霍夫直线变换
直线截据表达式:y = k*x+b
直线极坐标表达式:x*cosθ + y*sinθ = ρ,其中θ是直线垂足方向。
问题:已知直线上两点(xi,yi),(xj,yj),直线外一点(rx,ry)求直线外该点到直线的垂足坐标。即求两条直线的交点坐标
step1:直线的方向θ = atan2(yj-yi,xj-xi)
step2:theta_1 = θ + Π/2
step3:theta_2 = theta_1 + Π/2
step4:rho_1 = xi*cos(theta_1) + yi*sin(theta_1)
rho_2 = rx*cos(theta_2) + ry*sin(theta_2)
即求出了两条直线的θ和ρ。
step5:设垂足坐标为(x,y),那么:
x*cos(theta_1) + y*sin(theta_1) = rho_1
x*cos(theta_2) + y*sin(theta_2) = rho_2
step6:解得:
x=(rho_2*sin(theta_1)-rho_1*sin(theta_2))/(cos(theta_2)*sin(theta_1)-cos(theta_1)*sin(theta_2))
y=(rho_2*cos(theta_1)-rho_1*cos(theta_2))/(sin(theta_2)*cos(theta_1)-sin(theta_1)*cos(theta_2))