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Subject: Vitruvian Man - parts 1-6
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Wikimedia Commons - Da Vinci Vitruve (photo L. Viatour) 2,258 x 3,070 pixels, 5.81 MB:
https://upload.wikimedia.org/wikipedia/commons/2/22/Da_Vinci_Vitruve_Luc_Viatour.jpg
crude photo correction:
http://rawtherapee.com/
rotate -0.62
horizontal +0.7
vertical +0.5
save PNG
(35.56 MB)
import into inkscape:
https://inkscape.org/
skew +0.3h
width ~98.7
height 100.0
grid 20x20
square 16x16
"Vitruvius, the architect, says in his architectural work that the measurements
of man are in nature distributed in this manner, that is 4 fingers make a palm,
4 palms make a foot, 6 palms make a cubit, 4 cubits make a man, 4 cubits make
a footstep, 24 palms make a man and these measures are in his buildings. If you
open your legs enough that your head is lowered by 1/14 of your height and raise
your arms enough that your extended fingers touch the line of the top of your
head, let you know that the center of the ends of the open limbs will be the
navel, and the space between the legs will be an equilateral triangle"
+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+
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+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+
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+---+---+-x-----------------------------------------------------------x-+---+---+
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+---+---x---------------------------------------------------------------x---+---+
* 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)/(cos(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(10))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)+(2sin(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))-4sqrt(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)/6-4sqrt(3)sin(70)cos
(10)/3+sin(10)cos(10)+2(sin(70))^2-(sin(10))^2)+(sin(10))^2(csc(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,
leg 2(sin(10))^2/(sin(20)(cos(10)-sin(10))) = ~0.217376439936 cubits, altitude
(sin(10))^2/(cos(10)(cos(10)-sin(10))) = ~0.0377470226626 cubits, and area tan
(10)(sin(10))^2/(cos(10)-sin(10))^2 = ~0.00808065625672 square cubits; segment
"y" is also the diameter of a circle, area pi(sin(10))^2/(cos(10)-sin(10))^2 =
~0.143971899424 square cubits; this small circle is centered at the midpoint of
segment "y", i.e. between the drawing's center of circle and center of square;
* a radial grid of 36 10-degree sectors centered at each endpoint of segment "y"
highlights the many triangles and quadrangles evident in this geometric study;
a layer of about 20% opacity sector color fills makes distinguishing polygons
much easier; primary colors are red, orange, yellow, green, blue, violet, and
magenta; for zodiac equivalents, the following chart includes secondary colors
and polar angles measured in degrees from earth-sun ecliptic west at 0 scorpio:
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