28 Jun 04
Originally posted by jimmyb270Solid 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.
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)! 🙄
04 Jul 04
Originally posted by howzzatthe 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.????
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.
04 Jul 04
Originally posted by cosmic voiceErm, 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.
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.????
05 Jul 04
Originally posted by cosmic voiceccccosic voice, your result is equivalent to Acolytes'.
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.????
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.
05 Jul 04
Originally posted by rspoddar82Ignor my last comment. I was talking about faces and you about corners 😳.
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.
06 Jul 04
Originally posted by royalchickennever 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.
Ignor my last comment. I was talking about faces and you about corners 😳.
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).
06 Jul 04
Originally posted by rspoddar82I'll look at it. I haven't been the best at keeping up with the forums lately and forgot, but I'll try it.
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).
09 Jul 04
Originally posted by rspoddar82you earlier suggested that this is related to the solid angle at the corner of an equi-tetrahedron! How?..
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).
10 Jul 04
Originally posted by cheskmateimagine 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.
you earlier suggested that this is related to the solid angle at the corner of an equi-tetrahedron! How?..
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
10 Jul 04
Originally posted by rspoddar82Royalchiken's suggestion seems to be different. Are the two approaches equivalent?
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
13 Jul 04
Originally posted by rspoddar82Now perhaps , finally thatt's it ? . From ur eqn . the solid angle subtended at the corner of an equi-octahedron works out to be
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
equal to = 2*pi - 4 cos^-1(1/3).
16 Jul 04
Originally posted by rspoddar82There is perhaps another simpler way. Acolyte has worked out the angle between two adjacent faces of a regular tetrahedron is
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
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.