New SED Challenge

For those of us who have no access to a.b.s.e.: 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

-- "Librarians are hiding something." - Steven Colbert Website @

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

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

-- "Librarians are hiding something." - Steven Colbert Website @

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

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

Best regards, Spehro Pefhany

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 your website somewhere?

Thanks, Rich

Consider a tetrahedron with edge length of two, four spheres of unit radius centered at the vertices. 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 answer. I get the central sphere has a radius .3540064...

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

-- "Librarians are hiding something." - Steven Colbert Website @

Sure.

Tim

-- "Librarians are hid>

to

two

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

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

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:

Which produces the following results:

Get MacSpice updates, information, userguide, tutorials from:

Some useful Spice 3 commands:

source : 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 : 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:cdhw:Documents: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

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.

I drew a 3-D diagram, and came up with this:

Cheers! Rich

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

AAAAARRRRGGGGHHHHHH!

F*ck 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