Calling all geometers:
I want to make a tetrahedron out of square cross-section sticks. I want a
corner of the stick to define the edge of the tetrahedron - i.e., a plane
through the diagonal of the stick will pass through the center of the
tetrahedron. (I've already made one where a flat on the stick defines the
edge - the math is easy on that one).
What I can't figure out (and I've spent a frustrating day in the shop
trying) is:
What are the angles on the face of a square stick that work?
The critical information:
The central angle is 109.48 degrees. That is, the angle from the center of
the tetrahedron to any two vertices is 109.48.
Any help would be much appreciated!
Scott
"Elrond Hubbard" <[email protected]> wrote in message
news:[email protected]...
> "MikeWhy" <[email protected]> wrote in
> news:[email protected]:
>
>> "Elrond Hubbard" <[email protected]> wrote in message
>> news:[email protected]...
>>> "MikeWhy" <[email protected]> wrote in
>>> news:[email protected]:
>>>
>>>> "Elrond Hubbard" <[email protected]> wrote in message
>>>> news:[email protected]...
>>>>> Calling all geometers:
>>>>>
>>>>> I want to make a tetrahedron out of square cross-section sticks. I
>>>>> want a corner of the stick to define the edge of the tetrahedron -
>>>>> i.e., a plane through the diagonal of the stick will pass through
>>>>> the center of the tetrahedron. (I've already made one where a flat
>>>>> on the stick defines the edge - the math is easy on that one).
>>>>>
>>>>> What I can't figure out (and I've spent a frustrating day in the
>>>>> shop trying) is:
>>>>>
>>>>> What are the angles on the face of a square stick that work?
>>>
>>>> If you're asking about the compound miter at the vertices... :D That
>>>> will have to wait for the second pot of coffee.
>>>
>>> That's exactly what I'm asking :)
>>
>> If you haven't already, sketch it out on a cube as I described
>> earlier. The geometry at each vertex is surprisingly simple, and not
>> entirely obvious. Looking at one corner of the cube, the tetrahedron
>> edges form the prime, 45 degree diagnonals on each of the three faces
>> meeting at that corner.
>>
>> If you've already done that, I won't be able to help further. Compound
>> miters might as well be rocket science for me. I'm guessing, for a
>> radial arm saw, swing the arm 22.5 degrees to split the plan view 45
>> deg; tilt the blade 67.5 degrees (probably wrong) to split the 45
>> degree rise. Repeat for the other side. Maybe someone else can take it
>> from there...
>>
>> I'll lay it out in solidworks later, see if any more insight forms. It
>> sounds interesting enough to lay out on a small stick and whittle it.
>
> I think I've got it figured out. I did the mental exercise of making
> the original sticks then cutting off the corners to make the diagonally
> oriented one. I determined where the original cuts crossed the mid-line
> of the square stick. The midline of the square stick is the edge of the
> rotated stick - I can now calculate the angles, knowing the run and rise
> of those coordinates.
>
> I'll post the results tonight after I make a test piece when I come home
> from work.
I cut up some square sticks of cherry scrap and pocketed a marking knife and
3/4" butt chisel into the office. A few minutes of (completely,
therapeutically relaxing) whittling later, the first piece looks like it's
right on. Aside from the woodchips in the keyboard, I can't think of a
better way to break up a work day.
For those who might want to take their own whittle break, the dimensions for
1" square stock are 2.000" back along the top spine, and 1.692" back on the
two sides. Scale the numbers for your stock size, and mark and connect the
dots back to a sharp tip at the front end of the bottom edge. Picture in
your mind the two flat facets opening mostly upward and more gently to the
sides and forward, meeting in a straight spine connecting the top corner to
the pointy tip lying on the desk. Three sticks cut this way form a mitered
tetrahedron vertex.
(I'll have to find a better way to hold the workpiece. The corners are all
bruised from pressing it hard against the edge of the desk. I can now see a
need for a small tool chest in the desk drawer. ;)
Elrond Hubbard <[email protected]> wrote in
news:[email protected]:
> Jay Pique <[email protected]> wrote in
> news:f50c8142-e055-435b-8e1f-67d991602ee5@u18g2000pro.googlegroups.com:
>
>
>>>
>>> The central angle is 109.48 degrees. That is, the angle from the
>>> cente
>> r of
>>> the tetrahedron to any two vertices is 109.48.
>>>
>>> Any help would be much appreciated!
>>
>> I believe the answer is "42".
>
> That is the right answer - but just not to this particular question ;)
>
What is the question, and what are those two mice doing with Robatoy's
brain?
Puckdropper
--
If you're quiet, your teeth never touch your ankles.
To email me directly, send a message to puckdropper (at) fastmail.fm
"Elrond Hubbard" <[email protected]> wrote in message
news:[email protected]...
> Calling all geometers:
>
> I want to make a tetrahedron out of square cross-section sticks. I want a
> corner of the stick to define the edge of the tetrahedron - i.e., a plane
> through the diagonal of the stick will pass through the center of the
> tetrahedron. (I've already made one where a flat on the stick defines the
> edge - the math is easy on that one).
>
> What I can't figure out (and I've spent a frustrating day in the shop
> trying) is:
>
> What are the angles on the face of a square stick that work?
>
> The critical information:
>
> The central angle is 109.48 degrees. That is, the angle from the center
> of
> the tetrahedron to any two vertices is 109.48.
>
> Any help would be much appreciated!
>
> Scott
http://en.wikipedia.org/wiki/Tetrahedron
"Doug Miller" <[email protected]> wrote in message
news:[email protected]...
> In article <[email protected]>, Elrond Hubbard
> <[email protected]> wrote:
>
>>> If you haven't already, sketch it out on a cube as I described
>>> earlier. The geometry at each vertex is surprisingly simple, and not
>>> entirely obvious. Looking at one corner of the cube, the tetrahedron
>>> edges form the prime, 45 degree diagnonals on each of the three faces
>>> meeting at that corner.
>
> Ummm....no.
>
> There are no 45-degree angles anywhere on a tetrahedron.
You obviously haven't sketched it out, and clearly read too fast to catch
the import and meaning of "not entirely obvious" and "surprisingly".
Bucky Fuller spent the latter decades of his life noodling over just that
conundrum. For penance, scary sharpen a chisel and cut 45 degree chamfers on
two end grain and one long grain edge, and marvel at the isoceles triangle
that emerges. Repeat 3 more times just to be sure that there are no 45
degree angles in that triangle.
On Nov 9, 2:59=A0pm, Elrond Hubbard <[email protected]> wrote:
> "Ed Edelenbos" <[email protected]> wrote innews:[email protected]=
dual.net:
>
>
>
>
>
> > "Elrond Hubbard" <[email protected]> wrote in message
> >news:[email protected]...
> >> Calling all geometers:
>
> >> I want to make a tetrahedron out of square cross-section sticks. =A0I
> >> want a corner of the stick to define the edge of the tetrahedron -
> >> i.e., a plane through the diagonal of the stick will pass through the
> >> center of the tetrahedron. =A0(I've already made one where a flat on
> >> the stick defines the edge - the math is easy on that one).
>
> >> What I can't figure out (and I've spent a frustrating day in the shop
> >> trying) is:
>
> >> What are the angles on the face of a square stick that work?
>
> >> The critical information:
>
> >> The central angle is 109.48 degrees. =A0That is, the angle from the
> >> center of
> >> the tetrahedron to any two vertices is 109.48.
>
> >> Any help would be much appreciated!
>
> >> Scott
>
> >http://en.wikipedia.org/wiki/Tetrahedron
>
> Lots of info there, but none that answers my question. =A0I know what the
> mating surfaces of the sticks have to look like - my problem is projectin=
g
> those surfaces onto the faces of a square stick.
I've done similar projects with geometric solids. What really helped
me was to make very accurate models out of heavy cardboard. I made
them 3 to 4 inches on a side (some were as large as 6 inches on a
side); make yours to the same scale as your wooden one. Then you can
measure angles directly off of the model. You will probably need to
tweak the measurements slightly because your paper model won't be
absolutely perfect.
"Projecting onto faces of a square stick" is your problem.
The "sticks" would form the edges of the tetrahedron and, by
definition, the faces of those sticks would be the same as the faces
of the tetrahedron relative, one to the other. It would seem that you
would start off with "sticks" that were triangular, not square. The
image at: http://upload.wikimedia.org/wikipedia/commons/thumb/2/25/Tetrahedron.png/120px-Tetrahedron.png
shows one built out of "sticks" joined at little balls (as if some
sort of kids construction toy) and ignores the edges as such.
In your case, the end of your (triangular) stick, would be cut at the
same angle as the faces of the tetrahedron and each stick "end" would
form a third of the tetrahedron in miniature. Sticks of equal length,
glued together at end points so cut, should create the figure/shape
you asked about.
But square sticks won't cut it as the edges of the finished piece need
to be co-planer with the face(s) of the figure.
"Elrond Hubbard" <[email protected]> wrote in message
news:[email protected]...
> Calling all geometers:
>
> I want to make a tetrahedron out of square cross-section sticks. I want a
> corner of the stick to define the edge of the tetrahedron - i.e., a plane
> through the diagonal of the stick will pass through the center of the
> tetrahedron. (I've already made one where a flat on the stick defines the
> edge - the math is easy on that one).
>
> What I can't figure out (and I've spent a frustrating day in the shop
> trying) is:
>
> What are the angles on the face of a square stick that work?
Assuming that we know what a tetrahedron is, what do you mean "angles on the
face of a square stick"? (I understand square sticks, too, but the angles
are all 90 deg.)
The "simple" answer of the first part, the diagonal of the stick through the
center, is to first layout the tetrahedron on a cube. All edges of a
tetrahedron lie on the diagonals of its enclosing cube. It won't be easy to
describe in words, but here goes:
Say two squares stacked atop each other describes the cube.
Name the vertices of the top square a, b, c, and d.
Name the vertices of the other square e, f, g, and h.
(Sketch it up so the names go clockwise, e directly below a, f below b, etc.
Your sketch will then match mine.)
The cube encloses two tetrahedons.
The vertices a, c, h, and f are vertices of one tetrahedron described by the
cube.
The remaining vertices define the second tetrahedron.
The center of the cube is also the center of the tetrahedron(s).
The diagonals of your sticks, then, are perpendicular to the face of the
cube.
If you're asking about the compound miter at the vertices... :D That will
have to wait for the second pot of coffee.
(Hmmm. Here's a picture that matches:
http://en.wikipedia.org/wiki/Image:Compound_of_two_tetrahedra.png)
In article <[email protected]>, Elrond Hubbard <[email protected]> wrote:
>> If you haven't already, sketch it out on a cube as I described
>> earlier. The geometry at each vertex is surprisingly simple, and not
>> entirely obvious. Looking at one corner of the cube, the tetrahedron
>> edges form the prime, 45 degree diagnonals on each of the three faces
>> meeting at that corner.
Ummm....no.
There are no 45-degree angles anywhere on a tetrahedron.
"Elrond Hubbard" <[email protected]> wrote in message
news:[email protected]...
> "MikeWhy" <[email protected]> wrote in
> news:%[email protected]:
>> For those who might want to take their own whittle break, the
>> dimensions for 1" square stock are 2.000" back along the top spine,
>> and 1.692" back on the two sides. Scale the numbers for your stock
>> size, and mark and connect the dots back to a sharp tip at the front
>> end of the bottom edge. Picture in your mind the two flat facets
>> opening mostly upward and more gently to the sides and forward,
>> meeting in a straight spine connecting the top corner to the pointy
>> tip lying on the desk. Three sticks cut this way form a mitered
>> tetrahedron vertex.
>
> Nice!
>
> I solved the math today, and here's the solution:
>
> For those who want to lay the lines out with a protractor, the angles on
> two adjacent faces, rising from a common vertex, are 30.36 degrees.
> Continue on the other two faces at 73.67 degrees, and the lines will
> meet at a point on the opposite edge from where you started. The
> distance from your starting point, measured horizontally, is twice the
> thickness of your stick.
>
> Which is a convoluted way of saying exactly what Mike says above.
>
> A plane cut through each pair of lines (one long, one short) will make a
> 120 degree dihedral angle, three of which make one vertex of the
> tetrahedron.
I whittled a few more sticks, and gosh darn it, they don't quite make a
tetrahedron. That was earlier in the week. I figured at first it was just a
sloppy cut here or there; it really wasn't all that far off. But, while
figuring my way around Sketchup, I thought I would lay it out again, and see
what it has to say. This time, the distances came out a bit more rational,
although the geometry of it still escapes me. The distance I wrote earlier
as 1.692" is actually 1.707". That lays out to an exact 45 deg. miter, blade
tilt 35.264 degrees off horizontal. If you're using sine bars or a
TS-Aligner to set tilt, that works out to 0.707" per inch. (It was so much
easier to mark out and just whittle the 6 sticks than to set up a sledded
jig for the tablesaw. Depends on how many you're making, of course.)
There's really something magical about a tetrahedron's relationship to cubes
and 45 degree angles, as evidenced by the root 2 multiples everywhere you
look. It's easy to see how Fuller got trapped into this for so many years.
Puckdropper wrote:
> Elrond Hubbard <[email protected]> wrote in
> news:[email protected]:
>
>> Jay Pique <[email protected]> wrote in
>> news:f50c8142-e055-435b-8e1f-67d991602ee5@u18g2000pro.googlegroups.com:
>>
>>
>>>> The central angle is 109.48 degrees. That is, the angle from the
>>>> cente
>>> r of
>>>> the tetrahedron to any two vertices is 109.48.
>>>>
>>>> Any help would be much appreciated!
>>> I believe the answer is "42".
>> That is the right answer - but just not to this particular question ;)
>>
>
> What is the question, and what are those two mice doing with Robatoy's
> brain?
>
> Puckdropper
The true answer is: 9.
"Elrond Hubbard" <[email protected]> wrote in message
news:[email protected]...
> "MikeWhy" <[email protected]> wrote in
> news:[email protected]:
>
>> "Elrond Hubbard" <[email protected]> wrote in message
>> news:[email protected]...
>>> Calling all geometers:
>>>
>>> I want to make a tetrahedron out of square cross-section sticks. I
>>> want a corner of the stick to define the edge of the tetrahedron -
>>> i.e., a plane through the diagonal of the stick will pass through the
>>> center of the tetrahedron. (I've already made one where a flat on
>>> the stick defines the edge - the math is easy on that one).
>>>
>>> What I can't figure out (and I've spent a frustrating day in the shop
>>> trying) is:
>>>
>>> What are the angles on the face of a square stick that work?
>
>> If you're asking about the compound miter at the vertices... :D That
>> will have to wait for the second pot of coffee.
>
> That's exactly what I'm asking :)
If you haven't already, sketch it out on a cube as I described earlier. The
geometry at each vertex is surprisingly simple, and not entirely obvious.
Looking at one corner of the cube, the tetrahedron edges form the prime, 45
degree diagnonals on each of the three faces meeting at that corner.
If you've already done that, I won't be able to help further. Compound
miters might as well be rocket science for me. I'm guessing, for a radial
arm saw, swing the arm 22.5 degrees to split the plan view 45 deg; tilt the
blade 67.5 degrees (probably wrong) to split the 45 degree rise. Repeat for
the other side. Maybe someone else can take it from there...
I'll lay it out in solidworks later, see if any more insight forms. It
sounds interesting enough to lay out on a small stick and whittle it.
"Ed Edelenbos" <[email protected]> wrote in
news:[email protected]:
>
>
> "Elrond Hubbard" <[email protected]> wrote in message
> news:[email protected]...
>> Calling all geometers:
>>
>> I want to make a tetrahedron out of square cross-section sticks. I
>> want a corner of the stick to define the edge of the tetrahedron -
>> i.e., a plane through the diagonal of the stick will pass through the
>> center of the tetrahedron. (I've already made one where a flat on
>> the stick defines the edge - the math is easy on that one).
>>
>> What I can't figure out (and I've spent a frustrating day in the shop
>> trying) is:
>>
>> What are the angles on the face of a square stick that work?
>>
>> The critical information:
>>
>> The central angle is 109.48 degrees. That is, the angle from the
>> center of
>> the tetrahedron to any two vertices is 109.48.
>>
>> Any help would be much appreciated!
>>
>> Scott
>
> http://en.wikipedia.org/wiki/Tetrahedron
>
>
Lots of info there, but none that answers my question. I know what the
mating surfaces of the sticks have to look like - my problem is projecting
those surfaces onto the faces of a square stick.
Elrond Hubbard <[email protected]> wrote in
news:[email protected]:
> "Ed Edelenbos" <[email protected]> wrote in
> news:[email protected]:
>
>>
>>
>> "Elrond Hubbard" <[email protected]> wrote in message
>> news:[email protected]...
>>> Calling all geometers:
>>>
>>> I want to make a tetrahedron out of square cross-section
>>> sticks. I want a corner of the stick to define the edge
>>> of the tetrahedron - i.e., a plane through the diagonal
>>> of the stick will pass through the center of the
>>> tetrahedron. (I've already made one where a flat on
>>> the stick defines the edge - the math is easy on that
>>> one).
>>>
>>> What I can't figure out (and I've spent a frustrating day
>>> in the shop trying) is:
>>>
>>> What are the angles on the face of a square stick that
>>> work?
>>>
>>> The critical information:
>>>
>>> The central angle is 109.48 degrees. That is, the angle
>>> from the center of
>>> the tetrahedron to any two vertices is 109.48.
>>>
>>> Any help would be much appreciated!
>>>
>>> Scott
>>
>> http://en.wikipedia.org/wiki/Tetrahedron
>>
>>
>
> Lots of info there, but none that answers my question. I
> know what the mating surfaces of the sticks have to look
> like - my problem is projecting those surfaces onto the
> faces of a square stick.
>
Being geometricly (is that a word?) challenged myself, I would
draw it up in a CAD program and then let the software figure
it out. Just a thought...
Larry
Jay Pique <[email protected]> wrote in
news:f50c8142-e055-435b-8e1f-67d991602ee5@u18g2000pro.googlegroups.com:
>>
>> The central angle is 109.48 degrees. That is, the angle from the
>> cente
> r of
>> the tetrahedron to any two vertices is 109.48.
>>
>> Any help would be much appreciated!
>
> I believe the answer is "42".
That is the right answer - but just not to this particular question ;)
I agree with the cardboard model, and makig it wider to gain accuracy,
ut instead of cardboard, on our shop we use a lot of clear acetate
sheets of .020 "and .040" thickness. It breaks and peels off cleanly on
a knifed llne. Then after dry-fitting those pieces, you can trace the
angles onto your wood pieces. Thin styrene works well, too.
>> The critical information:
>>
>> The central angle is 109.48 degrees. That is, the angle from the
>> center of
>> the tetrahedron to any two vertices is 109.48.
> Do you want each corner to come out to a point?
Yes.
If it helps, here's another way to describe it -
Picture a square stick, viewed from the side. Roll the stick on its axis
so that it is now resting on one edge. One diagonal of the stick is now
vertical, one horizontal.
Make an angled cut at 35.26 degrees from the horizontal plane. This is one
corner of the triangle formed by the central angle and the two ends of the
stick.
Draw a line down the center of the plane formed by this cut, from the short
point on top to the long point on the bottom. Cut material away on both
sides of this line so that the resulting angle, with this line as the
vertex, is 120 degrees. In other words, cut a 30 degree slice from each
side, through this line. This completes one end of the stick.
The question I have is: What are the resulting angles, as measured on the
flats of the original square stick?
Puckdropper <puckdropper(at)yahoo(dot)com> wrote in news:000a9b1f$0$25939
[email protected]:
>
> What is the question, and what are those two mice doing with Robatoy's
> brain?
Going hungry, I suspect :P
"MikeWhy" <[email protected]> wrote in
news:[email protected]:
> "Elrond Hubbard" <[email protected]> wrote in message
> news:[email protected]...
>> Calling all geometers:
>>
>> I want to make a tetrahedron out of square cross-section sticks. I
>> want a corner of the stick to define the edge of the tetrahedron -
>> i.e., a plane through the diagonal of the stick will pass through the
>> center of the tetrahedron. (I've already made one where a flat on
>> the stick defines the edge - the math is easy on that one).
>>
>> What I can't figure out (and I've spent a frustrating day in the shop
>> trying) is:
>>
>> What are the angles on the face of a square stick that work?
>
> Assuming that we know what a tetrahedron is, what do you mean "angles
> on the face of a square stick"? (I understand square sticks, too, but
> the angles are all 90 deg.)
If you re-read my opening paragraph, you'll see that I know what a
tetrahedron is and how to make one.
What I want to do is make another one, with the sticks rotated 45 degrees
along their axes. The "angles on the face of the square stick" are what I
need to know to lay out the sticks.
"MikeWhy" <[email protected]> wrote in
news:[email protected]:
> "Elrond Hubbard" <[email protected]> wrote in message
> news:[email protected]...
>> Calling all geometers:
>>
>> I want to make a tetrahedron out of square cross-section sticks. I
>> want a corner of the stick to define the edge of the tetrahedron -
>> i.e., a plane through the diagonal of the stick will pass through the
>> center of the tetrahedron. (I've already made one where a flat on
>> the stick defines the edge - the math is easy on that one).
>>
>> What I can't figure out (and I've spent a frustrating day in the shop
>> trying) is:
>>
>> What are the angles on the face of a square stick that work?
> If you're asking about the compound miter at the vertices... :D That
> will have to wait for the second pot of coffee.
That's exactly what I'm asking :)
Dan Coby <[email protected]> wrote in
news:[email protected]:
> Elrond Hubbard wrote:
>> Calling all geometers:
>>
>> I want to make a tetrahedron out of square cross-section sticks. I
>> want a corner of the stick to define the edge of the tetrahedron -
>> i.e., a plane through the diagonal of the stick will pass through the
>> center of the tetrahedron. (I've already made one where a flat on
>> the stick defines the edge - the math is easy on that one).
>
> Hwy do any more math at all?
>
> Make a jig that holds the sticks rotated 45 degrees. (I.e. cut a V
> groove in a board.) Then use the jig to hold the sticks and cut them
> just like you did for your first tetrahedron.
That's a good idea, and would certainly work. I've also considered taking
one of my original sticks and cutting the corners off then measuring the
resulting angles - same result. But part of me would really like to get
there by calculation rather than by analog means.
"MikeWhy" <[email protected]> wrote in
news:[email protected]:
> "Elrond Hubbard" <[email protected]> wrote in message
> news:[email protected]...
>> "MikeWhy" <[email protected]> wrote in
>> news:[email protected]:
>>
>>> "Elrond Hubbard" <[email protected]> wrote in message
>>> news:[email protected]...
>>>> Calling all geometers:
>>>>
>>>> I want to make a tetrahedron out of square cross-section sticks. I
>>>> want a corner of the stick to define the edge of the tetrahedron -
>>>> i.e., a plane through the diagonal of the stick will pass through
>>>> the center of the tetrahedron. (I've already made one where a flat
>>>> on the stick defines the edge - the math is easy on that one).
>>>>
>>>> What I can't figure out (and I've spent a frustrating day in the
>>>> shop trying) is:
>>>>
>>>> What are the angles on the face of a square stick that work?
>>
>>> If you're asking about the compound miter at the vertices... :D That
>>> will have to wait for the second pot of coffee.
>>
>> That's exactly what I'm asking :)
>
> If you haven't already, sketch it out on a cube as I described
> earlier. The geometry at each vertex is surprisingly simple, and not
> entirely obvious. Looking at one corner of the cube, the tetrahedron
> edges form the prime, 45 degree diagnonals on each of the three faces
> meeting at that corner.
>
> If you've already done that, I won't be able to help further. Compound
> miters might as well be rocket science for me. I'm guessing, for a
> radial arm saw, swing the arm 22.5 degrees to split the plan view 45
> deg; tilt the blade 67.5 degrees (probably wrong) to split the 45
> degree rise. Repeat for the other side. Maybe someone else can take it
> from there...
>
> I'll lay it out in solidworks later, see if any more insight forms. It
> sounds interesting enough to lay out on a small stick and whittle it.
I think I've got it figured out. I did the mental exercise of making
the original sticks then cutting off the corners to make the diagonally
oriented one. I determined where the original cuts crossed the mid-line
of the square stick. The midline of the square stick is the edge of the
rotated stick - I can now calculate the angles, knowing the run and rise
of those coordinates.
I'll post the results tonight after I make a test piece when I come home
from work.
[email protected] (Doug Miller) wrote in
news:[email protected]:
>
>>> If you haven't already, sketch it out on a cube as I described
>>> earlier. The geometry at each vertex is surprisingly simple, and not
>>> entirely obvious. Looking at one corner of the cube, the tetrahedron
>>> edges form the prime, 45 degree diagnonals on each of the three
>>> faces meeting at that corner.
>
> Ummm....no.
>
> There are no 45-degree angles anywhere on a tetrahedron.
Read it again, but carefully, Doug. He's describing how a tetrahedron fits
into a cube. Each edge of a tetrahedron lies on one diagonal of each face
of a cube. And, yes indeedy, those diagonals are at 45 degrees to the
edges of the cube.
"MikeWhy" <[email protected]> wrote in
news:%[email protected]:
>
> I cut up some square sticks of cherry scrap and pocketed a marking
> knife and 3/4" butt chisel into the office. A few minutes of
> (completely, therapeutically relaxing) whittling later, the first
> piece looks like it's right on. Aside from the woodchips in the
> keyboard, I can't think of a better way to break up a work day.
>
> For those who might want to take their own whittle break, the
> dimensions for 1" square stock are 2.000" back along the top spine,
> and 1.692" back on the two sides. Scale the numbers for your stock
> size, and mark and connect the dots back to a sharp tip at the front
> end of the bottom edge. Picture in your mind the two flat facets
> opening mostly upward and more gently to the sides and forward,
> meeting in a straight spine connecting the top corner to the pointy
> tip lying on the desk. Three sticks cut this way form a mitered
> tetrahedron vertex.
Nice!
I solved the math today, and here's the solution:
For those who want to lay the lines out with a protractor, the angles on
two adjacent faces, rising from a common vertex, are 30.36 degrees.
Continue on the other two faces at 73.67 degrees, and the lines will
meet at a point on the opposite edge from where you started. The
distance from your starting point, measured horizontally, is twice the
thickness of your stick.
Which is a convoluted way of saying exactly what Mike says above.
A plane cut through each pair of lines (one long, one short) will make a
120 degree dihedral angle, three of which make one vertex of the
tetrahedron.
If I get a chance to make a model, I'll post pictures of both this
version and the flat-edge version on ABPW this weekend.
Thanks to everyone for the suggestions and help!
Scott
Hoosierpopi <[email protected]> wrote in
news:81920d03-ff81-4bbf-867d-077e47ba6f74@z28g2000prd.googlegroups.com:
> "Projecting onto faces of a square stick" is your problem.
>
> But square sticks won't cut it as the edges of the finished piece need
> to be co-planer with the face(s) of the figure.
Maybe your finished piece needs to have the faces co-planar with the edges,
but mine needs to be made from square sticks. The faces are not, nor do
they need to be, in the same plane as the triangle formed by the edges.
When I'm done, the faces of the sticks will be proud of the plane of the
edges. Not a problem - that's what I'm after.
"MikeWhy" <[email protected]> wrote in
news:[email protected]:
>
> I whittled a few more sticks, and gosh darn it, they don't quite make
> a tetrahedron. That was earlier in the week. I figured at first it was
> just a sloppy cut here or there; it really wasn't all that far off.
> But, while figuring my way around Sketchup, I thought I would lay it
> out again, and see what it has to say. This time, the distances came
> out a bit more rational, although the geometry of it still escapes me.
> The distance I wrote earlier as 1.692" is actually 1.707". That lays
> out to an exact 45 deg. miter, blade tilt 35.264 degrees off
> horizontal. If you're using sine bars or a TS-Aligner to set tilt,
> that works out to 0.707" per inch. (It was so much easier to mark out
> and just whittle the 6 sticks than to set up a sledded jig for the
> tablesaw. Depends on how many you're making, of course.)
>
> There's really something magical about a tetrahedron's relationship to
> cubes and 45 degree angles, as evidenced by the root 2 multiples
> everywhere you look. It's easy to see how Fuller got trapped into this
> for so many years.
The central angle of a tetrahedron - the angle formed by connecting the
ends of any edge to the center of the solid - is 109.472 degrees. If
you take your blade tilt angle of 35.264, multiply it by two, and add it
to the central angle, you get 180 degrees, which makes sense.
But that information is insufficient for cutting the sticks - it does
not give you the 120 degree dihedral that lets three ends come together
to a point.
The layout I gave in my previous post:
>>>For those who want to lay the lines out with a protractor, the angles
on two adjacent faces, rising from a common vertex, are 30.36 degrees.
Continue on the other two faces at 73.67 degrees, and the lines will
meet at a point on the opposite edge from where you started. The
distance from your starting point, measured horizontally, is twice the
thickness of your stick.<<<
allows you to cut the sticks in two passes (I used a bandsaw). But
the fly in the ointment that kept me from getting it right the first
time is that a correction must be made thusly:
http://www.auto-met.com/subtool/stcat/st_137.html
where 73.67 is angle A and 30.36 is angle B. The result is that the
table of the bandsaw gets tilted at 9.35 degrees. The model in the
picture I posted here:
http://www.flickr.com/photos/74619244@N00/3032662080/
was put together straight off the bandsaw, using that layout and
bandsaw setup.
And yes, it is tres cool that the 45 degree business keeps showing up in
a pyramid made of equilateral triangles!
Scott
In article <[email protected]>, "MikeWhy" <[email protected]> wrote:
>"Doug Miller" <[email protected]> wrote in message
>news:[email protected]...
>> In article <[email protected]>, Elrond Hubbard
>> <[email protected]> wrote:
>>
>>>> If you haven't already, sketch it out on a cube as I described
>>>> earlier. The geometry at each vertex is surprisingly simple, and not
>>>> entirely obvious. Looking at one corner of the cube, the tetrahedron
>>>> edges form the prime, 45 degree diagnonals on each of the three faces
>>>> meeting at that corner.
>>
>> Ummm....no.
>>
>> There are no 45-degree angles anywhere on a tetrahedron.
>
>You obviously haven't sketched it out, and clearly read too fast to catch
>the import and meaning of "not entirely obvious" and "surprisingly".
>
>Bucky Fuller spent the latter decades of his life noodling over just that
>conundrum. For penance, scary sharpen a chisel and cut 45 degree chamfers on
>two end grain and one long grain edge, and marvel at the isoceles triangle
Isosceles? Nope. Try again.
>that emerges. Repeat 3 more times just to be sure that there are no 45
>degree angles in that triangle.
On Nov 9, 8:14=A0pm, Robatoy <[email protected]> wrote:
> On Nov 9, 3:21=A0pm, Elrond Hubbard <[email protected]> wrote:
>
>
>
This PDF has formulas for making pyramid for "n" sides. I think a
tetrahedron would be a 3 sided pyramid. It has all the derivations,
but scroll to the end to see the answer.
www.swcp.com/~awa/images/pdf%20files/Angle.pdf
Mitch
On Nov 9, 3:21=A0pm, Elrond Hubbard <[email protected]> wrote:
> Calling all geometers:
>
> I want to make a tetrahedron out of square cross-section sticks. =A0I wan=
t a
> corner of the stick to define the edge of the tetrahedron - i.e., a plane
> through the diagonal of the stick will pass through the center of the
> tetrahedron. =A0(I've already made one where a flat on the stick defines =
the
> edge - the math is easy on that one).
>
> What I can't figure out (and I've spent a frustrating day in the shop
> trying) is:
>
> What are the angles on the face of a square stick that work?
>
> The critical information:
>
> The central angle is 109.48 degrees. =A0That is, the angle from the cente=
r of
> the tetrahedron to any two vertices is 109.48. =A0
>
> Any help would be much appreciated!
>
> Scott
Do you want each corner to come out to a point?
On Nov 9, 3:21=A0pm, Elrond Hubbard <[email protected]> wrote:
> Calling all geometers:
>
> I want to make a tetrahedron out of square cross-section sticks. =A0I wan=
t a
> corner of the stick to define the edge of the tetrahedron - i.e., a plane
> through the diagonal of the stick will pass through the center of the
> tetrahedron. =A0(I've already made one where a flat on the stick defines =
the
> edge - the math is easy on that one).
>
> What I can't figure out (and I've spent a frustrating day in the shop
> trying) is:
>
> What are the angles on the face of a square stick that work?
>
> The critical information:
>
> The central angle is 109.48 degrees. =A0That is, the angle from the cente=
r of
> the tetrahedron to any two vertices is 109.48. =A0
>
> Any help would be much appreciated!
I believe the answer is "42".
JP
"Doug Miller" <[email protected]> wrote in message
news:[email protected]...
> In article <[email protected]>, "MikeWhy"
> <[email protected]> wrote:
> ... cut 45 degree chamfers on
>>two end grain and one long grain edge, and marvel at the isoceles triangle
>
> Isosceles? Nope. Try again.
Fuller defined a geometry where perpendicular and orthogonal are 60 degrees,
not the 90 degrees of Euclidean geometry. I pay homage to his work by
nicking off the corner with one more cut, leaving an equilateral triangle
where the 3 cuts meet. The relevance? <shrug> Polyhedrons are more easily
defined in that system than Euclidean.
Elrond Hubbard wrote:
> Calling all geometers:
>
> I want to make a tetrahedron out of square cross-section sticks. I want a
> corner of the stick to define the edge of the tetrahedron - i.e., a plane
> through the diagonal of the stick will pass through the center of the
> tetrahedron. (I've already made one where a flat on the stick defines the
> edge - the math is easy on that one).
Hwy do any more math at all?
Make a jig that holds the sticks rotated 45 degrees. (I.e. cut a V groove in
a board.) Then use the jig to hold the sticks and cut them just like you did
for your first tetrahedron.
In article <[email protected]>, Elrond Hubbard <[email protected]> wrote:
>[email protected] (Doug Miller) wrote in
>news:[email protected]:
>
>>
>>>> If you haven't already, sketch it out on a cube as I described
>>>> earlier. The geometry at each vertex is surprisingly simple, and not
>>>> entirely obvious. Looking at one corner of the cube, the tetrahedron
>>>> edges form the prime, 45 degree diagnonals on each of the three
>>>> faces meeting at that corner.
>>
>> Ummm....no.
>>
>> There are no 45-degree angles anywhere on a tetrahedron.
>
>Read it again, but carefully, Doug. He's describing how a tetrahedron fits
>into a cube. Each edge of a tetrahedron lies on one diagonal of each face
>of a cube. And, yes indeedy, those diagonals are at 45 degrees to the
>edges of the cube.
>
Gotcha. I misunderstood his post the first time around, but, yes, that's
correct.