Originally posted by observantUno...your steps do not give the same value. It is different from the value as calculated on the lines suggested by the poser .
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 AP ...[text shortened]... pi/cos^-1(1/3)) = 2 cos^-1(1/3).
I hope this gives the same value for W8 as from your relation.
Originally posted by neverB4chessI think ,without any reference to equitetrahedron too, one can find the solid angle at a corner of an equi-octahedron. Take a=pi/2,
No, Acolyte's method will not work in this case.....
b=c=pi/3.This defines the triangular cone of a half-equi-octahedron.
The angles A,B & C between the three plane faces of the triangular cone can be found by the relations:
cos a = cos b cos c + sin b sin c cos A.
cos b = cos c cos a + sin c sin a cos B.& so on..Thus:
0 = (1/2)^2 + (((sqrt 3)/2)^2) cos A, hence cos A = -1/3.
1/2 = 0 + (sqrt(3)/2) cos B, hence cos B = 1/(sqrt(3) = cos C.
Solid angle ,thus,at the corner of an equioctahedron is
= 2(A+B+C-pi)=2{2cos^-1(1/sqrt(3))+cos^-1(-1/3) - pi}.
Thus it is possible to find the solid angle at a corner of an equi- octahedron without reference to that of an equi-tetrahedron.
Originally posted by howzzatThe value woked out by you and that worked out by cosmic voice are different from mine ,as given in my earlier post above...Check if we are talking about the samething..or??? Why should three different approaches give different analytical expressions or different values for the same thing? The poser should perhaps resolve it.
I think ,without any reference to equitetrahedron too, one can find the solid angle at a corner of an equi-octahedron. Take a=pi/2,
b=c=pi/3.This defines the triangular cone of a half-equi-octahedron.
The angles A,B & C between the three plane faces of the triangular cone can be found by the relations:
cos a = cos b cos c + sin b sin c cos A.
cos b = cos ...[text shortened]... solid angle at a corner of an equi- octahedron without reference to that of an equi-tetrahedron.
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Originally posted by howzzatConsider this yet simpler way..Let OABC be an equitetrahedron with length of each edge being a. With O as centre draw a sphere of radius a. Let POQ be the diameter of the great circle containing the equilateral triangle OAB such that PA=QB =a.. Now draw the great circle passing through P,Q and C. Clearly /_POC=/_QOC=pi/2. Also the tetrahedrons OPAC & OQBC are not equitetrahedrons but each is a quarter of an equioctahedron. The solid angle subtended at vertex O by each of these, is W8/2, where W8 is the solid angle at a corner of an equioctahedron. It can also be shown that the angle between the planes of the two great circles mentioned above is cos^-1(1/sqrt 3 ).
I think ,without any reference to equitetrahedron too, one can find the solid angle at a corner of an equi-octahedron. Take a=pi/2,
b=c=pi/3.This defines the triangular cone of a half-equi-octahedron.
The angles A,B & C between the three ...[text shortened]... equi- octahedron without reference to that of an equi-tetrahedron.
Thus the solid angle between these 2 planes = 2 cos^-1(1/sqrt 3 ).
Obviously, now W4+ 2*(W8/2) = 2 cosine inverse of 1/sqrt 3,
where W4 is the solid angle at a corner of an equitetrahedron...
This way too you get W8 if you know W4.
Originally posted by ranjan sinhaobservant..& Acolyte both have used the formula for the area of a spherical triangle to be expressed by A+B+C-pi, where A,B&C are the angles between the three planes forming the spherical triangle with its vertex at the centre of the sphere......But what is the proof of this?.
Consider this yet simpler way..Let OABC be an equitetrahedron with length of each edge being a. With O as centre draw a sphere of radius a. Let POQ be the diameter of the great circle containing the equilateral triangle OAB such that PA=QB =a.. Now draw the great circle passing through P,Q and C. Clearly /_POC=/_QOC=pi/2. Also the tetrahedrons OPAC ...[text shortened]... is the solid angle at a corner of an equitetrahedron...
This way too you get W8 if you know W4.
Originally posted by gavin75this thread is about solid angles. How come challenger gavin? If ur rating is really 9999, try solving this solid problem...try working out an analytical expression for the solid angle of a triangular cone whose vertex angles (of the 3 plane faces) are alpha, beta and gamma....s
All of you reading this that is looking for a challenger reply me your name so i could play you😀🙄[gid]realarcade[/gid]
Originally posted by rspoddar82In one of his posts above, Observant has already done it in the special case. I take ur challenge.. I think it can be done for the general case of yours too.... I will think.
this thread is about solid angles. How come challenger gavin? If ur rating is really 9999, try solving this solid problem...try working out an analytical expression for the solid angle of a triangular cone whose vertex angles (of the 3 plane faces) are alpha, beta and gamma....s
Originally posted by ranjan sinhaSeveral different results have been derived for the solid angle at the corner of a regular octahedron. All the apparently different expressions are basically the same....The poser's original suggestion that the solid angle (at the corner )of an equi -octahedron can not be worked out without finding that of an equi-tetrahedron & vice versa has been proved wrong in the post of howzzat above. Any comments mr. poser?
In one of his posts above, Observant has already done it in the special case. I take ur challenge.. I think it can be done for the general case of yours too.... I will think.
Originally posted by neverB4chessI agree with you. I had not visualised that half the solid angle at a corner of an equi-octahedron can be viewed as a triangular cone with the three angles of the three faces equal to pi/2, pi/3 and pi/3 respectively. Thank you for the new insight...
Several different results have been derived for the solid angle at the corner of a regular octahedron. All the apparently different expressions are basically the same....The poser's original suggestion that the solid angle (at the corner )of an equi -octahedron can not be worked out without finding that of an equi-tetrahedron & vice versa has been proved wrong in the post of howzzat above. Any comments mr. poser?
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Originally posted by cosmic voiceCan you clarify how did you do the actual integration?...I mean how did you set up the differetial area element? ur .....steps are not clear.
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 rspoddar82So? Finally you have conceded what I had been all along saying. See my earlier post regarding the unnecessary condition of linkage between tetrahedron and octahedron ...You should not assume that there is only one correct method. There are others who can find new and innovative ways of solving a puzzle...
I agree with you. I had not visualised that half the solid angle at a corner of an equi-octahedron can be viewed as a triangular cone with the three angles of the three faces equal to pi/2, pi/3 and pi/3 respectively. Thank you for the new insight...
Originally posted by cheskmateYes I 100 % agree. The most general method is that of direct integration. It was Howzzat perhaps who has done it for the particular case of an equitetrahedron. For a general (non-symmetric) tetrahedron ( which ,some of the respondent posters have called triangular cone) it should be possible too..
So? Finally you have conceded what I had been all along saying. See my earlier post regarding the unnecessary condition of linkage between tetrahedron and octahedron ...You should not assume that there is only one correct method. There are others who can find new and innovative ways of solving a puzzle...
Originally posted by ranjan sinhai think now it's all over...except the new challenge undertaken by you.
Yes I 100 % agree. The most general method is that of direct integration. It was Howzzat perhaps who has done it for the particular case of an equitetrahedron. For a general (non-symmetric) tetrahedron ( which ,some of the respondent posters have called triangular cone) it should be possible too..
..the general method for finding the solid angle of any arbitrary triangular cone will involve integration of an infinitesimally thin triangular cone with the three angles (of the three plane faces of the triangular cone) equal to phi, phi,and d(theta). What should be the solid angle of this thin strip triangular cone ? Will it be independent of phi? According to Acolyte's formula this should be
= (pi/2+pi/2+d(theta)-pi).
But this expression is independent of phi. Obviously this cannot be correct. What is the correct expression?