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CTK Exchange
Jeff Jencks
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Jun-25-04, 10:38 PM (EST) |
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"tetrahedron to sphere calculation"
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My primary question is: If you put a tetrahedron inside a sphere in such a way that the four points of the tetrahedron are on the surface of the sphere, what would be the lattitude on the sphere that the three points at the base of the tetrahedron would share. I could calculate this myself but I don't know the required formulas so here are some other questions about the formulas. I know that the surface formula for a sphere equals 4*p*r2. Is p the length of the perimeter (equator)? What are the formulas for angle and line length when we get into the third dimension. I'm not a mathematician. I'm a sci-fi author and these answers and formulas will help me with a book that I'm writing. |
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alexb
Charter Member
2185 posts |
Jun-27-04, 11:07 PM (EST) |
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2. "RE: tetrahedron calculation"
In response to message #0
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>My primary question is: If you put a tetrahedron inside a
>sphere in such a way that the four points of the tetrahedron
>are on the surface of the sphere, what would be the
>lattitude on the sphere that the three points at the base of
>the tetrahedron would share. I could calculate this myself
>but I don't know the required formulas so here are some
>other questions about the formulas. You can certainly do it yourself. Assume the tetrahedron has the base ABC and the apex S. Let Q be center of ABC, AQ extended intersect BC in K. Let O be the center of the sphere. If the side of the tetrahedron is 1, [BR>- AK = sqrt(3)/2,
- SK = sqrt(3)/2,
- in triangle SQK, Q is right,
- QK = AQ/3,
- use the pythagorean theorem in SQK to find SQ,
- SO = SQ·3/4,
- AO = SO,
- in triangle AOS, find angle O by the theorem of cosines.
>
>I know that the surface formula for a sphere equals 4*p*r2.
>Is p the length of the perimeter (equator)?
p is the famous constant p, which is approximately 3.14... |
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Jeff Jencks
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Jun-28-04, 04:20 PM (EST) |
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3. "RE: tetrahedron calculation"
In response to message #2
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Let me make sure I've got this. I don't understand "extended intersect" or "in K". I think what you are saying is that K is a point halfway between B and C and therefore along the same line as AQ if AQ was extended. Correct? How do you know that QK=AQ/3. Is that a known law or is there a formula that tells you that. I have the same question for SO=SQ*3/4. I've learned the theorem of cosines but I don't remember it off the top of my head. I've been known to get sine, cosine, cosecant and cotangent mixed up.
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alexb
Charter Member
2185 posts |
Jun-28-04, 04:29 PM (EST) |
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4. "RE: tetrahedron calculation"
In response to message #3
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>Let me make sure I've got this.
>
>I don't understand "extended intersect" or "in K". I think
>what you are saying is that K is a point halfway between B
>and C and therefore along the same line as AQ if AQ was
>extended. Correct? Yes, it is. >How do you know that QK=AQ/3. Triangle ABC is equilateral. The line AK is the median, the altitude, the angle (A) and the side (BC) bisector. Its being the median is important. In a sense, Q is also almost everything: barycenter (center fo gravity), circumcenter, incenter, orthocenter, etc. The important thing is that Q is the center of gravity of the points A, B, C with equal masses, say, 1. Then K is the center of gravity of points B and C. Now, place a mass of 2 = (1 + 1) in K, and think of the center of gravity of A with mass 1 and K with mass 2. It'll be 1/3 of the way from K. >Is that a known law or is
>there a formula that tells you that. I have the same
>question for SO=SQ*3/4. Exactly for the same reason, right. O is the center of gravity of SABC and could be found as the center of gravity of two masses: 1 at S and 3 at Q. >I've learned the theorem of cosines but I don't remember it
>off the top of my head. I've been known to get sine,
>cosine, cosecant and cotangent mixed up. Just search this site or the web. |
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francais
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Feb-11-08, 09:29 AM (EST) |
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5. "RE: tetrahedron calculation"
In response to message #4
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Hi my problem is a little different
i will be placing the center of the sphere on the tip
of a relatively large tetrahedron... so now the portion of the sides of tetrahedron that intersected
with sphere will be equal to the radius of sphere.. i want compute the volume of the interesected portion in the sphere.. someone please help me finding the solution
thanks francais5@rediffmail.com |
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