G Demonstration. CBD E inter- ameter CAD; To which draw the Diameter N at Right Angles. And let the MO, be projected into the Curve D FCG, in which draw F AG. Then is the Angle B. FA, a right one; and since D Á is perpendicular to B A, it will be perpendicular to FA another Line in the fame Plane FB A. Draw the Lines 1 M, NP, parallel to the Diameters FG, BE; Then is the Angle B F A = NIP, and both right ones ; also because N P, and P I, are parallel to B A, and FA ; the Angle B AF = NP I. Therefore the Triangles B E A, NIP, are Similar ; and FA:IP::BA:NP. Therefore it will be FAQ, : IPq:: BA : NPq. But it is B Aq: NPq:: CAXAD:CP XPD, (By 35 Prop. 3 Lib. Euclid.) Then F A9:1Pq:: CAXAD:CP XPD. But But this is a Property of the Circle and Ellipsis ; therefore the Figure CFDG must be one of them. But because the Circle is oblique to the Plane of the Projection, its Diameter B E is projected into a Line FG less than it felf, by Theorem 5. therefore fince FG is less than BE=DC, it follows the Curve FDGC is an Ellipfis, whose longer Diameter is DC, and the shorter FG. 2. E. D. G THEOREM' X. ; An Ellipsis standing at Right-Angles with the Plane of the Projection, is projected into a Right Line ; if it be parallel to the Plane of the Projection, 'cis projeted into an Ellipfis equal to it felf; if it be oblique to the said Plane, 'twill be projected into an Ellipsis less than it self ; except, lastly, when the longer Diameter is so far shore’ned by the Projection, that it be equal to the shorter Diameter, and then ’tis projected into a Circle. Demonstration. Self-evidence is the Demonstration of this Theorem in every of its Parts ; as is easy to conceive from what has been hitherto demonstrated of Lines and Circles. THEOREM XI.. The Quantity of any Right Line is to its Quantity when projected, as Radius to the Co-line of the Inclina· tion of that Right Line to the Plane of the Projection. VOL, II.-. D Demon t . Demonstration. Let the given Right Line be DC, its Projection (by Theo rem 4.) will be FC; The Angle of Inclina- A B tion is ACD, R whose Co-sine is DEE FC. Make the given Line Radius DC: Then, as the given Right Line DC : Is to its Projection FC=DE :: Radius : to the Co-line of ACD. 2. E. D. Corollary. Hence the Radius DC, the Arch DG, and its right Sine D E, are all evidently projected into the right Line FC = Sine. THEOREM XII. The Area of any Circle is to the Area of its Elliptic Projection, as the Radius to the Co-sine of its Inclination to the Plane of the Projection. Demonstration. Let the Quadrant of a Circle CGB- of an infinite Num- B And let each of thofe Lines go=Ca=Cd. tion of the Circle to Ву By similiar Triangles CDF, Cd f, : to the Co-fine of the Inclination CF; :: fo is the Area of the Quadrant CGB, : to the Area of its Projection CFH. And so is the Area of the whole Circle to its Pro jection. 2. E. D. CHA P. III. Problems of the Orthographical Projection of the Sphere, or Analemma. PROBLEM 1. B В Y the Sector to describe an Ellipsis, having the Transverse and Conjugate Diameters gis ven. VOL. II. D 2 PraEtice. Pralice, Let the Transverse Diameter be AB, 170 60 and the Conjugate Diameter CD; Take the half of each, QA 20 viz.. AE, CE, 10 and set them in the C! Sector from go to RE D D 90, in the Line of Sines ; rallel Sines of 10, 20, 30, &c. on each SemidiameB ter from Eto A, and Eto C. Then on each Division, raise the Perpendiculars 100, 200, 30a, &c. And thro' the Points of Intersection a, a, a, a, &c. Draw the Curvelineal Arch Aa, a, a, &c. C. After the same manner draw the other three Arches CB, BD, DA; so shall the Whole Ellipsis ACBD be compleat as required. 2. E. F. PROBLEM IT. To divide a right Circle in the Plane of the Projection into its proper Degrees. Practices |