To be specific I am talking about the steps used in this article: http://www.math.unl.edu/~bharbourne1/ST/sphericaltetrahedron.html

Many people don't know what a Reuleaux tetrahedron is, so here's a brief definition of the shape: the intersection of four balls of radius s centered at the vertices of a regular tetrahedron with side length s. The spherical surface of the ball centered on each vertex passes through the other three vertices, which also form vertices of the Reuleaux tetrahedron. Thus the center of each ball is on the surfaces of the other three balls.

I understand, so far, that the entire spherical cap which one face of the spherical tetrahedron rests on, can be calculated using $2\pi*r^2(1-\arccos(\theta))$, and $\theta$ equals 60 degrees so the area of the spherical cap is $\pi*r^2$. Then it goes on to calculate what fraction of that area contains the face, which is the dihedral angle of a tetrahedron divided by $2\pi$ multiplied by the total spherical cap area to get a fractional area. After this comes the issue. The article says it uses the spherical excess formula to calculate T (labelled in the diagram in the article, also attached as an image below) which is a formula saying that a curved triangle constructed from three points of a sphere has a surface area equal to the sum of its interior angles subtracted by $\pi$. However, the article not only sums the interior angles and subtracts $\pi$ from it, but it also multiplies it by $r^2$. Why does the author do that?

(Would also like to mention that multiplying by $r^2$ is not a mistake either. I double checked the final formula with the equation at Wolfram-Alpha https://mathworld.wolfram.com/ReuleauxTetrahedron.html by substituting in $r=1$, and both equations gave the exact same result.)

Labelled Sphere Diagram

  • $\begingroup$ The area is obviously proportional to $r^2$. The spherical excess formula in the form you report is valid only for a sphere with unit radius. $\endgroup$ Jan 27 at 22:53
  • $\begingroup$ @Intelligentipauca Oh I see! So the full formula for the area would actually be $r^2*(a+b+c-\pi)$? I have been for the longest time, under the impression that the formula was simply $a+b+c-\pi$. Thank you so much! $\endgroup$
    – 0x045f
    Jan 27 at 22:59
  • $\begingroup$ Yes, that is the correct formula. $\endgroup$ Jan 28 at 8:04


That's the same as looking for the solid angle of a tetrahedron


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