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Path: ...!news.misty.com!weretis.net!feeder6.news.weretis.net!usenet.blueworldhosting.com!diablo1.usenet.blueworldhosting.com!feeder.usenetexpress.com!tr2.iad1.usenetexpress.com!69.80.99.26.MISMATCH!local-2.nntp.ord.giganews.com!news.giganews.com.POSTED!not-for-mail NNTP-Posting-Date: Sat, 23 Mar 2024 22:38:04 +0000 Subject: Re: Vitruvian Man - parts 1-6 Newsgroups: soc.penpals References: <20240323.220408.d6a7b0af@mixmin.net> From: % <pursent100@gmail.com> Date: Sat, 23 Mar 2024 15:38:03 -0700 User-Agent: Mozilla/5.0 (Windows NT 10.0; WOW64; rv:91.0) Gecko/20100101 Firefox/91.0 SeaMonkey/2.53.18.1 MIME-Version: 1.0 In-Reply-To: <20240323.220408.d6a7b0af@mixmin.net> Content-Type: text/plain; charset=UTF-8; format=flowed Content-Transfer-Encoding: 7bit X-Antivirus: AVG (VPS 240323-4, 2024-3-23), Outbound message X-Antivirus-Status: Clean Message-ID: <ixudneobc-5RxGL4nZ2dnZfqnPudnZ2d@giganews.com> Lines: 229 X-Usenet-Provider: http://www.giganews.com X-Trace: sv3-99d+9S1yeTPWi0vbWjEDMUFOm7VEwZL+uOZwETcD/DlYI7K771B+c3MfrB0IDkwRC2xFJydaqOyC9vP!cUKuQr0mYA80UA0UYkbaLsTf80xBnd0IMDf06vA51NgAK8hfroiXJ5OVBDs1UZEwE0AKVAEutJ6E X-Complaints-To: abuse@giganews.com X-DMCA-Notifications: http://www.giganews.com/info/dmca.html X-Abuse-and-DMCA-Info: Please be sure to forward a copy of ALL headers X-Abuse-and-DMCA-Info: Otherwise we will be unable to process your complaint properly X-Postfilter: 1.3.40 Bytes: 18187 D wrote: > 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" > > +---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+ > | | | | | | | | | | - | | | | | | | | | | > +---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+ > | | | | | | | | | | | | | | | | | | | | | > +---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+ > | | | | | | | | | | | | | | | | | | | | | > +---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+ > | | | |<------------------------100 deg-------------------------->| | | | > +---+---+-x-----------------------------------------------------------x-+---+---+ > | | | \ | | | | | | | | | | | | | | | / | | | > +---+---|---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---|---+---+ > | | | | \ | | | | | | | | | | | | | / | | | | > +---+---|---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---|---+---+ > | | | | | \ | | | | | | | | | | | / | | | | | > +---+---|---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---|---+---+ > | | | | | | \ | | | | | | | | | / | | | | | | > +---+---|---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---|---+---+ > | | | | | | | \ | | | | | | | / | | | | | | | > +---+---|---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---|---+---+ > | | | | | | | | \ | | | | | / | | | | | | | | > +---+---|---+---+---+---+---+---+---+---.---+---+---+---+---+---+---+---|---+---+ > | | | | | | | | | \ | | | / | | | | | | | | | > +---+---|---+---+---+---+---+---+---+-------+---+---+---+---+---+---+---|---+---+ > | | | | | | | | | | \ | / | | | | | | | | | | > +---+---+---+---+---+---+---+---+---+---x---+---+---+---+---+---+---+---+---+---+ > | | | | | | | | | | / | \ | | | | | | | | | | > +---+---|---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---|---+---+ > | | | | | | | | | / | | | \ | | | | | | | | | > +---+---|---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---|---+---+ > | | | | | | | | / | | | | | \ | | | | | | | | > +---+---|---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---|---+---+ > | | | | | | | / | | | | | | | \ | | | | | | | > +---+---|---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---|---+---+ > | | | | | | / | | | | | | | | | \ | | | | | | > +---+---|---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---|---+---+ > | | | | | / | | | | | | | | | | | \ | | | | | > +---+---|---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---|---+---+ > | | | | / | | | | | | | | | | | | | \ | | | | > +---+---|---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---|---+---+ > | | | / | | | | | | | | | | | | | | | \ | | | > +---+---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 ========== REMAINDER OF ARTICLE TRUNCATED ==========