如何渲染尽可能少的顶点的圆？ [英] How to render a circle with as few vertices as possible?

问题描述

我想弄清楚如何决定需要多少顶点来使我的圈子尽可能平滑。

I'm trying to figure out how to decide how many vertices I need to have to make my circle look as smooth as possible.

这里是一个例子两个圆形，共有24个顶点：$ b​​ $ b

Here is an example of two circles, both having 24 vertices:

如你所见，圆圈越大，我需要隐藏直线的顶点越多。

As you see, the bigger the circle becomes, the more vertices I need to hide the straight lines.

刚开始我认为最小边缘上的一条线的长度应该是6px，但是当我增加圆的大小时，该方法失败：我有太多的顶点。我也想过计算角度，但我很快意识到角度在不同大小的圆圈上没有不同。我也检查了这个答案，但我没有一个线索如何将它转换为代码（和一些奇怪的东西有： th 使用自己计算自己）和我认为它甚至不工作，因为作者使用从一个切片到圆的中间的角度，如果圆圈变大则不改变。

At first I thought that the minimum length of one line on the edge should be 6px, but that approach failed when I increased the circle size: I got way too many vertices. I also thought about calculating the angles, but I quickly realised that angles doesn't differ on different sized circles. I also checked this answer, but I don't have a clue how to convert it into code (and some weird stuff there: th uses itself for calculating itself), and I think it doesn't even work, since the author is using the angle from one slice to the middle of circle, which doesn't change if the circle gets larger.

然后我意识到，也许解决方案是检查两个顶点之间的角度，在这样：

Then I realised that maybe the solution is to check the angle between two vertices at the edges, in this way:

如您所见，顶点越少，这些三角形的长度越大。所以这是必须的答案，我只是不知道如何使用这些信息计算顶点的数量。

As you see, the fewer vertices, the bigger the lengths are for those triangles. So this has to be the answer, I just don't know how to calculate the number of vertices by using this information.

推荐答案

你从这个答案所需要的决定性公式就是这个：

The decisive formula that you need from that answer is this one:

th = arccos(2 * (1 - e / r)^2 - 1)

这会告诉你两个顶点之间的角度，其中 r 是圆的半径， e 是您愿意容忍的最大误差，即您的多边形与圆的最大偏差 - 这是您的图表中标记的误差。例如，您可以选择将 e 设置为像素的0.5。

This tells you the angle between two vertices, where r is the radius of the circle and e is the maximum error you're willing to tolerate, i.e. the maximum deviation of your polygon from the circle -- this is the error marked in your diagram. For example, you might choose to set e to 0.5 of a pixel.

因为 th 以弧度表示，360度（整圆）等于以弧度表示的 2 * pi ，您需要的顶点数

Because th is measured in radians, and 360 degrees (a full circle) is equal to 2*pi in radians, the number of vertices you need is

num_vertices = ceil(2*pi/th)

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