CATransform3D-了解变换值 [英] CATransform3D - understanding the transform values

问题描述

图片显示了应用以下转换后的简单UIView：

The picture shows a simple UIView after applying the following transform:

- (CATransform3D) transformForOpenedMenu
{
  CATransform3D transform = CATransform3DIdentity;
  transform.m34 = -1.0 /450;
  transform = CATransform3DRotate(transform, D2R(40), 0, 1, 0);
  transform = CATransform3DTranslate(transform, 210, 150, -500);
  return transform;
}

我正在尝试用黑色突出显示的距离具有相等的长度。您能帮我理解值和计算背后的逻辑吗？

I'm trying to make the distances highlighted with black to have equal length. Could you please help me understand the logic behind the values and calculations?

干杯

外观就像删除3DTranslate一样使距离相等。我看到我可以使用图层的frame属性将旋转的视图重新定位到屏幕的左下角。尚不确定，但这实际上可能有效。

Looks like removing 3DTranslate keeps distances equal. I see I can use layer's frame property to reposition rotated view to the bottom left of the screen. Not yet sure, but this might actually work.

推荐答案

好的，所以最终解决我问题的是这个：

OK, so what eventually solved my question is this:

• 在Y轴上进行3D变换以像门一样摆动视图 transform = CATransform3DRotate（transform，D2R（40），0，1，0）;


• 在图层上设置Z锚点，将其移回 targetView.layer.anchorPointZ = 850;



• 调整图层位置，以使该视图稍微位于父视图的左下方：

• 3D transform on Y axis to swing the view like a door transform = CATransform3DRotate(transform, D2R(40), 0, 1, 0);
• set Z anchor point on a layer, to move it back targetView.layer.anchorPointZ = 850;

• adjust layer position so that the view is located slightly to the bottom left of the parent view:

newPosition.x + = 135 * positionDirection;
newPosition.y + = 70 * positionDirection;

此序列会调整 CATransform3DTranslate 的位置，并保持摆动效果不失真。

This sequence adjusts position without CATransform3DTranslate and keeps the 'swinged' effect not distorted.

谢谢大家！

这篇关于CATransform3D-了解变换值的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！