从一个点相对于该点的方向构造一个正方形 [英] Constructing a square from a point with respect to the direction of the point

问题描述

如下图所示，我在给定点周围/下方的正方形以给定的宽度(w)和长度(l)

I make a square around/below a given point with given width (w) and length (l) like shown below

但是现在我必须归纳出该点的方向(或可以这样说).我想创建一个朝向该方向的矩形，如下所示.我以前从未研究过标题，如果有人可以向我指出正确的方向，那将会有所帮助.

But now I have to induce the direction of the point (or heading so to say). I want to create a rectangle directed towards that direction like shown below. I have never worked with headings before, if anyone can point me in the right directions, that would be helpful.

----------------- EDIT ------------------- 谢谢@MBo& @HansHirse，我按照您的人员的建议实施了.为简单起见，我选择了该点在矩形的顶线上，而不是远离矩形.我的代码如下所示:

-----------------EDIT--------------- Thank you @MBo & @HansHirse, I implemented as suggested by your people. for simplicity, I chose the point to be on the top line of the rectangle rather to be away from the rectangle. my code is as shown below:

% Point coordinates
P = [5; 5];

% Direction vector
d = [1; -5];

%normalizing 
Len = sqrt(d(1)*d(1)+d(2)*d(2));
cs = d(1)/Len;
sn = d(2)/Len;

% Dimensions of rectangle
l = 200;
w = 100;

% Original corner points of rectangle

x1 = P(1) -w/2;
y1 = P(2);
x2 = P(1)+ w/2;
y2 = P(2);
x3 = P(1)+ w/2;
y3 = P(2) - l;
x4 = P(1) -w/2;
y4 = P(2) - l;
rect = [x1 x2 x3 x4; y1 y2 y3 y4];

%rotated rectangles coordinates:
x1_new = P(1)+ (x1 - P(1))* cs - (y1 - P(2))*sn;
y1_new = P(2) +(x1-P(1))*sn + (y1-P(2))*cs;
x2_new = P(1)+ (x2 - P(1))* cs - (y2 - P(2))*sn;
y2_new = P(2) +(x2-P(1))*sn + (y2-P(2))*cs;
x3_new = P(1)+ (x3 - P(1))* cs - (y3 - P(2))*sn;
y3_new = P(2) +(x3-P(1))*sn + (y3-P(2))*cs;
x4_new = P(1)+ (x4 - P(1))* cs - (y4 - P(2))*sn;
y4_new = P(2) +(x4-P(1))*sn + (y4-P(2))*cs;
new_rect = [x1_new x2_new x3_new x4_new; y1_new y2_new y3_new y4_new];


%plot:
figure(1);
plot(P(1), P(2), 'k.', 'MarkerSize', 21);  
hold on;
plot([P(1) P(1)+d(1)*10], [P(2) P(2)+d(2)*10], 'b');
patch(new_rect(1,:),new_rect(2,:), 'b', 'FaceAlpha', 0.2);
patch(rect(1,:),rect(2,:),'b')
hold off

旋转方式不是我想要的:我无法上传图片，imgur表现得很奇怪.您可以运行确切的代码，您将获得我得到的输出.

The way it is rotating is not what I wanted: I am not able to upload a pic, imgur is acting weird. You can run the exact code, you will get the output I am getting.

推荐答案

因此，您具有点P和矩形R(由坐标定义).

So you have point P, and rectangle R (defined by coordinates).

现在，您想围绕点P旋转矩形A角(据我了解)
每个顶点的新坐标为:

Now you want to rotate rectangle by angle A around point P (as far as I understand)
New coordinates for every vertex are:

NewV[i].X = P.X + (V[i].X - P.X) * Cos(A) - (V[i].Y - P.Y) * Sin(A)
NewV[i].Y = P.Y + (V[i].X - P.X) * Sin(A) + (V[i].Y - P.Y) * Cos(A)

如果您有方向矢量D = [d1, d2]，则无需使用trig进行操作.函数:仅利用归一化向量的成分(也许matlab包含用于归一化的函数):

If you have direction vector D = [d1, d2], you don't need to operate with trig. functions: just exploit components of normalized vector (perhaps matlab contains function for normalizing):

Len = magnitude(D) = sqrt(d1*d1+d2*d2)
//normalized (unit vector)
dx = d1 / Len
dy = d2 / Len
NewV[i].X = P.X + (V[i].X - P.X) * dx - (V[i].Y - P.Y) * dy
NewV[i].Y = P.Y + (V[i].X - P.X) * dy + (V[i].Y - P.Y) * dx

您还可以省略创建轴对齐矩形的坐标并立即计算顶点(错误的可能性更大)

Also you can omit creating coordinates of axis-aligned rectangle and calculate vertices immediately (perhaps more possibilities for mistakes)

//unit vector perpendicula to direction
perpx = - dy
perpy = dx

V[0].X = P.X - dist * dx  + w/2 * perpx
V[1].X = P.X - dist * dx  - w/2 * perpx
V[2].X = P.X - dist * dx  - w/2 * perpx - l * dx
V[3].X = P.X - dist * dx  + w/2 * perpx - l * dx

and similar for y-components

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