Methodus inveniendi lineas curvas

ELASTICIS. 291 $l \Phi = 4,7977381525$ subtr. $= 5,3144251332$ $= 9,4833130193$ ergo $\Phi = 0,3043077545$ add. $\frac{1}{2} \pi = 1,5707963268$ $\frac{a}{c} = 1,8751040813$ quo invento, erit $\frac{A}{B} = \text{tang. } \frac{1}{2} \Phi = 0,1533390624$. Reperitur ergo ratio constantium A & B, ex quibus & ratio reliquarum constantium C & D ad illas cognoscetur. 75. Restat adhuc prima aequatio $b = A+B+D$, quae; ob $D=A+B$, abit in $b=2A+2B$; ideoque $A+B = \frac{1}{2} b$: cum ergo fit $\frac{A}{B} = \text{tang. } \frac{1}{2} \Phi$, fiet $B(1+\text{tang. } \frac{1}{2} \Phi) [TRUNCATED]