1. The maximum length of the flag is twice its width; the minimum length is twice the altitude of the equilateral triangle.
2. Any side of the equilateral triangle is as long as the width of the flag.
3. (See accompanying illustration.) Solid golden sunburst without any markings―Sun with eight rays, equally spaced; Arc x with Sun ray = Free arc y; two opposite rays in horizontal axis and two in vertical axis; sun’s diameter D=w/5; each ray has one major beam, twice as broad as the minor beam on either side; length of major beam R=5/9 D; length of minor beam r = 4/5 R.
4. Three five-pointed golden stars of equal size, each star with one point directed to the vertex of the angle enclosing it; diameter of circumscribed circle of each star = 5/9 D diameter of inscribed circle of each star = 2/9 D; distance from each corner = D/2.
— relevant portion of text from Executive Order No. 23
Note: The text in the executive order does not provide the location for the centre of the sun, but the drawing provides enough information to calculate it.
Note: The government created another specification drawing in 1955; some of the listed dimensions appear to conflict with the 1936 executive order. It is assumed that the 1936 executive order takes precedence over the 1955 drawing.
For the purpose of these notes, a unit corresponds to the units used in this construction sheet. One unit is 1/90 the width of the flag or 1/180 the length of the flag.
the sun
Diameter of the central disk: D = 90/5 = 18 units.
Length of major rays: R = (5/9) × D = 10 units. Distance between two opposite tips is R+D+R = 10+18+10=38 units.
Length of minor rays: r = (4/5) × R = 8 units. Distance between two opposite minor tips is r+D+r = 8+18+8=34 units.
Distance from left edge of flag to left most ray on sun is D/2 = 9 units as shown on the 1936 drawing. Therefore distance from left edge of flag to sun center is D/2 + R + D/2 = 28 units.
It is apparent that the designers of the flag wanted the rays of the sun to appear as though they were clipped by two squares (or a square and a diamond... or an 8/2 regular star polygon). However, if the flag is drawn to spec, and one superimposes a square by connecting the tips of four opposite major rays, then the minor rays will protrude 0.0519779424 units outside of the square. This is about 0.06% of the width the flag and not generally noticeable to the naked eye. This construction sheet has a pair of red dashed squares connecting the major rays and a set of green dashed lines connecting the minor rays. It is necessary to zoom in to a high magnification level to see that these are actually distinct lines.
the stars
The three stars are all the same size. They are not 5/2 regular star polygons (pentagrams) and therefore they must be drawn using all 10 points.
Inner diameter of each star: (2/9) × D = 4 units
Outer diameter of each star: (5/9) × D = 10 units.
Each star is positioned such that one point is directed to the closest vertex of the white triangle.
Distance between each triangle vertex and the tip of the closest star: D/2 = 9. Therefore the distance from triangle vertex to star center is 9 + (10/2) = 14 units.
Lisanslama
Public domainPublic domainfalsefalse
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Bu görüntüde bir bayrak, arma, mühür veya başka bir resmi nişan gösterilmektedir. Bu sembollerin kullanımı bir çok ülkede kısıtlanmıştır. Bu kısıtlamalar telif hakkı durumundan bağımsızdır.
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