The values of the given distances in this picture are only approximate ratios / values!

- The ratio of the flag is officially defined to be 7:4 regardless of the designed flag ratio.
- the flag consists of three horicontal strips colored in green, white and red, which equally divide the height of the flag.
- the flag's width is made up of exactly 225 squares, which can be divided up into 3+11*(19+1)-1+3 squares, so that this defines eleven times the position of a "Allahu Akbar" pattern of width of 19 squares.
- Originally this 7x19 pattern fills out a certain fraction of the height of a (red and green) strip and so implicitly (distance of 7 squares is fixed) defined the flag's ratio to 1.7477...:1. Later this was effectively changed to exactly 1.75:1 with the consequence that the Tribar now occupies a bit more than one 7-th of a strip now.

Read it form top to bottom and from right to left! The upper two pictures show the construction of the basic 10-star Lower left shows the construction of the small cresent at the top Lower right shows the construction of the two larger cresents in the emblem |
Upper picture shows the construction of one half of the sword. Rotated by a quarter, the lower picture shows the precise placement and size of the Coat of Arms inside the flag. |

1) positive x-axis pointing right and positive y-axis pointing down. 2) circle with radius 1 given with center in (0,0), called the "unit-circle" Given a regular 10-star inscribing this circle. Its vertices are at Vi with Vi := ( sin(i*Pi/10), -cos(i*Pi/10) ) and i odd: V0 = ( 0 , -1 ) V1 = ( (sqrt(5)-1)/4 , -sqrt(10+2*sqrt(5))/4 ) V3 = ( (sqrt(5)+1)/4 , -sqrt(10-2*sqrt(5))/4 ) V4 = ( sqrt(10+2*sqrt(5))/4 , -(sqrt(5)-1)/4 ) V5 = ( 1 , 0 ) V7 = ( (sqrt(5)+1)/4 , sqrt(10-2*sqrt(5))/4 ) V8 = ( sqrt(10-2*sqrt(5))/4 , (sqrt(5)+1)/4 ) V9 = ( (sqrt(5)-1)/4 , sqrt(10+2*sqrt(5))/4 ) V10= ( 0, 1 ) V11= (-(sqrt(5)-1)/4 , sqrt(10+2*sqrt(5))/4 ) V13= (-(sqrt(5)+1)/4 , sqrt(10-2*sqrt(5))/4 ) V14= ( -sqrt(10+2*sqrt(5))/4 , (sqrt(5)-1)/4 ) V15= (-1 , 0 ) V17= (-(sqrt(5)+1)/4 , -sqrt(10-2*sqrt(5))/4 ) V19= (-(sqrt(5)-1)/4 , -sqrt(10+2*sqrt(5))/4 )

V15V3 cuts V17V9 in A A = ( -1/2, -sqrt(10-2*sqrt(5))*(5-sqrt(5))/40 ) V17V5 cuts V11V3 in A' A'= ( 1/2 , -sqrt(10-2*sqrt(5))*(5-sqrt(5))/40 ) Ay' = Ay = -sqrt(2-2/sqrt(5))*(sqrt(5)-1)/8 ~ -0.1624598481 Ax = -A'x AA' = RadiusA = 1 A is center and AA' is radius of a new circle * * cuts the unit-circle in D V1V13 cuts V3V15 in B B = (-3/4 + sqrt(5)/4 , (sqrt(10+2*sqrt(5))*(sqrt(5)-5) + sqrt(10-2*sqrt(5))*sqrt(5))/20 ) V9V17 cuts V11V19 in B ' B' = (1/4 - sqrt(5)/4 , (sqrt(10-2*sqrt(5))*(sqrt(5)-5) + sqrt(10+2*sqrt(5))*sqrt(5))/20 ) V3V11 cuts V7V15 in C' C' = ( sqrt(5)/2 - 1 , ( sqrt(10-2*sqrt(5))*(sqrt(5)-3)+sqrt(10+2*sqrt(5))*(sqrt(5)-1) ) / 8 ) C'x ~ 0.118033988 C'y ~ 0.3632712635 BB' cuts C'V17 in C C = ( (5-7*sqrt(5))/40 , ( sqrt(10+2*sqrt(5))*(3*sqrt(5)-5)+sqrt(10-2*sqrt(5))*(5*sqrt(5)-15) ) / 80 Cx ~ -0.2663118960 Cy ~ -0.03102707025 Dx = ((2*sqrt(5)-5)*sqrt(65+28*sqrt(5))-5)/20 Dy = ((sqrt(5)-5)*sqrt(10-2*sqrt(5))*(1-sqrt(65+28*sqrt(5))))/80 Dx ~ -0.5481496308 Dy ~ 0.8363802862 C is center and CD is radius of a new circle @ CD = RadiusC = sqrt(530-70*sqrt(5)+(220-100*sqrt(5))*sqrt(28*sqrt(5)+65))/20 CD ~ 0.9120460695 mirror D at the line AC to get the other intersection point D' D'x = (15*sqrt(5)-220 +(87*sqrt(5)-160)*sqrt(65+28*sqrt(5)))/620 D'y = sqrt(10-2*sqrt(5))*(163*sqrt(5)-355+(3*sqrt(5)-75)*sqrt(65+28*sqrt(5)))/2480 D'x ~ 0.3285435863 D'y ~ -0.7223844202 V9V17 cuts V5V13 in E E = ( -C'x , C'y ) Ex ~ -0.118033988 Ey ~ 0.3632712635 V3E cuts V5V15 in F F = ( sqrt(5)-2 , 0 ) Fx ~ 0.236067977 FV5 = RadiusF= 3-sqrt(5) F is the center and the distance FV5 is the radius of a new circle $ A'B' cuts V17C' in G V5V13 cuts horizontal (parallel to V5V15 ) through G in G' Gx = 5/22 - 2*sqrt(5)/11 Gy = ( sqrt(10-2*sqrt(5))*(sqrt(5)-15)+(7+sqrt(5))*sqrt(10+2*sqrt(5)) ) / 88 Gx ~ -0.1792850867 Gy ~ 0.05825325590 G'x = 27/22 - 2*sqrt(5)/11 G'y= Gy G'G = RadiusG = 1 G is the center and GG' is the radius of a new circle & F-circle and G-circle cuts in H & H' (Hx-Fx)^2+Hy^2=R3^2 (Hx-Gx)^2+(Hy-Gy)^2=1 Hx ~ 0.6183388412 Hy ~ 0.6614093320 H'x ~ 0.4216440014 H'y ~ -0.7410496960

V8V14 cuts V9V17 in J Jx = sqrt(10-2*sqrt(5))/4-1/2 Jy = (10*(1+sqrt(5))+(sqrt(5)-5)*sqrt(10-2*sqrt(5)))/40 Jx ~ 0.0877852522 Jy ~ 0.6465571461 V11V19 cuts V1V13 in K V1V9 cuts V7V19 in K' V3V15 cuts V7V19 in I Ix= sqrt(5)/2-1 ~ 0.118033988 Iy= -sqrt(10-2*sqrt(5))*(sqrt(5)-1)/8 Iy ~ -0.3632712640 KK' cuts vertical through I in L Lx = Ix; Ly = (sqrt(10-2*sqrt(5))*(5-sqrt(5))-sqrt(5)*sqrt(10+2*sqrt(5)))/20 Ly ~ -0.1004057078 IL' cuts V1V13 in M Mx = Ix; My = ((3*sqrt(5)-15)*sqrt(10+2*sqrt(5))+(3*sqrt(5)-5)*sqrt(10-2*sqrt(5)))/40 My ~ -0.6881909600 line from L to I ONLY above L (decrease y) until M line from L to J ONLY below L (increase y) until N V10V4 cuts LJ in N Nx ~ 0.07780924870 Ny ~ 0.8929047565 V9V17 cuts V7V15 in P P = (-3/4+sqrt(5)/4 , sqrt(10-2*sqrt(5))*(1+sqrt(5))/4/(5+sqrt(5)) ) V7V15 cuts V3E in P' IL cuts the unit-circle at Q RadiusQ = PP' ~ 0.1364082000 Q is the center and PP' is the radius of a new circle § Qx = Ix ~ 0.118033988 Qy = sqrt(4*sqrt(5)-5)/2 Qy ~ -0.9930095555 Sy = ( -44*sqrt(4*sqrt(5)-5)+sqrt(1319*sqrt(5)-2933)*sqrt(5+sqrt(5)) - sqrt(1319*sqrt(5)-2933)*sqrt(1+1/sqrt(5)) ) / 88 Sy ~ -0.9246343081 This circle intersects the unit-circle right from Q in T x^2+y^2 = 1 , (x-Qx)^2 + (y-Qy)^2 = R5^2 y = (Qx^2 + Qy^2 +1 - R5^2)/2/Qy - Qx/Qy*x 4*Qy^2*x^2 + (Qx^2 + Qy^2 +1 - R5^2 - 2*Qx*x)^2 = 4*Qy^2 Tx = (2317*sqrt(5)-4615+sqrt(221770*sqrt(5)+237175)*(3-sqrt(5)))/4840 Ty = -sqrt(18235720 +sqrt(5)*9222750 + (1130-1168*sqrt(5))*sqrt(237175+221770*sqrt(5)))*(3-sqrt(5))/4840 Tx ~ 0.2520750730 Ty ~ -0.9677076818 V0V4 cuts QI in R R is the center and RT is the radius of a new circle # RadiusR = RT ~ 0.1443102677 Rx = Ix ~ 0.118033988 Ry = (sqrt(10+2*sqrt(5))*(5*sqrt(5)-11)-8)/8 Ry ~ -0.914243288 S'y ~ -0.8312166012 T is the center and TU is the radius of a new circle % RadiusT = TU ~ 0.3099945325 this circle cuts off the upper line of the swords until the center-vertical in U Ux = 0; Uy ~ -0.7872782843

Considering the relative size and placement of the central emblem in the National flag: V1V5 cuts V0V10 at W' V5V9 cuts V0V10 at W W = ( 0 , sqrt(10+2*sqrt(5))/(5-sqrt(5)) ) W' = ( 0 , -sqrt(10+2*sqrt(5))/(5-sqrt(5)) ) ==> edge is at distance Wy ~ 1.37638192046 and thus this is the halve of one third of the flag's total height height of the flag ==> 3*(5+sqrt(5))*sqrt(10+2*sqrt(5))/10 For the precise height of the Takbir we need the distance between Z and Z': Z = ( (sqrt(5)-1)/4 , sqrt(10-2*sqrt(5))*(30-10*sqrt(5))/80 ) Zx ~ 0.3090169942 Zy ~ 0.2245139883 Z' = (Zx, -Zy) The distance ZZ' is the height of the Takbir, i.e. ZZ' = WV = W'V' in the picture. Mirroring the drawn halve of the central emblem and filling it with red color for the flag, we get together with the two Takbir strips:

7 squares = sqrt(10-2*sqrt(5))*(30-10*sqrt(5))/40 ~ 0.44902797658 ==> flag's width = 225 squares = 45*sqrt(10-2*sqrt(5))*(30-10*sqrt(5))/56 ~ 14.43304210434381Knowing the size of a square implies, the flag's width, which is made up of 225 squares, equals 45*sqrt(10-2*sqrt(5))*(30-10*sqrt(5))/56. Using the calculated relative flag-height we would get a flag-ratio of 75*(7*sqrt(5)-15)/28 = 1.7477031495496... = [1;1,2,1,26,2,5,1,15,...]. This irrational number has a periodic continued fraction whith period length 1660, starting after its very first coefficient. Comparing the officially defined 1.75 flag-ratio to its designed ratio, this means a distortion of about 0,13 %.

Remark: This page sprang of a private working paper which was not intended for publication. Sorry, for the plain layout.

First time calculated on 2012-06-18.

English translation (interpretation) of the official construction sheets Thanks to Juan Manuel Villascan.

Please send any comments or questions to webmaster.

Page created: 2012-06-19 12:00 UTC+2

Last updated: 2012-06-22 18:35 UTC+2

Last changes: 2012-06-24 14:20 UTC+2