 
2. “ To find the sun’s place in the ecliptic:” 
Find the day of the month in the calendar 
on the horizon, and right against it is the de- 
gree of the ecliptic, which the sun is in for 
that day. 
3. “ To find the sun’s declination.” Rectify 
the globe, bring the sun’s place in the eclip- 
tic to the meridian, and that degree which 
it cuts in the meridian is the declination re- 
quired. 
4. “ To find the sun’s right ascension.” 
Bring the sun’s place to the meridian, and 
the degree of the equinoctial cut by the 
meridian is the right ascension required. 
5. “ To find the sun’s amplitude.” Bring 
the sun’s place to the horizon, and the 
arch of the horizon intercepted between it 
and the east or west point, is the amplitude, 
north or south. 
6. “ To find the sun’s altitude for any 
given day and hour.” Bring the sun’s place 
to the meridian ; set the hour-index to the 
upper xu ; then turn the globe till the in- 
dex points to the given hour, where let it 
stand ; then screwing the quadrant of alti- 
tude in the zenith, lay it over the sun’s 
place, and the arch contained between it 
and the horizon, will give the degrees of al- 
titude required. 
7. “ To find the sun’s azimuth for any 
hour of the day.” Every thing being done 
as in the last problem, the arch of the hori- 
zon contained between the north point, and 
that where the quadrant of altitude cuts it, 
is the azimuth east or west, as required. 
