直线插补算法示例

p1=[-300,200,200];
p2=[300,200,400];
step=30;  
dx=(p2(1)-p1(1))/step;
dy=(p2(2)-p1(2))/step;
dz=(p2(3)-p1(3))/step;
x_i = zeros(step+1,1);y_i = zeros(step+1,1);z_i = zeros(step+1,1);
for t=0:1:step
    x_i(t+1)=p1(1)+dx*t;
    y_i(t+1)=p1(2)+dy*t;
    z_i(t+1)=p1(3)+dz*t;
end
figure1 = figure('Color',[1,1,1],'Position',[485,180,684,330]);
plot3(x_i,y_i,z_i,'.','color','k');
grid on;hold on;
plot3(p1(1),p1(2),p1(3),'o','color','m');
plot3(p2(1),p2(2),p2(3),'o','color','r');