Everyone knows that the solid angle subtended at the corner of a cube
is equal to pi/2. Can you work out the analytical value of the solid angle subtended at the corner of
(a) a regular tetrahedron,
and (b) a regular octagon?
Hint: You cant solve (a) without solving (b).
Originally posted by rspoddar82The answer to b) is 0, but I don't see what this has to do with a) 😛
Everyone knows that the solid angle subtended at the corner of a cube
is equal to pi/2. Can you work out the analytical value of the solid angle subtended at the corner of
(a) a regular tetrahedron,
and (b) a regular octagon?
Hint: You cant solve (a) without solving (b).
The angle in a) is equal to the area of an equilateral spherical triangle of side length pi/3.
*Acolyte reads some lecture notes*
cos a = cos b cos c + sin b sin c cos A
1/2 = 1/4 + 3/4 cos A
cos A = 1/3
Area of spherical trinagle = A + B + C - pi = 3 cos^-1(1/3) - pi
But what is cos^-1(1/3)?
Originally posted by Acolyteactually (b) should read "a regular octahedron" . My error . But without finding the solid angle at the corner of a regular octahedron U can't find the solid angle at the corner of a regular ettrahedronmi
The answer to b) is 0, but I don't see what this has to do with a) 😛
Originally posted by rspoddar82acolyte has found the value of the solid angle at the corner of a regular tetrahedron. But how is it related to the solid angle at the corner of a regular octehedron? AND WHAT IS ITS VALUE?
yes acolyte U have Got it. The answer is indeed
(a)3sec^-1(3) - pi which is equal to 3cos^-1(1/3) - pi.
But how could U solve it without solving for the solid angle at the corner of a regular octahedron?
This brings me to a funny story. My chemistry teacher told us some things about molecules and then that the bond angle in tetrahedral molecules was around 109.5 degrees. I told him:
There are five atoms in a tetrahedral molecule, at positions P1, P2, P3, P4, P5. Say P3 is the 'central' one. The vector P1P3 we'll call u, P2P3 v, P4P3 w, P5P3 x. The bond length is l = |u| = |v| = |w| = |x|. So, since the molecule is such that every interatomic distance is maximized relative to the others, any pair of atoms could be interchanged (rotation about P3) so that:
u + v + w + x = 0
Thus u*(u + v + w + x) = 0 (dot product here). This means:
l^2 + 3l^2 cos A = 0
where A is the bond angle (P3 to any of the others). Dividing by l^2 gives cos A = -1/3, so A is about 109.47 degrees (since obviously a bond angle is between 0 and 180 degrees).
The funny bit is that I very nearly failed chemistry.
(Acolyte's result can be got by taking a cross-section through any outer point and P3 and looking at one of the isoceles triangles, but his way doesn't involve the above rigamarole.)
Originally posted by royalchickenDid your chemistry teacher say, "Pardon??"
This brings me to a funny story. My chemistry teacher told us some things about molecules and then that the bond angle in tetrahedral molecules was around 109.5 degrees. I told him:
There are five atoms in a tetrahedral molecule, ...[text shortened]... es (since obviously a bond angle is between 0 and 180 degrees).
,
Originally posted by royalchickenthe story of ur chemistry teacher is interesting. by the way can ur method be applied in exactly similar way to find out the solid angle at the corner of a regular octahedron?/
This brings me to a funny story. My chemistry teacher told us some things about molecules and then that the bond angle in tetrahedral molecules was around 109.5 degrees. I told him:
There are five atoms in a tetrahedral molecule, at positions P1, P2, P3, P4, P5. Say P3 is the 'central' one. The vector P1P3 we'll call u, P2P3 v, P4P3 w, P5P3 x. ...[text shortened]... nd looking at one of the isoceles triangles, but his way doesn't involve the above rigamarole.)
Originally posted by cheskmateWill this be solved , by finding out the area of a spherical square(area on the surface of a sphere of unit radius) of side length equal to the magnitude of the plane angle pi/3 ?? This is the method used by one of the respondents above..He has found the area of a spherical equilateral triangle of side length = pi/3. Intutively , by finding the area of a spherical square of the same side length should give the solid angle of the octahedron.
the story of ur chemistry teacher is interesting. by the way can ur method be applied in exactly similar way to find out the solid angle at the corner of a regular octahedron?/
Can anybody say something definite on this and enlighten whether my intution is on the right track?
Originally posted by ranjan sinhaNo, Acolyte's method will not work in this case.....
Will this be solved , by finding out the area of a spherical square(area on the surface of a sphere of unit radius) of side length equal to the magnitude of the plane angle pi/3 ?? This is the method used by one of the respondents above..He has found the area of a spherical equilateral triangle of side length = pi/3. Intutively , by finding the area ...[text shortened]... anybody say something definite on this and enlighten whether my intution is on the right track?
Originally posted by cheskmatewell ,I guess the value of solid angle at the corner of an octahedron is pie/12..
acolyte has found the value of the solid angle at the corner of a regular tetrahedron. But how is it related to the solid angle at the corner of a regular octehedron? AND WHAT IS ITS VALUE?