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[其他] A Convenient Expression for Packed Circle Radius

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你们是我最耀眼的星盟 发表于 2016-10-16 07:53:51
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When you develop a solver for the Tammes problem you’re usually concerned with distributing points evenly on the sphere , ensuring they are equidistant from each other. The radius of the circles you place at those points is generally not considered:
   

A Convenient Expression for Packed Circle Radius

A Convenient Expression for Packed Circle Radius-1-技术控-concerned,solution,ensuring,general,between
   There are known solutions for a given number of circles but there is no known solution for any number of points. Relaxation can be used in the general case.
Once the unit sphere is packed with circles, finding the radius can start with two circles and the planes they lie within:
   

A Convenient Expression for Packed Circle Radius

A Convenient Expression for Packed Circle Radius-2-技术控-concerned,solution,ensuring,general,between
   If the circle center points $p_0$ and $p_1$ are on the unit sphere, they can also be considered the normals of the planes they lie within. The acute angle $\theta$ between these two planes is called the dihedral angle and is easily calculated as:
$$\theta = acos(p_0 \cdot p_1)$$
  The obtuse angle, $\alpha$, is called the anhedral angle and is thus:
$$\alpha = \pi - \theta$$
The adjacent/opposite sides of the triangle formed by the two planes and the line joining $p_0$ and $p_1$ will be the same size and can be considered the desired circle radius:
   

A Convenient Expression for Packed Circle Radius

A Convenient Expression for Packed Circle Radius-3-技术控-concerned,solution,ensuring,general,between
   Using the Law of Cosines an expression for $d$ and then $r$, can quickly be attained:
$$d^2 = r^2 + r^2 - 2r^2cos(\alpha)$$ $$d^2 = 2r^2(1 - cos(\alpha))$$ $$r = \sqrt{\frac{d^2}{2(1 - cos(\alpha))}}$$
  This works well enough to get the radius you want but the overuse of transcendentals feels like it can be squeezed some more. Following the Law of Sines makes things even worse but there’s a clue in the $cos(\alpha)$ term which expands to:
$$cos(\pi - \theta)$$
This can be reduced using a common trigonmetric identity:
$$cos(\pi - \theta) = cos(\pi)cos(\theta) + sin(\pi)sin(\theta)$$
Given that $cos(\pi)=-1$, $sin(\pi)=0$ and $\theta=acos(p_0 \cdot p_1)$, this immediately reduces to:
$$cos(\pi - \theta) = -(p_0 \cdot p_1)$$
Leaving the final substitution:
$$r = \sqrt{\frac{d^2}{2(p_0 \cdot p_1 + 1)}}$$



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廖程程 发表于 2016-10-16 09:34:54
我是火华哥,抢沙发专业户。。。。
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车小强 发表于 2016-10-17 01:39:25
赞一个!
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ㄗs_林 发表于 2016-10-17 22:28:29
别和我谈理想,戒了.
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narzo 发表于 2016-10-17 22:48:38
求沙发
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粉色美人鱼3 发表于 2016-10-18 02:04:57
不回帖,臣妾做不到啊!
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