There is no simple way, because the infinitesimal area is polar coordinates is
r dr dtheta where r is the radius and theta is the angle.
More intuitively, the farther you are from the center, the smaller the angle
that covers a fixed size pixel becomes.
All the best,
François
Le 3 sept. 2020 à 19:22 +0200, Roger Guay via use-livecode
<use-livecode@lists.runrev.com>, a écrit :
> Or to put it simply, how would one select random point (e.g. in a circle)
> using Polar Coordinates??
>
> Roger
>
> > On Sep 3, 2020, at 8:17 AM, Roger Guay via use-livecode
> > <use-livecode@lists.runrev.com> wrote:
> >
> > Jerry,
> >
> > You’ve done a very nice job of describing what’s actually(?) happening in
> > my code, but I think you missed the point of my question.
> > You agree that if you simply sample random pixels then the ratio of a
> > random pick inside the smaller circle will depend on the area of the
> > circles.
> > And, if I pick a random x and y within the concentric circles of radius R
> > and 2R, ¼ of the time they will lie in the smaller circle and ¾ of the time
> > in the bigger.
> > So, pick any random x and y and convert to radial coordinates. Everything
> > should work!
> > In my code I pick a random angle and a random radius (radial coordinates)
> > within the limits of the larger circle, thus picking random points within
> > the area of the larger circle, yet I get ½ (which you say is the right
> > answer).
> > My intent was to pick random points (using radial coordinates) for which
> > the result should be ¼!
> >
> > What’s wrong with my code?
> >
> > Thanks,
> >
> > Roger
> >
> > > On Sep 2, 2020, at 8:27 PM, Jerry Jensen via use-livecode
> > > <use-livecode@lists.runrev.com> wrote:
> > >
> > > 1/2 is the right answer.
> > >
> > > Take your drawing of the circles. Cut a verrrryy thin radial slice from
> > > the center to the outside circle. So thin that it is just a line.
> > >
> > > Now think of how likely a random point on that line will be in the part
> > > of the line that was in the smaller circle. The part that was from the
> > > smaller circle is HALF as long as the entire line.
> > >
> > > Now add up all the possible positions of that line. Why would that change
> > > the answer?
> > >
> > > Congratulations, you understand integrals!
> > > .Jerry
> > >
> > > > On Sep 2, 2020, at 7:38 PM, Roger Guay via use-livecode
> > > > <use-livecode@lists.runrev.com> wrote:
> > > >
> > > > Your chance to be Genius du Jour:
> > > >
> > > > If I construct 2 concentric circles, one being half the radius of the
> > > > larger, then simple math shows that the smaller circle has an area ¼
> > > > the area of the larger.
> > > > Now if I generate a random point within the radius of the larger
> > > > circle, I should expect that the probability of it landing in the
> > > > smaller circle to be ¼.
> > > > But, I must be doing something wrong because I get ½ !
> > > >
> > > > Here is my script:
> > > >
> > > > on mouseDown
> > > >
> > > > getStuff
> > > >
> > > > end mouseDown
> > > >
> > > >
> > > > local tR, tTheta, tX0, tY0, tX1, tY1, tTotCount, tL, tLongCount
> > > >
> > > > on getStuff
> > > >
> > > > put item 1 of the loc of grc OuterCircle into tx0
> > > >
> > > > put item 2 of the loc of grc OuterCircle into tY0
> > > >
> > > > put "" into tTotCount
> > > >
> > > > put "" into tLongCount
> > > >
> > > > emptyFlds
> > > >
> > > > end getStuff
> > > >
> > > >
> > > > on mouseUp
> > > >
> > > > lock screen
> > > >
> > > > repeat 1000
> > > >
> > > > put random(200) into tR -- 200 is half the width of the larger circle
> > > >
> > > > if tR > 1 then
> > > >
> > > > ## put random(2*pi) into tTheta1
> > > >
> > > > get random(360)
> > > >
> > > > put it*pi/180 into tTheta1
> > > >
> > > > put tR*cos(tTheta1) into tX1
> > > > put tR*sin(tTheta1) into tY1
> > > >
> > > > set the loc of grc Ptgrc to tX0 + tX1, tY0 - tY1 --- grc Ptgrc is a 2
> > > > pixle oval
> > > >
> > > > if intersect(grc Ptgrc, grc InnerCircle, "opaque Pixels") then add 1 to
> > > > tLongCount
> > > >
> > > > add 1 to tTotCount
> > > >
> > > > end if
> > > >
> > > > end repeat
> > > > put tTotCount into fld "totcountFld"
> > > >
> > > > put tLongCount into fld “LongCountFld"
> > > >
> > > > put tLongCount/tTotCount into fld "RatioFld"
> > > >
> > > > unlock screen
> > > >
> > > > end mouseUp
> > > >
> > > >
> > > > Apparently, this does not generate a random point within the larger
> > > > circle! Can someone please tell me what’s wrong here?
> > > >
> > > > Thanks,
> > > > Roger
> > > > _______________________________________________
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