# New SED Challenge

T

#### Tim Williams

Jan 1, 1970
0
Find the ratio of diameters (or radii, or whatever) of the large spheres to
the largest sphere which fits in the space between four adjecent large
spheres.

A couple of vertices (if not explicitly a dot, then an intersection of two
lines) and a radius are marked on the diagram. Remember this geometry
problem is three-dimensional.

Diagram is posted to alt.binaries.schematics.electronic.

Tim

K

#### kell

Jan 1, 1970
0
Find the ratio of diameters (or radii, or whatever) of the large spheres to
the largest sphere which fits in the space between four adjecent large
spheres.

A couple of vertices (if not explicitly a dot, then an intersection of two
lines) and a radius are marked on the diagram. Remember this geometry
problem is three-dimensional.

Diagram is posted to alt.binaries.schematics.electronic.

Tim

Are the large spheres of equal size and stacked in a tetrahedron (as
it were)?

T

#### Tim Williams

Jan 1, 1970
0
kell said:
Are the large spheres of equal size and stacked in a tetrahedron (as
it were)?

Yes.

Considering the simple base of the problem, I should've specified in text as
well

Tim

J

#### jasen

Jan 1, 1970
0
Are the large spheres of equal size and stacked in a tetrahedron (as
it were)?

spoiler follows.

So we have spheres centrered at a=(1,1,1) , b=(1,-1,-1),
c=(-1,-1,1) and d=(-1,1,-1) with radius measure(a,b)/2 = sqrt(8)/2
= sqrt(2)

and want to find the radius of the sphere at o=(0,0,0) that meets them
measure(o,a)/2 less the radius of the other sphere. (sqrt(3) - sqrt(2))

but you want it expressed in term of the 4 spheres so

so: (sqrt(3) - sqrt(2)) / sqrt(2)

so the answer is sqrt(3/2) - 1

that was easy.

Bye.
Jasen

T

#### Tim Williams

Jan 1, 1970
0
jasen said:

Odd, I got sqrt(5/3) - 1 (which is "close enough" by 6.7%, but a parsec
mathematically) with a different method. Another response (from the
original use of this problem) determined it experimentally as 0.224 times
smaller (or sqrt(2.9964 / 2) - 1), which is within 0.1% of your result.

So why is it that Jim screws with everyone and makes boastful threads miles
long, yet this plain intellectual challenge goes almost unnoticed? Probably
says something about the intellectual persuits of the rest of the group...

Tim

J

#### Jim Thompson

Jan 1, 1970
0
Odd, I got sqrt(5/3) - 1 (which is "close enough" by 6.7%, but a parsec
mathematically) with a different method. Another response (from the
original use of this problem) determined it experimentally as 0.224 times
smaller (or sqrt(2.9964 / 2) - 1), which is within 0.1% of your result.

So why is it that Jim screws with everyone and makes boastful threads miles
long, yet this plain intellectual challenge goes almost unnoticed? Probably
says something about the intellectual persuits of the rest of the group...

Tim

It's geometry, not circuits ;-)

Seriously, I'm up to my eyeballs in a chip design, so I only have a
few minutes here and there to comment.

...Jim Thompson

S

#### Spehro Pefhany

Jan 1, 1970
0
Odd, I got sqrt(5/3) - 1 (which is "close enough" by 6.7%, but a parsec
mathematically) with a different method. Another response (from the
original use of this problem) determined it experimentally as 0.224 times
smaller (or sqrt(2.9964 / 2) - 1), which is within 0.1% of your result.

So why is it that Jim screws with everyone and makes boastful threads miles
long, yet this plain intellectual challenge goes almost unnoticed? Probably
says something about the intellectual persuits of the rest of the group...

Tim

I would have tried it 'experimentally' (more like a Spice sim) with
Solidworks, but I recently found that program does not seem to
understand how to 'mate' spheres on tangents. It's not entirely a
theoretical question since machinists use tooling balls (almost
perfect metal spheres of known diameter) to form datums for measuring
tapers, so the designer has to specify the position of the tooling
ball, but I couldn't get the program to properly mate a sphere in an
assembly even in a simple round taper. Fortunately you can do it on a
2-D section, but then it's just a circle meeting a line at a tangent
and the math is easy for the program.

Best regards,
Spehro Pefhany

R

#### Rich Grise

Jan 1, 1970
0
Find the ratio of diameters (or radii, or whatever) of the large spheres to
the largest sphere which fits in the space between four adjecent large
spheres.

A couple of vertices (if not explicitly a dot, then an intersection of two
lines) and a radius are marked on the diagram. Remember this geometry
problem is three-dimensional.

Diagram is posted to alt.binaries.schematics.electronic.
I've been watching the thread, and this sounds kind of interesting,
given what the other posters have said, but I haven't seen the diagram;
it probably expired off my server. Could you re-post, or put it up on

Thanks,
Rich

K

#### kell

Jan 1, 1970
0
Find the ratio of diameters (or radii, or whatever) of the large spheres to
the largest sphere which fits in the space between four adjecent large
spheres.

A couple of vertices (if not explicitly a dot, then an intersection of two
lines) and a radius are marked on the diagram. Remember this geometry
problem is three-dimensional.

Diagram is posted to alt.binaries.schematics.electronic.

Tim

Consider a tetrahedron with edge length of two, four spheres of unit
My application of high school geometry, for better or ill, gives the
distance from the center of the tet to each vertex as the square root
of eleven sixths. Subtract one from that quantity, and you have your

T

#### Tim Williams

Jan 1, 1970
0
kell said:
My application of high school geometry, for better or ill, gives the
distance from the center of the tet to each vertex as the square root
of eleven sixths. Subtract one from that quantity, and you have your

Sure enough of that answer to give a proof? ;-)

So far I've heard...
sqrt(5/3) - 1 (mine)
sqrt(3/2) - 1
sqrt(6)/4 - 1/2 (half the proceeding value)
And now also sqrt(11/6) - 1.

Tim

T

Jan 1, 1970
0
K

#### kell

Jan 1, 1970
0
Sure enough of that answer to give a proof? ;-)

So far I've heard...
sqrt(5/3) - 1 (mine)
sqrt(3/2) - 1
sqrt(6)/4 - 1/2 (half the proceeding value)
And now also sqrt(11/6) - 1.

Tim

I'll give it a shot. I've defined the tetrahedron having edge length
of two. Looking at the base triangle and using the law of cosines to
find the distance from the center of the base to one of the corners of
that triangle, the distance equals the square root of four thirds.
The tetrahedron measures sqrt(8/3) high from the center of the base to
the top vertex (by the Pythagorean theorem).
The center of the tetrahedron lies at a point on this imaginary line
segment.
Now, the distance from the center of the tetrahedron's base to its
geometric center, again by Pythagorean th., equals
sqrt(x^2-(4/3))
The distance from the center of the base to the vertex
h = x + sqrt(x^2-(4/3))
that is,
h = sqrt(8/3) = x + sqrt(x^2-(4/3))
I'll skip typing in the algebra; I get x = sqrt(11/6)
result --> radius of central sphere = sqrt(11/6)-1

K

#### kell

Jan 1, 1970
0
I'll give it a shot. I've defined the tetrahedron having edge length
of two. Looking at the base triangle and using the law of cosines to
find the distance from the center of the base to one of the corners of
that triangle, the distance equals the square root of four thirds.
The tetrahedron measures sqrt(8/3) high from the center of the base to
the top vertex (by the Pythagorean theorem).
The center of the tetrahedron lies at a point on this imaginary line
segment.
Now, the distance from the center of the tetrahedron's base to its
geometric center, again by Pythagorean th., equals
sqrt(x^2-(4/3))
The distance from the center of the base to the vertex
h = x + sqrt(x^2-(4/3))
that is,
h = sqrt(8/3) = x + sqrt(x^2-(4/3))
I'll skip typing in the algebra; I get x = sqrt(11/6)
result --> radius of central sphere = sqrt(11/6)-1- Hide quoted text -
I made a mistake in the algebra. x = sqrt(3/2)
r = sqrt(3/2)-1 = .22474487...

J

#### Jonathan Kirwan

Jan 1, 1970
0
I made a mistake in the algebra. x = sqrt(3/2)
r = sqrt(3/2)-1 = .22474487...

And I got 3/sqrt(6)-1 by somewhat different logic, but where it's the
same result.

Jon

C

#### [email protected]

Jan 1, 1970
0
So far I've heard...
sqrt(5/3) - 1 (mine)
sqrt(3/2) - 1
sqrt(6)/4 - 1/2 (half the proceeding value)
And now also sqrt(11/6) - 1.

Tim

I've used a prototype version of MacSpice 3f5 2.9p34 to solve this
problem. This version 2.9p34 has a built-in optimizer that uses a
simplex method. Using default settings to minimize and n-parameter
cost function, it starts by constructing a regular n-dimensional unit
simplex with its centroid at the origin. I can extract these and solve
the problem fairly easily, using this command file:

*Gap between four spheres calculation
..control
echo
echo "Get a regular simplex with four vertices on on a unit sphere:"
echo
optimize 'let cost = 1' npar 3 report -1 maxeval 3
foreach v 0 1 2 3
print line vertices[$v] end echo echo "Rescale it so each point is separated by 2 units from the others:" echo let delta = (vertices[2]-vertices[3]) let separation = sqrt(length(delta)*mean(delta*delta)) let rescaled = 2*vertices/separation foreach v 0 1 2 3 print line rescaled[$v]
end
echo
echo "The distance from the origin to any of the rescaled vertices
is:"
echo
let d = sqrt(length(rescaled[0])*mean(rescaled[0]*rescaled[0]))
print d
echo
echo "Hence the largest sphere to fit into the void has radius:"
echo
set numdgt = 15
print (d-1.0)
echo
echo "Which is the same as:"
echo
print sqrt(3/2)-1.0
..endc

Which produces the following results:

**************
*** MacSpice - 3f5 v2.9 PATCHLEVEL: 34
*** Carbon Version by CDHW
*** Date Created: Apr 7 2007
**************

Get MacSpice updates, information, userguide, tutorials from:

<http://newton.ex.ac.uk/teaching/CDHW/MacSpice/>

Some useful Spice 3 commands:

source <filename> : loads the named file as the new circuit.
run : executes the specified analyses.
edit : edits source with helper application.
display : displays the list of output-vectors.
plot <plotargs> : draws the results.
set : displays all internal variables.
rusage all : prints the resource usage.
help : on-line help information (obsolete).
quit : quit spice.

MacSpice 1 -> source Primary:Users:cdhwocuments:MacSpice:4-
spheres:four-spheres.src

Get a regular simplex with four vertices on on a unit sphere:

vertices[0] = ( -8.16497e-01 -4.71405e-01 -3.33333e-01 )
vertices[1] = ( 8.164966e-01 -4.71405e-01 -3.33333e-01 )
vertices[2] = ( 0.000000e+00 9.428090e-01 -3.33333e-01 )
vertices[3] = ( 0.000000e+00 0.000000e+00 1.000000e+00 )

Rescale it so each point is separated by 2 units from the others:

rescaled[0] = ( -1.00000e+00 -5.77350e-01 -4.08248e-01 )
rescaled[1] = ( 1.000000e+00 -5.77350e-01 -4.08248e-01 )
rescaled[2] = ( 0.000000e+00 1.154701e+00 -4.08248e-01 )
rescaled[3] = ( 0.000000e+00 0.000000e+00 1.224745e+00 )

The distance from the origin to any of the rescaled vertices is:

d = 1.224745e+00

Hence the largest sphere to fit into the void has radius:

(d-1.0) = 2.247448713915894e-01

Which is the same as:

sqrt(3/2)-1.0 = 2.247448713915889e-01
MacSpice 2 ->

Happy Easter

Charles

J

#### jasen

Jan 1, 1970
0
Sure enough of that answer to give a proof? ;-)

So far I've heard...
sqrt(5/3) - 1 (mine)
sqrt(3/2) - 1
sqrt(6)/4 - 1/2 (half the proceeding value)
And now also sqrt(11/6) - 1.

after a slow start this is shaping up to prove as controversial as
the other challenges...

Bye.
Jasen

K

#### kell

Jan 1, 1970
0
after a slow start this is shaping up to prove as controversial as
the other challenges...

Bye.
Jasen

You found an elegant approach using analytical geometry.
I used euclidean geometry, some elementary trig & algebra.
Another guy used computer simulation of some sort.
All in good fun, with no flames.

T

#### Tim Williams

Jan 1, 1970
0
The top and left spheres do not appear to be tangent, though it could just
be the crummy way drawn (there isn't a good way to draw the surfaces while
seeing everything in this problem).

Tim

R

#### Rich Grise

Jan 1, 1970
0
The top and left spheres do not appear to be tangent, though it could just
be the crummy way drawn (there isn't a good way to draw the surfaces while
seeing everything in this problem).

AAAAARRRRGGGGHHHHHH!

**** me seven ways from Sunday - the "crummy" way it's drawn is that I
didn't even get the fundamental equilateral tetrahedron right!

Well, like they used to say, "Back to the ol' drawing board!"

Thanks for not trashing me for such a horrendous blunder.

Thanks,
Rich

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