Calculate bevel edge Icosphere












5












$begingroup$


I have an Icosphere with 80 faces, 120 edges. Now i am looking to find out what the angle is of the bevel between all the faces. With the bevel i mean the following see the image below:



enter image description here



So i am looking on how to find the angle shown in pink.

Also would i be correct to assume that the bevel angle is the same on all sides of each individual triangle. For all the 80 triangles in the icosphere?



One more side question would i be correct to also state that the angles of the actual triangle would all be 120 degrees? See image below:
enter image description here

So that A1 = A2 = A3 = 60 degrees



after the comment of @Blue i thought maybe the shape is a combination of compound elements. A element consisting out of 5 triangles and then singles to fill the spaces between the compound elements that exist of 5 triangles. See Image below:



enter image description here
After doing some counting i came to the conclusion that it is possible to fit 12 compound shapes consisting of 5 triangles on the Icosphere. resulting in a leftover of 20 single triangles.



I am an art student so i have little knowledge when it comes to math. So if any clarification of my problem is needed i am happy to try and provide it!






geometry triangle angle







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    I din't think the angle will be the same on all pairs of triangles. Each vertex sees either 5 or 6 triangles, and some adjacent vertices both see 6, while other pairs of adjacent vertices have one see 5 and the other see 6. The bevel on the edge between the former pair of vertices is a larger angle (closer to 180 degrees) than the bevel on the latter type of edge.
    $endgroup$
    – Adam Cartisano
    Dec 10 '18 at 15:13










  • $begingroup$
    For those (such as myself) unfamiliar with the term, can you confirm that an "icosphere" is a geodesic polyhedron?
    $endgroup$
    – Blue
    Dec 10 '18 at 15:14












  • $begingroup$
    @Blue yeah looking at your link it is indeed a geodesic polyhedron
    $endgroup$
    – FutureCake
    Dec 10 '18 at 15:15






  • 1




    $begingroup$
    thanks i will edit that in the question!
    $endgroup$
    – FutureCake
    Dec 10 '18 at 15:19






  • 1




    $begingroup$
    The face-angles cannot all be $60^circ$ (you have a typo, above the second image, of $120^circ$), since there are groups of six around some vertices. (If all angles were $60^circ$, then these groups would combine to $360^circ$ at the vertex, making the figure flat; but it's not flat.) Whether all the triangles are even identical depends upon exactly which polyhedron this is and what its symmetries are. (I haven't looked very hard, but I suspect the fact that there are groups of five and six about different vertices implies that the triangles are not identical.)
    $endgroup$
    – Blue
    Dec 10 '18 at 15:35


















5












$begingroup$


I have an Icosphere with 80 faces, 120 edges. Now i am looking to find out what the angle is of the bevel between all the faces. With the bevel i mean the following see the image below:



enter image description here



So i am looking on how to find the angle shown in pink.

Also would i be correct to assume that the bevel angle is the same on all sides of each individual triangle. For all the 80 triangles in the icosphere?



One more side question would i be correct to also state that the angles of the actual triangle would all be 120 degrees? See image below:
enter image description here

So that A1 = A2 = A3 = 60 degrees



after the comment of @Blue i thought maybe the shape is a combination of compound elements. A element consisting out of 5 triangles and then singles to fill the spaces between the compound elements that exist of 5 triangles. See Image below:



enter image description here
After doing some counting i came to the conclusion that it is possible to fit 12 compound shapes consisting of 5 triangles on the Icosphere. resulting in a leftover of 20 single triangles.



I am an art student so i have little knowledge when it comes to math. So if any clarification of my problem is needed i am happy to try and provide it!






geometry triangle angle







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    I din't think the angle will be the same on all pairs of triangles. Each vertex sees either 5 or 6 triangles, and some adjacent vertices both see 6, while other pairs of adjacent vertices have one see 5 and the other see 6. The bevel on the edge between the former pair of vertices is a larger angle (closer to 180 degrees) than the bevel on the latter type of edge.
    $endgroup$
    – Adam Cartisano
    Dec 10 '18 at 15:13










  • $begingroup$
    For those (such as myself) unfamiliar with the term, can you confirm that an "icosphere" is a geodesic polyhedron?
    $endgroup$
    – Blue
    Dec 10 '18 at 15:14












  • $begingroup$
    @Blue yeah looking at your link it is indeed a geodesic polyhedron
    $endgroup$
    – FutureCake
    Dec 10 '18 at 15:15






  • 1




    $begingroup$
    thanks i will edit that in the question!
    $endgroup$
    – FutureCake
    Dec 10 '18 at 15:19






  • 1




    $begingroup$
    The face-angles cannot all be $60^circ$ (you have a typo, above the second image, of $120^circ$), since there are groups of six around some vertices. (If all angles were $60^circ$, then these groups would combine to $360^circ$ at the vertex, making the figure flat; but it's not flat.) Whether all the triangles are even identical depends upon exactly which polyhedron this is and what its symmetries are. (I haven't looked very hard, but I suspect the fact that there are groups of five and six about different vertices implies that the triangles are not identical.)
    $endgroup$
    – Blue
    Dec 10 '18 at 15:35
















5












5








5





$begingroup$


I have an Icosphere with 80 faces, 120 edges. Now i am looking to find out what the angle is of the bevel between all the faces. With the bevel i mean the following see the image below:



enter image description here



So i am looking on how to find the angle shown in pink.

Also would i be correct to assume that the bevel angle is the same on all sides of each individual triangle. For all the 80 triangles in the icosphere?



One more side question would i be correct to also state that the angles of the actual triangle would all be 120 degrees? See image below:
enter image description here

So that A1 = A2 = A3 = 60 degrees



after the comment of @Blue i thought maybe the shape is a combination of compound elements. A element consisting out of 5 triangles and then singles to fill the spaces between the compound elements that exist of 5 triangles. See Image below:



enter image description here
After doing some counting i came to the conclusion that it is possible to fit 12 compound shapes consisting of 5 triangles on the Icosphere. resulting in a leftover of 20 single triangles.



I am an art student so i have little knowledge when it comes to math. So if any clarification of my problem is needed i am happy to try and provide it!






geometry triangle angle







share|cite|improve this question











$endgroup$




I have an Icosphere with 80 faces, 120 edges. Now i am looking to find out what the angle is of the bevel between all the faces. With the bevel i mean the following see the image below:



enter image description here



So i am looking on how to find the angle shown in pink.

Also would i be correct to assume that the bevel angle is the same on all sides of each individual triangle. For all the 80 triangles in the icosphere?



One more side question would i be correct to also state that the angles of the actual triangle would all be 120 degrees? See image below:
enter image description here

So that A1 = A2 = A3 = 60 degrees



after the comment of @Blue i thought maybe the shape is a combination of compound elements. A element consisting out of 5 triangles and then singles to fill the spaces between the compound elements that exist of 5 triangles. See Image below:



enter image description here
After doing some counting i came to the conclusion that it is possible to fit 12 compound shapes consisting of 5 triangles on the Icosphere. resulting in a leftover of 20 single triangles.



I am an art student so i have little knowledge when it comes to math. So if any clarification of my problem is needed i am happy to try and provide it!






geometry triangle angle




geometry triangle angle






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 10 '18 at 16:01







FutureCake

















asked Dec 10 '18 at 15:05









FutureCakeFutureCake

1285




1285








  • 1




    $begingroup$
    I din't think the angle will be the same on all pairs of triangles. Each vertex sees either 5 or 6 triangles, and some adjacent vertices both see 6, while other pairs of adjacent vertices have one see 5 and the other see 6. The bevel on the edge between the former pair of vertices is a larger angle (closer to 180 degrees) than the bevel on the latter type of edge.
    $endgroup$
    – Adam Cartisano
    Dec 10 '18 at 15:13










  • $begingroup$
    For those (such as myself) unfamiliar with the term, can you confirm that an "icosphere" is a geodesic polyhedron?
    $endgroup$
    – Blue
    Dec 10 '18 at 15:14












  • $begingroup$
    @Blue yeah looking at your link it is indeed a geodesic polyhedron
    $endgroup$
    – FutureCake
    Dec 10 '18 at 15:15






  • 1




    $begingroup$
    thanks i will edit that in the question!
    $endgroup$
    – FutureCake
    Dec 10 '18 at 15:19






  • 1




    $begingroup$
    The face-angles cannot all be $60^circ$ (you have a typo, above the second image, of $120^circ$), since there are groups of six around some vertices. (If all angles were $60^circ$, then these groups would combine to $360^circ$ at the vertex, making the figure flat; but it's not flat.) Whether all the triangles are even identical depends upon exactly which polyhedron this is and what its symmetries are. (I haven't looked very hard, but I suspect the fact that there are groups of five and six about different vertices implies that the triangles are not identical.)
    $endgroup$
    – Blue
    Dec 10 '18 at 15:35
















  • 1




    $begingroup$
    I din't think the angle will be the same on all pairs of triangles. Each vertex sees either 5 or 6 triangles, and some adjacent vertices both see 6, while other pairs of adjacent vertices have one see 5 and the other see 6. The bevel on the edge between the former pair of vertices is a larger angle (closer to 180 degrees) than the bevel on the latter type of edge.
    $endgroup$
    – Adam Cartisano
    Dec 10 '18 at 15:13










  • $begingroup$
    For those (such as myself) unfamiliar with the term, can you confirm that an "icosphere" is a geodesic polyhedron?
    $endgroup$
    – Blue
    Dec 10 '18 at 15:14












  • $begingroup$
    @Blue yeah looking at your link it is indeed a geodesic polyhedron
    $endgroup$
    – FutureCake
    Dec 10 '18 at 15:15






  • 1




    $begingroup$
    thanks i will edit that in the question!
    $endgroup$
    – FutureCake
    Dec 10 '18 at 15:19






  • 1




    $begingroup$
    The face-angles cannot all be $60^circ$ (you have a typo, above the second image, of $120^circ$), since there are groups of six around some vertices. (If all angles were $60^circ$, then these groups would combine to $360^circ$ at the vertex, making the figure flat; but it's not flat.) Whether all the triangles are even identical depends upon exactly which polyhedron this is and what its symmetries are. (I haven't looked very hard, but I suspect the fact that there are groups of five and six about different vertices implies that the triangles are not identical.)
    $endgroup$
    – Blue
    Dec 10 '18 at 15:35










1




1




$begingroup$
I din't think the angle will be the same on all pairs of triangles. Each vertex sees either 5 or 6 triangles, and some adjacent vertices both see 6, while other pairs of adjacent vertices have one see 5 and the other see 6. The bevel on the edge between the former pair of vertices is a larger angle (closer to 180 degrees) than the bevel on the latter type of edge.
$endgroup$
– Adam Cartisano
Dec 10 '18 at 15:13




$begingroup$
I din't think the angle will be the same on all pairs of triangles. Each vertex sees either 5 or 6 triangles, and some adjacent vertices both see 6, while other pairs of adjacent vertices have one see 5 and the other see 6. The bevel on the edge between the former pair of vertices is a larger angle (closer to 180 degrees) than the bevel on the latter type of edge.
$endgroup$
– Adam Cartisano
Dec 10 '18 at 15:13












$begingroup$
For those (such as myself) unfamiliar with the term, can you confirm that an "icosphere" is a geodesic polyhedron?
$endgroup$
– Blue
Dec 10 '18 at 15:14






$begingroup$
For those (such as myself) unfamiliar with the term, can you confirm that an "icosphere" is a geodesic polyhedron?
$endgroup$
– Blue
Dec 10 '18 at 15:14














$begingroup$
@Blue yeah looking at your link it is indeed a geodesic polyhedron
$endgroup$
– FutureCake
Dec 10 '18 at 15:15




$begingroup$
@Blue yeah looking at your link it is indeed a geodesic polyhedron
$endgroup$
– FutureCake
Dec 10 '18 at 15:15




1




1




$begingroup$
thanks i will edit that in the question!
$endgroup$
– FutureCake
Dec 10 '18 at 15:19




$begingroup$
thanks i will edit that in the question!
$endgroup$
– FutureCake
Dec 10 '18 at 15:19




1




1




$begingroup$
The face-angles cannot all be $60^circ$ (you have a typo, above the second image, of $120^circ$), since there are groups of six around some vertices. (If all angles were $60^circ$, then these groups would combine to $360^circ$ at the vertex, making the figure flat; but it's not flat.) Whether all the triangles are even identical depends upon exactly which polyhedron this is and what its symmetries are. (I haven't looked very hard, but I suspect the fact that there are groups of five and six about different vertices implies that the triangles are not identical.)
$endgroup$
– Blue
Dec 10 '18 at 15:35






$begingroup$
The face-angles cannot all be $60^circ$ (you have a typo, above the second image, of $120^circ$), since there are groups of six around some vertices. (If all angles were $60^circ$, then these groups would combine to $360^circ$ at the vertex, making the figure flat; but it's not flat.) Whether all the triangles are even identical depends upon exactly which polyhedron this is and what its symmetries are. (I haven't looked very hard, but I suspect the fact that there are groups of five and six about different vertices implies that the triangles are not identical.)
$endgroup$
– Blue
Dec 10 '18 at 15:35












1 Answer
1






active

oldest

votes


















4












$begingroup$

This "icosphere" of yours should be a geodesic polyhedron, created from an icosahedron whose faces are divided into four equal triangles. New vertices (i.e. midpoints of icosahedron edges) are then projected onto the embedding sphere.



There are two kinds of dihedral angles, depending on the common edge. A quick construction with GeoGebra shows that (your bevel angle is half dihedral angle, I suppose):




  • angles having a common edge with an endpoint which is the center of a pentagon, measure about $157.54108°$;


  • angles having a common edge with endpoints which aren't the center of a pentagon, measure about $161.9709°$.



And there are two kinds of triangles:




  • triangles having a vertex which is the center of a pentagon, have angles of $55.56901°$, $55.56901°$, $68.86198°$.


  • triangles having no vertex which is the center of a pentagon, have three equal angles of $60°$.







share|cite|improve this answer











$endgroup$













  • $begingroup$
    Hi thanks for the answer much appreaciated! I just found a function in a piece of software. that gave me about the same results the only difference i found was that the angles of a triangle with the center of a pentagon have angles the following: 55.56 , 55.56 , 68.86 could this be correct.
    $endgroup$
    – FutureCake
    Dec 10 '18 at 17:56












  • $begingroup$
    Yes, of course you are right about the angles: I just copy-pasted the wrong number. I'll correct in a moment.
    $endgroup$
    – Aretino
    Dec 10 '18 at 18:08










  • $begingroup$
    One more question if you dont mind, if i need to know the bevel angle of one triangle lets say of the edge with a common endpoint in the center of a pentagon i need to do (180 - 157.54) / 2 right? since two angles connect there? or should the "bevel" of each edge of the individual triangle be 180 - 157.54?
    $endgroup$
    – FutureCake
    Dec 10 '18 at 18:16










  • $begingroup$
    Unfortunately I'm not familiar with "bevel angle": from your question I thought it was just a half of dihedral angle, but I'm not sure. I think you could read this Wikipedia entry about dihedral angle and find yourself the answer.
    $endgroup$
    – Aretino
    Dec 10 '18 at 19:00











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1 Answer
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active

oldest

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1 Answer
1






active

oldest

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active

oldest

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active

oldest

votes









4












$begingroup$

This "icosphere" of yours should be a geodesic polyhedron, created from an icosahedron whose faces are divided into four equal triangles. New vertices (i.e. midpoints of icosahedron edges) are then projected onto the embedding sphere.



There are two kinds of dihedral angles, depending on the common edge. A quick construction with GeoGebra shows that (your bevel angle is half dihedral angle, I suppose):




  • angles having a common edge with an endpoint which is the center of a pentagon, measure about $157.54108°$;


  • angles having a common edge with endpoints which aren't the center of a pentagon, measure about $161.9709°$.



And there are two kinds of triangles:




  • triangles having a vertex which is the center of a pentagon, have angles of $55.56901°$, $55.56901°$, $68.86198°$.


  • triangles having no vertex which is the center of a pentagon, have three equal angles of $60°$.







share|cite|improve this answer











$endgroup$













  • $begingroup$
    Hi thanks for the answer much appreaciated! I just found a function in a piece of software. that gave me about the same results the only difference i found was that the angles of a triangle with the center of a pentagon have angles the following: 55.56 , 55.56 , 68.86 could this be correct.
    $endgroup$
    – FutureCake
    Dec 10 '18 at 17:56












  • $begingroup$
    Yes, of course you are right about the angles: I just copy-pasted the wrong number. I'll correct in a moment.
    $endgroup$
    – Aretino
    Dec 10 '18 at 18:08










  • $begingroup$
    One more question if you dont mind, if i need to know the bevel angle of one triangle lets say of the edge with a common endpoint in the center of a pentagon i need to do (180 - 157.54) / 2 right? since two angles connect there? or should the "bevel" of each edge of the individual triangle be 180 - 157.54?
    $endgroup$
    – FutureCake
    Dec 10 '18 at 18:16










  • $begingroup$
    Unfortunately I'm not familiar with "bevel angle": from your question I thought it was just a half of dihedral angle, but I'm not sure. I think you could read this Wikipedia entry about dihedral angle and find yourself the answer.
    $endgroup$
    – Aretino
    Dec 10 '18 at 19:00
















4












$begingroup$

This "icosphere" of yours should be a geodesic polyhedron, created from an icosahedron whose faces are divided into four equal triangles. New vertices (i.e. midpoints of icosahedron edges) are then projected onto the embedding sphere.



There are two kinds of dihedral angles, depending on the common edge. A quick construction with GeoGebra shows that (your bevel angle is half dihedral angle, I suppose):




  • angles having a common edge with an endpoint which is the center of a pentagon, measure about $157.54108°$;


  • angles having a common edge with endpoints which aren't the center of a pentagon, measure about $161.9709°$.



And there are two kinds of triangles:




  • triangles having a vertex which is the center of a pentagon, have angles of $55.56901°$, $55.56901°$, $68.86198°$.


  • triangles having no vertex which is the center of a pentagon, have three equal angles of $60°$.







share|cite|improve this answer











$endgroup$













  • $begingroup$
    Hi thanks for the answer much appreaciated! I just found a function in a piece of software. that gave me about the same results the only difference i found was that the angles of a triangle with the center of a pentagon have angles the following: 55.56 , 55.56 , 68.86 could this be correct.
    $endgroup$
    – FutureCake
    Dec 10 '18 at 17:56












  • $begingroup$
    Yes, of course you are right about the angles: I just copy-pasted the wrong number. I'll correct in a moment.
    $endgroup$
    – Aretino
    Dec 10 '18 at 18:08










  • $begingroup$
    One more question if you dont mind, if i need to know the bevel angle of one triangle lets say of the edge with a common endpoint in the center of a pentagon i need to do (180 - 157.54) / 2 right? since two angles connect there? or should the "bevel" of each edge of the individual triangle be 180 - 157.54?
    $endgroup$
    – FutureCake
    Dec 10 '18 at 18:16










  • $begingroup$
    Unfortunately I'm not familiar with "bevel angle": from your question I thought it was just a half of dihedral angle, but I'm not sure. I think you could read this Wikipedia entry about dihedral angle and find yourself the answer.
    $endgroup$
    – Aretino
    Dec 10 '18 at 19:00














4












4








4





$begingroup$

This "icosphere" of yours should be a geodesic polyhedron, created from an icosahedron whose faces are divided into four equal triangles. New vertices (i.e. midpoints of icosahedron edges) are then projected onto the embedding sphere.



There are two kinds of dihedral angles, depending on the common edge. A quick construction with GeoGebra shows that (your bevel angle is half dihedral angle, I suppose):




  • angles having a common edge with an endpoint which is the center of a pentagon, measure about $157.54108°$;


  • angles having a common edge with endpoints which aren't the center of a pentagon, measure about $161.9709°$.



And there are two kinds of triangles:




  • triangles having a vertex which is the center of a pentagon, have angles of $55.56901°$, $55.56901°$, $68.86198°$.


  • triangles having no vertex which is the center of a pentagon, have three equal angles of $60°$.







share|cite|improve this answer











$endgroup$



This "icosphere" of yours should be a geodesic polyhedron, created from an icosahedron whose faces are divided into four equal triangles. New vertices (i.e. midpoints of icosahedron edges) are then projected onto the embedding sphere.



There are two kinds of dihedral angles, depending on the common edge. A quick construction with GeoGebra shows that (your bevel angle is half dihedral angle, I suppose):




  • angles having a common edge with an endpoint which is the center of a pentagon, measure about $157.54108°$;


  • angles having a common edge with endpoints which aren't the center of a pentagon, measure about $161.9709°$.



And there are two kinds of triangles:




  • triangles having a vertex which is the center of a pentagon, have angles of $55.56901°$, $55.56901°$, $68.86198°$.


  • triangles having no vertex which is the center of a pentagon, have three equal angles of $60°$.








share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 10 '18 at 18:08

























answered Dec 10 '18 at 17:23









AretinoAretino

23.7k21443




23.7k21443












  • $begingroup$
    Hi thanks for the answer much appreaciated! I just found a function in a piece of software. that gave me about the same results the only difference i found was that the angles of a triangle with the center of a pentagon have angles the following: 55.56 , 55.56 , 68.86 could this be correct.
    $endgroup$
    – FutureCake
    Dec 10 '18 at 17:56












  • $begingroup$
    Yes, of course you are right about the angles: I just copy-pasted the wrong number. I'll correct in a moment.
    $endgroup$
    – Aretino
    Dec 10 '18 at 18:08










  • $begingroup$
    One more question if you dont mind, if i need to know the bevel angle of one triangle lets say of the edge with a common endpoint in the center of a pentagon i need to do (180 - 157.54) / 2 right? since two angles connect there? or should the "bevel" of each edge of the individual triangle be 180 - 157.54?
    $endgroup$
    – FutureCake
    Dec 10 '18 at 18:16










  • $begingroup$
    Unfortunately I'm not familiar with "bevel angle": from your question I thought it was just a half of dihedral angle, but I'm not sure. I think you could read this Wikipedia entry about dihedral angle and find yourself the answer.
    $endgroup$
    – Aretino
    Dec 10 '18 at 19:00


















  • $begingroup$
    Hi thanks for the answer much appreaciated! I just found a function in a piece of software. that gave me about the same results the only difference i found was that the angles of a triangle with the center of a pentagon have angles the following: 55.56 , 55.56 , 68.86 could this be correct.
    $endgroup$
    – FutureCake
    Dec 10 '18 at 17:56












  • $begingroup$
    Yes, of course you are right about the angles: I just copy-pasted the wrong number. I'll correct in a moment.
    $endgroup$
    – Aretino
    Dec 10 '18 at 18:08










  • $begingroup$
    One more question if you dont mind, if i need to know the bevel angle of one triangle lets say of the edge with a common endpoint in the center of a pentagon i need to do (180 - 157.54) / 2 right? since two angles connect there? or should the "bevel" of each edge of the individual triangle be 180 - 157.54?
    $endgroup$
    – FutureCake
    Dec 10 '18 at 18:16










  • $begingroup$
    Unfortunately I'm not familiar with "bevel angle": from your question I thought it was just a half of dihedral angle, but I'm not sure. I think you could read this Wikipedia entry about dihedral angle and find yourself the answer.
    $endgroup$
    – Aretino
    Dec 10 '18 at 19:00
















$begingroup$
Hi thanks for the answer much appreaciated! I just found a function in a piece of software. that gave me about the same results the only difference i found was that the angles of a triangle with the center of a pentagon have angles the following: 55.56 , 55.56 , 68.86 could this be correct.
$endgroup$
– FutureCake
Dec 10 '18 at 17:56






$begingroup$
Hi thanks for the answer much appreaciated! I just found a function in a piece of software. that gave me about the same results the only difference i found was that the angles of a triangle with the center of a pentagon have angles the following: 55.56 , 55.56 , 68.86 could this be correct.
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– FutureCake
Dec 10 '18 at 17:56














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Yes, of course you are right about the angles: I just copy-pasted the wrong number. I'll correct in a moment.
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– Aretino
Dec 10 '18 at 18:08




$begingroup$
Yes, of course you are right about the angles: I just copy-pasted the wrong number. I'll correct in a moment.
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– Aretino
Dec 10 '18 at 18:08












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One more question if you dont mind, if i need to know the bevel angle of one triangle lets say of the edge with a common endpoint in the center of a pentagon i need to do (180 - 157.54) / 2 right? since two angles connect there? or should the "bevel" of each edge of the individual triangle be 180 - 157.54?
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– FutureCake
Dec 10 '18 at 18:16




$begingroup$
One more question if you dont mind, if i need to know the bevel angle of one triangle lets say of the edge with a common endpoint in the center of a pentagon i need to do (180 - 157.54) / 2 right? since two angles connect there? or should the "bevel" of each edge of the individual triangle be 180 - 157.54?
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– FutureCake
Dec 10 '18 at 18:16












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Unfortunately I'm not familiar with "bevel angle": from your question I thought it was just a half of dihedral angle, but I'm not sure. I think you could read this Wikipedia entry about dihedral angle and find yourself the answer.
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– Aretino
Dec 10 '18 at 19:00




$begingroup$
Unfortunately I'm not familiar with "bevel angle": from your question I thought it was just a half of dihedral angle, but I'm not sure. I think you could read this Wikipedia entry about dihedral angle and find yourself the answer.
$endgroup$
– Aretino
Dec 10 '18 at 19:00


















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