数学问题:根据外角半径/厚度确定内边界的角半径

问题描述

这是数学高手(不是我最擅长的科目)的数学/几何问题.这是针对 WPF 的，但无论如何都应该足够通用以解决问题:

Here's a math/geometry problem for the math whizzes (not my strongest subject). This is for WPF, but should be general enough to solve regardless:

我有两个嵌入的边框元素，外边的元素具有一定的圆角半径，R 和边框厚度，T.给定这两个值，内边框的圆角半径应该是多少，R' 应该设置为两个角边缘不重叠或没有孔?

I have two embedded Border elements, with the outer one having a certain corner radius, R and border thickness, T. Given these two values, what should the corner radius of the inner Border, R' be set to such that the two corner edges meet with no overlap or holes?

到目前为止，我一直在关注它，但如果有人能给我一个合适的公式，那就太好了.如果可以的话，请尊重点！;)

So far I've just been eyeballing it, but if someone can give me a proper formula that would be great. Respect points if you can!! ;)

推荐答案

T'/2 + R` = R - T/2

T'/2 + R` = R - T/2

或

对于给定的 T、R 和 T'，则 R' = R - T/2 - T'/2

For a given T, R and T' then R' = R - T/2 - T'/2

因此，例如，对于 (T=10) 的外边框厚度和 (R=8) 的半径，以及 (T'=4) 的内边框厚度，您需要 1 的内边框半径.

So for instance for an outer border thickness of (T=10), and radius of (R=8), and an inner border thickness of (T'=4), you'd need an inner border radius of 1.

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