Top #circlechallenge dweets

function u(t) {

c.width=1920 // clear the canvas
for(i=0;i<9;i++)
x.fillRect(400+i*100+S(t)*300,400,50,200) // draw 50x200 rects

} // 122/140

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    u(t) is called 60 times per second.
    t: Elapsed time in seconds.
    S: Shorthand for Math.sin.
    C: Shorthand for Math.cos.
    T: Shorthand for Math.tan.
    R: Function that generates rgba-strings, usage ex.: R(255, 255, 255, 0.5)
    c: A 1920x1080 canvas.
    x: A 2D context for that canvas.

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Current theme challenge: #ScaryMovies

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donbright

function u(t) {

限=199
for(m=-限;m<限;m+=(2+t))for(n=-限;n<限;n+=(3+t)){
g=2*m*n
b=m*m+n*n
r=m*m-n*n
x.fillRect(1920/2+frame*g/b,1080/2+frame*r/b,1,1)}

Nov 2 2021 3:12 PM UTC

}// 130/140

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As part of #circlechallenge I go back to the old rational parametrization of the circle from Wildberger's "chromogeometry", which has elements of Euclid. Each frame draws rational points on a circle, and each frame the radius gets bigger. All of the patterns are a kind of optical illusion.

u/donbright

changing '2' and '3' to different numbers will give different patterns

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rep_movsd

function u(t) {

t||(M=O=.018,N=P=.9998,r=540)
if(t&&t<3.1)with(x){for(i=200,K=t*r,beginPath();--i;lineTo(K*B+r,K*A+r))A=O*N+P*M,B=P*N-O*M,O=A,P=B;stroke()}

May 15 2021 10:22 AM UTC

}// 139/140

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Look Ma - no trig!

u/rep_movsd

#Circlechallenge

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Xen

remix of d/18454 by u/ersagun

function u(t) {

t?x.rotate(!x.fillRect(0,0,5,399)):x.translate(960,540)

Apr 23 2020 2:37 PM UTC

}// 55/140

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#circlechallenge more Minimal

u/ersagun

just another way to exploit irrational numbers
u/Xen

Reality is chaos, Chaos is a fractal, All mostly infinite if you ask me

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New_Core

function u(t) {

x.font='6cm a'
for(i=7-~~t%8;i--;)x.beginPath(),x.setTransform(C(a=i/1.1+t)*2,S(a)*2,S(a)*2,-C(a)*2,960,540),x.fillText("🍕",-40,-30)

Apr 22 2020 2:09 PM UTC

}// 132/140

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the best circles are served warm #circlechallenge

u/donbright

lol

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KilledByAPixel

remix of d/18368 by u/veubeke

function u(t) {

x.strokeRect(960-(W=C(t)*300),540-(H=S(t)*300),2*W,2*H)

Apr 21 2020 7:54 AM UTC

}// 55/140

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u/sigveseb

#circlechallenge

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BriSeven

remix of d/18368 by u/veubeke

function u(t) {

x.fillStyle="#0001"
x.fillRect(960,540,C(100*t|0)*300,S(100*t|0)*300)

Apr 21 2020 7:17 AM UTC

}// 69/140

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u/sigveseb

#circlechallenge

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magna

remix of d/24304 by u/cantelope

function u(t) {

x.fillText(x.font='20in ●',-4e3,1e3)

Nov 11 2021 6:10 AM UTC

}// 36/140

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#circlechallenge

u/rep_movsd

5e3

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katkip

function u(t) {

x.font=`420cm"`+x.fillText(`.`,-264,944)

Apr 24 2020 10:23 AM UTC

}// 40/140

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#circlechallenge

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iznax

function u(t) {

for(r=240,X=1e3,t?0:c.width=X;X--;x.fillRect(X,Y,1,(X-c.width/2)**2+(Y-562/2)**2>r*r))Y=t*60

Apr 22 2020 5:04 AM UTC

}// 92/140

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Inverse #circlechallenge

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DaSpider

function u(t) {

A=Array(s=2e3).fill().map(Math.random)
for(i=s;i--;(a-.5)**2+(b-.27)**2>.05?0:x.fillRect(a*s,b*s,5,5))a=A[i],b=A[s-i]

Apr 21 2020 11:42 PM UTC

}// 117/140

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#circlechallenge Monte Carlo Method

u/DaSpider

Using only good old Pythagoras. This method can be used to estimate π ≈ 4*(number of points in the circle / total number of points) // t?1:d=g=0;A=Array(s=6e3).fill().map(Math.random);for(i=s;i--;(a**2+b**2)**.5<=1?d++:1)a=A[i],b=A[s-i];g+=s;throw(4*d/g)

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