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Thro' the Center C,
G and (always) the remo
D test of the two given
В Points P, 0, draw the Diameter AB, which cross with a fe
cond Diameter DE, at Right Angles ; join PD;
A on the End of which D, erect the Perpendicular D I, cutting the first Diameter produced at pleasure in 1; then thro' the three Points P, 0, 1, describe the Circle FOGI; and it is the Great Circle required. See Euclid, Lib. 3. Prop. 25.
Note, If the Diameter FG, passeth thro' the Cena ter C, the Circle is drawn true ; otherwise not.
To Measure any Spherical Angle in the Projection.
In this Problem are four Cafes.
and on the Line of Chords
F you will find it contain
H 50° 30', the Quantity of the Angle
В С Н, , as required.
B А Case 2. When the Angle
C is at the Top of the Periphery of the Primi
tive E. Let the Angle to be mea
sured be AED; Find the Pole a, of the Oblique Circle EDF; then draw the Lines Ed, Eg; so shall the Arch Fg = Ad, measured on the Line of Chords, give the Angle
AED = 44° 30' ; and thus the Arch d F, will give the Angle DEC
a R Primitive ; but at P,
t6 Thus ; let the Angle to be
measured be B P Q.; from the Angular Point
D lay a Ruler from the Pole
C, of the Oblique Circle B O H, and it will cut the Primitive in h; But the Point D, is the Pole of the Right Circle Q R ; wherefore the Arch Db,
is the Measure of the Angle B'P Q,
as required. Cafe 4. When the Angle, as A, is contained under two Circles oblique to the Primi
tive, as BOH, DFE, Thus ; to measure the Angle B AD, find the Poles of the Oblique Circles a, C; lay a Ruler from rhe Oblique Point A, to each Pole, and it will cut the Primitive in, b, e ; wherefore the Distance of the Arch be, applied to the Chords, gives the Angle BAD = 35°.;
as was required.
To lay off any Number of Degrees on any Great Circle.
In this Problem are three Cases. Cafe 1. To divide the Pri
mitive A EBC this is done by a Line of
xh Chords thus ; fit the Sector
n to Radius AF;
HE set off the Chord of 46° 30° A A from A to e,
S IF IDB and it is done, Case 2. To lay any Number of Degrees on a Right Circle EFC.
e This is performed by a Line of Half-Tangents thus ;
from the Center F,
and it is done ; for the Arch - Ff = 46° 30'.
To lay any Number of Degrees on an Oblique Circle, as
EDC. First, find the Pole of that Circle then from the Line of Chords, lay off 50° 30', on the Primitive, from
B to bi then lay a Ruler on the Points and it will cut the Oblique Circle in so fhall the Arch
Dn"= 5030; as was 'required. Note, That to measure any part of a great Circle, is only
the Reverse of this problem in each Case ; and cannot be difficult to any one, who hath orderly come thus far.
To draw a Parallel Circle.
Practice. . In this Problem are three Cases. Cafe 1. To draw a Paral·lel to the Primitive Circle
AEDK. With the Half - Tangent L.
с FEC B, of its Distance from the Pole,
AB of the Primitive, describe
the Circle BGFH; and that is the Parallel required.
N to e,
Cafe 2. To draw .a Parallel to any Right Circle,
A D. Set the Half-Tangent
CN, of its Distance, ( suppose from the Right Circle ) from
C to N; Then fet the Tangent of
60°, (the Complement of its Distance from in the Diameter produced to On that Point, as a Center, draw
LNM, and it shall be the Parallel Circle required. Cafe 3. To draw a Paral
lel to any Oblique Cir-
46° 30'; First, find the Pole of the D Oblique Circle
B Then measure the Dis
h on the Half-Tangents, and
it will be 32° 30' ; which fubftract from
43° 30', ( the Complement of the given Distance 46.. 30') and there will remain
II?:oo' set the Half-Tangent of this from and set the Half-Tangent of,
439 30', from the Point
C to fi then draw the Line
ef; bisect this Line in on which, as a Center, describe
and that will be the Parallel Circle required.
C to es
By this Problem, the Path of the Vertex of any Place is described in Astronomical Schemes,