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I've tried calculating the exact dihedral angle of a Disdyakis Triacontahedron, with no success. I cannot seem to find it online either. What is the correct approach to trying to figure out this value?

Thank you.

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    $\begingroup$ I would suggest finding an explicit coordinate set and doing the whole thing 'manually' - find the normals for two adjacent faces from their coordinates and find the angle between them from a dot or cross product. $\endgroup$ Commented Aug 6, 2017 at 16:04
  • $\begingroup$ On polyhedra like the Triakis Icosahedron, it is possible to select a height for the "pyramids" on the faces such that the entire solid has the same dihedral angle everywhere. Can this not be done on the Disdyakis Dodecahedron and Triacontahedron? $\endgroup$
    – Disousa
    Commented Aug 6, 2017 at 16:14
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    $\begingroup$ @Disousa: As for calculating the vertices, if $\gamma = \frac{1}{2}(1 + \sqrt{5})$ denotes the golden ratio, the twelve points $(\pm\gamma, \pm1, 0)$ (all four choices of sign) and their cyclic permutation constitute vertices of a regular icosahedron. Take three vertices of one face, form their arithmetic mean, and scale to get a vector on the same sphere as the existing vertices. $\endgroup$ Commented Aug 6, 2017 at 16:14
  • $\begingroup$ (It can indeed be done - I deleted my mistaken comment about dihedral angles.) $\endgroup$ Commented Aug 6, 2017 at 16:15
  • $\begingroup$ According to Mathematica the angle is: $\pi -\arccos\left(\frac{1}{241} \left(179+24 \sqrt{5}\right)\right)$ $\endgroup$ Commented Aug 6, 2017 at 16:21

1 Answer 1

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I am learning MathJax yet, i could not find any tool online to convert text into MathJax as desmos.com does but that format is not supported here..

Apologies, but i will convert it to required format in some time.


First an expression for the dihedral angle is must for this problem.

enter image description here

w.r.t the above figure

Cos(Dihedral angle) = 1 - 2*((sin(Dihedral Angle/2))^2) = 1 - 2*((Cos pi/n)/( cos theta))^2 = 2*asin((Cos pi/n)/( cos theta))

or

sin(Dihedral Angle/2) = (cos pi/n)/(cos theta)

Its a long derivation to show this, but to cut short the steps, lets start from this

In the below figure showing a disdyakis tricontahedron, let the angle of the unitary scalen triangle which goes up in making this be as below

enter image description here

now there can be three dihedral angles, each calculated taking one of the vertices of this scalen triangle as point 'O' of figure 1.

cos (Dihedral angle from A) = 1 - 2((Cos pi/10)/( cos angle A/2))^2, as n = 10 around A --(i)

cos (Dihedral angle from B) = 1 - 2((Cos pi/6)/( cos angle B/2))^2, as n = 6 around B

cos (Dihedral angle from C) = 1 - 2((Cos pi/4)/( cos angle C/2))^2 as n = 4 around C

since all of the three dihehdral angles are equal, we get the folowing equation

sin(Dihedral Angle/2) = (cos pi/10)/(cos angle A/2) = (cos pi/6)/(cos angle B/2) = (cos pi/4)/(cos angle C/2)

But angle (A + B + C) = pi & we have cos pi/10 = ((5^0.5)/(2(5^0.5 - 1)) )^0.5 (=a say) & cos pi/6 = (3^0.5)/2 (=b say) & cos pi/4 = 1/(2^0.5) (=c say)

from the above equations and after following long process of simplification, we get

cos(angle A/2) = ((((b+c)^2 - a^2)(a^2 - (b-c)^2))^0.5)/(2bc) = ((33 + (125^0.5))/48)^0.5

So, from (i) --> Cos(Dihedral angle) = 1 - 2((Cos pi/10)/( cos angle A/2))^2 = - acos((179 + (24*(5^0.5)))/241) on substitution of above values

or Dihedral angle = pi - acos((179 + (24*(5^0.5)))/241)

Also all the angles of the scalen triangle i.e. angles "A, B & C" can be found

as angle A = 2*acos(((33 + (125^0.5))/48)^0.5) = 32.7702785 degrees

and substituting this value in above equations we get

angle B = 2*acos((241/(280 + (1280^0.5)))^0.5) = 58.2379196 degrees &

angle C = 2*acos((241/(420 + (2880^0.5)))^0.5) = 88.9918019 degrees

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