18 



U. S. COAST AND GEODETIC SURVEY. 



But the right-hand member of this equation is equal to 

 one-half the arc of the parallel of latitude (p from X = to 

 the value X. If then in figure 3 we lay off the distance MN 

 on the tangent to the parallel drawn from the point where 

 it crosses the central meridian and take it equal in length 

 to one-half the arc of this parallel up to the given longitude 

 X, the angle MCN wiU be equal to one-hau of 6. To de- 

 termine th*e point of intersection, from N £is center with a 

 radius iV^if construct an arc intersecting the parallel at M^. 

 The point M^ is then the intersection of the meridian X 

 with the parallel ip. 



This projection has been much used by the English War 

 Office for the construction of maps. 



Fig. 3. — Construction of arc of parallel on rectangular polyconic projection. 



We can easily determine the radius of curvature of the 

 meridians in this projection. In figure 2 



M'M=(ds cos e-dp), 



since in this cas.e cos ^=1. 



l-tan2 



cos = 



1— 7sm' <p 



e x^ . 



1 + tan^^ 1+4 sm^ (p 



The angle between two successive radii of curvature is the 

 Angle between the tangents to the parallels of <p and <p + d(p 



