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Proceedings of the Royal Society 
and consequently 
_ (sin |-A)2 - (sin 
” (sin 
( 6 ) 
or 
/ sin 
X‘^ \sin JA/ 
If, from 0 as a centre, we describe the semicircle BC6 (fig. 7), and 
draw P/iQ parallel to the axis, we have = OB^ - HP^ = X^ - 
wherefore OC : /zQ : : sin JA : sin Ja j and hence if, having made 
Oe equal to sin JA to the radius OB, we describe the semi-ellipse 
must be ; and if we make rn equal to hr, the radius On must be 
perpendicular to the straight line touching the curve at P, or, in 
other words, the direction of the circumference at n is parallel to 
that of the curve at P. 
Similarly, if a parallel to B5 were drawn through c, and the arc 
intercepted from h doubled, we should get the limiting position X, 
where the circumference is parallel to the curve at A, that is to say, 
the angle &0X is equal to the maximum inclination of the curve. 
Thus it is easy to obtain, by calculation or by construction, the 
value of X corresponding to a given inclination, or the inclination 
corresponding to a given value of x, when OB and the maximum 
inclination A are known. 
We may thus obtain an approximation to the true form of the 
curve by a graphic process. A number of lines parallel to the axis, 
and crossing OB, having been drawn, we ascertain the angle at 
which each of them should be crossed by the curve, and draw a 
series of short connected lines with their inclinations. In this 
operation we have our choice among three processes. We may 
divide the ordinate OB equally, computing the corresponding inclina- 
tions ; we may divide the inclinations equally, computing the succes- 
sive values of x ; or else we may graduate the quadrant BQC equally. 
The last-named arrangement is the most convenient of the three. 
A still closer approximation may be made by considering the 
curvature. From equation (5) we at once get the value of that 
is, of the product rx, so that the radius of curvature is easily com- 
puted ; and we are able to compose a series of short circular arcs to 
Bc&, the portion liq intercepted by it must be equal to sin Ja to the 
same radius. Hence, if we draw Qr parallel to B&, the angle hOr 
