Side length of a regular tetrahedron given the radius of the sphere tangent to its edges
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I am working on a physics project where we have to build a container to protect a glass ornament when dropped from a high place. My design involves building a tetrahedron out of straws and putting the glass ornament inside of it. I'm not sure what length to cut each side of the straw tetrahedron. I did some measurements and I estimated the radius of the sphere to be about 1.3. The tetrahedron has no "walls" where the ornament will touch at a single point. It is simply an empty frame made out of regular McDonald's straws, and the ornament will protrude out of each side a little bit, thus there are no faces, only edges. How can I find the best side length?
geometry 3d
edited Jan 1 at 2:57
Blue
49.2k870157
asked Dec 31 '18 at 23:30
user22333user22333
362
$endgroup$
add a comment |
$begingroup$
I am working on a physics project where we have to build a container to protect a glass ornament when dropped from a high place. My design involves building a tetrahedron out of straws and putting the glass ornament inside of it. I'm not sure what length to cut each side of the straw tetrahedron. I did some measurements and I estimated the radius of the sphere to be about 1.3. The tetrahedron has no "walls" where the ornament will touch at a single point. It is simply an empty frame made out of regular McDonald's straws, and the ornament will protrude out of each side a little bit, thus there are no faces, only edges. How can I find the best side length?
geometry 3d
edited Jan 1 at 2:57
Blue
49.2k870157
asked Dec 31 '18 at 23:30
user22333user22333
362
$endgroup$
2
$begingroup$
Tetrahedron - Wikipedia Has all the formulae you need. I think your ornament will be midsphere of straw tetrahedron
$endgroup$
– Daniel Mathias
Dec 31 '18 at 23:38
$begingroup$
The best I have seen was on an episode of "Modern Family." I think it was an egg being dropped. The smart daughter made a little parachute... en.wikipedia.org/wiki/Egg_Drop I guess the scene with Alex and the parachute was an extra, after most of the show was over.
$endgroup$
– Will Jagy
Dec 31 '18 at 23:56
add a comment |
$begingroup$
I am working on a physics project where we have to build a container to protect a glass ornament when dropped from a high place. My design involves building a tetrahedron out of straws and putting the glass ornament inside of it. I'm not sure what length to cut each side of the straw tetrahedron. I did some measurements and I estimated the radius of the sphere to be about 1.3. The tetrahedron has no "walls" where the ornament will touch at a single point. It is simply an empty frame made out of regular McDonald's straws, and the ornament will protrude out of each side a little bit, thus there are no faces, only edges. How can I find the best side length?
geometry 3d
edited Jan 1 at 2:57
Blue
49.2k870157
asked Dec 31 '18 at 23:30
user22333user22333
362
$endgroup$
I am working on a physics project where we have to build a container to protect a glass ornament when dropped from a high place. My design involves building a tetrahedron out of straws and putting the glass ornament inside of it. I'm not sure what length to cut each side of the straw tetrahedron. I did some measurements and I estimated the radius of the sphere to be about 1.3. The tetrahedron has no "walls" where the ornament will touch at a single point. It is simply an empty frame made out of regular McDonald's straws, and the ornament will protrude out of each side a little bit, thus there are no faces, only edges. How can I find the best side length?
geometry 3d
geometry 3d
edited Jan 1 at 2:57
Blue
49.2k870157
asked Dec 31 '18 at 23:30
user22333user22333
362
edited Jan 1 at 2:57
Blue
49.2k870157
asked Dec 31 '18 at 23:30
user22333user22333
362
edited Jan 1 at 2:57
Blue
49.2k870157
edited Jan 1 at 2:57
Blue
49.2k870157
edited Jan 1 at 2:57
Blue
49.2k870157
49.2k870157
asked Dec 31 '18 at 23:30
user22333user22333
362
asked Dec 31 '18 at 23:30
user22333user22333
362
asked Dec 31 '18 at 23:30
user22333user22333
362
362
2
$begingroup$
Tetrahedron - Wikipedia Has all the formulae you need. I think your ornament will be midsphere of straw tetrahedron
$endgroup$
– Daniel Mathias
Dec 31 '18 at 23:38
$begingroup$
The best I have seen was on an episode of "Modern Family." I think it was an egg being dropped. The smart daughter made a little parachute... en.wikipedia.org/wiki/Egg_Drop I guess the scene with Alex and the parachute was an extra, after most of the show was over.
$endgroup$
– Will Jagy
Dec 31 '18 at 23:56
add a comment |
2
$begingroup$
Tetrahedron - Wikipedia Has all the formulae you need. I think your ornament will be midsphere of straw tetrahedron
$endgroup$
– Daniel Mathias
Dec 31 '18 at 23:38
$begingroup$
The best I have seen was on an episode of "Modern Family." I think it was an egg being dropped. The smart daughter made a little parachute... en.wikipedia.org/wiki/Egg_Drop I guess the scene with Alex and the parachute was an extra, after most of the show was over.
$endgroup$
– Will Jagy
Dec 31 '18 at 23:56
2
2
$begingroup$
Tetrahedron - Wikipedia Has all the formulae you need. I think your ornament will be midsphere of straw tetrahedron
$endgroup$
– Daniel Mathias
Dec 31 '18 at 23:38
$begingroup$
Tetrahedron - Wikipedia Has all the formulae you need. I think your ornament will be midsphere of straw tetrahedron
$endgroup$
– Daniel Mathias
Dec 31 '18 at 23:38
$begingroup$
The best I have seen was on an episode of "Modern Family." I think it was an egg being dropped. The smart daughter made a little parachute... en.wikipedia.org/wiki/Egg_Drop I guess the scene with Alex and the parachute was an extra, after most of the show was over.
$endgroup$
– Will Jagy
Dec 31 '18 at 23:56
$begingroup$
The best I have seen was on an episode of "Modern Family." I think it was an egg being dropped. The smart daughter made a little parachute... en.wikipedia.org/wiki/Egg_Drop I guess the scene with Alex and the parachute was an extra, after most of the show was over.
$endgroup$
– Will Jagy
Dec 31 '18 at 23:56
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Your inscribed sphere is touching edges of tetrahedron, not its faces which do not exist in reality. The radius of the sphere touching edges of terahedron is known to be:
$$r=frac{a}{sqrt8}iff a=2rsqrt2approx2.82r$$
Proving the formula is a fairly simple exercise.
https://en.wikipedia.org/wiki/Tetrahedron#Formulas_for_a_regular_tetrahedron
answered Jan 1 at 0:33
OldboyOldboy
8,97111138
$endgroup$
$begingroup$
is 'a' the side length of the tetrahedron?
$endgroup$
– user22333
Jan 1 at 0:42
1
$begingroup$
@user22333 Exactly.
$endgroup$
– Oldboy
Jan 1 at 6:32
add a comment |
$begingroup$
Added: in the construction below, the point on an edge closest to the origin is at $(0,0,1),$ radius for this is exactly $1,$ therefore the edge length divided by $sqrt 8$
===================================================================
one way to place a regular tetrahedron is at every other vertex of the cube with all vertices $(pm1,pm1,pm1).$ For example
$$ (-1,-1,-1) ; , ; ; ; (-1,1,1) ; , ; ; ; (1,-1,1) ; , ; ; ; (1,1,-1) ; . ; ; ; $$
The edges are all length $sqrt 8$
One triangle side is $$ x+y+z=1. $$
The closest point to the origin in that plane is
$$ left( frac{1}{3}, frac{1}{3}, frac{1}{3} right) $$
with distance from the origin
$$ frac{sqrt 3}{3} = frac{1}{sqrt 3},$$
which is the radius of the inscribed sphere. Thus the radius is the edge length divided by $$ sqrt{24} $$
answered Jan 1 at 0:17
Will JagyWill Jagy
104k5102201
$endgroup$
1
$begingroup$
OP said: "...and the ornament will protrude out of each side a little bit". You have calculated the radius of inscribed sphere touching all faces of tetrahedron, not the radius of insphere touching its edges.
$endgroup$
– Oldboy
Jan 1 at 0:35
$begingroup$
@Oldboy I see what you mean. Added a note at the beginning.
$endgroup$
– Will Jagy
Jan 1 at 0:40
add a comment |
Your Answer
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2 Answers
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active
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2 Answers
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$begingroup$
Your inscribed sphere is touching edges of tetrahedron, not its faces which do not exist in reality. The radius of the sphere touching edges of terahedron is known to be:
$$r=frac{a}{sqrt8}iff a=2rsqrt2approx2.82r$$
Proving the formula is a fairly simple exercise.
https://en.wikipedia.org/wiki/Tetrahedron#Formulas_for_a_regular_tetrahedron
answered Jan 1 at 0:33
OldboyOldboy
8,97111138
$endgroup$
$begingroup$
is 'a' the side length of the tetrahedron?
$endgroup$
– user22333
Jan 1 at 0:42
1
$begingroup$
@user22333 Exactly.
$endgroup$
– Oldboy
Jan 1 at 6:32
add a comment |
$begingroup$
Your inscribed sphere is touching edges of tetrahedron, not its faces which do not exist in reality. The radius of the sphere touching edges of terahedron is known to be:
$$r=frac{a}{sqrt8}iff a=2rsqrt2approx2.82r$$
Proving the formula is a fairly simple exercise.
https://en.wikipedia.org/wiki/Tetrahedron#Formulas_for_a_regular_tetrahedron
answered Jan 1 at 0:33
OldboyOldboy
8,97111138
$endgroup$
$begingroup$
is 'a' the side length of the tetrahedron?
$endgroup$
– user22333
Jan 1 at 0:42
1
$begingroup$
@user22333 Exactly.
$endgroup$
– Oldboy
Jan 1 at 6:32
add a comment |
$begingroup$
Your inscribed sphere is touching edges of tetrahedron, not its faces which do not exist in reality. The radius of the sphere touching edges of terahedron is known to be:
$$r=frac{a}{sqrt8}iff a=2rsqrt2approx2.82r$$
Proving the formula is a fairly simple exercise.
https://en.wikipedia.org/wiki/Tetrahedron#Formulas_for_a_regular_tetrahedron
answered Jan 1 at 0:33
OldboyOldboy
8,97111138
$endgroup$
Your inscribed sphere is touching edges of tetrahedron, not its faces which do not exist in reality. The radius of the sphere touching edges of terahedron is known to be:
$$r=frac{a}{sqrt8}iff a=2rsqrt2approx2.82r$$
Proving the formula is a fairly simple exercise.
https://en.wikipedia.org/wiki/Tetrahedron#Formulas_for_a_regular_tetrahedron
answered Jan 1 at 0:33
OldboyOldboy
8,97111138
edited Jan 1 at 0:38
answered Jan 1 at 0:33
OldboyOldboy
8,97111138
answered Jan 1 at 0:33
OldboyOldboy
8,97111138
answered Jan 1 at 0:33
OldboyOldboy
8,97111138
8,97111138
$begingroup$
is 'a' the side length of the tetrahedron?
$endgroup$
– user22333
Jan 1 at 0:42
1
$begingroup$
@user22333 Exactly.
$endgroup$
– Oldboy
Jan 1 at 6:32
add a comment |
$begingroup$
is 'a' the side length of the tetrahedron?
$endgroup$
– user22333
Jan 1 at 0:42
1
$begingroup$
@user22333 Exactly.
$endgroup$
– Oldboy
Jan 1 at 6:32
$begingroup$
is 'a' the side length of the tetrahedron?
$endgroup$
– user22333
Jan 1 at 0:42
$begingroup$
is 'a' the side length of the tetrahedron?
$endgroup$
– user22333
Jan 1 at 0:42
1
1
$begingroup$
@user22333 Exactly.
$endgroup$
– Oldboy
Jan 1 at 6:32
$begingroup$
@user22333 Exactly.
$endgroup$
– Oldboy
Jan 1 at 6:32
add a comment |
$begingroup$
Added: in the construction below, the point on an edge closest to the origin is at $(0,0,1),$ radius for this is exactly $1,$ therefore the edge length divided by $sqrt 8$
===================================================================
one way to place a regular tetrahedron is at every other vertex of the cube with all vertices $(pm1,pm1,pm1).$ For example
$$ (-1,-1,-1) ; , ; ; ; (-1,1,1) ; , ; ; ; (1,-1,1) ; , ; ; ; (1,1,-1) ; . ; ; ; $$
The edges are all length $sqrt 8$
One triangle side is $$ x+y+z=1. $$
The closest point to the origin in that plane is
$$ left( frac{1}{3}, frac{1}{3}, frac{1}{3} right) $$
with distance from the origin
$$ frac{sqrt 3}{3} = frac{1}{sqrt 3},$$
which is the radius of the inscribed sphere. Thus the radius is the edge length divided by $$ sqrt{24} $$
answered Jan 1 at 0:17
Will JagyWill Jagy
104k5102201
$endgroup$
1
$begingroup$
OP said: "...and the ornament will protrude out of each side a little bit". You have calculated the radius of inscribed sphere touching all faces of tetrahedron, not the radius of insphere touching its edges.
$endgroup$
– Oldboy
Jan 1 at 0:35
$begingroup$
@Oldboy I see what you mean. Added a note at the beginning.
$endgroup$
– Will Jagy
Jan 1 at 0:40
add a comment |
$begingroup$
Added: in the construction below, the point on an edge closest to the origin is at $(0,0,1),$ radius for this is exactly $1,$ therefore the edge length divided by $sqrt 8$
===================================================================
one way to place a regular tetrahedron is at every other vertex of the cube with all vertices $(pm1,pm1,pm1).$ For example
$$ (-1,-1,-1) ; , ; ; ; (-1,1,1) ; , ; ; ; (1,-1,1) ; , ; ; ; (1,1,-1) ; . ; ; ; $$
The edges are all length $sqrt 8$
One triangle side is $$ x+y+z=1. $$
The closest point to the origin in that plane is
$$ left( frac{1}{3}, frac{1}{3}, frac{1}{3} right) $$
with distance from the origin
$$ frac{sqrt 3}{3} = frac{1}{sqrt 3},$$
which is the radius of the inscribed sphere. Thus the radius is the edge length divided by $$ sqrt{24} $$
answered Jan 1 at 0:17
Will JagyWill Jagy
104k5102201
$endgroup$
1
$begingroup$
OP said: "...and the ornament will protrude out of each side a little bit". You have calculated the radius of inscribed sphere touching all faces of tetrahedron, not the radius of insphere touching its edges.
$endgroup$
– Oldboy
Jan 1 at 0:35
$begingroup$
@Oldboy I see what you mean. Added a note at the beginning.
$endgroup$
– Will Jagy
Jan 1 at 0:40
add a comment |
$begingroup$
Added: in the construction below, the point on an edge closest to the origin is at $(0,0,1),$ radius for this is exactly $1,$ therefore the edge length divided by $sqrt 8$
===================================================================
one way to place a regular tetrahedron is at every other vertex of the cube with all vertices $(pm1,pm1,pm1).$ For example
$$ (-1,-1,-1) ; , ; ; ; (-1,1,1) ; , ; ; ; (1,-1,1) ; , ; ; ; (1,1,-1) ; . ; ; ; $$
The edges are all length $sqrt 8$
One triangle side is $$ x+y+z=1. $$
The closest point to the origin in that plane is
$$ left( frac{1}{3}, frac{1}{3}, frac{1}{3} right) $$
with distance from the origin
$$ frac{sqrt 3}{3} = frac{1}{sqrt 3},$$
which is the radius of the inscribed sphere. Thus the radius is the edge length divided by $$ sqrt{24} $$
answered Jan 1 at 0:17
Will JagyWill Jagy
104k5102201
$endgroup$
Added: in the construction below, the point on an edge closest to the origin is at $(0,0,1),$ radius for this is exactly $1,$ therefore the edge length divided by $sqrt 8$
===================================================================
one way to place a regular tetrahedron is at every other vertex of the cube with all vertices $(pm1,pm1,pm1).$ For example
$$ (-1,-1,-1) ; , ; ; ; (-1,1,1) ; , ; ; ; (1,-1,1) ; , ; ; ; (1,1,-1) ; . ; ; ; $$
The edges are all length $sqrt 8$
One triangle side is $$ x+y+z=1. $$
The closest point to the origin in that plane is
$$ left( frac{1}{3}, frac{1}{3}, frac{1}{3} right) $$
with distance from the origin
$$ frac{sqrt 3}{3} = frac{1}{sqrt 3},$$
which is the radius of the inscribed sphere. Thus the radius is the edge length divided by $$ sqrt{24} $$
answered Jan 1 at 0:17
Will JagyWill Jagy
104k5102201
edited Jan 1 at 0:39
answered Jan 1 at 0:17
Will JagyWill Jagy
104k5102201
answered Jan 1 at 0:17
Will JagyWill Jagy
104k5102201
answered Jan 1 at 0:17
Will JagyWill Jagy
104k5102201
104k5102201
1
$begingroup$
OP said: "...and the ornament will protrude out of each side a little bit". You have calculated the radius of inscribed sphere touching all faces of tetrahedron, not the radius of insphere touching its edges.
$endgroup$
– Oldboy
Jan 1 at 0:35
$begingroup$
@Oldboy I see what you mean. Added a note at the beginning.
$endgroup$
– Will Jagy
Jan 1 at 0:40
add a comment |
1
$begingroup$
OP said: "...and the ornament will protrude out of each side a little bit". You have calculated the radius of inscribed sphere touching all faces of tetrahedron, not the radius of insphere touching its edges.
$endgroup$
– Oldboy
Jan 1 at 0:35
$begingroup$
@Oldboy I see what you mean. Added a note at the beginning.
$endgroup$
– Will Jagy
Jan 1 at 0:40
1
1
$begingroup$
OP said: "...and the ornament will protrude out of each side a little bit". You have calculated the radius of inscribed sphere touching all faces of tetrahedron, not the radius of insphere touching its edges.
$endgroup$
– Oldboy
Jan 1 at 0:35
$begingroup$
OP said: "...and the ornament will protrude out of each side a little bit". You have calculated the radius of inscribed sphere touching all faces of tetrahedron, not the radius of insphere touching its edges.
$endgroup$
– Oldboy
Jan 1 at 0:35
$begingroup$
@Oldboy I see what you mean. Added a note at the beginning.
$endgroup$
– Will Jagy
Jan 1 at 0:40
$begingroup$
@Oldboy I see what you mean. Added a note at the beginning.
$endgroup$
– Will Jagy
Jan 1 at 0:40
add a comment |
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$begingroup$
Tetrahedron - Wikipedia Has all the formulae you need. I think your ornament will be midsphere of straw tetrahedron
$endgroup$
– Daniel Mathias
Dec 31 '18 at 23:38
$begingroup$
The best I have seen was on an episode of "Modern Family." I think it was an egg being dropped. The smart daughter made a little parachute... en.wikipedia.org/wiki/Egg_Drop I guess the scene with Alex and the parachute was an extra, after most of the show was over.
$endgroup$
– Will Jagy
Dec 31 '18 at 23:56