( 3?? i 
1. Prepare a Rular A B {Fig. z>) of a convenient Length, 
in which let 5 0 be equal to the Radius of the intended 
Proje&ion. To the Point 0 as a Center (on the narrow* 
er Edge of the Rular) fallen a little PI at e- Wheel, t? A tight 
to the Rular, and of a Diameter a little more than the 
thicknefsof the Rular. Let KR {Fig. 3.) reprefent ano- 
ther long Rular, to \vhich AR is a perpendicular Line- 
Place the Rular 4 B upon the Line A R, with the Center 
of the Wbeelat^. Then with one Hand holding fall 
the Rular K R, with the other Hand Hide the end B of 
the Rular A B by the Edge of K R 3 fo will the little 
Wheels h defcribe on the Paper a Curve Line AC B, to 
be continued a$ far as is convenient. 
2. Having drawn the Curve ACB , draw a flreight 
Line K R by the Edge of the Rular K R : which Line is 
the Meridian to be divided, and alfo an Afymptote to the 
Curve A CB. 
3. fn this Meridian, (accounting R to be the Point of its 
Inrerfedlion with the Equator,) the Point anf\yering to 
any Degree of Latitude is thus Found. In the perpendi- 
cular A R, make R 6 equal to the Ccfine of Latitude 
(Radius being A R,) and from G draw G C parallel to 
K R, and interfering the Curve in C. With Center C 
and Radius C M=z A /?, llrike an Arc cutting the Meri- 
dian at M; fo is M the Point defir’d. 
4. In the Curve A C , let c be a Point infinitely near to 
C, and cm, (= CM,) a Tangent to the Curve at c, ma- 
king the little Angle M C to which let the Angle R Ar 
beequab fo is Rr = Md (a Perpendicular from M to cm.) 
Draw C D equal and parallel ,to r /4 R, interfering k R in 
S. With' Center Can d Radius CD draw the Arc D M, 
and its Tangent D E and Secant C E. 
5. Becaufeof the like Triangles C DE, Md m ; CD : 
C £ : :Md : Mm, that is, as Radius to , Recant. of the 
Arc D M, ( whole Cofine is C S — G R, ) : : fo is M d 
Ggg (- 
