如何在透视模式下计算旋转角度以使宽度适合所需尺寸?
How to calculate angle of rotation to make width fit desired size in perspective mode?
我正在尝试想出一种方法,通过 CSS 绕 Y 轴透视旋转图像,以便最终可见宽度等于所需的像素数。
例如,我可能想要旋转 300px
图像,以便在应用旋转和透视后,图像的宽度现在为 240px
(原始图像的 80%)。通过反复试验我知道我可以设置 transform: perspective(300) rotateY(-12.68)
并将左上角点放在 -240px
(这是使用图像的右侧作为原点)
我不太明白如何对此进行逆向工程,以便对于任何给定的图像宽度、透视图和所需宽度,我都可以计算出必要的旋转。
例如。对于同一张 300px
图像,我现在希望它在旋转后宽度为 150px
- 获得必要角度所需的计算是什么?
这里有一个 playground,可以让您了解我在寻找什么,我已经复制了透视和旋转变换所做的数学计算来计算最左边点的最终位置,但我没有鉴于矩阵数学和涉及的多个步骤,我无法弄清楚如何求解角度。
https://repl.it/@BenSlinger/PerspectiveWidthDemo
const calculateLeftTopPointAfterTransforms = (perspective, rotation, width) => {
// convert degrees to radians
const rRad = rotation * (Math.PI / 180);
// place the camera
const cameraMatrix = math.matrix([0, 0, -perspective]);
// get the upper left point of the image based on middle right transform origin
const leftMostPoint = math.matrix([-width, -width / 2, 0]);
const rotateYMatrix = math.matrix([
[Math.cos(-rRad), 0, -Math.sin(-rRad)],
[0, 1, 0],
[Math.sin(-rRad), 0, Math.cos(-rRad)],
]);
// apply rotation to point
const rotatedPoint = math.multiply(rotateYMatrix, leftMostPoint);
const cameraProjection = math.subtract(rotatedPoint, cameraMatrix);
const pointInHomogenizedCoords = math.multiply(math.matrix([
[1, 0, 0 / perspective, 0],
[0, 1, 0 / perspective, 0],
[0, 0, 1, 0],
[0, 0, 1 / perspective, 0],
]), cameraProjection.resize([4], 1));
const finalPoint = [
math.subset(pointInHomogenizedCoords, math.index(0))
/ math.subset(pointInHomogenizedCoords, math.index(3)),
math.subset(pointInHomogenizedCoords, math.index(1))
/ math.subset(pointInHomogenizedCoords, math.index(3)),
];
return finalPoint;
}
<div id="app"></div>
<script crossorigin src="https://unpkg.com/react@16/umd/react.development.js"></script>
<script crossorigin src="https://unpkg.com/react-dom@16/umd/react-dom.development.js"></script>
<script src="https://cdnjs.cloudflare.com/ajax/libs/babel-standalone/6.26.0/babel.js"></script>
<script type="text/babel" data-plugins="transform-class-properties" >
// GOAL: Given the percentage defined in desiredWidth, calculate the rotation required for the transformed image to fill that space (shown by red background)
// eg: With desiredWidth 80 at perspective 300 and image size 300, rotation needs to be 12.68, putting the left point at 300 * .8 = 240.
// How do I calculate that rotation for any desired width, perspective and image size?
// factor out some styles
const inputStyles = { width: 50 };
const PerspDemo = () => {
const [desiredWidth, setDesiredWidth] = React.useState(80);
const [rotation, setRotation] = React.useState(25);
const [perspective, setPerspective] = React.useState(300);
const [imageSize, setImageSize] = React.useState(300);
const [transformedPointPosition, setTPP] = React.useState([0, 0]);
const boxStyles = { outline: '1px solid red', width: imageSize + 'px', height: imageSize + 'px', margin: '10px', position: 'relative' };
React.useEffect(() => {
setTPP(calculateLeftTopPointAfterTransforms(perspective, rotation, imageSize))
}, [rotation, perspective]);
return <div>
<div>
<label>Image size</label>
<input
style={inputStyles}
type="number"
onChange={(e) => setImageSize(e.target.value)}
value={imageSize}
/>
</div>
<div>
<label>Desired width after transforms (% of size)</label>
<input
style={inputStyles}
type="number"
onChange={(e) => setDesiredWidth(e.target.value)}
value={desiredWidth}
/>
</div>
<div>
<label>Rotation (deg)</label>
<input
style={inputStyles}
type="number"
onChange={(e) => setRotation(e.target.value)}
value={rotation}
/>
</div>
<div>
<label>Perspective</label>
<input
style={inputStyles}
type="number"
onChange={(e) => setPerspective(e.target.value)}
value={perspective}
/>
</div>
<div>No transforms:</div>
<div style={boxStyles}>
<div>
<img src={`https://picsum.photos/${imageSize}/${imageSize}`} />
</div>
</div>
<div>With rotation and perspective:</div>
<div style={boxStyles}>
<div style={{ display: 'flex', position: 'absolute', height: '100%', width: '100%' }}>
<div style={{ backgroundColor: 'white', flexBasis: 100 - desiredWidth + '%' }} />
<div style={{ backgroundColor: 'red', flexGrow: 1 }} />
</div>
<div style={{
transform: `perspective(${perspective}px) rotateY(-${rotation}deg)`,
transformOrigin: '100% 50% 0'
}}>
<img src={`https://picsum.photos/${imageSize}/${imageSize}`} />
</div>
</div>
<div>{transformedPointPosition.toString()}</div>
</div>;
};
ReactDOM.render(<PerspDemo />, document.getElementById('app'));
</script>
<script src="https://cdnjs.cloudflare.com/ajax/libs/mathjs/6.0.4/math.min.js"></script>
非常感谢任何帮助!
我会考虑用一种不同的方法来找到没有矩阵计算的公式1以获得以下内容:
R = (p * cos(angle) * D)/(p - (sin(angle) * D))
其中p
是视角,angle
是角度旋转,D
是元素宽度,R
是我们要搜索的新宽度。
如果我们有 -45deg
的角度和等于 100px
的视角以及初始宽度 200px
那么新的宽度将是:58.58px
.box {
width: 200px;
height: 200px;
border: 1px solid;
background:
linear-gradient(red,red) right/58.58px 100% no-repeat;
position:relative;
}
img {
transform-origin:right;
}
<div class="box">
<img src="https://picsum.photos/id/1/200/200" style="transform:perspective(100px) rotateY(-45deg)">
</div>
如果我们有 -30deg
的角度和等于 200px
的视角以及初始宽度 200px
那么新的宽度将为 115.46px
.box {
width: 200px;
height: 200px;
border: 1px solid;
background:
linear-gradient(red,red) right/115.46px 100% no-repeat;
position:relative;
}
img {
transform-origin:right;
}
<div class="box">
<img src="https://picsum.photos/id/1/200/200" style="transform:perspective(200px) rotateY(-30deg)">
</div>
1 为了更好地理解公式让我们考虑下图:
想象一下,我们正在从顶部看一切。红线是我们旋转的元素。大黑点是我们的视角,与场景的距离等于 p
(这是我们的视角)。由于 transform-origin
是右边的,因此将此点放在右边是合乎逻辑的。否则,它应该在中心。
现在,我们看到的是R
设计的宽度,W
是我们没有透视看到的宽度。很明显,有大视角我们看到的几乎和没有视角一样
.box {
width: 200px;
height: 200px;
border: 1px solid;
}
img {
transform-origin:right;
}
<div class="box">
<img src="https://picsum.photos/id/1/200/200" style="transform: rotateY(-30deg)">
</div>
<div class="box">
<img src="https://picsum.photos/id/1/200/200" style="transform:perspective(9999px) rotateY(-30deg)">
</div>
并且在小视角下我们看到的宽度很小
.box {
width: 200px;
height: 200px;
border: 1px solid;
}
img {
transform-origin:right;
}
<div class="box">
<img src="https://picsum.photos/id/1/200/200" style="transform: rotateY(-30deg)">
</div>
<div class="box">
<img src="https://picsum.photos/id/1/200/200" style="transform:perspective(15px) rotateY(-30deg)">
</div>
如果考虑图中O
标注的角度我们可以写出如下公式:
tan(O) = R/p
和
tan(O) = W/(L + p)
所以我们将有 R = p*W /(L + p)
与 W = cos(-angle)*D = cos(angle)*D
和 L = sin(-angle)*D = -sin(angle)*D
这将给我们:
R = (p * cos(angle) * D)/(p - (sin(angle) * D))
为了找到角度,我们可以将公式转换为:
R*p - R*D*sin(angle) = p*D*cos(angle)
R*p = D*(p*cos(angle) + R*sin(angle))
那么按照here1我们可以得到如下等式:
angle = sin-1((R*p)/(D*sqrt(p²+R²))) - tan-1(p/R)
如果你想要 190px
的视角和 150px
的 R 和 200px
的 D
你需要 -15.635deg
的旋转
.box {
width: 200px;
height: 200px;
border: 1px solid;
background:
linear-gradient(red,red) right/150px 100% no-repeat;
position:relative;
}
img {
transform-origin:right;
}
<div class="box">
<img src="https://picsum.photos/id/1/200/200" style="transform:perspective(190px) rotateY(-15.635deg)">
</div>
1 感谢 https://math.stackexchange.com 社区帮助我确定了正确的公式
我正在尝试想出一种方法,通过 CSS 绕 Y 轴透视旋转图像,以便最终可见宽度等于所需的像素数。
例如,我可能想要旋转 300px
图像,以便在应用旋转和透视后,图像的宽度现在为 240px
(原始图像的 80%)。通过反复试验我知道我可以设置 transform: perspective(300) rotateY(-12.68)
并将左上角点放在 -240px
(这是使用图像的右侧作为原点)
我不太明白如何对此进行逆向工程,以便对于任何给定的图像宽度、透视图和所需宽度,我都可以计算出必要的旋转。
例如。对于同一张 300px
图像,我现在希望它在旋转后宽度为 150px
- 获得必要角度所需的计算是什么?
这里有一个 playground,可以让您了解我在寻找什么,我已经复制了透视和旋转变换所做的数学计算来计算最左边点的最终位置,但我没有鉴于矩阵数学和涉及的多个步骤,我无法弄清楚如何求解角度。
https://repl.it/@BenSlinger/PerspectiveWidthDemo
const calculateLeftTopPointAfterTransforms = (perspective, rotation, width) => {
// convert degrees to radians
const rRad = rotation * (Math.PI / 180);
// place the camera
const cameraMatrix = math.matrix([0, 0, -perspective]);
// get the upper left point of the image based on middle right transform origin
const leftMostPoint = math.matrix([-width, -width / 2, 0]);
const rotateYMatrix = math.matrix([
[Math.cos(-rRad), 0, -Math.sin(-rRad)],
[0, 1, 0],
[Math.sin(-rRad), 0, Math.cos(-rRad)],
]);
// apply rotation to point
const rotatedPoint = math.multiply(rotateYMatrix, leftMostPoint);
const cameraProjection = math.subtract(rotatedPoint, cameraMatrix);
const pointInHomogenizedCoords = math.multiply(math.matrix([
[1, 0, 0 / perspective, 0],
[0, 1, 0 / perspective, 0],
[0, 0, 1, 0],
[0, 0, 1 / perspective, 0],
]), cameraProjection.resize([4], 1));
const finalPoint = [
math.subset(pointInHomogenizedCoords, math.index(0))
/ math.subset(pointInHomogenizedCoords, math.index(3)),
math.subset(pointInHomogenizedCoords, math.index(1))
/ math.subset(pointInHomogenizedCoords, math.index(3)),
];
return finalPoint;
}
<div id="app"></div>
<script crossorigin src="https://unpkg.com/react@16/umd/react.development.js"></script>
<script crossorigin src="https://unpkg.com/react-dom@16/umd/react-dom.development.js"></script>
<script src="https://cdnjs.cloudflare.com/ajax/libs/babel-standalone/6.26.0/babel.js"></script>
<script type="text/babel" data-plugins="transform-class-properties" >
// GOAL: Given the percentage defined in desiredWidth, calculate the rotation required for the transformed image to fill that space (shown by red background)
// eg: With desiredWidth 80 at perspective 300 and image size 300, rotation needs to be 12.68, putting the left point at 300 * .8 = 240.
// How do I calculate that rotation for any desired width, perspective and image size?
// factor out some styles
const inputStyles = { width: 50 };
const PerspDemo = () => {
const [desiredWidth, setDesiredWidth] = React.useState(80);
const [rotation, setRotation] = React.useState(25);
const [perspective, setPerspective] = React.useState(300);
const [imageSize, setImageSize] = React.useState(300);
const [transformedPointPosition, setTPP] = React.useState([0, 0]);
const boxStyles = { outline: '1px solid red', width: imageSize + 'px', height: imageSize + 'px', margin: '10px', position: 'relative' };
React.useEffect(() => {
setTPP(calculateLeftTopPointAfterTransforms(perspective, rotation, imageSize))
}, [rotation, perspective]);
return <div>
<div>
<label>Image size</label>
<input
style={inputStyles}
type="number"
onChange={(e) => setImageSize(e.target.value)}
value={imageSize}
/>
</div>
<div>
<label>Desired width after transforms (% of size)</label>
<input
style={inputStyles}
type="number"
onChange={(e) => setDesiredWidth(e.target.value)}
value={desiredWidth}
/>
</div>
<div>
<label>Rotation (deg)</label>
<input
style={inputStyles}
type="number"
onChange={(e) => setRotation(e.target.value)}
value={rotation}
/>
</div>
<div>
<label>Perspective</label>
<input
style={inputStyles}
type="number"
onChange={(e) => setPerspective(e.target.value)}
value={perspective}
/>
</div>
<div>No transforms:</div>
<div style={boxStyles}>
<div>
<img src={`https://picsum.photos/${imageSize}/${imageSize}`} />
</div>
</div>
<div>With rotation and perspective:</div>
<div style={boxStyles}>
<div style={{ display: 'flex', position: 'absolute', height: '100%', width: '100%' }}>
<div style={{ backgroundColor: 'white', flexBasis: 100 - desiredWidth + '%' }} />
<div style={{ backgroundColor: 'red', flexGrow: 1 }} />
</div>
<div style={{
transform: `perspective(${perspective}px) rotateY(-${rotation}deg)`,
transformOrigin: '100% 50% 0'
}}>
<img src={`https://picsum.photos/${imageSize}/${imageSize}`} />
</div>
</div>
<div>{transformedPointPosition.toString()}</div>
</div>;
};
ReactDOM.render(<PerspDemo />, document.getElementById('app'));
</script>
<script src="https://cdnjs.cloudflare.com/ajax/libs/mathjs/6.0.4/math.min.js"></script>
非常感谢任何帮助!
我会考虑用一种不同的方法来找到没有矩阵计算的公式1以获得以下内容:
R = (p * cos(angle) * D)/(p - (sin(angle) * D))
其中p
是视角,angle
是角度旋转,D
是元素宽度,R
是我们要搜索的新宽度。
如果我们有 -45deg
的角度和等于 100px
的视角以及初始宽度 200px
那么新的宽度将是:58.58px
.box {
width: 200px;
height: 200px;
border: 1px solid;
background:
linear-gradient(red,red) right/58.58px 100% no-repeat;
position:relative;
}
img {
transform-origin:right;
}
<div class="box">
<img src="https://picsum.photos/id/1/200/200" style="transform:perspective(100px) rotateY(-45deg)">
</div>
如果我们有 -30deg
的角度和等于 200px
的视角以及初始宽度 200px
那么新的宽度将为 115.46px
.box {
width: 200px;
height: 200px;
border: 1px solid;
background:
linear-gradient(red,red) right/115.46px 100% no-repeat;
position:relative;
}
img {
transform-origin:right;
}
<div class="box">
<img src="https://picsum.photos/id/1/200/200" style="transform:perspective(200px) rotateY(-30deg)">
</div>
1 为了更好地理解公式让我们考虑下图:
想象一下,我们正在从顶部看一切。红线是我们旋转的元素。大黑点是我们的视角,与场景的距离等于 p
(这是我们的视角)。由于 transform-origin
是右边的,因此将此点放在右边是合乎逻辑的。否则,它应该在中心。
现在,我们看到的是R
设计的宽度,W
是我们没有透视看到的宽度。很明显,有大视角我们看到的几乎和没有视角一样
.box {
width: 200px;
height: 200px;
border: 1px solid;
}
img {
transform-origin:right;
}
<div class="box">
<img src="https://picsum.photos/id/1/200/200" style="transform: rotateY(-30deg)">
</div>
<div class="box">
<img src="https://picsum.photos/id/1/200/200" style="transform:perspective(9999px) rotateY(-30deg)">
</div>
并且在小视角下我们看到的宽度很小
.box {
width: 200px;
height: 200px;
border: 1px solid;
}
img {
transform-origin:right;
}
<div class="box">
<img src="https://picsum.photos/id/1/200/200" style="transform: rotateY(-30deg)">
</div>
<div class="box">
<img src="https://picsum.photos/id/1/200/200" style="transform:perspective(15px) rotateY(-30deg)">
</div>
如果考虑图中O
标注的角度我们可以写出如下公式:
tan(O) = R/p
和
tan(O) = W/(L + p)
所以我们将有 R = p*W /(L + p)
与 W = cos(-angle)*D = cos(angle)*D
和 L = sin(-angle)*D = -sin(angle)*D
这将给我们:
R = (p * cos(angle) * D)/(p - (sin(angle) * D))
为了找到角度,我们可以将公式转换为:
R*p - R*D*sin(angle) = p*D*cos(angle)
R*p = D*(p*cos(angle) + R*sin(angle))
那么按照here1我们可以得到如下等式:
angle = sin-1((R*p)/(D*sqrt(p²+R²))) - tan-1(p/R)
如果你想要 190px
的视角和 150px
的 R 和 200px
的 D
你需要 -15.635deg
的旋转
.box {
width: 200px;
height: 200px;
border: 1px solid;
background:
linear-gradient(red,red) right/150px 100% no-repeat;
position:relative;
}
img {
transform-origin:right;
}
<div class="box">
<img src="https://picsum.photos/id/1/200/200" style="transform:perspective(190px) rotateY(-15.635deg)">
</div>
1 感谢 https://math.stackexchange.com 社区帮助我确定了正确的公式