Bouncing balls in a heptagon · mimo-v2.6-flash

Let me build the file. Plan: canvas, heptagon rotating at 2π/5 s, walls as line segments with velocity due to rotation. Ball-wall collision: compute closest point on segment, if overlap, resolve position and reflect velocity relative to wall point velocity, with restitution and friction.

Ball-ball: circle overlap resolution + impulse.

Balls numbered, drawn with rotation.

Let me write the code carefully.

Physics details:
- dt fixed timestep, e.g., substeps: dt = 1/60 with 4 substeps.
- gravity 900 px/s².
- Ball radius: heptagon circumradius R = 340 (canvas 900x700? heptagon needs fully visible). Canvas 1000x700 maybe. Let's use canvas 960x720, heptagon centered at (480,360), circumradius 320. Inradius = R cos(π/7) ≈ 320*cos(25.71°)=320*0.901=288. So balls must fit inside; inner space diameter ~576.

20 balls: choose radius 16 → area 20*π*256=16085, inradius region area of heptagon ≈ (7/2)R² sin(2π/7) = 3.5*102400*0.7818=280,000. Plenty of space. But initial compact arrangement: place in a grid near center. e.g., 5 columns x 4 rows spacing 34 (2*16+2). Width 5*34=170, height 4*34=136, fits within inradius.

Wall rotation: angular velocity ω = 2π/5 rad/s.

Ball-wall collision with moving wall: wall point velocity v_w = ω × r (perpendicular): vx = -ω*(py-cy), vy = ω*(px-cx).

Relative velocity: v_rel = v_ball - v_w. If penetrating and v_rel·n < 0 (n pointing inward from wall to ball... define n as normal from closest point to ball center, which points inward-ish), reflect: v_rel' = v_rel - (1+e)(v_rel·n)n, plus friction tangential damping: tangential component multiplied by (1 - μ). Then v_ball = v_w + v_rel'.

Positional correction: move ball along n to resolve overlap (with slop/split).

Actually for stability, also add positional projection: ball.pos += n * overlap * 0.8.

Ball-ball: standard impulse with restitution 0.6, positional correction 0.8 split by mass (equal mass).

Damping: global linear damping 0.999 per substep, angular damping 0.995.

Angular motion: ball angular velocity from friction in collisions (tangential relative velocity → torque), simpler: set angular velocity change from tangential impulse: Δω = -tangential_rel_v / r * factor. Also initial small random spin.

Number rotates with ball: draw number rotated by ball.angle.

Walls friction: apply tangential friction to relative velocity.

Also prevent unbounded speed: clamp speed to e.g., 1500 px/s.

Wall thickness: draw heptagon as polygon; collision segments on the polygon edges (the inner faces). To keep balls contained, use polygon vertices at circumradius R_inner maybe reduced slightly for drawing offset. Simplest: collision polygon = heptagon with circumradius R_col = 315, drawn polygon with line width 10 centered at R_draw=320... just draw the same polygon with thick stroke, stroke centered means inner edge at ~R - 5. Balls contained inside polygon edges means fine since we detect on segment; ball could visually overlap stroke slightly but ok. Use polygon vertices at R, stroke width 12 → inner face at R-6 along normal. Collision at R polygon plane means ball center can be up to R (distance) — actually ball constrained to inside of polygon edges; stroke straddles so ball overlaps stroke by ~6px visually. Acceptable, or set collision radius R-6. Let's use collision vertices at R_c = 314, draw polygon at same vertices with stroke width 12 (inner edge at 308, outer at 320). Ball collides at 314 → 6px inside of stroke inner edge... wait ball surface touches the segment line at 314 which is 6 px outside the stroke inner edge (308) i.e., within stroke. Hmm: stroke covers radius 308..320. Collision surface at 314 = middle of stroke. Fine, looks like ball touches the wall line's center. Good enough — draw collision polygon as the visual polygon.

Containment concern: fast-moving balls may tunnel. Use substeps (e.g., 8 substeps of dt/8) and continuous check? With speed clamp 1500 px/s, per substep at 60fps with 8 substeps: 1500/(60*8)=3.1px per substep < radius 16. Fine.

Heptagon vertices: angle offset so a flat or vertex at top — rotate by base angle.

Reset: restore positions, velocities=0, angles, wall angle=0, time=0.

Also balls shouldn't drift to overlap persistently: positional correction handles.

Initial arrangement "let them fall": start slightly above center? "near the enclosure center" — place grid centered slightly above center, say center y - 40, zero velocity, gravity pulls down.

Wait: heptagon rotating at 2π/5; with gravity, balls pile at bottom of rotating container. Fine.

Let's also handle ball-ball overlap in initial arrangement: spacing 34 with radius 16 → gap 2px. Good.

Draw: number in white/black, rotate ball.angle + text.

Also add Pause/Resume button, Reset button, maybe a status (angle). Keep it simple.

Sub-stepping loop: accumulate real time, fixed dt = 1/240, max steps per frame.

Let's write it.

Collision resolution detail for wall, per contact:
```
const c = closestPointOnSegment(p, a, b);
const dx = p.x - c.x, dy = p.y - c.y;
let d = Math.hypot(dx,dy);
if (d < R) {
  n = (dx,dy)/d (if d very small, use polygon inward normal)
  overlap = R - d
  wall point velocity: vw = ω × (c - center) = (-ω*(c.y-cy), ω*(c.x-cx))
  vrel = v - vw
  vn = vrel·n
  if (vn < 0) {
     // restitution with bounce
     const e = 0.55;
     vrel -= (1+e)*vn*n
     // friction tangential
     t = vrel - (vrel·n)n
     vrel -= friction * t  (friction 0.25)
  }
  v = vw + vrel
  // positional
  p += n * overlap  (maybe 1.0 for walls to prevent sinking, use 0.9)
}
```
Issue: pushing along n which is direction from wall point to ball — for a ball near a vertex, closest point at vertex, normal radial-ish. OK.

Also need to handle case ball outside polygon (e.g., due to rotation carrying it out through... rotating wall pushes ball; if penetration deep, closest point normal still works). But if ball center goes outside polygon beyond wall, d computed from closest point still pushes it away from segment, which would push it further out! Need normal inward: ensure n points toward polygon interior. Compute n = (p - c)/d; if dot(n, inwardNormalAtClosest) < 0, flip? Actually the correct resolution pushes the ball back inside. If ball center is outside the polygon, we want to push it inward, i.e., n should be the polygon interior direction. Simplest: after computing closest point, take n as from c to p; if the ball center lies outside the polygon (test point-in-polygon or dot with segment inward normal), flip n. Let's compute segment inward normal: for polygon vertices in CCW order... let's just define polygon vertices in clockwise order on screen? Simpler: compute polygon centroid (center cx,cy) and inward direction at closest point = normalize(center - c). If dot(n, inward) < 0 then n = -n? Hmm, that's not geometrically right generally but close enough: n from c to p; if p is outside polygon, then c to p points outward → flip to -n to push inward. Determine "p outside polygon" via point-in-polygon test (ray casting) — cheap enough (7 edges × 20 balls × 8 substeps = 1120 checks/frame, fine).

Actually if p is outside, correct resolution: push p inside along... the shortest way is along (c - p) direction = -n. So n = -n. And then vn relative: if vn<0 reflect. Since ball moving outward, vrel·(-n) >0 → no impulse? Hmm. Let's think: ball outside, normal pointing inward = -n. Ball velocity moving outward means vrel·n_inward = positive → separating, so no bounce impulse, but positional correction pushes it inside. That's acceptable (it's like tunneling recovery).

Edge case d≈0: n = inward normal from segment midpoint.

Also multiple walls contact: fine.

Speed clamp and also energy damping to prevent gaining speed from rotating walls (rotating wall can pump energy). Wall speed: ω*R = (2π/5)*320 ≈ 402 px/s — that's fast-ish. Balls pushed by wall get up to ~400 px/s plus gravity. Clamp at 1400. Damping 0.999/substep.

Ball-ball with equal mass:
```
n = (pb - pa)/d; overlap = ra+rb-d
positional: each moves overlap/2 * 0.8 along n
vrel = vb - va; vn = vrel·n
if vn < 0: j = -(1+e)*vn/2; va -= j*n... 
```
Standard: impulse magnitude J = -(1+e)*vn / (1/ma + 1/mb) with m=1 → J = -(1+e)*vn/2. va -= J*n*(1/ma)... let's: vrel = vb - va; if vrel·n<0 (approaching since n from a to b): impulse scalar j = -(1+e)*(vrel·n)/(1/ma+1/mb); va -= j/ma * n; vb += j/mb * n. With ma=mb=1: j = -(1+e)*vn/2, va -= j*n, vb += j*n. Check: vn negative → j positive → a moves -n (away from b), b moves +n. Correct.

Friction & spin for ball-ball: apply tangential friction: t = vrel - vn*n; jt = -|t|... simpler: relative tangential vt = vrel·t; impulse jt = -vt * friction / (1/ma+1/mb) clamp to |jt|<=|j|*μ... Simplify: apply jt = -vt/(2)*0.3 and add angular change: Δωa = -(jt * r) / I? For visible spin, use: angular velocity change from tangential impulse: ωa += (jt_cross) ... Let's do 2D: torque impulse about center for ball a: Δω_a = -(jt * r_a) / I_a where I = 0.5 m r². Actually tangential impulse jt applied at contact on a in direction t: torque = r_vec × impulse. Contact point on a is +n*ra direction from center a. torque_z = (ra*n_x)*(jt*t_y) - (ra*n_y)*(jt*t_x) = ra*jt*(n_x t_y - n_y t_x) = ra*jt*(cross(n,t)). For 2D cross(n,t)= n.x*t.y - n.y*t.x. Since t perpendicular to n, cross(n,t)= ±1. Δω = torque/I.

I'll implement with signs and verify visually (spin visible regardless of exact sign).

For wall friction similarly, apply angular change.

Simpler approach for spin that's robust: track contact tangential relative velocity and apply spin impulse. I'll implement carefully.

Let me define for a ball, at contact with normal n (pointing from surface toward ball center... let's be careful). For wall: n points from wall point to ball center (ball pushed along +n). Impulse on ball from wall: J = -(1+e)*vn*n + jt*t where vn = vrel·n (<0). The tangential contact velocity of ball relative to wall: vt = vrel·t. Friction opposes it: jt = -vt * μ_eff (with clamping to |jt| <= μ|Jn|). Torque on ball: contact point r_vec from ball center to contact = -n*ra. torque_z = r_vec.x * J.y - r_vec.y * J.x = (-ra n.x)(J.y) - (-ra n.y)(J.x) = ra (n.y J.x - n.x J.y) = -ra * cross(n, J) where cross(n,J)= n.x J.y - n.y J.x.

Δω = torque / I, I = 0.5 m r². With J tangential magnitude jt: cross(n, jt*t) = jt*(n.x t.y - n.y t.x).

I'll just code it and check that spin appears (visually numbers rotate). Not critical to exact sign.

Then update: v += J/m (J is impulse applied to ball). For wall: ball velocity gets J added (wall infinite mass, but we already accounted for wall velocity in vrel).

Let me define per-contact algorithm (applies to both wall and ball-ball, with proper masses):

resolveContact(ball, contactPoint, wallVel, n, isStatic...) — I'll write two functions.

Ball-ball:
```
n from a to b, contact at pa + n*ra
vrel = (vb + wb?) ... ignore surface spin for normal; include tangential relative surface velocity:
 surface point velocities: va_pt = va + ωa × (-n*ra) = va + (-ωa * (-n.y*ra)? )
 2D cross: ω × r = (-ω*r.y, ω*r.x). r_a_contact = -n*ra → spinVel_a = (-ωa*(-n.y*ra), ωa*(-n.x*ra)) = (ωa*n.y*ra, -ωa*n.x*ra)
 similarly r_b_contact = n*rb → spinVel_b = (-ωb*n.y*rb, ωb*n.x*rb)
 vrel = (vb + spinVel_b) - (va + spinVel_a)
 vn = vrel·n; if vn<0 apply normal impulse
 tangential: t = perpendicular of n; vt = vrel·t; jt = -vt * (2/ (1/ma+1/mb + ...)) — simplify: jt = -vt / (1/ma+1/mb) * friction, clamped by μ*|j|.
```
Effective inverse mass for tangential including rotation: 1/ma + 1/mb + (ra²/Ia + rb²/Ib) = 1/m + 1/m + 2 + 2 = 6 for m=1, I=0.5mr² → ra²/I = 2. So denominator = 1+1+2+2=6. jt = -vt/6 * μ... Actually full effective inverse mass k_t = 1/ma+1/mb+ra²/Ia+rb²/Ib = 6. jt = -vt / k_t, clamp |jt| <= μ|j|.

Normal effective mass k_n = 1/ma+1/mb = 2.

Then apply: va -= J/ma... J applied on a is -impulse_on_b. Let impulse on b = P (vector). P_n = j*n where j = -(1+e)*vn/k_n... Let's derive: vb_new - va_new along n = -(1+e) vn. With impulse P on b, -P on a: vn_new = vn + (P·n)(1/mb + 1/ma) = vn + p*k_n where p=P·n. Set = -(1+e)vn → p = -(1+e+1)*vn/k_n... wait: vn + p*k_n = -(1+e)*vn → p*k_n = -(2+e)*vn → p = -(2+e)*vn/k_n. Hmm standard: j = -(1+e)*vn / (1/ma+1/mb) applied with relative approach... Standard formula: j = -(1+e)*vrel·n / (1/ma+1/mb) where n from a to b and impulse on b = j*n, on a = -j*n. Check: vn_new = vn + j*n·(n/mb) ... change in vrel·n = j/mb + j/ma = j*k_n. So vn_new = vn + j*k_n = vn - (1+e)*vn = -e*vn. Yes correct. So j = -(1+e)*vn/k_n, with vn<0 gives j>0, b pushed along +n. Good.

Angular: P applied at contact on b at r = n*rb: Δω_b = (r × P)/I_b = (n.x*(j*n.y + jt*t.y) - n.y*(j*n.x+jt*t.x))/I = jt*(n.x t.y - n.y t.x)/I_b. Similarly on a: impulse -P at r = -n*ra: Δω_a = ((-n)×(-P))/I_a = (n×P)/I_a = same cross(n,P)/I_a. Note both same cross value. cross(n,t) where t = (-n.y, n.x): n.x*n.x - n.y*(-n.y) = 1. So t = perp(n) = (-ny, nx) gives cross=1.

So Δω = crossval * jt / I where crossval = 1 for this t. For a: Δω_a = jt/I_a (positive jt means... jt = impulse tangential on b; on a it's -jt tangential). Let's recompute a: P_a = -P_t = -jt*t. r_a = -n*ra. r × P_a = (-n.x)(-jt*t.y) - (-n.y)(-jt*t.x) = jt*(n.x t.y - n.y t.x)... careful: (-n.x)*(-jt t.y) = jt n.x t.y; (-n.y)*(-jt t.x) = +jt n.y t.x; subtract: jt n.x t.y - jt n.y t.x = jt * cross(n,t) = jt. So Δω_a = jt/I_a. And Δω_b: r_b = n*rb, P_b = jt*t: cross = n.x*jt t.y - n.y*jt t.x = jt. Δω_b = jt/I_b. Both same sign? Hmm that seems wrong but check physically: ball a's contact surface moves... let's test: two balls, a left, b right, n = (1,0), t = (0,1). Suppose a moving right faster than b: vrel = vb-va = (-u, 0), vn=-u<0 → normal impulse. Tangential: suppose a moving down relative? Take va=(0, -u) (a moving up? screen y down). Let's just take vt = vrel·t = vb.y - va.y (+spin terms). If a slides down relative to b (va.y > vb.y i.e., a moving +y, b still): vrel·t = vb.y - va.y = -u. jt = -vt/k_t > 0. Δω_a = jt/I > 0 → a spins clockwise? In screen coords with y down, positive ω rotates... we define ω such that velocity = (-ω ry, ω rx), positive ω = clockwise on screen? With y down, positive angle is clockwise visually, and our draw rotates by angle with canvas rotate (which is clockwise for positive in y-down). Point at (1,0) with ω>0 gets velocity (0, ω) → moving +y (down) → yes positive ω = clockwise on screen.

a moving down (+y) at contact with b below-right... a's bottom surface moving down relative to b → friction on a should oppose: push a's contact point up → a rotates counter-clockwise (negative ω). But we got positive. Let's recheck sign of jt application: I applied P_b = jt*t with jt = -vt/k_t... but friction impulse on b should oppose relative tangential motion of b relative to a: if b relative to a moves in -t direction (vt<0), friction on b pushes +t → jt>0 on b: b surface pushed down(+y)?? Wait t = (0,1) = +y (down). vt = vrel·t = vb.y - va.y = -u (b moving up relative to a). Friction on b: opposes b's relative motion → push b in +t (down): jt>0 on b, so Δω_b = jt/I >0 → b spins clockwise: b's contact point (left side, n=(1,0) so contact on b is at (-rb,0), left side) velocity from ω clockwise... spinVel_b = (-ω*n.y*rb, ω*n.x*rb) with n=(-1,0) for b's contact (r_b = n*rb = (-rb,0)): spinVel = (-ω*0, ω*(-1)) = (0, -ω rb). With ω>0, contact point moves up (-y). But friction pushed b's contact point down. Contradiction → sign error.

Let's redo: torque on b: r_b = (-rb, 0), P_b = jt*(0,1) = (0, jt). torque_z = r.x*P.y - r.y*P.x = (-rb)(jt) - 0 = -rb*jt. I earlier wrote cross(n,P) with n from a to b = (1,0): cross(n,P) = n.x P.y - n.y P.x = jt. But r_b = n*rb?? No! r_b = contact point relative to b's center = -n*rb (contact is on b's side facing a, which is at -n direction from b's center). I had it backwards. Let's fix: contact point on b = pb + n_from_b_to_contact... n points a→b, contact on b is at pb - n*rb. Contact on a is at pa + n*ra.

Redo: r_b = -n*rb, P_b = jt*t: torque = (-n.x*rb)(jt t.y) - (-n.y*rb)(jt t.x) = -rb*jt*(n.x t.y - n.y t.x) = -rb*jt. Δω_b = -rb*jt/I_b = -jt/(0.5 rb) (since I=0.5 m r² → rb/I = 2/(m rb)... units whatever). With jt>0 → Δω_b < 0 → counterclockwise: b's contact (left side) moves... ω<0: spinVel at r=(-rb,0): (-ω*0, ω*(-rb)) = (0, -ω rb) = (0, +|ω| rb) → down (+y). Friction pushed b's contact down ✓. 

r_a = +n*ra, P_a = -jt*t: torque = (n.x ra)(-jt t.y) - (n.y ra)(-jt t.x) = -ra*jt*cross(n,t) = -ra*jt. Δω_a = -jt ra/I_a <0?? Wait we need a to rotate: a's contact point at (ra, 0) right side, friction on a is -t (up): torque should make contact move up: spinVel = (-ω*0, ω*ra) = (0, ω ra); want negative (up) → ω<0 ✓. So Δω_a = -jt ra/I_a = also negative. Hmm both same sign: with t=(0,1), cross(n,t)= n.x t.y - n.y t.x = 1*1-0=1. Δω_a = (r_a × P_a)/I = (n*ra × (-jt t))/I = -ra jt cross(n,t)/I = -jt ra/I. Yes both negative.

Physical check: a moving down relative to b, friction drags a upward at contact (slows a's downward motion) and drags b downward. a: contact at right side pushed up → a rotates counterclockwise (ω<0) ✓. b: contact at left side pushed down → b rotates ... left side moving down = clockwise? Point at left (-r,0) moving down (+y): for rotation ω, v = (-ω y, ω x) = (0, ω*(-r)) → v_y = -ω r; want +y → ω<0 → counterclockwise. Hmm, counterclockwise: left point moves down? In standard CCW (math, y up), left point moves down-in-math = up on screen. Ugh, y-flip confusion. In screen coords (y down), ω<0 is counterclockwise visually. Left point under CCW (visually) moves down? Visually CCW: top→left→bottom→right. Left point moving CCW goes toward bottom → down ✓. And friction on b: b's left surface pushed down ✓ consistent. And a's right surface pushed up: a's right surface under CCW goes toward top ✓ (right→top is CCW). Consistent! Great, both ω decrease. OK formulas: Δω_a = -ra*jt*cross(n,t)/I_a... in general Δω = (r × P)/I computed properly. I'll just compute cross products numerically rather than simplify — safer.

So general code:
```
// ball a
ra_vec = {x: n.x*ra, y: n.y*ra}
Pa = {x: -P.x, y: -P.y}
wa += (ra_vec.x*Pa.y - ra_vec.y*Pa.x)/Ia
// ball b
rb_vec = {x: -n.x*rb, y: -n.y*rb}
Pb = P
wb += (rb_vec.x*Pb.y - rb_vec.y*Pb.x)/Ib
```
where P = j*n + jt*t applied to b.

For wall: treat as ball b with infinite mass: v_wall point velocity vw used in vrel; impulse P applied to ball only; ball's contact point r_a = n*ra (n from wall point to ball). Δω = (r × P)/I.

Normal impulse for wall: j = -(1+e)*vn / (1/ma) = -(1+e)*vn (since 1/ma=1). Wait derived: vn_new = vn + j*k_n where k_n includes ball's inverse mass = 1 → j = -(1+e)*vn. Good.

Tangential effective mass for wall: k_t = 1/ma + ra²/I = 1 + 2 = 3. jt = -vt/3 clamped to μ|j|.

Position correction: ball-ball: separate by overlap/2 each along n (mass weighted equal). Wall: move ball by overlap*n (full, maybe with n outward-corrected).

Now spin damping: angular velocity *= 0.995 each substep; linear *= 0.9995.

Let's also cap angular speed.

Numbers: draw with ctx.rotate(ball.angle).

Initial layout: 5 cols × 4 rows, spacing 2r+3=35, centered at (cx, cy-30).

Let's write the file now.

Buttons: Pause/Resume toggle, Reset. Also maybe show wall angle.

Canvas responsive? Fixed 960x720 with CSS scaling maybe. Keep fixed size, centered.

Heptagon fully visible: circumradius 314 + stroke 12 → max extent 320 from center; canvas 960x720 center (480,360): 360-320=40 margin bottom/top ✓, 480-320=160 ✓.

Ball radius 16, 20 balls. Space: area fine.

Gravity: 1100 px/s².

Let me write.

Drawing order: background, heptagon walls (stroke polygon), balls.

Also draw a subtle inner fill.

Rotation: wallAngle += ω*dt during substeps (even when paused? no — pause stops everything).

Reset: wallAngle = 0, balls back to initial, time to 0.

Point-in-polygon test for outward normal flip: implement.

Also handle the case where ball center is outside polygon (should be rare): also apply strong positional correction inward.

Closest point on segment function.

Let me also consider balls escaping at vertices when wall rotates fast: wall point speed 400px/s, ball near wall... substep dt=1/240 → wall point moves 1.7px per substep. Fine.

Write code.

Substeps: use fixed dt = 1/240, run up to some max (e.g., accumulate real delta clamped to 0.05).

Let me write the full HTML now.

Also verification: run browser, screenshot at a few moments, check console, click pause/reset, compare screenshots.

For comparing reset arrangement: take screenshot right after load (t≈0 before falling? it falls immediately). Maybe take screenshot quickly after reload vs after reset+pause. Better: reset then immediately screenshot (paused? reset resumes running). I could pause then reset? Requirement: Reset restores initial arrangement and wall orientation. To compare, I'll screenshot right after reset (fast) vs initial. To make comparison easy, maybe Reset also pauses? Not required; but I can make reset keep running state. I'll capture quickly.

Alternatively add a "hold" — simplest: in browser, click Reset then immediately screenshot; motion in first 100ms is tiny. Good.

Let's write the file.

Draw ball: circle with radial gradient? Keep flat colors with slight gradient, dark border, number in bold.

Ball colors: distinct hues via hsl(i*18, 70%, 55%).

Number: white text with dark shadow, font size ~ r*1.0.

Also draw a small marker dot on ball to visualize spin? The number rotating suffices.

Write code:

```js
const canvas=..., ctx=...;
const CX=480, CY=360, R=314, N=7;
const OMEGA = 2*Math.PI/5;
let wallAngle = 0;
function polyVerts(angle){ verts = for i in 0..6: a = angle + i*2π/7 + base; x=CX+R cos a ...}
```
Base offset: start with vertex at angle -π/2 (top). 

Gravity 1100. Restitution wall 0.5, ball-ball 0.6. Friction wall 0.35, ball-ball 0.3.

Clamp speed 1600, angular 30 rad/s.

Now write everything.

Also initial: give tiny random velocities? "let them fall under gravity" — start with v=0. Fine.

Status display: show t, wall angle, ball count maybe.

Let's write the file.

Let me design the simulation.

Canvas: 900x700 maybe. Heptagon centered, fully visible: circumradius ~ 260. Rotating at 1 rev / 5s → ω = 2π/5 ≈ 1.2566 rad/s.

20 balls, radius ~ 26. Initial arrangement: compact grid near center, non-overlapping: e.g., 5 columns × 4 rows spacing 2r+2 = 54. Width 5*54 = 270, might exceed heptagon inradius? inradius = R cos(π/7) = 260*cos(25.7°)=260*0.901=234. So inradius 234; a 270-wide box fits inside only near center... half width 135, half height 108 — distance to heptagon boundary fine at center. Actually the box must be inside: max distance from center to corner = sqrt(135²+108²)=173 < inradius 234? The heptagon interior contains all points within inradius from center only along the apothem direction; points at distance 173 in any direction are within inradius 234 so inside. Good.

Physics implementation:

Walls: heptagon edges as line segments rotating with angle θ. For each ball, for each edge: compute closest point on segment to ball center; if dist < r, resolve. Wall velocity at contact point: v_wall = ω × r_perp = ω * (-(py-cy), (px-cx)) i.e. tangential velocity. Relative velocity approach: compute normal n (pointing inward from wall to ball), penetration = r - dist. Correct position: push ball along n by penetration. Compute contact velocity of wall point, relative velocity vrel = v_ball - v_wall. If vrel·n < 0, reflect: vrel' = vrel - (1+e)(vrel·n)n, plus friction on tangential component: vt' = vt * (1 - friction). Then v_ball = v_wall + vrel'. Also apply angular: spin from tangential friction: Δω_ball = -tangential_rel / r * factor.

But careful: normal should point from wall into interior (toward ball). If ball is exactly at center of segment-line outside... Handle degenerate.

Also, ball might be pushed outside if deep. Use closest point approach with clamping, plus for robustness do multiple substeps.

Ball-ball: standard circle collision: positional correction (split overlap), impulse with restitution and friction, angular impulse.

Damping: linear damping per step (multiply velocity by 0.999 per substep?) plus restitution e.g. 0.6 ball-ball, 0.55 wall. Also gravity ~ 900 px/s².

Rotation of numbers: draw number rotated by ball.angle.

Also apply angular velocity decay and rolling friction between ball and wall: when ball contacts wall, tangential relative velocity reduces spin.

Containment: heptagon fully visible — canvas 900x700, center (450,350), circumradius 260 → bounding: vertices reach 260 in some direction; max extent of heptagon with vertex up: horizontal extent = R*cos(π/7)... let's just check all vertices within canvas: with circumradius 260, worst-case x extent ~ ±260*cos(small angle). Vertex angles: starting at -90° + k*360/7. cos values: for k=0: -90° → x=0, y=-260. k=1: -90+51.4=-38.6° → x=202, y=-162. k=2: 12.9° → x=253, y=59. k=3: 64.3° → x=112, y=234. So max |x| = 253+? vertices at ±: also k=... symmetric. Max |x| ≈ 253 < 450 ✓. Max |y| = 260 ✓ < 350. Plus line stroke width 6, fine. Balls contained inside, so all visible.

Ball radius 24, 20 balls. Grid 5×4 spacing 50 → 250×200 area, fits.

Initial arrangement near center, let fall: place with some margin above center? "compact, non-overlapping arrangement near enclosure center and let them fall" — place slightly above center maybe, zero velocity, gravity pulls down.

Stability: substeps — use fixed dt = 1/60 with e.g. 4 substeps per frame. Also speed clamp.

Rotating wall can push balls; with ω=1.2566 rad/s, wall point speed at r=260 is ~327 px/s — fine.

Potential issue: balls escaping through walls when wall moves fast — do position correction fully and clamp max speed (e.g., 1500 px/s).

Also need to check ball inside polygon after correction: if the ball's center is outside the polygon (pushed out), project it back. Simplest robust approach: for each edge compute signed distance d = n·(p - a) where n is inward normal; if d < r, push along n by (r - d). This handles infinite line; for segments, edges are convex polygon so infinite-line test works for interior points (as long as ball center remains inside polygon). Ball center outside the polygon → all edge tests... could fail. Add fallback: if center outside polygon (all/any signed distance negative significantly), pull back toward center.

Use signed distance approach per edge (7 edges) — cheap and robust for convex polygon containing the ball.

Wall velocity at the foot of perpendicular: contact point p_c = p - d*n... actually p_c = p - n*d (closest point on line). Wall velocity at p_c: v = ω × (p_c - center) = ω * (-(y), x) for (x,y)=p_c-center.

Ball-ball: iterate over pairs, positional correction with 0.8 factor split, impulse.

Angular motion: gravity gives no torque; spin comes from wall/ball friction. Also "give each ball angular motion" — initialize small random angular velocities, and friction generates spin on wall contact. Also could add a tiny rolling coupling: when touching wall, tangential friction drives spin toward rolling: ω → v_tangential / r.

Numbers rotate with ball: draw text rotated by ball.angle.

UI: Pause/Resume button, Reset button. Maybe show wall angle. Style nicely.

Reset: restore initial positions, velocities, angles, angular velocities, wall angle = 0, time = 0, and running state? Keep running (or keep current pause state). Reset should restore arrangement and wall orientation — do that and continue running if running.

Draw: heptagon with thick stroke, maybe also inner fill. Balls with radial gradient, number in center, and a marker dot near edge to show spin? Number itself rotating suffices; adding a small dot could enhance, but number rotated is required.

Let me write the code.

Key physics detail for moving wall bounce:

```
function collideWalls(b) {
  for each edge (a, e1):
    inward normal n (unit)
    d = dot(n, b.p - a)
    if (d < R) {
      // contact point on wall line
      const cx = b.p.x - n.x*d, cy = b.p.y - n.y*d;
      const rx = cx - cx0, ry = cy - cy0;
      const wvx = -omega*ry, wvy = omega*rx;
      let rvx = b.v.x - wvx, rvy = b.v.y - wvy;
      const vn = rvx*n.x + rvy*n.y;
      // positional correction
      b.p += n * (R - d)   (full, maybe 1.0)
      if (vn < 0) {
        // reflect normal
        rvx -= (1+e_wall)*vn*n.x; rvy -= (1+e_wall)*vn*n.y;
        // friction tangential
        tx = -n.y, ty = n.x;
        let vt = rvx*tx + rvy*ty;
        rvx -= friction*vt*tx; rvy -= friction*vt*ty;
        // spin: contact point velocity relative tangential change
        // torque from friction impulse J_t = -friction*vt*m
        // Δω = (r × J)/I ; r = contact - center = -n*d... 
      }
      b.v = wv_at_contact? no: b.v = rv + wv_contact
    }
}
```

Hmm but careful: wall velocity varies along wall; using contact point velocity is right.

For spin: contact vector from ball center to contact point: rvec = (cx - b.p.x, cy - b.p.y) = -n*d (pointing from ball center to wall). Friction impulse on ball: J = -friction*vt * t * m (m=1). Torque = rvec × J (z-component) = rvec.x*J.y - rvec.y*J.x. Δω = torque / I, I = 0.5 m R² for disc.

After positional correction, rvec changes but approximate fine.

Let's compute sign: Suppose ball resting at bottom of polygon (not rotating). n points up (0,-1) in screen coords (y down). Ball slides right (vx>0). Contact point is below ball, rvec = (0, +R) (down). t = (-n.y, n.x) = (1, 0). vt = rv·t = vx > 0. J = -friction*vt*t = (-f*vx, 0). Torque = rvec.x*J.y - rvec.y*J.x = 0 - R*(-f*vx) = R*f*vx > 0. Δω > 0 → positive ω in screen coords (y down) means clockwise rotation... A ball sliding right on the floor should spin clockwise (in normal orientation), which in y-down screen coords is positive? In screen coords (y down), positive angle rotation is clockwise visually. Rolling right = clockwise visually = positive ω. ✓ matches. Good.

Ball-ball impulse with friction and spin:

```
n = (pb - pa)/dist; overlap = Ra+Rb - dist
positional: pa -= n*overlap*0.5*corr; pb += n*overlap*0.5*corr
vrel = vb - va
vn = dot(vrel, n); if vn < 0:
  j = -(1+e)*vn / (1/ma + 1/mb)  (m=1 → /2)
  va -= j*n; vb += j*n  ... wait sign: impulse applied to b is +j*n along n; to a is -j*n.
  tangential friction: t = perp(n); vt = dot(vrel_after_normal, t)
  effective inverse mass for tangential with spin: 1/ma+1/mb + (ra²/Ia + rb²/Ib) = 1+1+ (R²/(0.5R²) each = 2) → 1+1+2+2 = 6
  jt = -vt / 6, clamp |jt| <= friction * |j|
  apply: va -= jt*t... impulse on b = +jt*t, on a = -jt*t
  spin: Δω_a = torque from (-jt*t) applied at contact point r_a = n*ra... contact point relative to a: r_a = n * ra (from a's center toward b). Torque_a = r_a × (-jt*t) = ra*n.x*(-jt*t.y) - ra*n.y*(-jt*t.x) = -jt*ra*(n.x*t.y - n.y*t.x). With t = (-n.y, n.x): n.x*t.y - n.y*t.x = n.x*n.x - n.y*(-n.y) = 1. So Torque_a = -jt*ra. Δω_a = -jt*ra/I_a.
  For b: r_b = -n*rb; force +jt*t: Torque_b = (-rb n.x)(jt t.y) - (-rb n.y)(jt t.x) = -jt*rb*(n.x t.y - n.y t.x)... compute: (-rb n.x)(jt t.y) - (-rb n.y)(jt t.x) = -jt rb n.x t.y + jt rb n.y t.x = -jt rb (n.x t.y - n.y t.x) = -jt*rb. Hmm same sign? Δω_b = -jt*rb/I_b.

Hmm, check with simple case: two balls, a left, b right, n = (1,0), t = (0,1) (pointing down in screen). Suppose a moving down relative to b: va.y = +5, vb.y=0 → vrel = vb - va = (0,-5); vt = dot(vrel,t) = -5. jt = -vt/6 = +0.83 (positive). Impulse on b = jt*t = (0, +0.83) → b pushed down. Impulse on a = (0,-0.83) → a pushed up. Correct (friction resists relative motion: a moves down relative to b, so friction drags b down and a up ✓).

Spin: Δω_a = -jt*ra/I = negative → counterclockwise (screen). Ball a sliding down past b, friction at its right side pulling up... contact point on a is at right side (r_a = (ra, 0)), force up (0,-0.83): torque = r × F = ra*(-0.83) - 0 = negative ✓ counterclockwise. Physically: a moving down, contact on right side feels upward friction → torque = x*Fy - y*Fx = ra*(-F) <0 → counterclockwise ✓. Good.

Δω_b = -jt*rb/I_b also negative: contact on b at left (r_b=(-rb,0)), force down (0, jt): torque = (-rb)*(jt) - 0 = -jt*rb < 0 ✓ same as formula. Good.

Wait both same sign — that's fine.

Wall friction spin formula: rvec = from ball center to contact = -n*d... but after correction, the ball center is at distance R from the wall line, so rvec = -n*R. Torque = rvec × J where J = -friction*vt*t... Let me recompute general: torque_z = rvec.x*J.y - rvec.y*J.x with rvec = -n*R: = (-n.x R)(J.y) - (-n.y R)(J.x) = R(-n.x J.y + n.y J.x) = -R * cross(n, J) where cross(n,J)=n.x J.y - n.y J.x.

With J = jt*t, jt = -friction*vt: cross(n,t) with t=(-n.y,n.x) = n.x*n.x - n.y*(-n.y) = 1. So torque = -R*jt = R*friction*vt. Verify earlier example: floor: n = (0,-1) up; ball sliding right vt>0; torque = R*f*vt >0 → clockwise ✓ matches earlier. Good.

I = 0.5*m*R² → Δω = torque / I = 2*friction*vt / R. Hmm for a single wall contact, full rolling friction would give Δω such that contact relative velocity → 0. Actually tangential relative velocity change: Δv_t = J_t/m = friction*vt (reduces vt by friction*vt — note friction ∈[0,1] acts as Coulomb-ish coefficient with implicit normal impulse; fine). And Δω changes contact point velocity: contact point tangential velocity v_cp_t = v_t + ω × rvec ... in 2D: contact point velocity = v + ω_perp*rvec where ω cross rvec = (-ω*rvec.y, ω*rvec.x). Tangential component with t: t·(ω*(-rvec.y, rvec.x)) = ω(-rvec.y t.x + rvec.x t.y) = ω * cross(rvec, t)... let rvec = -nR: = ω*(n.y... compute: rvec.x t.y - rvec.y t.x = (-n.x R)(n.x) - (-n.y R)(-n.y) = -R(n.x² + n.y²) = -R. So contact point velocity tangential = v_t - ω*R.

After applying friction: v_t' = v_t + jt = v_t - f*vt (with vt = current relative tangential = v_t - wv_t - ω*R... careful with wall tangential velocity too).

ω' = ω + Δω, Δω = -R*jt/I = -R*jt/(0.5 R²) = -2*jt/R (using torque = -R*jt from above... earlier torque = R*f*vt = -R*jt since jt = -f*vt). So Δω = -2jt/R.

New contact relative tangential: (v_t - f... ) let's just: rel_t' = v_t' - wv_t - ω'*R = (v_t - jt) - wv_t - (ω - 2jt/R)*R = rel_t - jt + 2jt = rel_t + jt = rel_t - f*rel_t = (1-f)*rel_t ✓. So with f=1 it fully sticks tangentially (rolling). Good; use f_wall ≈ 0.35 so contact slides but damps, plus spin decay.

Hmm but with f=0.35 each contact frame reduces relative tangential by 35%, that's strong damping — fine, keeps stability.

Also apply general linear damping: v *= (1 - 0.15*dt) and ω *= (1 - 0.8*dt) as air drag/spin decay.

Gravity: 800 px/s².

Ball radius 24, mass 1 (or π r² for scaling; uniform anyway).

Max speed clamp: 2000 px/s.

Timestep: use requestAnimationFrame with accumulator, fixed h = 1/240 s, up to some max steps per frame (e.g., 8 substeps → covers 1/30 s). Actually 1/240 with 4-8 steps per frame at 60fps = fine. Wall motion per step at h=1/240: point velocity 327 px/s * 1/240 = 1.4 px — good.

Collision iterations: within each substep, run wall + ball collisions maybe 2 iterations for positional stability.

Also need to handle the polygon rotating: recompute vertices each substep from θ.

Note: as polygon rotates, balls outside? Not expected.

Also to keep balls contained: after all corrections, if ball center outside polygon, clamp: find deepest edge penetration and push inside; plus pull toward center. Add safety: if outside, project point to polygon interior (move along inward normal of most-negative edge by penetration + R).

Edge case: signed distance for a ball near a vertex — using all 7 edge lines and pushing along each inward normal handles vertices okay-ish for convex polygons as long as center is inside.

Escape prevention from high-speed wall: substeps handle.

Initial arrangement: 5 cols × 4 rows, spacing 52 (2*24+4), starting x = cx - 2*52, y = cy - 1.5*52 - 40 (slightly above center so they fall). Actually "near the enclosure center and let them fall" — place centered slightly above center: rows y = cy-90..., let's compute: 4 rows spacing 52 → total height 156. Start y0 = cy - 78 - 30 = cy - 108. Width 4*52=208, x0 = cx-104.

Non-overlapping ✓ (spacing 52 > 48).

Check containment: max distance from center = sqrt(104² + 108² + r?) point at (104,108) from center, plus radius 24 → need inside heptagon: inradius 234 (R=260, cos(π/7)=0.9009 → 234). Corner distance 150 < 234-24=210 ✓.

Ordering: number ball i with label i+1: fill row by row 1..20.

Reset: θ=0, positions to initial, velocities 0, ω=0 (maybe tiny random spin initially? "give each ball angular motion" — during fall they get spin from contacts; give initial ω = small random to have visible spin at start? Better: initial ω random in [-1.5,1.5] so numbers visibly rotate as they fall. But reset must restore initial arrangement — angles too: set angle=0 initially? If ω random, angle 0 at t=0. I'll set initial ω small random and angle 0... but reset should restore identical view: set angle = 0, ω = small deterministic random (seeded) — simpler: use a fixed array of initial spins. I'll generate deterministic spins from index: ω0 = ((i*37)%11 - 5)*0.6.

Hmm, is random spin needed? Ball-ball and wall friction will generate spin during the fall anyway. But during first instants in free fall no spin. Requirement: "Give each ball angular motion and rotate its number with the ball so spin is visible." Deterministic initial spin makes it clearly visible. I'll include modest initial spin.

Pause/Resume and Reset buttons. Reset while paused keeps paused state (restores state). 

Draw heptagon: stroke polygon with lineWidth 8, color; also draw vertices dots maybe. Draw faint fill.

Also display small HUD: wall angle, time.

Let's also ensure canvas fits: use CSS max-width. Canvas 900x700.

Now write index.html.

Verification: run in browser, sample screenshots at moments, check console, click pause/reset, compare.

Also add optional: ball numbering font size ~20, white with contrast; ball colors varying hue.

Let me write the file.

Additional numeric stability: positional correction with slop? Simple: correct fully but that can inject energy with wall moving; use correction factor 1.0 for wall (needed for containment) and 0.6 for ball-ball.

Also after wall positional correction, we don't add velocity (only impulse when approaching). But rotating wall pushing into ball: positional correction moves ball; the normal impulse computed from relative velocity handles bounce. There's risk of energy pumping from position correction (moving wall drags ball) — that's physical-ish (wall does work), damped by restitution and friction. Should be fine with damping.

Also clamp: if ball ends up with huge speed from repeated wall impulses, cap speed to e.g. 1200.

One more: gravity plus rotation, balls settle at bottom of heptagon rotating container, swept around — nice.

Write code now.

Draw order: background, heptagon (fill + stroke), balls, HUD.

For number rotation: ctx.translate(ball.x, ball.y); ctx.rotate(ball.angle); draw circle (no rotation needed) then text rotated: draw text with rotate applied.

Actually draw circle then rotate then text. Also add a small dot/stripe to reinforce spin visualization? I'll add a small dot on the rim maybe — optional, keeps numbers as main indicator. I'll add a subtle radial marker (small circle near edge) — helps see spin for balls where number orientation ambiguous... number is unambiguous. Skip extra marker? A light highlight that doesn't rotate could conflict. I'll rotate everything including a small rim dot.

Code:

```js
const canvas=..., ctx=...
const W=900,H=700, CX=450, CY=350, R_POLY=260, SIDES=7;
const OMEGA = 2*Math.PI/5;
const G=800, BALL_R=24, N_BALLS=20;
const H_STEP=1/240;
const E_WALL=0.5, E_BALL=0.6, FRIC_WALL=0.35, FRIC_BALL=0.3;
const LIN_DAMP=0.12, ANG_DAMP=0.9;
```

state: theta, balls array, paused, time.

polyVerts(theta): angle0 = -Math.PI/2 + theta.

edge data: for i: a=v[i], b=v[(i+1)%7]; edge dir, normal inward: For CCW polygon in standard math coords... our polygon: vertices at angles -90° + k*51.4°, going counterclockwise in math sense (increasing angle) which on screen (y down) appears clockwise. Inward normal: n = perpendicular to (b-a) pointing toward center: compute mid = (a+b)/2, n = normalize(center - mid). Simple and robust ✓.

Wall collision per ball:
```
for each edge:
  n = normal (unit, toward center)
  d = dot(n, p - a)
  if d < R:
     penetration = R - d
     contact = p - n*d
     rel to center: rx=contact.x-CX, ry=contact.y-CY
     wv = (-thetaDot*ry, thetaDot*rx)
     // position
     p += n*penetration
     rv = v - wv
     vn = dot(rv,n)
     if vn<0:
        // normal impulse
        jn = -(1+E)*vn   (m=1, inverse mass 1)
        v += n*jn  (apply to v: v = rv + n*jn + wv)
        recompute rv
     tangential...
```
Let's structure: rv = v - wv (before). if vn<0: rv' = rv + jn*n with jn = -(1+E)*vn. Then t=(-n.y,n.x); vt = dot(rv',t); jt = -FRIC*vt; rv'' = rv' + jt*t. Then spin: Δω = -R*jt / I... wait earlier: torque = -R*jt (derived with rvec=-nR) → Δω = torque/I = -R*jt/(0.5*R²) = -2jt/R. Hmm let me double check that derivation sign with concrete floor example: floor bottom of screen, n=(0,-1) (up toward center). t = (-n.y, n.x) = (1, 0). Ball v=(5,0) moving right, wall static (wv maybe nonzero if rotating; use static). rv=(5,0), vt=5, jt=-FRIC*5 (negative). Impulse applied = jt*t = (-f*5, 0) → ball slows ✓. Δω = -2*jt/R = +2*f*5/R > 0 → clockwise ✓ (rolling right). Great: Δω = -2*jt/R.

But wait — should spin update use contact point based on d (actual distance) rather than R? After correction d=R. Fine.

Then v = rv'' + wv.

Hmm, careful: position correction pushes ball by penetration along n; also should I add the wall's positional movement? The wall edge moves between substeps; since we recompute geometry each substep, ball positions correction relative to current wall handles it implicitly (wall sweeping into ball → penetration → push out). The pushed-out direction imparts no velocity directly but subsequent impulses handle bounce. To account moving wall velocity better, we could add the wall's normal velocity contribution when there's penetration even if vn≥0? Actually vn already includes wall velocity (rv = v - wv). If wall moves into ball, rv·n > 0 means separating relative to wall... no: if wall pushes toward ball, wv has component along n (n points from wall to ball... n points inward toward center, ball is inside; wall moving inward means wv·n < 0 ... hmm wv is tangential for rotation! For a rotating infinite line... wait the wall's velocity at contact is tangential to... no. The heptagon edge line: its motion due to rotation at a point is tangential to the rotation, which generally has a component normal to the edge. wv = ω×r is perpendicular to r (radial vector), and the edge normal n is roughly radial-ish. So yes, wv·n ≠ 0 generally. Good — so rv·n<0 captures the wall slamming into the ball. 

Ball-ball loop: standard.

Order per substep: 
1. integrate velocity (gravity), damping
2. integrate position
3. resolve collisions (2 iterations: walls, then pairs; or iterate a few times)
4. spin decay

Actually do: apply gravity to v; then positions; then collision resolution (positions + impulses) ×2; then clamp.

Angle += ω*dt.

Clamp: speed max 1500, |ω| max 40.

Let's write the HTML now.

UI: buttons top, canvas, hint text. Style: dark background, nice gradient balls.

Let me write the file.

I need to finalize quickly. There are some issues in the code: a CSS typo `#2b3away`, `#2b3purpose` (I already added corrected lines after them, but the typo lines should be removed as they're invalid — browsers ignore invalid declarations, and I put corrected versions after, so it's fine, but better to clean). Also the pairCollide function has messy dead code and an unused variable `avx`, `maxJt`. Also the wall friction torque sign — let me not overthink; but check stability via browser quickly.

Let me quickly clean the obvious dead code and typos, then run browser check.

Wait — the friction relative velocity after the normal impulse: I recomputed rv using updated velocities (A/B velocities were updated by normal impulse), so that's fine. But note the spin velocity sign: contact point on A is rA = +n*R from A's center (toward B). Spin velocity at that point = w × r = (-w*r.y, w*r.x) = (-A.w*rAy, A.w*rAx). My code: (A.vx - A.w*rAy, A.vy + A.w*rAx). ✓. For B, rB = -n*R: spin vel = (-B.w*rBy, B.w*rBx) = (B.vx - B.w*rBy, B.vy + B.w*rBx). ✓.

Wall torque: contact point relative to ball center = -n*d ≈ -n*R. Friction impulse J = jt*t applied to ball. Torque = r × J where r = -n*R... wait r = contact - center = -n*R (d≈R after correction). Torque_z = r.x*J.y - r.y*J.x = (-n.x*R)*(jt*ty) - (-n.y*R)*(jt*tx) = -R*jt*(n.x*ty - n.y*tx). With t = (-ny, nx): n.x*ty - n.y*tx = n.x*nx - n.y*(-ny) = 1. So torque = -R*jt. Code: b.w += (-BALL_R*jt)/BALL_I ✓.

Now quickly run and verify in browser.

Syntax check via pipe failed for environment reasons. Let's just load in the browser.

Initial state looks good: 20 balls 1-20 in compact 5x4 grid, heptagon visible, wall rotating (5° at t=0.07s... wait OMEGA*0.07 = 0.088 rad = 5° ✓).

Now check a couple moments during fall/collisions, check spin (numbers rotated), wall rotation over time, pause/reset.

Simulation is running: balls fell, settled toward bottom-left, wall rotated to ~99°, numbers rotated (many are upside down — spin visible). All 20 balls contained, no visible overlaps. Numbers clearly rotated with the balls.

Wait, balls seem to be piled to the left/bottom, but the wall rotated 99° — physically plausible as the rotating floor sweeps balls. Containment holds.

Check at another moment for overlaps/speed, then test Pause and Reset. Let's pause, check time doesn't advance, then reset and compare with initial layout.

Paused works. All 20 balls contained, no persistent overlaps visible. Now click Reset while paused — then the arrangement should be instantly the initial grid with wall at 0°.

Reset restored: t=0.00, wall=0°, vertex straight up (same orientation as initial screenshot), 5×4 grid identical to initial. 

Now resume and let it run a bit longer to verify no unbounded speed/escape over a longer period. Let's click Resume, wait ~4s, screenshot and check console.

Resumed from reset — falling starts again, matches initial view. Now let it run ~6s more and screenshot to check long-run stability (containment, no explosion).

At t=12.33s, wall=168° (12.33/5*360 = 887.7° → mod 360 = 167.7° ✓ correct rotation rate). All 20 balls contained, no visible persistent overlaps, numbers rotated at various angles (spin visible). No console errors.

Everything verified. One more quick check: rotation rate exactness confirmed by math above. Save final file (already saved). Maybe quickly verify ball count = 20 in the frame — count balls in screenshot: 7,8,16,14,13,19,2,3,5,12,1,17,11,15,4,14?... visible ones: 7,8,16,14,13,19,2,3,5,12,1,17,11,15,4,12?,6,18,10,9 — that's 20. Yes appears all 20 present.

Done. Report.