Berechnung des Seitenverhältnisses der Nationalfahne von Nepal


Geometric / Algebraic Construction of the Flag

Let A= (0,0), B= (3,0)
=> AB = 3
=> C= (0,4), D= (0,3)
=> AD = 3, AC = 4
=> BE = 3
E= (3-3/Sqrt(2) , 3/Sqrt(2))

=> FG = 3
F= (0, 3/Sqrt(2))
=> AF = 3/Sqrt(2) , CF = 4- 3/Sqrt(2)
G= (3, 3/Sqrt(2))) AH = 3/4
H= (3/4 , 0)
I= (3/4, 3 + 3/4/Sqrt(2))

J= (0, 2 + 3/2/Sqrt(2))
K= (3/2, 2 + 3/2/Sqrt(2))
L= (3/4, 2 + 3/2/Sqrt(2))

M= (3/4 , 2 + 3/2/Sqrt(2) - 1/4*(4- 3/Sqrt(2))/2)

x= point where IH cuts DB = ( 3/4 , 3-3/4)
xM = -3/4 + 15/8/Sqrt(2)
MN = xM/Sqrt(2) = 15/16-3/4/Sqrt(2)
N = (3/4 , 9/16 + 21/8/Sqrt(2))
O = (0, 3/2 + 15/8/Sqrt(2))
LN= 23/16 - 9/8/Sqrt(2)

circle around L with radius LN
Q = (3/4 + Sqrt(609-366*Sqrt(2))/16 , 3/2 + 15/8/Sqrt(2))
P = (3/4 - Sqrt(609-366*Sqrt(2))/16 , 3/2 + 15/8/Sqrt(2))

circle around N with radius MN
both circles cut each others at R and S at height h
T = (3/4, h)
TS = MN*Sqrt(1-(MN/2/LN)^2)

consider the triangle (L,N,S) with lengths of sides  LN, LN, MN.

Bisecting it into two rectangle triangles 

we get:  NS^2 + NT^2 = MN^2  
and      NS^2 + LT^2 = LN^2

=>  MN^2-NT^2 = LN^2-LT^2
moreover we have LT+NT = LN , thus we can eliminate NT getting
MN^2-(LN-LT)^2 = LN^2-LT^2
or
MN^2-LN^2 +2*LT*LN = LN^2
or
LT =  (2*LN^2-MN^2)/(2*LN) = LN-MN^2/(2*LN)

now we can calculate h starting at the y-coordinate of L as

h = 2 + 3/2/Sqrt(2) - LT
  = 2 + 3/2/Sqrt(2) -LN+MN^2/2/LN
  = 9/16 + 21/8/Sqrt(2) + MN^2/LN/2
  = 9/16 + 21/8/Sqrt(2) + (15/16-3/4/Sqrt(2))^2/(23/16 - 9/8/Sqrt(2))/2
  
= 9/16 +21/8/Sqrt(2) + (225/256+9/32-90/64/Sqrt(2))/(23/16 - 9/8/Sqrt(2))/2
= 9/16 +21/8/Sqrt(2) + (297/512-45/64/Sqrt(2))/(23/16 - 9/8/Sqrt(2))
= 9/16 +21/8/Sqrt(2) + (297/512-45/64/Sqrt(2))*(23/16 + 9/8/Sqrt(2))/(529/256-81/128)
= 9/16 +21/8/Sqrt(2) + (297/512-45/64/Sqrt(2))*(23/16 + 9/8/Sqrt(2))/(367/256)
= 9/16 +21/8/Sqrt(2) + (297/32-45/4/Sqrt(2))*(23 + 18/Sqrt(2))/367
= 9/16 +21/8/Sqrt(2) + (297*23/32-45*23/4/Sqrt(2)+297*18/32/Sqrt(2)-45/4*18/2)/367
= 9/16 +21/8/Sqrt(2) + (6831/32-8280/32/Sqrt(2)+5346/32/Sqrt(2)-810/8)/367
= 9/16 +21/8/Sqrt(2) + (3591/32-2934/32/Sqrt(2))/367

= 10197/367/32 + 27894/367/32/Sqrt(2)  ~ 2.54777218616

and using again  LT+NT = LN  we get

TN = NT = MN^2/(2*LN) = (3591-2934/Sqrt(2))/32/367
TS = MN*Sqrt(1-(MN/2/LN)^2)
   = 3*Sqrt(2467 - 1476*Sqrt(2))*(-5+2*Sqrt(2)) / 32 / (-23+9*Sqrt(2))
   = (237-3*Sqrt(2))*Sqrt(2467-1476*Sqrt(2))/11744   ~ 0.38615562772035130798
Nationl flag of Nepal
AU = AF/2
W= (3/4, 3/2/Sqrt(2))
m := TN = NT = MN^2/(2*LN) ~ 0.1291168855584698
lower left corner angel at A = 90 = Pi/2
lower right corner angel d = 45 = Pi/4
middle inner corner angel c = -45 = -Pi/4
Pi/2 - upper left corner angle a = right outer corner angle b =: w
National Flag of Nepal used since at least December 1962
t := tan(w) = 4/3 - 1/Sqrt(2)
w = arctan(4/3 - 1/Sqrt(2))   ~ 0.1780882838*Pi

length of bottom line:  m+3+m+m*Sqrt(2)  ~ 3.44083262177

tan(w) = x/(m+m/sin(w)) = x/m/(1+Sqrt(1+tan^2(w))/tan(w))
x = m*(tan(w)+Sqrt(1+tan^2(w)))

x= m*(1+1/sin(w))*(4/3-1/Sqrt(2)) = m*(tan(w)+1/cos(w))
length of left vertical line:  m+4+x  ~ 4.3623180847787892670

z = m/tan(w/2) = m*(1+cos(w))/sin(w) = m*(1/sin(w)+1/tan(w))
= m*(Sqrt(1+tan^2(w))+1)/tan(w)
length of horicontal middle peak:  m+3+z   ~ 3.5785734947467747880

ratio resp. Verhältnis = (m+4+x)/(m+3+z)   ~ 1.219010337829452
  
=  (3591-2934/Sqrt(2))/367/32+4+(3591-2934/Sqrt(2))/367/32*(t+Sqrt(1+t^2))/
   ((3591-2934/Sqrt(2))/367/32+3+(3591-2934/Sqrt(2))/367/32*(1+Sqrt(1+t^2))/t)

=  (3591-2934/Sqrt(2))+4*367*32+(3591-2934/Sqrt(2))*(t+Sqrt(1+t^2))/
   ((3591-2934/Sqrt(2))+3*367*32+(3591-2934/Sqrt(2))*(1+Sqrt(1+t^2))/t)

1+t^2 = 1+(4/3-1/Sqrt(2))^2 = 1+ 16/9+1/2-8/3/Sqrt(2) = 32/18+27/18-8/3/Sqrt(2)
Sqrt(1+t^2) = Sqrt(59/18-8/3/Sqrt(2)) = Sqrt(59/2-24/Sqrt(2))/3

(m+4+x)= (Sqrt(118-48*Sqrt(2))*(1197-489*Sqrt(2))-10437*Sqrt(2)+113644)/23488

(m+3+z)= (Sqrt(118-48*Sqrt(2))*(19926-963*Sqrt(2))-73260*Sqrt(2)+1905414)/540224

(m+4+x)/(m+3+z)= 23*(Sqrt(118-48*Sqrt(2))*(1197-489*Sqrt(2))-10437*Sqrt(2)+113644)*(Sqrt(118-48*Sqrt(2))*(19926-963*Sqrt(2))+73260*Sqrt(2)-1905414)/(255505506840*Sqrt(2)-3590582103648)
 = (Sqrt(118-48*Sqrt(2))*(16146755136*Sqrt(2)-1560707136)+591939501840*Sqrt(2)-4924217880984)/(255505506840*Sqrt(2)-3590582103648)
 = (Sqrt(118-48*Sqrt(2))*(702032832*Sqrt(2)-67856832)+25736500080*Sqrt(2)-214096429608)/(11108935080*Sqrt(2)-156112265376)
= (Sqrt(118-48*Sqrt(2))*(29251368*Sqrt(2)-2827368)+1072354170*Sqrt(2)-8920684567)/(462872295*Sqrt(2)-6504677724)
= (6136891429688-306253616715*Sqrt(2)-Sqrt(118-48*Sqrt(2))*(934861968+20332617192*Sqrt(2)))/4506606337686

= 1.219010337829452184570024869930988566...

is the value of the sides-ratio of the circumscribing rectangle of the flag.
   r - s*Sqrt(2) - Sqrt(118-48*Sqrt(2)) * (u+v*Sqrt(2)) 
= ------------------------------------------------------
                           w

 with

r = 6136891429688
s = 306253616715
u = 934861968
v = 20332617192
w = 4506606337686


corner coordinates of the flag (A:B:C = 0:3:4) :

at lower left (near A): (-m,-m)
at upper left (near C): (-m,4+x)
at lower right (near B): (3+m+m*Sqrt(2),-m)
at middle inner (near E): (3-3/Sqrt(2)+m+m*Sqrt(2),3/Sqrt(2)-m)
at middle outer (near G): (3+z,3/Sqrt(2)-m)

with m = 9*(399-163*Sqrt(2))/11744
     x = m*(8-3*Sqrt(2)+Sqrt(118-48*Sqrt(2)))/6
     z = m*(3+Sqrt(59/2-24/Sqrt(2)))/(4-3/Sqrt(2))


the area of the flag is ~ 9.05889579867594   or
   A = (43590392214+7537121109*Sqrt(2)+(1176214419-509719833*Sqrt(2))*Sqrt(118-48*Sqrt(2)))/6344390656

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

First time calculated at 2012-04-13 18:38 UTC+2

References

Constitution of The Kingdom of Nepal, Article 5, Shedule 1 translated in english.
All geometric sheets for the Nepalesian Flag at FOTW Thanks to Sammy Kanadi.
Further background information on Nepal's Flag Thanks to
Please send any comments or questions to webmaster.
Page created: 2012-05-11
Last updated: 2012-06-08 20:23 UTC+2