PHP中的坐标旋转 [英] Coordinate Rotation in PHP

问题描述

(这个问题是针对 PHP 的，我知道在其他语言中讨论过这个问题，但是我在用 PHP 实现它时遇到了问题.)

(This question is specific to PHP, I know this is discussed in other languages, but I'm having trouble with implementing it in PHP.)

我正在尝试旋转 x &要放置在旋转图像上的特征的 y 坐标.

I'm attempting to rotate the x & y coordinates of a feature which is to be placed on a rotated image.

$x &$y 是图像旋转前块的原始 x,y 坐标.

$x & $y are the original x,y coordinates of the block before the image was rotated.

$width2 &$height2 是旋转中心(也就是图片的中心).

$width2 & $height2 are the center of rotation (which is the center of the image).

$sin &$cos 是正弦和余弦，由 sin($radians) 和cos($radians) 关于(背景)图像旋转的角度(以弧度为单位)

$sin & $cos are the sine & cosine, which are obtained with sin($radians) and cos($radians) on the degree of rotation the (background) image was rotated by (in radians)

function RotatePoints($x,$y,$width2,$height2,$sin,$cos)
    {
    // translate point back to origin:
    $x -= $width2;
    $y -= $height2;

    // rotate point
    $x = $x * $cos - $y * $sin;
    $y = $x * $sin + $y * $cos;

    // translate point back:
    $x += $width2;
    $y += $height2;

    return array($x,$y);
    }

据说这个函数应该给我块的新坐标，并考虑到旋转.但是定位比较远.

Supposedly this function should give me the new coordinates of the block, with the rotation taken into account. But the positioning is quite far off.

我做错了什么?

推荐答案

在计算旋转时，您应该在代码中使用其他变量:

You should use other variables when you compute the rotation, in your code:

$x = $x * $cos - $y * $sin;
$y = $x * $sin + $y * $cos;

$x 被第一个方程修改，然后您在第二个方程中使用了错误的 $x 值.

$x is modified by the first equation, then you're using wrong value of $x in the second.

更改为:

$temp_x = $x * $cos - $y * $sin;
$temp_y = $x * $sin + $y * $cos;

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