![]() ![]() Separating the components, resolving dist and squaring both sides, we have: I just realized that there are 101 circles. Tile this 4 times, and you see that yes, 100 half-radius circles can cover the whole rectangle. Let coxy = oldxy - cornerxy, for brevity. Heres the trick: If 25 circles can cover the full rectangle, then 25 circles of half-radius can cover the half-scale rectangle (i.e. We also want t to be the lowest possible value, as long as it is between 0 and 1, so pick up the lowest solution that is between 0 and 1. That is, solve for t in this equation: dist(oldxy + t * vxy, cornerxy) = circle.rad. Best fit strategy The original algorithm does not produce the smallest rectangle to hold all the circles (it simply tries to fit a bunch of circles into a given rectangle). We have a point, oldxy, and a vector, vxy = xy - oldxy we want to calculate the fraction of this vector, t, such that oldxy + t * (xy - oldxy) results in the circle touching the corner. Each of these triangles that this squares divided into, these are all equal. The algorithm simply starts with the four bounding circles and adds one circle at a time, using the greedy heuristic lambda parameter to pick the 'best' location. So, all that is left is knowing how to calculate cx and cy (the centre of the circle as it collides with the corner). This is pseudocode, not code you need to separate xy into circle.x and circle.y, oldxy into circle.oldx and circle.oldy, cxy into cx and cy, and tgxy into tgx and tgy, and do the operations separately for each component. In vector form, If cxy is the position of the centre of the circle at the instant of the collision, oldxy is the old position, xy is the new position, and tgxy is the normalized tangent vector, the formula for finding the new, bounced oldxy and xy points is:Ĭode: Select all oldxy = cxy + tgxy * dot(oldxy - cxy, tgxy) This tangent needs to be normalized before using the reflection formula if done correctly, its length is the radius of the circle, so just divide it by the radius. exchange x and y and then put a minus sign in front of one of them for example: tgx, tgy = ny, -nx). For the reflection formula, you actually need the tangent vector, which is the normal rotated 90° in any direction (i.e. The crossing line is the line that you need to bounce against, and n is the normal vector of the collision. The red cross represents the centre of the circle at the time of the collision. That's the first point along the path from old to new, where the distance from said point to the collided corner equals the radius of the circle. You first need the position of the centre of the circle at the time of the collision. Two circles are inscribed such that they touch each other (circles are adjacent to each other) and also their circumferences touch 2 sides of the rectangle. Couldn't we just use velocity values? Thank you for your interest.Ĭode: Select all circle.oldx = circle.oldx + 2 * (rect.x + rect.w + circle.rad - circle.oldx)Ĭircle.x = circle.oldx + 2 * (rect.x + rect.w + circle.rad - circle.x)īouncing against a corner is a lot trickier. It looks like oldx and oldy variables are used like velocity properties. I also had some understanding issues with Verlet Integration. Graphics.circle("fill", circle.x, circle.y, circle.rad) Graphics.rectangle("fill", rect.x, rect.y, rect.w, rect.h) All Math Words Encyclopedia - How to Circumscribe a Circle About a Rectangle: How to circumscribe a circle about a rectangle using a compass and a straight. ![]() If circle.x + circle.rad > graphics.getWidth() thenĬircle.x = graphics.getWidth() - circle.radĮlseif circle.x graphics.getHeight() thenĬircle.y = graphics.getHeight() - circle.rad Vx, vy = (circle.x - circle.oldx) * friction, (circle.y - circle.oldy) * frictionĬircle.oldx, circle.oldy = circle.x, circle.yĬircle.x, circle.y = circle.x + vx, circle.y + vy + gravity circle.x, circle.y = mouse.getPosition() My whole Lua code looks like this (both intersection codes included): Number of circles ( n)Ģ + 2 2 + 6 2 ĭense packings of circles in non-square rectangles have also been the subject of investigations.Local distance = math.sqrt(distX ^ 2 + distY ^ 2) The obvious square packing is optimal for 1, 4, 9, 16, 25, and 36 circles (the six smallest square numbers), but ceases to be optimal for larger squares from 49 onwards. Solutions (not necessarily optimal) have been computed for every N ≤ 10,000. To convert between these two formulations of the problem, the square side for unit circles will be L = 2 + 2 / d n. Equivalently, the problem is to arrange n points in a unit square aiming to get the greatest minimal separation, d n, between points. ![]() The formula for circumference is C 2 r The formula for area of a circle is A r 2 Remember, that we approximate with 3. Circle packing in a square is a packing problem in recreational mathematics, where the aim is to pack n unit circles into the smallest possible square. Circumference is the perimeter of a circle. ![]()
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