erect the Perpendicular D I,

G

D

B

E

cutting the first Diameter produced at pleasure in 1; then thro' the three Points P, O, I,

defcribe the Circle FO GI;

and it is the Great Circle required. See Euclid, Lib. Prop. 25.

3.

Note, If the Diameter FG, paffeth thro' the Center C, the Circle is drawn true; otherwife not.

PROBLEM V.

To Measure any Spherical Angle in the Projection.

Practice.

In this Problem are four Cafes.

Cafe 1. When the Angle is at the Center of the Pri

mitive C,

As the Angle BCH ;

Take the Arch B H,

and

Ad,

of the Oblique Circle EDF;

then draw the Lines E d, Eg;

fo fhall the Arch Fg

measured on the Line of Chords, give the Angle

AED = 44° 30';

and thus the Arch d F,

will give the Angle DEC = 45° 30'.

Cafe 3. When the Angle is neither in the Center, nor Periphery of the Primitive; but at P, Thus; let the Angle to be measured be BP2;

from the Angular Point

P,

lay a Ruler from the Pole

C,

of the Oblique Circle BO H,

and it will cut the Primitive in b; But the Point D,

is the Pole of the Right Circle QR; wherefore the Arch Db,

is

is the Measure of the Angle BP 2,

and will be found on the Chords to contain 659 15' as required.

Cafe 4. When the Angle, as A,

is contained under two Circles oblique to the Primitive, as BOH, DFE,

Thus; to measure the Angle BAD,

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 = 35o; as was required.

PROBLEM VI.

To lay off any Number of Degrees on any Great Circle.

Practice.

In this Problem are three Cafes.

Cafe 1. To divide the Pri

mitive A E B C

this is done by a Line of Chords thus ; fit the Sector to Radius AF;

E

xh

e

n

fet off the Chord of 46° 30′

A

from A to e,

g FDB

This is performed by a Line of Half-Tangents thus ;

from the Center F,

fet the Half-Tangent of 46° 30' to f,

in the Right Circle FE,

and

and it is done; for the Arch:

Ff46° 30°. Cafe 3. To lay any Number of Degrees on an Ob

lique Circle, as

Firft, find the Pole of that Circle

then from the Line of Chords, lay off

on the Primitive, from

then Tay a Ruler on the Points

ED C. g;

50° 30%

B to bi

Dn = 50° 30′;

and it will cut the Oblique Circle in

fo fhall the Arch

as was required.

Note, That to measure any Part of a great Circle, is only the Reverse of this Problem in each Cafe; and cannot be difficult to any one, who hath orderly come thus far.

Cafe 2.

as

To draw a Parallel to any Right Circle,

Set the Half-Tangent

of its Distance, (fuppofe

from the Right Circle) from

Then fet the Tangent of

(the Complement of its Distance) from in the Diameter produced to

On that Point, as a Center, draw

AD.

CN,

30%

C to N;

60°,

N to e,

e;

LNM,

E

g

af B

and it fhall be the Parallel Circle required. Cafe 3. To draw a Paral

lel to any Oblique Cir

EDG;

it will be

32° 30';

G

which fubftract from

(the Complement of the given Distance

and there will remain

fet the Half-Tangent of this from

from the Point

and fet the Half-Tangent of

then draw the Line

bifect this Line in

on which, as a Center, defcribe

and that will be the Parallel Circle required.

By this Problem, the Path of the Vertex of any

Place is defcribed in Aftronomical Schemes,

PRO