已知圆中,一段弧长为C,对应的弦长为L,求弧高H?
弧半径为R.
Rn+1=(1+(L-2*Rn*SIN(C/(2*Rn)))/(L-C*COS(C/(2*Rn))))*Rn
取R0初始值,求得R1；再以R1代入求得R2；.;最后得Rn+1与Rn很接近,得弧半径R=Rn
R^2=(R-H)^2+(L/2)^2
R^2=R^2-2*R*H+H^2+L^2/4
H^2-2*R*H+L^2/4=0
H1=R+(R^2-L^2/4)^0.5
H2=R-(R^2-L^2/4)^0.5