3d_design

David Cary feedback.html d.cary@ieee.org. · superappliance

Batteries
finishing nails
wire
WD-40
superglue
Duct Tape
masking tape
scotch tape (writable with pencil)
clear contact paper
paint
spare light bulbs
ziplock bags
trash bags

Screwdriver
Tape measure
hammer
scissors
pliers (needle-nose)
flashlight
Dentist's Mirror
toolbox
extension cord
power strip

http://web.shorty.com/geeks/96/sep/msg00159.html

how to mismake a soccer ball

     Subject: how to mismake a soccer ball
     From: (Anton Sherwood)
     Date: Tue, 3 Sep 1996 23:24:53 -0700

I wrote a little C program to enumerate the ways to make a closed ball
from 12 pentagons and up to 20 hexagons.  (I don't care whether the
faces are regular, because I got on this track by wondering about
the varieties of buckyballs.)

I thought I'd have to build (and debug) a representation of the polygons
and their connections.  Easier to do the search by hand... But eventually
it hit me that all I need is a bit string representing the convex and
concave vertices of the boundary of the incomplete ball.

        Solutions      = 526363
        Dead ends      = 21982613
        Face overflows = 19186580
        Edge overflows = 410

The output is also a bit string: 1 for a pentagon, 0 for a hexagon; and
they're added in a strict spiral sequence, so even though the program
knows nothing about the shape, it's easy for a human to build it up
from its code.  Each solution appears in the tree 120/G times, where
G is the size of its symmetry group.

And what's my point in doing this?  Well, if you should ever want to
build a dome house with a square floor, let me know!

Anton Sherwood   *\\*   +1 415 267 0685   *\\*  

"NO!!!  Not THAT button!!!"

	RealEarth GlobeMaps
	Universal Marketing Company
	1013 Barrington Oaks Place
	Roswell GA  30075-4790

        Graphite whiskers       961 km
        Al2O3 whiskers          527 km
        Iron whiskers           162 km
        Si3N4 whiskers          455 km
        SiC whiskers            704 km
        Si whiskers             337 km

        Bulk aluminum             10 km
        Bulk iron                 11 km
        Bulk steel                40 km
        Nylon                     88 km
        Fiberglass                98 km
        Kevlar                   195 km

From: "Scott Jensen" 
Subject: Re: Space Elevator
Date: 03 Nov 1999 00:00:00 GMT
Newsgroups: sci.nanotech
...
It was
first conceived by Yuri Artsutanov in 1960 in Russia.  Arthur C. Clarke
wrote about it in his book "Ascent To Orbit: A Scientific Autobiography"
(John Wiley & Sons, 1984) in chapter twenty-one "The Space Elevator -- And
Beyond".  The chapter is nothing more than a short introduction for a paper
that appeared in "Advances in Earth Orientated Applications of Space
Technology" (Vol. I, no. 1, 1981, pp. 39-48), but I'm guessing you'll have a
better chance of getting hold of Clarke's book thus why I gave it.

...

The material Artsutanov suggested for using only existed in microscopic
quantities.  On the bright side, we'd only need 900 tons of it.  Sorry, I
forget what was the material he proposed using.  However, if the
manufacturing of it could be solved, then yes the space elevator could be
easily constructed.

...

> ...and is not likely to in the near
>  future.  The only materials even theoretically capable of doing the job
>  are things like single-crystal (gem quality) diamond 80 thousand miles
long!

What?  Carbon crystals have breaking lengths of up to 3000kms.  A
"to-sea-level" space elevator would "only" need to be 5000km long.

...

> Not to mention the
>  transportation system to get a trillion tons into geosynchronous orbit.

No one proposes using Earth materials.  It is either asteriods or lunar.
Probably lunar.  Oh, as a side note, we can make a space elevator for the
Moon using mere steel.  What many propose is first using mass drivers to
create the lunar space elevator then use the lunar space elevator to send up
the material to make the Earth space elevator.

Scott

From: dietz@Computerese.Information.Network (Paul Dietz)
Newsgroups: sci.space.science
Subject: Re: Ion drive
Date: 24 Nov 1995 03:50:48 GMT
Organization: Computerese Information Network (CIN.net )
Lines: 20
Return-Path: news@news.cin.net
To: sci-space-science@uunet.uu.net

Joshua B Hopkins (jbhopkin@ux4.cso.uiuc.edu) wrote:

: In choosing a propellant for an ion drive, you want to look at two major
: issues.  The energy required to ionize the propellant must be fairly low,
: otherwise you spend too much energy ionizing the fuel and not as much
: accelerating it.  Generally, heavier atoms/molecules are desireable,
: because while they decrease the Isp, they increase the thrust.

Heavier atoms/molecules do not necessarily decrease Isp, because you can
just increase the voltage in proportion to the mass/charge ratio of the
ions.

The real reason to go to high mass/charge ratio particles is the effect on
thrust density.  There is a limit to the current density in ion engines
due to the space charge of the ions. At fixed exhaust velocity, thrust
density goes as the square of the mass/charge ratio.

	Paul

        0

o       1

 o       3
o o

  o       6
 o o
o o o

   o       10
  o o
 o o o
o o o o

    o       15
   o o
  o o o
 o o o o
o o o o o

     o       21
    o o
   o o o
  o o o o
 o o o o o
o o o o o o

      o       28
     o o
    o o o
   o o o o
  o o o o o
 o o o o o o
o o o o o o o

       o     36
      o o
     o o o
    o o o o
   o o o o o
  o o o o o o
 o o o o o o o
o o o o o o o o

       o     36
      o o
     o o o
    o o o o
   o o o o o
  o o o o o o
 o o o o o o o
o o o o o o o o

       x
      x x
     x x x
    o o o x
   o o o x x
  o o o x x x
 o o o x x x x
o o o x x x x x

5*3 = 15 remaining oranges.

            +
           / \
          / b \    b = orange on bottom layer
         +-----+
        / \ t / \  t = orange from top layer
       / b \ / b \     squashed down into the gaps on the bottom layer.
      +-----+-----+
     / \ t / \ t / \
    / b \ / b \ / b \
   +-----+-----+-----+

            +
           / \
          / b \
         +-----+
  length = 1   total: 1 orange
            +
           / \
          / b \
         +-----+
        / \ t / \
       / b \ / b \
      +-----+-----+
  length = 2   total: 4 oranges

            +
           / \
          / b \
         +-----+
        / \ t / \
       / b \ / b \
      +-----+-----+
     / \ t / \ t / \
    / b \ / b \ / b \
   +-----+-----+-----+
  length = 3   total: 9 oranges
               +
              / \
             / b \
            +-----+
           / \ t / \
          / b \ / b \
         +-----+-----+
        / \ t / \ t / \
       / b \ / b \ / b \
      +-----+-----+-----+
     / \ t / \ t / \ t / \
    / b \ / b \ / b \ / b \
   +-----+-----+-----+-----+
  length = 4  total: 16 oranges
   ...
  length = x  total: x^2 oranges

            +
           / \
          /   \  original trangle made up of any number of oranges
         +-----+
	    +
           / \
          /   \
         +-----+  new bigger triangle
        / \   / \
       /   \ /   \
      +-----+-----+

c == [ (a+b)^2 + 3*a + b ]/2
c == [ a^2 + b^2 + 2*a*b + 3*a + b ]/2
c == [ a^2 + b^2 + 2*a*b + 3*a + b ]/2

we recognize the pattern "(x^2 + x)/2 = t(x)",
usually a clue that triangles will be helpful,
and convert to triangle notation:

c == [ (a^2 + a)/2 + 2*a/2 + (b^2 + b)/2 + 2*a*b/2 ]
c ==   (a^2 + a)/2 +   a   + (b^2 + b)/2 +   a*b
c ==   t(a) +          a   + t(b) +          a*b

Next we substitute the definition of multiply,
  a*b = t(a+b) - ( t(a) + t(b) )
to get
c == t(a) + a + t(b) + t(a+b) - ( t(a) + t(b) )

Lots of terms cancel out, and we are left with

c == a + t(a+b).

To recover the original 2 values from c,
we merely find the largest triangle
we can build from c oranges,
which has sides of length L = (a+b).
Then we count the "a" leftover oranges,
and we calculate b = L-a.

We know that c must have come from
a single unique pair of
whole integers (a,b),
since
  a < a+b+1
  a + t(a+b) < t(a+b) + a+b+1
  a + t(a+b) < t(a+b+1)
.
In other words,
no matter how many "leftover" oranges a we have,
it's never enough to build a triangle
even 1 row larger.

Unfortunately,
there are multiple (a,b) pairs
where b is a integer, but a is some real number,
that map to the same value c.
For example,
b = 1, a = 1.772, c = a+t(a+b) = 7
b = 2, a = 1.000, c = a+t(a+b) = 7

            c c c
          c c c c
    a a a c c c c e e e
  a a a a c c c e e e e
    a a a a D D D e e e e
    a a a D D D D e e e
    b b b D D D D f f f
  b b b b D D D f f f f
    b b b b       f f f f
    b b b         f f f

          c c
        c c c c
    a a c c c c e e
  a a a a c c e e e e
  a a a a D D e e e e
    a a D D D D e e
    b b D D D D f f
  b b b b D D f f f f
  b b b b     f f f f
    b b         f f

  ...

    b b
  b b b b
    b b
...
          c c
    a a c c c c e e
  a a a a c c e e e e
    a a X D D X e e
    b b D D D D f f
  b b b b D D f f f f Doesn't quite tile properly.
    b b         f f

...

      c c
  a a c c e e
  a a D D e e
  b b D D f f
  b b     f f

    c
  a c e
  a D e
  b D f
  b   f

         c c c c
         c c c c
   a a a c c c c e e e
   a a a a c c e e e e
   a a a a D D e e e e
   a a a D D D D e e e
   b b b D D D D f f f
 b b b b D D D D f f f f
 b b b b         f f f f
   b b b         f f f

 Every region is the same shape and size (14 pixels)
 (but rotated in all 4 directions)
 , no single-pixel protuberances.

   a c c c e
 a a a c e e e
 b b b D f f f
   b D D D f ...

        +
       / \
      / a \     ====    +--+
     +-----+            | a|
    / \ c / \   ==== +--+--+--+
   / b \ / d \       |b | c| d|
  +-----+-----+      +--+--+--+

      a
    a a a c
    b a c c c e
  b b b D c e e e
    b D D D f e
        D f f f
            f

    b   b
      b c   c
  a b a b c
    a x c x c
  a   a x
      x   x

Similar to slash code, but each blob encodes
2 bits (rather than 1 bit):

the first bit encodes horizontally,

    x x
  a * * b
  b * * a
    x x

where a = not(b) = encoded bit.

The other bit of this blob encodes vertically,

    c d
  x * * x
  x * * x
    d c

where c = not(d) = other encoded bit.

From: spyknife@aol.com (SpyKnife)
Subject: Re: Space Elevator
Date: 19 Nov 1999 00:00:00 GMT
Approved: josh@discuss.foresight.org
X-Complaints-To: abuse@verio.net
Organization: AOL http://www.aol.com
NNTP-Posting-Date: Fri, 19 Nov 1999 21:09:12 GMT
Newsgroups: sci.nanotech
X-Psw: Cadmium 0.01

Diamondoid Solids and other nanomaterials would probably be more reliable than
systems composed of active moving parts full of ratchets and computers.

[Presumably this is in the context of large structures?  Maybe, maybe not.
 Large structures that last a long time have typically overpowered the
 problem (forces handled, erosion, etc) with mass, like the pyramids.
 That's not an option for skyhooks and the like.
 Inert matter, no matter how strong, is subject to wear and tear;
 self-repairing systems can last much longer.  Consider living organisms
 (ones not programmed to die) such as Galapagos tortoises and Sequoia trees.
 A stone spire the height of a redwood wouldn't last a fraction of the
 thousands of years they can attain.
 --JoSH]

shape	size in unit tetras (or unit triangles)	size in unit cubes (or unit squares)
equilateral triangle with sides of length s (see #t(x) for details)	s^2	s^2 * sqrt(3) / 4
equilateral triangle with sides of length 2*s	4*s^2	s^2 * sqrt(3)
square with sides of length s	s^2 * 4 / sqrt(3)	s^2
square with diagonal of length s	s^2 * 2 / sqrt(3)	s^2 / 2
area of hexagon inscribed inside a circle of radius r	r^2 * 6	r^2 * 3 * sqrt(3) / 2
area of hexagon with a circle inscribed of radius r (it's pretty easy to see that once you chop out the inscribed hexagon, the 6 remaining fragments can be assembled into 2 more triangles the same size as the 6 interior trangles)	r^2 * 8	r^2 * 6 / sqrt(3)
area of a circle of radius r (obviously more than the 6 triangles of the inscribed hexagon ... the little leftovers between that hexagon and the circle add up to a bit more than 1 more triangle) (obviously more than the 2 squares of the inscribed square with diagonal of length 2*r ... the little leftovers add up to a bit more than 1 more square)	4πr^2 / sqrt(3) =~= 7.26*r^2	pir^2 = πr^² =~= 3.14*r^2
area of pentagon with sides of length s	s^2 * sqrt( 25 + 10*sqrt(5) ) / sqrt(3)	s^2 * sqrt( 25 + 10*sqrt(5) ) / 4


area of sphere (compare the unit dodecahedron, which has a surface of 20 unit triangles ...)	16pir^2 / sqrt(3) =~= (29.02)*r^2	4pir^2 =~= (12.566)*r^2
volume of a sphere http://mathworld.wolfram.com/Sphere.html /* was sphere */ = 2/3 the volume of cylinder circumscribed around it		4pir^3/3
volume of cylinder circumscribed around unit sphere		2pir^2
the Platonic solids http://mathworld.wolfram.com/PlatonicSolid.html
volume of regular tetrahedon with edges of length s	s^3	s^3 / ( 6 * sqrt(2) )
volume of regular tetrahedon with edges of length sqrt(2)	2*sqrt(2)	1/3
volume of cube with edges of length s	s^3 * 6 * sqrt(2)	s^3
volume of cube with face diagonal of length s	s^3 * 3	s^3 * / ( 2 * sqrt(2) )
volume of regular octahedron with edges of length s	s^3 * 4	s^3 * sqrt(2) / 3
volume of regular dodecahedron with edges of length s	s^3 * ( 15 + 7 * sqrt(5) ) * sqrt(2) * 3 / 2	s^3 * ( 15 + 7 * sqrt(5) ) / 4
volume of regular icosahedron with edges of length s	s^3 * ( 3 + sqrt(5) ) * sqrt(2) * 5 / 2	s^3 * ( 3 + sqrt(5) ) * 5 / 12

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