(24)
| ῶ = ½ ∫ | G′ − Gz | dt | = ∫ z3 | (G′ − Gz) / 2An | dz | , | ||
| E − z | A | E − z | √ (2Z) |
(25)
an elliptic integral, of the third kind, with pole at z = E; and then
ῶ − ψ = KCH = tan−1 KH/CH
| = tan−1 | A sin θ dθ/dt | = tan−1 | √ (2Z) | , |
| G′ − G cos θ | (G′ − Gz) / An |
(26)
which determines ψ.
Otherwise, from the geometry of fig. 4,
C′K sin θ = OC − OC′ cos θ,