Correct answer is Kashubian btw

this is not meant to be taken seriously, obviously this is insane, but I was just wondering - what if it was taken to an absolute extreme.

also do not ask me what any of the flags are, I made this ages ago

Blue representing the Jewish population, Green the Muslim population and a golden sapling representing peace. Something to the effect of "Levantine Republic" to avoid specific religious naming or symbols. Inspired by the Irish and Cypriot flags.

I was curious what the Colorado flag would look like with each letter of the alphabet.

Built this concept around a simple idea:

Before nationality, politics, religion, race, or borders — we are all human, we all bleed red and we all carry a heartbeat.

The heartbeat line represents humanity as one continuous system made up of every race and culture across Earth.

Curious what you guys think about the symbolism and overall concept.

What I think is the most likely outcome of a UK flag without Scotland.

Other versions of this i've seen have used super low quality rips of the design, and weird/bad colors. The blue is inspired by the UN, as well as the ocean.

It shows the ancient negative handprints our ancestors created tens of thousands of years ago. I think this is the only kind of flag that is universal and shows off all of humanity.

I was inspired by u/Cumohgc, however wanted it to slightly more traditional with the blue, I also believe its particularly good for a future interstellar/space faring humanity because the hand is outstretched reaching towards the sun/ a star.

Examining the symbolism of the pride flag.

Stick figure graphics from Michelle Dahlmanns via Wikimedia Commons, available under Creative Commons Attribution 4.0 International

I was wondering one day, "how many stars could the United States add to its flag (preserving the current star size) before it runs out of room?" Turns out, a lot! Solving this problem led to some interesting math, which I will share below.

First, I had to figure out the sizes of the stars in relation to the rest of the flag. Executive Order 10834 (1959) specifies these ratios with the length of the hoist (height) of the flag as unity.

Hoist (height) of the flag: A = 1

Fly (width) of the flag: B = 19/10

Hoist (height) of the canton: C = A * 7/13 = 7/13

Fly (width) of the canton: D = B × 2/5 = 19/25

Width of stripe: L = A/13 = 1/13

Diameter of star: K = L × 4/5 = 4/65

We'll ignore E, F, G, and H in the diagram in slide two of this post as the stars are going to get *very* close together in the new version...

We'll need the ratios of the measurements within the pentagram for our calculation later, so I'll use this helpful diagram from Wolfram MathWorld (slide two) to calculate the new values. Wolfram put things in relation to the length from one tip to another connected tip as unity, but we'll need to convert that to match the flag code's definition of the star in terms of the diameter of the inscribing circle (K). It's as simple as multiplying the measurements we'll need by K and ratio of 2*rho = 1 and 2a + b = 1. As it turns out, we'll only need a, b, y, rho, and R:

a = (sqrt(5-2*sqrt(5))/2) * K

b = ((sqrt((25-11*sqrt(5))/2)/2) * K

y = ((3*sqrt(5) - 5)/8) * K

rho = 1/2 * K

R = ((3-sqrt(5)/4) * K

We'll use these later when we calculate how many stars fit into the canton.

Next, I had to figure out the optimal way to arrange the most number of stars (pentagrams in math jargon) into the canton, a rectangle. This kind of puzzle is called a packing problem, which is a whole class of optimization problems in math that involves placing objects into containers as densely as possible (or to pack as many objects into a given container as possible, as this puzzle will frame it.)

I am no expert in this kind of problem, and, according to Wikipedia, the "packing of irregular objects," which a pentagram certainly is, "is a problem not lending itself well to closed form solutions." I figured that I would need to break the stars up into rectilinearly-tilable groups so I can fit them into a rectangular container (lol, autocorrect wanted “rectilinearly-tilable” to be "rectally-intelligible" 💀).

The optimal packing for a regular pentagon is called a "double-lattice packing," basically where stacks of pentagons alternate facing up and down. Pentagons and pentagrams share a lot of symmetries, so I figured their optimal packings must be related too. The third slide in the post shows a diagram of how I derived the star packing. Notice that the star in the top left corner touches three of its legs to the edges of the box—this is important because if the top row of stars was instead red (i.e. if only the top is touched by one leg of the stars), the stars take up more room for the same number of stars.

Now, when we go to fit these stars into the container that is the canton, we'll need to test both star-stacks going up-and-down and left-to-right. At this point, there is no way to tell whether orienting the stars vertically or horizontally will maximize the number of stars as it depends on the ratio of the height and width of the stars and the height and width of the canton. These values are essentially random and unrelated, so we’ll need to check both cases.

Ok, now we need to figure out the formula for the height and width of the star-stacks, which is not as simple as multiplying the height or width of one star by the number of stars in a stack. Because the stars are concave, they interlock in a way that makes it a little more difficult to calculate. Even more strangely, the formula for the height of a stack depends on whether there are an even or odd number of stars in the stack. There is another special case where it may be that the number of stars in every other column of stars differs by one. This occurs when there's not enough room for the bottom two legs of the bottom-most row of stars (see slide four).

Let's define some terms. I'm calling M the number of rows of stars and N the number of columns of stars. We'll want to divide both the canton height and width by both the height formula and the width formula and take the floor to determine M and N. Again, M depends on whether an even or odd number of stars fits, so we will just manually check at the end.

The height and width formulae are as follows (visually explained on slide four).

C or D = y + M(rho + R) + y(mod(M, 2))

D or C = N(a + b) + a

Therefore,

M = floor(([C or D] - [y or 2y])/(rho + R))

N = floor(([D or C] - a)/(a+b))

Then, to find the maximum number of stars, we find the maximum of either M(C)*N(D) or M(D)*N(C).

Ok, now we can just plug the values in for a, b, y, rho, and R as defined above, and we can get our answer. (Thank you Mathematica for dealing with all those sqrts!)

Let's start with the stars going left-to-right (i.e. the "tops" of the stars pointing left and right alternating in rows). Plugging in our values, we get N = 14 and M = 17, giving 238 stars.

Now, for the grand finale... Stars going top-to-bottom. Plugging in the values, we get N = 20 and M = 12... 240!! But wait! We need to check if that last row of stars *actually* fits (remember, there is the possibility of every other row of stars having 11 stars, bringing the count down from 240 to 230!) Basically, we need to find if C - (12(rho + R) + y >= y. Miraculously, that last row of stars fits with 0.0019 the height of the flag's room to spare! So, 240 > 238, and we have our final count.

I hope that was a bit of fun, but seriously, I do not support the Trump administration's latest neoimperialist proclivities. Ultimately, I intend this as a satire... There are around 200-ish countries in the entire world, so the joke is that the United States would have to take over every country to run out of room! I respect the sovereignty of Canada, Greenland, Venezuela, Cuba, Ir4n, and P4lestine and mean this flag as a bit of math fun and political critique. Thanks for reading!

Made this for Kerbal Space Program, lol.

It's not as good as the current flag, but I think it still looks decent.

yes thats the hungarian flag.

This is excluding the pawn.