186 PHILOSOPHICAL TKANS ACTIONS. [aNNO 1715. 



AB by the edge of kr; so will the little wheel wh describe on the paper a curve 

 line ACB, to be continued as far as is convenient. 



2. Having drawn the curve acb, draw a straight line kr by the edge of the 

 ruler kr, which line is the meridian to be divided, and also an asymptote to the 

 curve ACB. 



3. In this meridian, accounting r to be the point of its intersection with the 

 equator, the point answering to any degree of latitude is thus found. In the 

 perpendicular ar, make rg equal to the cosine of latitude, radius being ar, and 

 from G draw gc parallel to kr, and intersecting the curve inc. With centre c 

 and radius cm = ar, strike an arc cutting the meridian at m, so is m the point 

 desired. 



4. In the curve ac, let cbe a point infinitely near to c, and cm, (= cm) a 

 tangent to the curve at c, making the little angle Mcm, to which let the angle 

 RAr be equal: so is Rr = Md a perpendicular from m to cm. Draw cd equal 

 and parallel to ar, intersecting kr in s. With centre c and radius cd draw the 

 arc DM, and its tangent de and secant ce. 



5. Because of the like triangles cde, Mdm; gd : ce :: Md : Mm, that is, as 

 radius to secant of the arc dm, whose cosine is cs = gr, :: so is Md, = Rr a 

 degree or particle of the equator : to Mm the fluxion or correspondent particle 

 of the meridian line rm. Whence, and from what is premised concerning the 

 nature of this nautical projection, it is evident that rm is the meridional part 

 answering to the latitude whose cosine is gr. Or thus, with centre r and radius 

 AR describe the quadrant Axa, in which let the arc Ax be equal to the given lati- 

 tude. From X. draw xc parallel to kr, and intersecting the curve in c, so is cx 

 the meridional part desired, being equal to rm, as is easy to show. 



As to the other properties of this curve, it is evident, from its construction, 

 that its tangent, as cm, is a constant line, every where equal to ar; the curve 

 being generated by the motion of the wheel at the end of the ruler, which is its 

 tangent. And from hence the curve acb may, for distinction, be called the 

 equitangential curve. 



7. The fluxion of the area armc is the little sector or triangle Mcd, which is 

 also the fluxion of the sector cdm : whence the areas armc, cdm, are equal, 

 and the whole area acb &c, kmr being infinitely continued, is equal to the 

 quadrant ARa. 



8. To find the radius of curvature of any particle, as cc, from c draw an 

 indefinite line ct perpendicular to cm, on the concave side of the curve, and 

 Irom c another line perpendicular to cm, which lines, because of the inclination 

 of cm to CM, will somewhere meet, as at t, making an angle ctc= Mcm. 



