which is to be equated to Ln, so that
−An2 sin θk cos θk − Cn (1 − cos θk) sin θk + Kn sin θk
−Ta cos θk (tan αk+1 + tan αk) + 2Tα sin θk = 0.
(5)
In addition
Mn2xk + T (tan αk+1 − tan αk) = 0,
(6)
with the geometrical relation
xk+1 − xk − a (sin θk+1 + sin θk) − 2l sin αk+1 = 0.
(7)
When the polygon is nearly coincident with Oz, these equations can be replaced by