由网友(th,well)分享简介：下面是数学学有专长（不是我的最强主体）数学/几何问题。这是WPF，但应足够全面解决不分：Here's a math/geometry problem for the math whizzes (not my strongest subject). This is for WPF, but should be gene...

下面是数学学有专长（不是我的最强主体）数学/几何问题。这是WPF，但应足够全面解决不分：

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:

我有两个内嵌边框元素，与外层具有一定角半径，研究和边框厚度， T 。鉴于这两个值，又该的内边框的圆角半径， R'设置为使得两个拐角边没有重叠或孔满足？

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，我刚刚被目测，但如果有人可以给我一个正确的公式，将是巨大的。如果你能尊重点！ ;）

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，对于一个外边框的厚度（T = 10），和半径（T'= 4）（R = 8），内边框的厚度，你需要为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.