you may solve this solid problem

Originally posted by jimmyb270
Everyone knows?!? I don't even know what you mean by that, let alone know what it is (well actually, I do know that much now, it's pi/2)! 🙄

Solid angle is measured by the area on the surface of a unit sphere (a sphere of unit radius) by the solid angle, that is the equivalent of an angle in three dimensions.

Originally posted by howzzat
Solid angle is measured by the area on the surface of a unit sphere (a sphere of unit radius) by the solid angle, that is the equivalent of an angle in three dimensions.

the answers posted earlier by acolyte and others seem to be wrong. By actual integration , it comes out that the solid angle at the corner of a regular tetrahedron is = cosine inverse of 23/27. The poser too has endorsed the seemingly wrong answers posted earlier.????

Originally posted by cosmic voice
the answers posted earlier by acolyte and others seem to be wrong. By actual integration , it comes out that the solid angle at the corner of a regular tetrahedron is = cosine inverse of 23/27. The poser too has endorsed the seemingly wrong answers posted earlier.????

Erm, you've probably found the solid angle as described earlier, while I found the angle between the two faces. I dunno if you're talking about the same thing though, as you'd expect the 'solid angle' subtended by a face of the tetrahedron to be just pi, a quarter of the whole of space, which actual integration also bears out.

Originally posted by cosmic voice
the answers posted earlier by acolyte and others seem to be wrong. By actual integration , it comes out that the solid angle at the corner of a regular tetrahedron is = cosine inverse of 23/27. The poser too has endorsed the seemingly wrong answers posted earlier.????

ccccosic voice, your result is equivalent to Acolytes'.
in fact had you carried your calculations only a little further you would have discovered that acolyte's value i.e.3cos^-1(1/3)- pi is indeed exactly equal to cos^-1(23/27).
So no contradiction there.

Originally posted by rspoddar82
ccccosic voice, your result is equivalent to Acolytes'.
in fact had you carried your calculations only a little further you would have discovered that acolyte's value i.e.3cos^-1(1/3)- pi is indeed exactly equal to cos^-1(23/27).
So no contradiction there.

Ignor my last comment. I was talking about faces and you about corners 😳.

Originally posted by royalchicken
Ignor my last comment. I was talking about faces and you about corners 😳.

never mind. in ur earlier post U correctly calculated the angle between two bonds of a tetrahedral molecule. And suggested that acolytes result can be obtained from that by taking projections.
By the way why not attempt the second part of the puzzle?....Finding the solid angle at the corner of a regular octahedron
(equi-octahedron).

Originally posted by rspoddar82
never mind. in ur earlier post U correctly calculated the angle between two bonds of a tetrahedral molecule. And suggested that acolytes result can be obtained from that by taking projections.
By the way why not attempt the second part of the puzzle?....Finding the solid angle at the corner of a regular octahedron
(equi-octahedron).

I'll look at it. I haven't been the best at keeping up with the forums lately and forgot, but I'll try it.

Make a cube with vertices at the center of each face of your octahedron. Can you see how my earlier method can be applied from there?

Originally posted by royalchicken
Make a cube with vertices at the center of each face of your octahedron. Can you see how my earlier method can be applied from there?

How does that help?...please make it more clear.

Originally posted by rspoddar82
never mind. in ur earlier post U correctly calculated the angle between two bonds of a tetrahedral molecule. And suggested that acolytes result can be obtained from that by taking projections.
By the way why not attempt the second part of the puzzle?....Finding the solid angle at the corner of a regular octahedron
(equi-octahedron).

you earlier suggested that this is related to the solid angle at the corner of an equi-tetrahedron! How?..

Originally posted by cheskmate
you earlier suggested that this is related to the solid angle at the corner of an equi-tetrahedron! How?..

imagine a cube. it has 12 edges. imagine all the mid-points of all the 12 edges, joined to the mid points of all neighbouring edgees. thus you have 8 equilateral triangles & 6 square faces,surrounding the body centre of the cube from all sides.
The body centre of the cube is the vertex of 8 equitetrahedrons, and the vertex of 6 half-octahedrons(hence of 6 equi-octahedrons). Let the solid angle at a corner of an equitetrahedron be W4, and let the solid angle at the corner of an equitetrahedron be W8. then obviously we have,
6 x W8 + 8 x W4 = 4 x pi.
This gives W8 in terms of W4. The value of W4 has already been found in the earlier posts. Thus W8 can be worked out..........
and s

Originally posted by rspoddar82
imagine a cube. it has 12 edges. imagine all the mid-points of all the 12 edges, joined to the mid points of all neighbouring edgees. thus you have 8 equilateral triangles & 6 square faces,surrounding the body centre of the cube from all sides.
The body centre of the cube is the vertex of 8 equitetrahedrons, and the vertex of 6 half-octahedrons(hence ...[text shortened]... W4 has already been found in the earlier posts. Thus W8 can be worked out..........
and s

Royalchiken's suggestion seems to be different. Are the two approaches equivalent?

Originally posted by royalchicken
Make a cube with vertices at the center of each face of your octahedron. Can you see how my earlier method can be applied from there?

No... at least I cant see. pl demystify the trick if it really works.

Originally posted by rspoddar82
imagine a cube. it has 12 edges. imagine all the mid-points of all the 12 edges, joined to the mid points of all neighbouring edgees. thus you have 8 equilateral triangles & 6 square faces,surrounding the body centre of the cube from all sides.
The body centre of the cube is the vertex of 8 equitetrahedrons, and the vertex of 6 half-octahedrons(hence ...[text shortened]... W4 has already been found in the earlier posts. Thus W8 can be worked out..........
and s

Now perhaps , finally thatt's it ? . From ur eqn . the solid angle subtended at the corner of an equi-octahedron works out to be
equal to = 2*pi - 4 cos^-1(1/3).

Originally posted by rspoddar82
imagine a cube. it has 12 edges. imagine all the mid-points of all the 12 edges, joined to the mid points of all neighbouring edgees. thus you have 8 equilateral triangles & 6 square faces,surrounding the body centre of the cube from all sides.
The body centre of the cube is the vertex of 8 equitetrahedrons, and the vertex of 6 half-octahedrons(hence ...[text shortened]... W4 has already been found in the earlier posts. Thus W8 can be worked out..........
and s

There is perhaps another simpler way. Acolyte has worked out the angle between two adjacent faces of a regular tetrahedron is
cos^-1(1/3).
In a unit sphere,with center at O, let AOB be a vertical diameter. Now draw two planespassing through this diameter, and inclined at angle cosine inverse of 1/3, cut the sphere surface in the semi-great-circles APQB & ARSB. The points P, Q, R, & S are so chosen on the semi-circles that the straight lines OA=AP=PQ=QB=AR=RS=SB=OQ=OS=OR=PR=Qs.
Thus the 3-d figure OAPR is an equitetrahedron. So is OBQS. The 3-d figure OPQSR is a semi octahedron with vertex at O.
Now in your notation a new relation between solid angles W4, & W8 is found ....
W8 + 2*w4 = 4*pi/(2*pi/cos^-1(1/3)) = 2 cos^-1(1/3).
I hope this gives the same value for W8 as from your relation.