Having drawn the Semi-Circle HER, let it represent the Northern half of the Earth’s illuminate Disk (because the Sun is in *Gemini*) projected upon the said Plane, the Sun it’s Center, the Point therein opposite to the Sun, H☉︎R an Arc of the Ecliptick passing through it. Upon ☉︎ raise GE, perpendicular to the Ecliptick HR, and the Point E wherein it intersects the Limb of the Disk, will be the Pole of the Ecliptick, and ☉︎E it’s Axis.

Again; Make ☉︎E equal to the Radius of a Line of Chords (by *Use III. of the Line of Chords*), from which taking the Chord of 23 Deg. 30 Min. (the constant Distance of the two Poles) set it off both ways from E to B and C, draw the Line BC, in which the Northern Pole of the World shall be found.

Make BA equal to AC, the half of this Line, the Radius of a Line of Sines, and therein set off the Sine of the Sun’s Distance from the solstitial Colure 19 Deg. 19 Min. 30 Sec. from A to P, on the Left-Hand of the Axis of the Ecliptick (because the Sun is in *Gemini*), and draw the Line OP, which will be the Axis of the Earth, and P the Place of the North-Pole in the illuminate Hemisphere of the Disk.

Or the Angle E☉︎I, which the Axis of the Earth and Ecliptick make with each other, may be more accurately determined by Calculation. For,

As Radius—to the Sine of the Sun’s Distance from the Solstitial Colure | 90° 00′ 00″ | 10,000000 |

19° 19′ 30″ | 9,519731 | |

So is the Tangent of the Sun’s greatest Declination to the Tangent of the Inclination of the Axis | 23° 30′ 00″ | 9,637956 |

8° 10′ 54″ | 9,157687 |

Count the said 8 Deg. 10 Min. 54 Sec. in the Limb of the Disk from E to I, on the Left-hand, and draw the Line ☉︎I, this shall be the Axis; and the Point P wherein it intersects the Line BC, the Place of the Pole in the illuminate Disk.

The next thing required will be the Sun’s Distance from the Pole, or the Complement of his Declination, which will be found 67 Deg. 57 Min. 48 Sec. this added to the Distance of the Vertex from the Pole 38 Deg. 32 Min. makes 106 Deg. 29 Min. 48 Sec. and the same 38 Deg. 32 Min. taken from 67 Deg. 51 Min. 48 Sec. gives 29 Deg. 25 Min. 48 Sec. the Meridional Distance of the Sun from the Vertex.

Make ☉︎E the Radius of the Disk, to be the Radius of a Line of Sines, from which take the Sine of 73 Deg. 30 Min. 12 Sec. (the Complement of 106 Deg. 29 Min. 48 Sec. to a Semi-Circle) and set it off in the Axis from ☉︎ to 12; it there gives the Meridional Intersection of the Nocturnal Arc of the Path with the Axis.

Take the Sine of 29 Deg. 25 Min. 48 Sec. from the same Line of Sines, and set it off the same way from to M, and it there gives the Intersection of the diurnal Arc of the Path with the Meridian. Whence M12 will be the conjugate Diameter of the Path, it being the Difference of the Sines of 70 Deg. 30 Min. 12 Sec. and 29 Deg. 25 Min. 48 Sec. that is, the Difference of the Sines of the Sum, and Difference of the Distances of the Path and Sun from the Pole, which will be the conjugate Diameter of any Path.

Bisect 12M in C, and through it draw 6C6 at right Angles to the Axis of the Globe; and then taking the Sine of 38 Deg. 32 Min. the Distance of the Pole from the Vertex, set it off from C both ways to 6 and 6; then the Line 66 will be the Tranverse-Diameter of the Path, and C6 the Semi-Diameter.

Making C6 equal to the Radius of a Line of Sines, if from the same you take the right Sines of 15, 30, 45, 60, 75 Degrees, and set them off severally both ways from C in the Tranverse-Diameter, and from the Points so found erect Perpendiculars, *a*11, *a*1, *a*10, *a*2, &c. equal to the Co-Sines of the said Arcs, taken from a Line of Sines, whose Radius shall be C12, equal to CM, you will have twenty-four Points given, through which the Ellipsis representing the Path shall pass, which shall also shew the Place of the Vertex at every Hour of the Day. In the same manner may the Parts of an Hour be pricked down in the Path, in laying off the Sine of the Degrees and Minutes corresponding thereto from C towards 6, and then raising Perpendiculars from the Points so found in the Semi-Transverse, and setting off from the said Semi-Transverse each way upon the Perpendiculars, the Sines of the Complements of the Degrees and Minutes corresponding to the aforesaid Parts of an Hour. As, for Example; To denote half an Hour past 11 and 12, cake the Sine of 7 Deg. 30 Min. and lay it off on both sides from C to *b* and *b*; then take the Co-sine of 7 Deg. 30 Min. and having raised the Perpendiculars *b*\(\frac{1}{2}\), lay off the laid Sine-Complement from *b* to \(\frac{1}{2}\), and you will have the Points in the Periphery of the Ellipsis, for half an Hour past 11, and half an Hour past 12; and in this manner may the Path be divided into Minutes, if the Ellipsis be large enough.

Take this for another Example; Suppose I would represent upon the Plane of the Earth’s Disk, the Path of *Gibraltar*, whose Latitude is 35 Deg. 32 Min. North, and the Sun’s Place is in 15 Deg. 45 Min. of *Leo*.