[continued from previous message]   
      
    * angle between extended middle finger tips tangent circle-square   
   intersections   
      is actually 100 degrees; navel at center of circle is 1 1/2 times higher   
   than   
      1/14 of man's height; angle between raised legs at calf muscle is actually   
   60   
      degrees as measured from the center of the square; also, angle between   
   raised   
      legs at center of ball of foot is 60 degrees as measured from center of   
   circle;   
      
    * the line segment between center of circle and center of square is the   
   opposite   
      side of a right triangle, with adjacent side the horizontal circle radius,   
   and   
      hypotenuse from the center of the square to the end of that same circle   
   radius,   
      the angle of which is 80 degrees; the center of square is 2 cubits above   
   floor   
      line, and its base is tangent to the base of circle at the vertical   
   centerline;   
      thus solving for "y": y/(y + 2) = tan 10; y = ~0.428148 cubits; 4 cubits/14   
   is   
      ~0.285714, for a ratio of ~1.49852; very nearly 1 1/2 times higher than   
   "1/14";   
      
    * circle radius 2 + y = ~2.428148 cubits; circle diameter 2y + 4 = ~4.856296   
   cu-   
      bits; circle area (2 + y)^2 * pi = ~18.522525 square cubits; top of circle   
   is   
      2y = ~0.856296 cubits above square, segment chord 4 * sqrt(2y) = ~3.701451   
   cu-   
      bits, central angle is 2 * arctan (2 * sqrt(2y)/(2 - y)) = ~99.316396   
   degrees   
      (inside edge extended middle finger tips); 1 finger is 1/24 cubit =   
   ~0.041667   
      cubits; 1 palm is 1/6 cubit = ~0.166667 cubits; 1 foot 4/6 = ~0.666667   
   cubits;   
      
    * simplifying the value of "y", y/(y+2)=tan(10): y = 2sin(10)/(   
   os(10)-sin(10));   
      circle chord at top of square = 8sqrt(sin(10)/(cos(10)-sin(10))) =   
   ~3.701451;   
      2arcTan((cos(10)-sin(10))sqrt(sin(10)/(cos(10)-sin(10)))/(cos   
   10)/2-sin(10)))   
      is central angle of top circle sector ~99.316396 degrees; top circle sector   
      area = pi(2sin(10)/(cos(10)-sin(10))+2)^2arcTan((cos(10)-sin(   
   0))sqrt(sin(10)   
      /(cos(10)-sin(10)))/(cos(10)/2-sin(10)))/180 = ~5.109973 square cubits; top   
      circle segment area = pi(2sin(10)/(cos(10)-sin(10))+2)^2arcTan((cos(10)-sin   
      (10))sqrt(sin(10)/(cos(10)-sin(10)))/(cos(10)/2-sin(10)))/180+(8sin(10)/(cos   
      (10)-sin(10))-8)sqrt(sin(10)/(cos(10)-sin(10))) = ~2.200907 square cubits;   
      
    * circle chord at side of square = 2sqrt((2sin(10)/(cos(10)-sin(10))+2)^2-4) =   
      ~2.753836; central angle of side circle sector = 2arcTan(sqrt((2sin(10)/(cos   
      (10)-sin(10))+2)^2-4)/2) = ~69.091629 degrees; side circle sector area = pi   
      (2sin(10)/(cos(10)-sin(10))+2)^2arcTan(sqrt((2sin(10)/(cos(10)-sin(10))+2)^2   
      -4)/2)/180 = ~3.554865 square cubits; area of side circle segment = pi(2sin   
      (10)/(cos(10)-sin(10))+2)^2arcTan(sqrt((2sin(10)/(cos(10)-sin(10))+2)^2-4)/   
      2)/180-2sqrt((2sin(10)/(cos(10)-sin(10))+2)^2-4) = ~0.801029 square cubits;   
      
    * circle chord at bottom of square = 4 cubits; central angle of bottom circle   
      sector = 2arcTan(2/sqrt((2sin(10)/(cos(10)-sin(10))+2)^2-4)) = ~110.908371   
      degrees; bottom circle sector area = pi(2sin(10)/(cos(10)-sin(10))+2)^2arc   
      Tan(2/sqrt((2sin(10)/(cos(10)-sin(10))+2)^2-4))/180 = ~5.706397 square cu-   
      bits; bottom circle segment area = pi(2sin(10)/(cos(10)-sin(10))+2)^2arcTan   
      (2/sqrt((2sin(10)/(cos(10)-sin(10))+2)^2-4))/180-2sqrt((2sin(10)/(cos(10)-   
      sin(10))+2)^2-4) = ~2.952562 square cubits;   
      
    * area of bottom square corner outside circle = -pi(2sin(10)/(cos(10)-sin(10))   
      +2)^2arcTan(2/sqrt((2sin(10)/(cos(10)-sin(10))+2)^2-4))/360-sqrt((2sin(10)/   
      (cos(10)-sin(10))+2)^2-4)+4sin(10)/(cos(10)-sin(10))+4 = ~0.626179 square   
      cubits; circle chord at top corner of square = sqrt((-sqrt((2sin(10)/(cos   
      (10)-sin(10))+2)^2-4)-2sin(10)/(cos(10)-sin(10))+2)^2+(-4sqrt(sin(10)/(cos   
      (10)-sin(10)))+2)^2) = ~0.245524 cubits; central angle of top square corner   
      circle sector = -arcTan(sqrt((2sin(10)/(cos(10)-sin(10))+2)^2-4)/2)-arcTan   
      ((cos(10)-sin(10))sqrt(sin(10)/(cos(10)-sin(10)))/(cos(10)/2-sin(10)))+90   
      = ~5.795988 degrees; top square corner circle sector area = (2sin(10)/(cos   
      (10)-sin(10))+2)^2(-piarcTan(sqrt((2sin(10)/(cos(10)-sin(10))+2)^2-4)/2)/   
      360-piarcTan((cos(10)-sin(10))sqrt(sin(10)/(cos(10)-sin(10)))/(cos(10)/2-   
      sin(10)))/360+pi/4) = ~0.298212 square cubits; top square corner circle   
      segment area = -sqrt(((-sqrt((2sin(10)/(cos(10)-sin(10))+2)^2-4)-2sin(10)   
      /(cos(10)-sin(10))+2)^2+(-4sqrt(sin(10)/(cos(10)-sin(10)))+2)^2)(-(-sqrt   
      ((2sin(10)/(cos(10)-sin(10))+2)^2-4)-2sin(10)/(cos(10)-sin(10))+2)^2/4-   
      (-4sqrt(sin(10)/(cos(10)-sin(10)))+2)^2/4+(2sin(10)/(cos(10)-sin(10))+2)   
      ^2))/2+(2sin(10)/(cos(10)-sin(10))+2)^2(-piarcTan(sqrt((2sin(10)/(cos(10)   
      -sin(10))+2)^2-4)/2)/360-piarcTan((cos(10)-sin(10))sqrt(sin(10)/(cos(10)-   
      sin(10)))/(cos(10)/2-sin(10)))/360+pi/4) = ~0.000508 square cubits; area   
      of top square corner outside circle = sqrt(((-sqrt((2sin(10)/(cos(10)-sin   
      (10))+2)^2-4)-2sin(10)/(cos(10)-sin(10))+2)^2+(-4sqrt(sin(10)/(cos(10)-sin   
      (10)))+2)^2)(-(-sqrt((2sin(10)/(cos(10)-sin(10))+2)^2-4)-2sin(10)/(cos(10)   
      -sin(10))+2)^2/4-(-4sqrt(sin(10)/(cos(10)-sin(10)))+2)^2/4+(2sin(10)/(cos   
      (10)-sin(10))+2)^2))/2+(2sin(10)/(cos(10)-sin(10))+2)^2(piarcTan(sqrt((2sin   
      (10)/(cos(10)-sin(10))+2)^2-4)/2)/360+piarcTan((cos(10)-sin(10))sqrt(sin(10)   
      /(cos(10)-sin(10)))/(cos(10)/2-sin(10)))/360-pi/4)+(-2sqrt(sin(10)/(cos(10)   
      -sin(10)))+1)(-sqrt((2sin(10)/(cos(10)-sin(10))+2)^2-4)-2sin(10)/(cos(10)-   
      sin(10))+2) = ~0.014041 (= 0.0140410224358...) square cubits;   
      
    * line segment "y" is also the shortest side of a scalene triangle, with   
   longest   
      side the circle radius, and adjacent side "a" 100 degrees from vertical   
   center-   
      line to the end of that same circle radius; thus solving for "a": a =   
   sqrt((-8   
      sin(10)cos(10)cos(70)+4(sin(10))^2)/(-2sin(10)cos(10)+1)+(2si   
   (10)/(cos(10)-sin   
      (10))+2)^2) = ~2.316912 cubits (2.31691186136...); area of triangle =   
   sin(10)cos   
      (10)cos(20)/(-sin(10)cos(10)+1/2) = ~0.488455 (0.488455385956...) square   
   cubits;   
      
    * segment "y" is shortest side of yet another, slightly smaller scalene   
   triangle   
      with adjacent side "a" 110 degrees from vertical centerline, and longest   
   side   
      60 degrees from the same vertical centerline; thus solving for "a": a =   
   sqrt(3)/   
      (cos(10)-sin(10)) = ~2.135278 (2.13527752148...) cubits; longest side =   
   2sin(70)   
      /(cos(10)-sin(10)) = ~2.316912 (2.31691186136...) cubits, which extends   
   2sin(70)   
      /(cos(10)-sin(10))-4sqrt(3)/3 = ~0.00751078 cubits beyond intersection   
   w/square;   
      area of triangle = sqrt(3)sin(10)sin(70)/(cos(10)-sin(10))^2 = ~0.429540   
   square   
      cubits (0.429540457576...); the tiny fraction of this triangle outside   
   square is   
      described by shortest side = sin(10)(2/(cos(10)-sin(10))-4sqr   
   (3)/(3sin(70))) =   
      ~0.00138794 cubits (0.00138793689527...); longest side = 2sin   
   70)/(cos(10)-sin   
      (10))-4sqrt(3)/3 = ~0.00751078 (0.00751078459977...) cubits; adjacent side   
   "a":   
      a = sqrt(3)(1/(cos(10)-sin(10))-2sqrt(3)/(3sin(70))) = ~0.00692198 cubits   
   (0.00   
      692197652921...); area of tiny triangle = (sqrt(3)sin(10)(csc   
   70))^2(sqrt(3)sin   
      (10)/3-sqrt(3)cos(10)/3+sin(70)/2)(5sqrt(3)sin(10)sin(70)/6-4   
   qrt(3)sin(70)cos   
      (10)/3+sin(10)cos(10)+2(sin(70))^2-(sin(10))^2)+(sin(10))^2(c   
   c(70))^2(-sqrt(3)   
      (cos(10)-sin(10))+3sin(70)/2)(-sqrt(3)(cos(10)-sin(10))/3+sin   
   70)/2))/(cos(10)-   
      sin(10))^2 = ~0.00000451394 square cubits (0.0000045139387711...);   
      
    * segment "y" is the base of an isosceles triangle with vertex angle 160   
   degrees,   
      
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   --- SoupGate-Win32 v1.05   
    * Origin: you cannot sedate... all the things you hate (1:229/2)