Bouncing balls in a heptagon · laguna-s-2.1

Let me analyze this task. I need to build a 2D physics simulation of numbered balls inside a rotating heptagon. Let me break down the requirements:

1. 20 balls numbered 1-20, placed in compact non-overlapping arrangement near center, falling under gravity
2. Rotating heptagon - one revolution every 5 seconds, fully visible
3. Ball-ball and ball-wall collision detection implemented myself
4. Ball-wall collision should account for moving-wall velocity in bounce response
5. Damping/friction
6. Each ball has angular motion, numbers rotate with balls
7. Reasonable sizes and physical constants for stability
8. Pause/Resume and Reset buttons
9. Reset restores initial ball arrangement and wall orientation

Let me think about the physics:

**Heptagon**: A 7-sided polygon. Center at some point, radius R. Rotating at constant angular velocity ω = 2π/5 rad/s.

**Walls of heptagon**: Each wall is a line segment. For a heptagon centered at origin with radius R (distance from center to vertices), the walls are the edges between consecutive vertices.

**Ball-wall collision**: Need to detect when a ball penetrates a wall and push it out, and reflect velocity considering the wall's movement at the contact point.

**Ball-ball collision**: Detect overlap, push apart, reflect velocities with some restitution and damping.

Let me think about the coordinate system. I'll use canvas coordinates with y pointing down for rendering, but use y-up physics or handle sign carefully.

Let me set up:
- Canvas centered, heptagon centered at canvas center
- Gravity pointing down (in canvas, increasing y)
- Heptagon rotates with angular velocity

For the heptagon walls in world space: vertices are at (cx + R*cos(θ + 2πk/7), cy + R*sin(θ + 2πk/7)) for k=0..6, where θ is the current rotation angle.

Each wall is the edge between vertex k and vertex k+1.

The inward normal of each wall points toward the center of the heptagon.

For ball-wall collision: For each wall, compute distance from ball center to wall line. If ball center is outside the heptagon (penetrating the wall), push it back. Actually, the ball should stay inside, so we check if the ball center is within R_eff = R - ballRadius of the center, but that's only for circular approximation. For polygonal walls, we need distance to each wall.

Let me compute the distance from a point to each wall line. The wall line passes through two vertices. The normal pointing inward (toward center) can be computed. The signed distance from center = dot(point - vertex, normal). If this distance is less than ballRadius (i.e., ball center is within ballRadius of the wall, meaning it has penetrated past the wall inward... wait, need to think about sign).

Let me define: normal points inward. signed distance d = dot(ballCenter - wallPoint, normal). If d < ballRadius, the ball is too close to or past the wall (penetrating), and we need to push it inward by (ballRadius - d).

The wall velocity at contact point: The wall is part of a rigid rotating body. At a point P on the wall, velocity = ω × (P - center) where ω is the angular velocity vector. In 2D, if rotating at angular speed ω (scalar), the velocity at point P relative to center C is: v = ω * (-(P.y - C.y), P.x - C.x) (for counterclockwise rotation). Actually the tangential velocity = ω × r where r is the vector from center. In 2D with ω out of page, v = ω * (-r_y, r_x).

So at the contact point (projection of ball center onto wall), the wall velocity is ω × (contactPoint - heptagonCenter).

For the bounce response: relative velocity of ball w.r.t. wall at contact. Normal component: if ball is moving into wall (relative normal velocity > 0 into wall... need sign convention), reflect with restitution.

Let me define normal pointing inward. The ball's velocity along normal: v_n = dot(ballVel, normal). The wall velocity along normal: w_n = dot(wallVel, normal). If the ball is moving toward the wall in the inward direction faster than the wall... hmm, this is getting complex. Let me think clearly.

The condition for collision: ball center within ballRadius of wall (d < ballRadius). 

Relative velocity at contact point along normal: the ball's velocity component along normal vs wall's velocity along normal. Let v_rel_n = dot(ballVel, normal) - dot(wallVel, normal).

If v_rel_n < 0, the ball is moving away from the wall (in the inward direction, meaning toward interior). Wait, normal points inward. If ball moves inward, dot(ballVel, normal) > 0. Hmm, I think I'm confusing myself.

Let me re-set: Let me use a clearer convention. Let the wall have an outward normal (pointing outside the heptagon, away from center). Actually, let me think about what matters.

Standard approach: The wall line. Normal pointing into the polygon (inward). d = signed distance to the line, positive inside. Collision when d < r (ball radius), ball penetrates.

Penetration depth = r - d (positive when overlapping).

To resolve: move ball center along inward normal by penetration depth.

For velocity: The wall at contact point moves with velocity v_wall. The component of v_wall along the inward normal. The ball's velocity along inward normal is v_ball_n = dot(ballVel, n_in).

For the ball to stay inside, it should not be moving inward into the wall beyond the wall's motion. The relative velocity into the wall: if the wall is moving inward (into the polygon), it can "push" the ball. 

Relative velocity along inward normal: v_rel = v_ball_n - v_wall_n. If v_rel < 0, the ball is moving toward the wall faster than the wall moves away (i.e., ball is penetrating the wall from inside = ball heading to exit). Hmm, no.

OK here's the thing: the inward normal points into the polygon (interior direction). A ball at the wall wants to stay inside, so its position is constrained to have d >= r. If the ball is moving so that d decreases (moving toward the wall), dot(ballVel, n_outward) would be... ugh, let me just pick outward normal for clarity in the collision logic, since it's about "leaving the polygon."

Let me use outward normal n (pointing away from center, outside polygon). d = dot(ballCenter - wallPoint, n) is positive when ball center is outside the wall line. Actually for a convex polygon, the interior is where d <= 0 for all walls (if n is outward). 

Hmm, let me just carefully use inward normal and be consistent.

Inward normal n_in points toward center.
d = dot(ballCenter - anyPointOnWall, n_in). This is the distance from the wall line, measured inward. Interior points have positive d. The ball is contained when d >= r for all walls... no wait. d is distance from wall, large values are deep inside. When ball is right at the wall, d ≈ 0. The ball penetrates the wall when d < r (ball center is closer to wall than its radius, so the ball crosses the wall).

Actually, d = distance from wall to ball center, measuring how far inside. When ball is exactly at wall, d = 0. Ball stays inside if the ball doesn't cross, i.e., the ball's closest approach... the constraint is d >= r? No. If d = 0 the ball center is on the wall. The ball (radius r) would then be half-in half-out. To keep fully inside, d >= r. Yes.

So collision/penetration: d < r. Penetration depth = r - d.

Now n_in points inward. v_ball_n = dot(ballVel, n_in) is how fast the ball moves inward (toward center / away from wall). If v_ball_n is large positive, ball moves deeper inside (away from wall) - fine. If v_ball_n is negative, ball moves toward the wall (d decreases) - heading to exit.

For a static wall, we want to prevent the ball from exiting, i.e., keep d >= r. If the ball is moving toward the wall (v_ball_n < 0) and penetrating or about to, we reflect: v_ball_n -> -e * v_ball_n (with restitution e), making it move inward.

Now with a moving wall: the wall moves with velocity v_wall at the contact point. The wall velocity component along n_in: v_wall_n = dot(v_wall, n_in). This is how fast the wall itself moves inward. If the wall moves inward (into polygon), it can push the ball inward.

Relative velocity of ball w.r.t. wall along n_in: v_rel_n = v_ball_n - v_wall_n. If v_rel_n < 0, ball is moving toward wall faster than wall moves away (relative approach). We want to reflect the relative velocity: the new relative velocity should be -e * v_rel_n (bouncing off, moving inward). 

So new v_ball_n = v_wall_n - e * v_rel_n = v_wall_n - e*(v_ball_n - v_wall_n) = v_wall_n*(1+e) - e*v_ball_n.

Wait, reflecting relative velocity: if v_rel_n < 0 (approaching), after bounce v_rel_n' = -e * v_rel_n > 0 (moving apart). v_ball_n' = v_wall_n + v_rel_n' = v_wall_n - e*(v_ball_n - v_wall_n).

Yes that's the standard approach. Only apply when v_rel_n < 0 (approaching).

But there's subtlety: the wall is rotating, so the contact point velocity. Let me use the contact point = the closest point on the wall line to the ball center (projection). The wall velocity there = ω × (contactPoint - heptagonCenter).

This is the key requirement: "Account for moving-wall velocity in bounce response."

For ball-ball collisions: standard, with restitution and damping (maybe use average or separate). Also positional correction for overlap.

Angular motion of balls: each ball can rotate (angular velocity ω_ball). When ball-ball collision has offset from centers line, apply torque. But simpler: just give each ball some spin and let it rotate, update angle, render number rotated. To make spin visible, I can apply spin based on collisions or just give initial spin. The requirement says "Give each ball angular motion and rotate its number with the ball so spin is visible." So just have each ball with its own angular velocity, update it, and render the number at that rotation. Angular velocity can change through collisions (friction/torque).

Let me keep it reasonably physical but stable. I'll apply some torque during ball-wall and ball-ball collisions based on relative tangential velocity (friction), so spins develop naturally. That makes it more interesting and "visible."

Let me think about stability. With 20 balls in a heptagon, gravity, rotation, collisions... I need small timestep, maybe substeps. Let me use fixed timestep with multiple substeps per frame.

Let me think about sizes:
- Canvas: say 800x800, center (400, 400).
- Heptagon radius (vertex to center): R = 350. So vertices at distance 350 from center.
- Wall distance from center (apothem): a = R * cos(π/7) = 350 * cos(25.7°) = 350 * 0.9009 ≈ 315.
- Ball radius: need 20 balls to fit near center. If balls radius ~12, diameter 24. A 5x4 grid = 20 balls needs width ~5*24=120, height ~4*24=96. Fits near center. Let me use r = 14 or so. Let me reconsider for stability—bigger balls might be too heavy for fast rotation. Let me use r = 12.

Actually, with rotation and gravity, the balls will fill the heptagon. 20 balls of radius 12, packing in a heptagon of apothem 315 — plenty of room for a couple layers. Should be fine.

Physical constants:
- g = 600 px/s² (or use a gravity). Let me use g = 800.
- Angular velocity of heptagon: ω = 2π/5 ≈ 1.2566 rad/s.
- Ball mass: proportional to area, m = density * π r². Use density = 1, so m = π r² ≈ 452.
- Wall restitution e_w = 0.8, ball-ball e_b = 0.7.
- Friction/damping: apply velocity damping each frame (multiply velocity by 0.98 or use linear damping based on dt). Or apply friction in collisions.

For angular motion development, I'll add friction at contact: impulse along tangent with some coefficient.

Hmm, this is getting complex. Let me write a reasonably complete simulation. Let me consider using a simpler but effective approach.

Actually, let me think about whether to use impulse-based or position-based dynamics. For stability with many balls, position-based (PBD) or sequential impulses work well. But the requirement emphasizes collision response with moving walls. Let me use an impulse-based approach with substeps and positional correction, plus damping to keep stable.

Let me think about the simplest stable approach:

Each frame:
1. Determine dt. Substep into n sub-steps.
2. For each substep:
   a. Apply gravity: v += g * dt_sub.
   b. For collisions: iterate a few times (Jacobi/sequential) to resolve ball-ball and ball-wall.
   c. Update positions: x += v * dt_sub.
   d. Apply damping: v *= (1 - damping*dt_sub).

Let me structure collision resolution as: for each pair/contact, compute relative velocity, apply impulse if approaching, plus positional correction.

Actually, let me use the "find collision, apply impulse, positional correction" but to be stable with 20 balls, I'll do multiple iterations. Since the balls also can spin, I'll include angular.

Let me define ball state: x (pos vector), v (vel), angle, av (angular velocity), r (radius), m (mass). Also compute moment of inertia I = (1/2) m r² for solid circle (in 2D, I = 0.5 m r²).

Ball-ball collision:
- d = dist(b1.x, b2.x); if d < r1 + r2 (overlap):
  - normal n = (b2.x - b1.x)/d
  - penetration = r1+r2 - d; push apart by penetration/2 each (weighted by mass)
  - relative velocity at contact: v_rel = b1.v - b2.v (+ angular contributions: ω × contactOffset). For 2D: v_contact1 = b1.v + b1.av * (-(c1.y), c1.x)... let me think.
  - Let me compute contact points on each ball: c1 = b1.x + n*r1 (the point on b1 toward b2), c2 = b2.x - n*r2.
  - velocity of contact point 1 = b1.v + b1.av × (c1 - b1.x), where × in 2D: av (scalar) cross vector (rx,ry) = av*(-ry, rx). So v1_contact = b1.v + av1*(-(c1-b1).y, (c1-b1).x).
  - Similarly v2_contact.
  - v_rel_vec = v1_contact - v2_contact; v_rel_n = dot(v_rel_vec, n). If v_rel_n < 0 (approaching):
    - Apply normal impulse: j = -(1+e) * v_rel_n / (1/m1 + 1/m2 + (cross terms from inertia)). 
    - The inertia term: for ball, the normal impulse contribution = (r1² * I1_inv) for the cross... Let me recall: the denominator = 1/m1 + 1/m2 + (r_cross_n)² * (1/I1 + 1/I2) where r_cross_n = (contactOffset × n) scalar = in 2D, for offset r_vec and normal n: r_vec × n (scalar) = r_vec.x * n.y - r_vec.y * n.x. For ball1, contact offset = c1 - b1.x = n * r1 (pointing along n). r_vec × n = (n*r1) × n = 0! Because the contact point is along the normal, the cross product with normal is zero. So angular doesn't affect normal impulse magnitude for circles. Good, simplifies: j = -(1+e) v_rel_n / (1/m1+1/m2).
    - Apply: b1.v -= j/m1 * n; b2.v += j/m2 * n.
    - Friction (tangential): t = perpendicular to n. v_rel_t = dot(v_rel_vec, t). Apply friction impulse to develop spin: jt = -v_rel_t / (1/m1+1/m2 + (r × t)² (1/I1+1/I2)). r_vec × t for offset n*r1 and tangent t: = (n*r1) × t = r1 * (n × t). |n × t| = 1 (since perpendicular). sign depends. This develops angular velocity. Apply jt along t, with torque.
    
This is getting involved but doable. Let me write it.

Actually, for friction between balls, I'll keep a small tangential friction to induce spin but not overcomplicate. Let me include it.

Ball-wall collision:
- For each wall (inward normal n_in, wall point p0):
  - d = dot(ball.x - p0, n_in). If d < r: penetration = r - d.
    - Push ball along n_in by penetration.
    - wall contact point = ball.x - n_in * d (projection onto wall line, since d is signed dist along n_in, the foot = ball.x - d*n_in... wait: ball.x = foot + d*n_in, so foot = ball.x - d*n_in). Hmm, d = dot(ball.x-p0, n_in), foot = p0 + (ball.x-p0) projection onto line = ball.x - d*n_in. Yes.
    - wall velocity at foot: v_wall = ω_hept × (foot - center). In 2D, ω_hept is scalar, v_wall = ω_hept * (-(foot-center).y, (foot-center).x).
    - v_rel = ball.v - v_wall (at contact; for circle against infinite line, contact offset is along normal so angular doesn't affect... actually the wall is a line, contact point on ball = ball.x - n_in*r? No. The ball touches the wall at the point closest. contact point on ball = ball.x - n_in * 0 ... hmm. For a line, the ball's contact point relative to center is along n_in direction to the wall: offset = -n_in * r (toward wall/outward). n_in points inward, wall is outside relative... wait.
    
    Let me re-examine. n_in points inward (toward center). The wall is at the boundary. The ball is inside. The ball touches the wall at its outward side, i.e., along -n_in direction from center. So contact point on ball = ball.x - n_in * r (move from center toward outside = opposite of inward normal = -n_in). 
    
    The offset from ball center to contact point = -n_in * r. Angular velocity cross this offset: av × (-n_in * r). For normal impulse, cross of offset with n_in: offset × n_in = (-n_in*r) × n_in = 0 (parallel). So normal impulse unaffected by spin, as expected.
    
    - v_rel_n = dot(v_rel, n_in). If < 0 (ball moving toward wall/outward relative to wall motion): apply impulse j = -(1+e_w)*v_rel_n / (1/m) (only ball has mass; wall is infinite). v_ball' = v_ball - (j/m)*n_in... wait sign. 
    - Let me be careful with signs. We want: if ball approaches wall (moving outward faster), reflect so it moves inward. 
    - The relative normal velocity v_rel_n = dot(ball.v - v_wall, n_in). n_in inward. If ball moves outward (away from center, toward wall), ball.v has negative component along n_in, so v_rel_n < 0 when ball moving toward wall. 
    - Reflect: v_rel_n' = -e_w * v_rel_n. So new ball.v along n_in = v_wall_n + (-e_w * v_rel_n) = v_wall_n - e_w*(dot(ball.v,n_in) - v_wall_n). 
    - Impulse magnitude j = -(1+e_w)*v_rel_n / (1/m)... let me just compute impulse approach: 
      j = -(1+e) * v_rel_n / (1/m1 + 1/m2 + ...). Here m2 = ∞ (wall), so 1/m2 = 0, inertia term for wall = 0. So j = -(1+e) v_rel_n * m (since 1/m1 = 1/m). Actually j = -(1+e)*v_rel_n / (1/m) = -(1+e)*v_rel_n * m.
      - Apply to ball: ball.v += (j/m) * n_in? Let me check sign. Impulse along n_in on the ball... We have the formula v1' = v1 + (j/m1) * n for collision where n points from 2 to 1. Here the wall is "object 2", n_in points from wall toward ball interior (inward = from wall toward ball). So n = n_in (points from wall to ball, i.e., inward). Standard: v1' = v1 + j/m1 * n (n from 2 to 1). 
      - So ball.v += (j/m) * n_in. With j = -(1+e)*v_rel_n * m where v_rel_n = dot(ball.v - v_wall, n_in). Since n points from wall(2) to ball(1)... and v_rel_n computed with ball - wall. Let me just verify: v_rel = v1 - v2 = ball.v - v_wall. v_rel_n = dot(v_rel, n) = dot(ball.v - v_wall, n_in). Approaching means v_rel_n < 0. j = -(1+e) v_rel_n m > 0. ball.v += (j/m) n_in = (1+e)(-v_rel_n) n_in = (1+e)(-dot(ball.v-v_wall,n_in)) n_in. New ball.v_n = dot(ball.v,n_in) + (1+e)(-v_rel_n) = ball.v_n + (1+e)(-ball.v_n + v_wall_n) = ball.v_n(1 - (1+e)) + (1+e) v_wall_n = -e*ball.v_n + (1+e) v_wall_n. And v_rel_n' = new_ball_v_n - v_wall_n = -e*ball.v_n + (1+e)v_wall_n - v_wall_n = -e*ball.v_n + e*v_wall_n = -e*(ball.v_n - v_wall_n) = -e * v_rel_n. ✓. 
    - Also apply friction (tangent) to the wall to develop spin: t = tangent along wall. v_rel_t = dot(v_rel, t). Apply friction impulse to change ball's tangential velocity and spin.
    - Positional correction: push ball along n_in by penetration.

    Also need to make sure penetration resolution with moving wall doesn't cause issues; positional correction is separate from velocity.

Ball rendering: draw circle, then draw number text rotated by ball.angle. The number rotates with spin.

Angular velocity for balls: I'll let collisions generate spin via friction. Also could add a tiny constant. Let me make sure spin is visible—maybe add a bit of initial spin or rely on friction. Let me add friction in collisions (both ball-wall and ball-ball) with a reasonable coefficient so spins develop.

Let me also add global damping to keep energy bounded: linear velocity damping and angular damping. The rotating heptagon continuously injects energy though (moving walls). Hmm, that could cause unbounded speeds. The damping/friction must balance energy input. The requirement says "keep them stable, contained, do not persistently overlap or gain unbounded speed." So damping is important. Let me add velocity damping (e.g., multiply v by 0.99 per second, or use linear damping force = -k*v). 

Energy injection from rotating walls: when a ball is at a wall and the wall pushes it (wall moves inward), the ball gains energy. This is like a rotating drum. With damping, it should reach equilibrium. Let me tune damping to be enough.

Let me set:
- gravity g = 900 (px/s²), downward.
- heptagon ω = 2π/5 ≈ 1.2566 rad/s.
- ball r = 14, mass m = r² (proportional).
- e_ball = 0.5, e_wall = 0.6 (lower to reduce energy).
- friction coefficient μ such that friction reduces tangential speed. 
- damping: linear damping coefficient cl = 0.5 (force = -cl * v), angular damping ca = 0.3.

I'll tune.

Let me reconsider the ball initial arrangement. "compact, non-overlapping arrangement near the enclosure center." Let me place them in a small cluster: maybe a 5x4 grid offset to one side, or a pentagon/hexagonal packing of 20 near center. Let me do a small diamond/cluster. Simplest: arrange in a small circle or grid. Let me place them in a tight hexagonal-ish cluster near center so they're compact and non-overlapping. But "let them fall under gravity"—they start near center and fall. If I cluster them exactly at center, gravity pulls down and they spread. Let me place the cluster slightly above center so they fall into the lower part. Actually "near the enclosure center" and "let them fall under gravity" — place near center, gravity pulls down, they drop to bottom of heptagon and interact. Let me place cluster centered slightly above the geometric center, e.g., center - (0, 40), so they fall downward.

Let me make initial positions: a compact arrangement. I'll generate positions on a small triangular grid within a radius. Or simpler: place balls on concentric-ish pattern. Let me just do a 5 wide × 4 tall grid with spacing 2.2*r, centered at (400, 350). 5*2.2*14 = 5*30.8 = 154 wide, 4*30.8 = 123 tall. That fits. They'll fall from y=350 downward (gravity pulls +y in canvas). Good.

Actually for "compact," triangular packing is more compact. But grid is fine and clearly non-overlapping. Let me do a grid but maybe offset rows for hexagonal packing to be more compact and stable. Let me do hexagonal: rows offset, 5 per row, 4 rows. Spacing 2.2r horizontally, vertical spacing ~1.9r.

Hmm, let me keep it simple with a square grid offset slightly. Non-overlapping guaranteed by spacing.

Let me now also handle the Reset: store initial state (positions, velocities=0, angles, angular velocities). Reset restores positions, zeroes velocities, restores heptagon angle to initial (θ=0), and zeroes ball velocities/angle. Actually "wall orientation" restored — heptagon angle back to 0. And ball arrangement back to initial positions with zero velocity. Let me also reset ball angles/spins to 0 for clean comparison? The requirement: "Reset restores the initial ball arrangement and wall orientation." So arrangement (positions) and wall orientation. Velocities should be zero (so they fall fresh). Let me reset velocities to 0 too. Angles—let me reset to 0 to make comparison clean, but it says arrangement and wall orientation. I'll reset positions, zero velocities, reset heptagon angle to 0. For ball angles, resetting to 0 is reasonable (initial had 0 spin). Let me reset ball angle and av to 0.

Let me track initial positions array.

Visualization: I'll draw the heptagon as filled/outline rotating. Draw balls as circles with numbers. Number rotates with ball.angle. Use canvas.

Let me also draw the heptagon edges. Use ctx. Rotate canvas by heptagon angle around center, draw heptagon, then draw balls in world coordinates (not rotating). The heptagon is in world space; I compute wall normals each frame based on current rotation.

For collision detection, I compute walls in world space each frame using current heptagon angle θ.

Let me write the code now.

Let me define:
- canvas size W=800, H=800.
- center cx=400, cy=400.
- R = 340 (vertex radius), apothem = R cos(π/7) ≈ 306.
- Ball r = 13. 20 balls.
- g_y = 900.

Walls: 7 walls. For wall i (between vertex i and i+1), compute midpoint or use vertex i as point. Inward normal: average of inward normals of the two vertices? Easier: The inward normal of wall i (edge from vertex i to vertex i+1) points toward center. Compute as: n_in = normalize(center - midpoint). For a regular polygon, the apothem direction (from edge midpoint toward center) is the inward normal. Good, that's exact for regular polygon.

midpoint_i = (v_i + v_{i+1})/2. n_in_i = normalize(center - midpoint_i). d_i = dot(ball.x - midpoint_i, n_in_i). Containment d_i >= r.

Wall velocity at foot: foot = ball.x - d_i * n_in_i (closest point on the infinite line). v_wall = ω_hept × (foot - center), 2D cross scalar ω: v = ω * perp, where perp of (dx,dy) = (-dy, dx). So v_wall = ω_hept * (-(foot.y - cy), foot.x - cx).

Let me code it up.

For performance, 20 balls, 7 walls, substeps ~10, iterations ~4. That's fine.

Let me write the simulation loop with requestAnimationFrame, compute real dt, cap it, then substep.

Actually, for stability and predictability, I'll use a fixed internal timestep (e.g., dt = 1/120 ≈ 0.00833) and accumulate real elapsed time, doing as many fixed steps as needed (with a cap to avoid spiral of death). Subdivide each fixed step into sub-steps.

Hmm, but if I fix dt and real framerate varies, the simulation time advances in real time correctly (fixed timestep with accumulator). That's good practice. Let me do fixed dt = 1/120, substeps = 1 (or 2)? Let me use dt = 1/180 with 2 sub-substeps? Let me just use fixed physics step 1/120 and run multiple per frame if needed, capped at like 5.

Actually simpler: I'll compute dt from elapsed, clamp to 0.02, then do substeps = ceil(dt / sub_dt) with sub_dt = 1/180. This keeps it stable.

Let me write.

Let me define the impulse-based collision with friction.

```
function resolveBallBall(b1, b2) {
  let dx = b2.x - b1.x, dy = b2.y - b1.y;
  let dist = Math.hypot(dx, dy);
  let rsum = b1.r + b2.r;
  if (dist < rsum && dist > 0) {
    let nx = dx/dist, ny = dy/dist;
    // normal points from b1 to b2
    let pen = rsum - dist;
    // positional correction (mass weighted)
    let m1sum = 1/b1.m, m2sum = 1/b2.m;
    let total = m1sum + m2sum;
    let c1 = pen * m2sum/total, c2 = pen * m1sum/total;
    b1.x -= nx * c1; b1.y -= ny * c1;
    b2.x += nx * c2; b2.y += ny * c2;
    // relative velocity at contact
    // contact points: c1p on b1 = center + n*r1 (toward b2), c2p on b2 = center - n*r2
    let v1x = b1.vx + b1.av*(-0*... ) ...
```
For angular velocity cross offset: v_contact = v + av × offset, where av × (ox,oy) = av*(-oy, ox). 
For b1, offset to contact = (nx*r1, ny*r1) = n*r1. av1 × offset = av1*(-ny*r1, nx*r1). 
For b2, offset to contact = (-nx*r2, -ny*r2). av2 × offset = av2*(ny*r2, -nx*r2).
v_rel = (v1_contact - v2_contact).
v_rel_n = nx*v_rel_x + ny*v_rel_y.
if v_rel_n < 0:
  // normal impulse (inertia term zero for circles since offset along n)
  let j = -(1+e)*v_rel_n / (1/b1.m + 1/b2.m);
  b1.vx -= j*nx/b1.m; b1.vy -= j*ny/b1.m;
  b2.vx += j*nx/b2.m; b2.vy += j*ny/b2.m;
  // friction
  let tx = -ny, ty = nx;
  let v_rel_t = v_rel_x*tx + v_rel_y*ty;
  let r_cross_t1 = (n*r1) × t = ... scalar = (offset1.x*t.y - offset1.y*t.x). offset1=(nx*r1,ny*r1), t=(-ny,nx). scalar = nx*r1*nx - ny*r1*(-ny) = r1*(nx²+ny²)=r1. So (r × t) = r1. Similarly for b2 offset2=(-nx*r2,-ny*r2), t=(-ny,nx): scalar = offset2.x*t.y - offset2.y*t.x = -nx*r2*nx - (-ny*r2)*(-ny)= -r2*(nx²+ny²) = -r2. 
  // denominator friction: 1/m1+1/m2 + r1²/I1 + r2²/I2
  let denom_t = 1/b1.m + 1/b2.m + r1²/b1.I + r2²/b2.I; // note (r×t)² = r²
  let jt_max = Math.abs(j) * friction; // clamp normal impulse by friction coeff
  let jt = -v_rel_t / denom_t;
  jt = clamp(jt, -jt_max, jt_max);
  b1.vx -= jt*tx/b1.m; b1.vy -= jt*ty/b1.m;
  b2.vx += jt*tx/b2.m; b2.vy += jt*ty/b2.m;
  b1.av -= jt*r1/b1.I;  // torque = offset × impulse; for b1, τ = r1 * jt (sign)
  b2.av += jt*r2/b2.I;
```

Hmm the friction sign for angular needs care. Let me think. The friction impulse on b1 is along -t direction if jt... Actually I applied b1.v -= jt*tx/m. Let me reconsider. The standard: impulse j along normal n from 1 to 2 means b2 gets +j*n, b1 gets -j*n. For friction tangent t, impulse magnitude jt, b2 gets +jt*t, b1 gets -jt*t. Then angular impulse on b1 = (offset1) × (-jt*t) ... the torque vector (2D scalar) = offset × impulse = offset1.x*(-jt*t.y) - offset1.y*(-jt*t.x) = -jt*(offset1.x*t.y - offset1.y*t.x) = -jt*(r1) (we computed r1). So τ1 = -jt*r1, av1 += τ1/I1 = -jt*r1/I1. I had b1.av -= jt*r1/b1.I ✓. For b2, impulse = +jt*t, offset2=(-nx*r2,-ny*r2), τ2 = offset2 × (jt*t) = jt*(offset2.x*t.y - offset2.y*t.x) = jt*(-r2). av2 += -jt*r2/I2. I had b2.av += jt*r2/b2.I ✗ (sign). Let me fix: b2.av -= jt*r2/b2.I. Wait let me recompute: τ = r × F (scalar in 2D = rx*Fy - ry*Fx). For b2: offset2 = (-nx*r2, -ny*r2). Force2 = jt*(tx,ty) = jt*(-ny, nx). τ2 = offset2.x*Force2.y - offset2.y*Force2.x = (-nx*r2)*(jt*nx) - (-ny*r2)*(jt*(-ny)) = -r2*jt*(nx²) - (-ny*r2)*(jt*(-ny)) = -r2*jt*nx² - (ny²*r2*jt) [since -ny*r2 * -jt*ny = +jt*r2*ny²... wait let me be careful: (-ny*r2)*(jt*(-ny)) = (-ny*r2)*(-jt*ny)= +jt*r2*ny²]. So τ2 = -r2*jt*nx² + jt*r2*ny²? That's not symmetric. Hmm I think I made an error. Let me redo carefully.

t = (-ny, nx) (tangent perpendicular to n=(nx,ny), rotated -90°... actually (-ny,nx) is n rotated +90°). 
offset2 = contact point on b2 relative to center = center_to_contact = -n*r2 = (-nx*r2, -ny*r2).
Force on b2 from friction impulse = jt * t = jt*(-ny, nx).
τ2 (scalar, positive = counterclockwise) = rx*Fy - ry*Fx where (rx,ry)=offset2, (Fx,Fy)=Force.
= (-nx*r2)*(jt*nx) - (-ny*r2)*(jt*(-ny))
= -nx*r2*jt*nx - (-ny*r2)*(-jt*ny)
= -r2*jt*nx² - (ny*r2)*(jt*ny)   [because (-ny*r2)*(-jt*ny) = +jt*r2*ny², so minus that = -jt*r2*ny²]

Wait: -(-ny*r2)*(-jt*ny). Let me compute (-ny*r2)*(-jt*ny) = jt*r2*ny² (product of two negatives = positive, times jt). Then there's a minus sign in front: - jt*r2*ny². So τ2 = -r2*jt*nx² - r2*jt*ny² = -r2*jt*(nx²+ny²) = -r2*jt.

So av2 += τ2/I2 = -jt*r2/I2. I had b2.av -= jt*r2/b2.I ✓ (matches -jt*r2/I2). Good, my original was right; my "fix" was wrong. Let me keep b2.av -= jt*r2/b2.I. Wait I need to recheck my original code: I wrote `b2.av += jt*r2/b2.I;` then said ✗. The correct is av2 += -jt*r2/I2 = b2.av -= jt*r2/b2.I. Good.

Similarly τ1: offset1 = n*r1 = (nx*r2... = (nx*r1,ny*r1). Force on b1 = -jt*t = -jt*(-ny,nx) = jt*(ny,-nx). τ1 = offset1.x*Force.y - offset1.y*Force.x = (nx*r1)*(-jt*nx) - (ny*r1)*(jt*ny) = -jt*r1*nx² - jt*r1*ny² = -jt*r1. av1 += -jt*r1/I1. I wrote b1.av -= jt*r1/b1.I ✓.

Wait, Force on b1: friction impulse on b1 is -jt*t (since b2 gets +jt*t, b1 gets -jt*t). -jt*t = -jt*(-ny,nx) = (jt*ny, -jt*nx). Then τ1 = offset1.x*Fy - offset1.y*Fx = (nx*r1)*(-jt*nx) - (ny*r1)*(jt*ny) = -jt*r1*nx² - jt*r1*ny² = -jt*r1. ✓. Good.

OK so signs consistent. Let me double check the v_rel sign usage. v_rel = v1_contact - v2_contact. v1_contact = b1.v + av1×offset1. The impulse reduces relative velocity. Standard collision: if v_rel_n < 0 (approaching, since n from b1 to b2, b1 moving toward b2 means... n points b1->b2; b1 moving in +n direction toward b2 => v_rel_n>0 means approaching?). 

Hmm sign of approach. n points from b1 to b2. If b1 moves toward b2, b1.v has component along +n >0. If b2 moves toward b1, b2.v has component along -n, so along n <0, making v_rel_n = v1n - v2n... if b2 moves toward b1, v2n <0, so v_rel_n = v1n - v2n = v1n - (negative) > v1n. So approaching = v_rel_n > 0? Wait that means if they approach, v_rel_n >0. Let me reconsider: v_rel = v1 - v2. v_rel_n = dot(v1-v2, n). If b1 approaches b2 (b1 moves +n) and b2 stationary: v_rel_n >0. So approaching corresponds to v_rel_n > 0? But the relative position d = b2 - b1 moving... they're overlapping and approaching means distance decreasing. distance derivative = dot(v1-v2, (b2-b1)/dist) = dot(v1-v2, n) = v_rel_n. If decreasing, v_rel_n < 0. 

Contradiction! distance decreasing => d(dist)/dt < 0 => dot(v1-v2, n) < 0 => v_rel_n < 0. So approaching = v_rel_n < 0. Let me recheck the first: if b1 moves toward b2 (b1 moves +n) and b2 stationary, then v1-v2 = v1 (positive n component), dot with n >0 => v_rel_n>0. But distance b2-b1: b1 moving toward b2 decreases distance, so should be <0. 

Ah, I see the confusion: if b1 moves in +n direction (toward b2), that decreases the distance, so d(dist)/dt should be negative. d(dist)/dt = (b1-b2 unit)·(v1-v2) = -n·(v1-v2) = -v_rel_n. So d(dist)/dt = -v_rel_n. If b1 moves toward b2, dist decreases => d(dist)/dt<0 => -v_rel_n<0 => v_rel_n>0. 

OK so I had it: approaching (dist decreasing) => v_rel_n > 0. Wait but that contradicts "distance derivative = dot(v1-v2, n)". Let me recompute distance derivative. dist = |b2 - b1|. d(dist)/dt = (b2-b1)/|b2-b1| · (v2 - v1) = n · (v2 - v1) = -n·(v1-v2) = -v_rel_n. Right! So d(dist)/dt = -v_rel_n. Approaching (dist decreasing) => d(dist)/dt<0 => -v_rel_n<0 => v_rel_n>0.

So approaching => v_rel_n > 0, NOT <0. I had it backwards. Let me fix: apply impulse if v_rel_n > 0 (approaching). 

Hmm wait, but many references define n from 2 to 1 (pointing from second to first) and v_rel_n<0 for approaching. It depends on convention. Let me just be careful and consistent with my n pointing b1->b2.

With n from b1 to b2: v_rel_n = dot(v1-v2, n). Approaching => v_rel_n > 0. So condition: if v_rel_n > 0, they approach, apply impulse to separate.

j = -(1+e) v_rel_n / (1/m1+1/m2). Since v_rel_n>0, j<0. b1 gets -j*n/m1... wait. Impulse: b2 += j*n/m2, b1 -= j*n/m1 (n from 1 to 2, standard gives v1 += j*n/m1 with n from 2 to 1...). Ugh. Let me just derive from the restitution.

We want post-approach velocity v_rel_n' = -e * v_rel_n (reverse and reduce). Change in v_rel_n = -(1+e) v_rel_n. The total impulse along n changes v_rel_n by -j*(1/m1+1/m2) [if b1 gets -j*n/m1 and b2 gets +j*n/m2, then v1_n changes by -j/m1, v2_n by +j/m2, v_rel_n = v1_n - v2_n changes by -j/m1 - j/m2 = -j(1/m1+1/m2)]. We want change = -(1+e)v_rel_n. So -j(1/m1+1/m2) = -(1+e)v_rel_n => j = (1+e)v_rel_n/(1/m1+1/m2). Since v_rel_n>0, j>0. b2 += j*n/m2 (b2 moves +n, away from b1 — correct, separates). b1 -= j*n/m1 (b1 moves -n, away from b2 — correct). 

So: j = (1+e)*v_rel_n/(1/m1+1/m2), b1.v -= j*n/m1, b2.v += j*n/m2. Good, fix sign from my earlier draft.

Let me re-examine wall. n_in points inward (from wall toward interior, i.e., from wall toward ball side). Equivalent to "from wall(2) to ball(1)". With n = n_in from wall to ball. v_rel_n = dot(ball.v - v_wall, n_in). Approaching wall => ball moving toward wall (outward) => dot(ball.v, n_in) <0 (moving opposite to inward normal). v_rel_n = (negative) - v_wall_n. If wall stationary, v_rel_n <0 => approaching. So for wall, approaching => v_rel_n < 0 (since n points into polygon, ball moving outward is "toward wall"). 

Hmm so the sign convention differs because n_in points from wall to ball (interior). Approaching the wall means moving outward (away from interior, toward wall), i.e., opposite to n_in. So v_rel_n = dot(ball.v - v_wall, n_in); approaching => v_rel_n < 0.

Apply if v_rel_n < 0. j should push ball inward (+n_in direction). Impulse: ball.v += (j/m)*n_in with j>0. Change in v_rel_n: v_rel_n = dot(ball.v, n_in) - v_wall_n. Adding (j/m)*n_in to ball.v increases dot(ball.v,n_in) by j/m, so v_rel_n increases by j/m. We want post = -e*v_rel_n_initial. Change = v_rel_n' - v_rel_n_initial = -e*v_rel_n - v_rel_n = -(1+e)v_rel_n. We need change = j/m. So j/m = -(1+e)v_rel_n => j = -(1+e)v_rel_n * m. Since v_rel_n<0, j>0. ✓. ball.v += (j/m) n_in = -(1+e)v_rel_n * n_in. 

Let me double check final: ball.v_n_new = ball.v_n + j/m = ball.v_n + (-(1+e)v_rel_n) = ball.v_n - (1+e)(ball.v_n - v_wall_n) = ball.v_n - (1+e)ball.v_n + (1+e)v_wall_n = -e*ball.v_n + (1+e)vwall_n. Then v_rel_n_new = ball.v_n_new - vwall_n = -e*ball.v_n + (1+e)vwall_n - vwall_n = -e*ball.v_n + e*vwall_n = -e(ball.v_n - vwall_n) = -e*v_rel_n. ✓.

Great. So wall: condition v_rel_n < 0, impulse ball.v += -(1+e)v_rel_n * n_in * (... wait j/m). Let me just write j_scalar = -(1+e) * v_rel_n (this is the impulse per mass? No). Let me define impulse magnitude P = -(1+e)*v_rel_n * m (since 1/m2=0). Then ball.v += (P/m) * n_in = -(1+e)*v_rel_n * n_in. So effectively I can skip computing P and m: ball.v += -(1+e)*v_rel_n * n_in. But if I want friction and spin, I need the actual impulse and m for friction direction and angular. Let me keep P.

Actually for friction with the wall, the wall is fixed-mass-inertia ∞, so friction only affects the ball. Friction impulse along tangent t (along wall): limited by μ*P (normal impulse magnitude). jt = -v_rel_t/denom_t where denom_t = 1/m + (r×t)²/I, (r×t) for offset = -r*n_in... wait offset2? For wall, ball contact offset = -n_in*r? The contact point on ball is ball.x - n_in*r (toward wall). offset from center = -n_in*r = (-nx*r, -ny*r) using n_in=(nx,ny). (offset × t) scalar: offset.x*t.y - offset.y*t.x. t = tangent along wall, say t = (ny, -nx)? Need t perpendicular to n_in. n_in=(nx,ny). Two perpendiculars: (-ny,nx) and (ny,-nx). Let me pick t = (-ny, nx) (rotate n by +90°). offset=(-nx*r,-ny*r). offset×t = (-nx*r)*(nx) - (-ny*r)*(-ny) = -r*nx² - r*ny² = -r. So (r×t)² = r². Good, denom_t = 1/m + r²/I.

Friction impulse magnitude jt, clamped by μ*|P|. Apply ball.v += (jt/m)*t, av += (τ)/I where τ = offset × (jt*t) = jt*(offset×t) = jt*(-r). So av += -jt*r/I. Hmm sign; let me just compute τ = offset.x*(jt*t.y) - offset.y*(jt*t.x). With t=(-ny,nx), jt*t = (jt*(-ny), jt*nx). τ = (-nx*r)*(jt*nx) - (-ny*r)*(jt*(-ny)) = -r*jt*nx² - (-ny*r)*(-jt*ny) = -jt*r*nx² - jt*r*ny² = -jt*r. So av += τ/I = -jt*r/I. 

I'll apply friction to make spin develop when ball slides along wall.

This is a lot. Let me just code it carefully. Given time constraints, let me write clean code.

Actually, to keep within budget and ensure stability, maybe I should reduce complexity of friction-induced spin and just include it but keep coefficients moderate.

Let me also reconsider: the moving-wall velocity at the contact point matters for the bounce. The wall velocity v_wall = ω_hept × (foot - center). foot = ball.x - d*n_in where d = dot(ball.x - midpoint, n_in). 

Let me now also make sure the heptagon is drawn rotated and collision uses the rotated vertices. Since the heptagon rotates, vertices rotate. midpoint and n_in rotate accordingly. I'll recompute each frame.

Now let me think: gravity direction. Canvas y+ is down. Let me use y+ down for physics too (simpler rendering). Then gravity g > 0 in +y. n_in toward center. The wall velocity v_wall = ω × (foot - center) with ω being a scalar; for counterclockwise visual rotation, ω>0. In screen coords (y down), the 2D cross of scalar ω with vector r = (rx,ry): the rotation ω×r in standard math (y-up) = (-ω*ry, ω*rx). But in y-down, a positive scalar ω meaning CCW visually... Let me define rotation angle θ(t) = ω*t increasing, and visually CCW in standard math orientation, but canvas y is down so CCW in math = CW visually? Ugh.

Let me just decide: heptagon rotates visually clockwise or counterclockwise, doesn't matter as long as consistent. Let me define θ increases with time, and vertices computed with standard parametric: vertex = center + R*(cos(θ+2πk/7), sin(θ+2πk/7)) but in canvas y, sin gives y; if θ increases, vertex moves counterclockwise in standard math, which is clockwise visually (since y down flips). That's fine—visually clockwise rotation. To make it visually counterclockwise, use -sin. Doesn't matter. Let me use sin/cos normally; the visual will be clockwise. Fine.

For wall velocity in canvas coords: v_wall = ω × (foot-center) where × for scalar ω: I'll define v = ω * (-(foot.y-center.y), foot.x-center.x)? Let me derive from the vertex motion. Position of a point at angle: p(t) = center + R*(cos(θ), sin(θ)) with θ=ω*t. dp/dt = R*(-sin(θ)*ω, cos(θ)*ω) = ω*(-R*sinθ, R*cosθ) = ω*(-(py-center.y), (px-center.x)). So v_wall = ω*(-(foot.y-cy), (foot.x-cx)). Yes. So v_wall = ω * (-(foot.y-cy), foot.x-cx).

Good, consistent with the ω×r in y-down convention giving CCW... anyway consistent.

Now gravity: g_y = +900 (down). balls fall down.

Let me now write everything.

I'll set up:
- W=820,H=820, margin. Let me use 800x800.
- cx=400, cy=400.
- R_verts = 340.
- apothem ≈ 306.2.
- ball r = 13. 20 balls.
- Number scale for drawing.

Let me compute initial positions: hexagonal close packed near center but offset up a bit. Let me place in a small hexagon pattern of 20. Actually a 5x4 grid is simplest and clearly non-overlapping. Spacing sx = 2.3r = 30, sy = 2.3r = 30, with row offset. 5 columns × 4 rows = 20. Center the cluster at (cx, cy-30) so they fall down into the heptagon. Let me center at (400, 370). With 5 cols, total width = 4*30=120, so leftmost x = 370-60=310, rightmost 430. rows y from 370-1.5*30=325 to 370+1.5*30=415. Fine.

Actually "compact, non-overlapping arrangement near center." Grid spacing 2.3r ensures non-overlap (2r=26 < 30). Good. Let me even do hexagonal (offset every other row by sx/2, vertical spacing sy*0.87) for compactness. But then need 5 cols ×4 rows; offset rows. Let me just do square grid, it's fine and clearly non-overlapping.

Hmm, but a square grid with spacing 2.3r: when they fall, the first ones hit bottom wall and stack. Should be stable enough.

Let me set number drawing: ctx font, fillText with centered alignment, rotated by ball.angle. Draw ball circle, then save ctx, translate to ball center, rotate by angle, draw number, restore. But also need to handle that the number shouldn't be upside down weirdly; rotating with ball is fine—"rotate its number with the ball so spin is visible."

Let me also give the numbers color and make them readable (white on colored balls).

Let me also draw the heptagon: fill with dark color, stroke. Maybe make it a frame (walls). Let me draw filled heptagon with dark background and a bright stroke, so balls (lighter colors) are visible inside.

Let me set heptagon fill = dark gray/black, stroke = white. Balls = colored circles with numbers.

Let me write the HTML.

For performance, 20 balls is trivial.

Let me think about damping to prevent runaway from rotating walls. The heptagon rotation at ω=1.256 rad/s; at radius ~306, wall speed ≈ 1.256*306 ≈ 384 px/s at the wall. That's significant—balls near wall get pushed. With gravity g=900, terminal-ish. Damping needed. Let me add linear damping acceleration = -cl * v with cl ~ 3-5 (so damping force grows with speed). And angular damping. Let me tune: maybe cl = 4. Velocity damping reduces energy input. Also e_wall=0.5 reduces bounce energy.

Actually, energy can be pumped by the rotating wall (like a paddle). But damping dissipates. Let me pick moderate values and test, then tune.

Let me also cap maximum velocity to prevent explosions (safety): if speed > Vmax (say 2000), scale down. Or rely on substeps. Let me add a soft cap as safety.

Let me write the code and test, then tune.

Let me consider the substep dt. gravity g=900, r=13. Free fall distance in heptagon ~600 px. Time to fall ~ sqrt(2*600/900)=1.15s. dt_frame ~16ms. Sub_dt=1/180≈5.5ms. Number of sub-steps per frame = dt_frame/sub_dt ≈ 3. With collision iterations 3 per substep. 20*7 + 20*20/2 = 140+190=330 collision checks per substep × 3 iterations × 3 substeps = ~3000 ops, trivial. Good.

Let me write.

I'll structure:
```
const canvas, ctx.
const balls = [{x,y,vx,vy,angle,av,r,m,I}].
let theta=0 (heptagon angle), initialTheta=0.
const omega = 2π/5.
gravity.
state: running/paused.
initBalls(), save initial positions.
function reset(){ restore positions, zero velocities/angles, theta=0. }
function step(dt){ subdivide; } 
function collide(){ }
function draw(){ }
loop via rAF.
```

Let me write the full file now.

For numbers: draw text with font size ~ 0.5*r? r=13, font ~ 14px. Centered. Use black text on light balls, or white. Let me make balls with gradient and a border, number in contrasting color.

Let me also add wall thickness visually? Not needed for physics (infinitely thin walls), but visually I can make the heptagon border thick.

Let me write.

Let me also handle the contact point / foot for wall velocity. foot = ball.x - d*n_in where d = dot(ball.x - mid, n_in). Good.

Edge case: dist between balls =0; guard with dist>1e-6 else n = (1,0).

Let me code ball-wall: iterate 7 walls, for each check, resolve, then iterate again for multiple walls? Since a ball could be near a corner (two walls). I'll do multiple passes. Actually I'll do collision iterations that include both ball-wall and ball-ball.

Let me write the resolve functions and a collision loop.

Now, regarding "moving-wall velocity in bounce response"—I include v_wall in v_rel_n and friction. Good.

Let me also ensure balls don't escape: after collision resolution, do positional correction; the constraint d >= r enforced. With substeps and iterations should keep contained.

Let me also add that if a ball is outside (d<0 significantly), clamp.

Let me write the code.

I think I have a clear plan. Let me also add a small tangential spin to make numbers rotate visibly even early. The friction during collisions will create spin, but the initial fall has no collisions until they hit walls. Once they bounce, spin develops. To ensure visible spin early, maybe give a tiny initial av. But requirement says give angular motion (from physics). Friction will handle it. Let me also add rolling friction / a small spin-up. I'll test.

Let me write the file.

One more: rendering the rotating heptagon with rotating numbers inside—balls are in world coords (not rotating), only the heptagon rotates. Numbers rotate with each ball's own angle. Good.

Let me also make the heptagon semi-transparent fill so we see the background, but better solid dark.

Let me write now.

For the number rotation: since the whole canvas isn't rotated (only the heptagon shape is drawn rotated via ctx), the balls and their numbers are drawn in world space. Ball.angle is its own spin. Good.

Let me write the heptagon drawing using a transformed path (rotate about center) each frame.

Let me write:

```
function heptagonVertices(theta){
  const V=[];
  for(let k=0;k<7;k++){
    const a = theta + k*2π/7;
    V.push({x:cx+R*Math.cos(a), y:cy+R*Math.sin(a)});
  }
  return V;
}
```

Walls: for i in 0..6: v_i, v_{i+1 mod 7}. mid = avg. n_in = normalize(c - mid) (points to center). Store.

Draw: ctx.save, ctx.translate(cx,cy), ctx.rotate(theta), draw heptagon centered at origin, ctx.restore. But walls computed in world coords from V (which already in world coords). Either way fine. Let me compute V in world coords and use for both drawing and collision.

Draw heptagon:
```
ctx.beginPath();
for vertices: ctx.lineTo(v.x,v.y)
ctx.closePath(); fill, stroke.
```

Let me set R large enough: canvas 800, cx cy 400, R=340, top vertex y = 400 - 340 = 60, bottom = 740, left = 60, right=740. Fits with margin ~60. Good. Actually for a heptagon oriented with a vertex up, top vertex at cy-R=60. Good, fully visible.

Let me write the code now. I'll then run and inspect, tune.

Let me write a fairly complete, careful implementation.

I'll also add velocity/angle display? Not required. Keep minimal UI: Pause/Resume, Reset buttons. Maybe a speed note.

Let me write.

Let me think about the friction-induced spin magnitude. With μ~0.3 and impulses, spins develop. Good.

Let me also add: when balls rest on bottom wall, the rotating wall friction could spin them and the wall lifts them (centrifugal-like). Damping keeps stable.

Let me code and test.

I realize I should double-check the angular inertia term and friction denominator for walls. For wall, the "other body" infinite, so denom_t = 1/m + (offset×t)²/I = 1/m + r²/I. With I = 0.5*m*r², r²/I = r²/(0.5 m r²)= 2/m. So denom_t = 1/m + 2/m = 3/m. jt = -v_rel_t/(3/m) = -m*v_rel_t/3. Clamped by μ*P. P = -(1+e)v_rel_n*m. For v_rel_n<0, P>0. Clamp |jt| <= μ*P.

For ball-ball, denom_t = 1/m1+1/m2 + r1²/I1+r²/I2 = 1/m1+1/m2+2/m1+2/m2 = 3/m1+3/m2 = 3(1/m1+1/m2).

OK.

Let me define I = 0.5*m*r² for each ball. m = r² (density 1 area-ish; area=πr², but constant density cancels, just use m ∝ r²). Let me set m = r*r (so proportional). I = 0.5*m*r² = 0.5*r⁴. Fine.

Let me now write the file with all this.

Let me write helper for clamp.

Let me write the full thing.

Actually, let me reconsider gravity and damping for stability and nice motion. I'll start with:
g = 900, omega=2π/5, R=340, r=13, e_ball=0.6, e_wall=0.5, friction μ=0.25, cl=3 (linear damping coeff, a_damp = -cl*v), ca=2 (angular damping), Vmax=1500 (cap). sub_dt=1/180, iterations=3.

Let me write and test, then tune damping if runaway.

Let me write the HTML file.

I need to be careful with the requestAnimationFrame time and pause.

Let me write:

```
let last=0, paused=true initially? No, start running.
```

Start running after init.

Buttons: Pause/Resume toggles; Reset restores.

Let me write now.

Let me also draw numbers with a stroke outline for readability: fillText and strokeText.

Number font color: let me use black text with a whiteish stroke (strokeStyle = lighter). Balls colored per ball. Let me give each ball a hue based on index, fill style = hsl(hue,70%,55%), stroke dark. Number: white with black stroke.

Let me code.

Let me also add: ball slightly smaller than r so gap visible, no.

Write it.

I'll also include a small instruction text.

Let me write the file.

For the spin visibility, let me ensure angular velocities get reasonable magnitudes. If not visible, I'll scale up friction or initial.

Let me write and then test in browser.

Let me also reconsider: to make spin clearly visible from the start, I could give each ball a small initial angular velocity. But "give each ball angular motion" might be satisfied by physics-driven spin. I'll let physics drive it but also the friction at wall will spin them. To be safe and make numbers rotate visibly, I'll add a small initial av (e.g., 1 rad/s, random sign) so even before many collisions there's visible rotation. Actually that might look odd. Let me instead ensure friction produces spin and test. If numbers barely rotate, increase μ. I'll test.

Let me write the code.

Let me also make the rendering smooth.

Now writing.

Let me carefully write resolve functions with correct signs (using the derived formulas: ball-ball applying j when v_rel_n>0; wall applying when v_rel_n<0 with impulse along n_in).

Wait, I need to double-check ball-ball condition and signs once more with a clean convention to avoid bugs.

Ball-ball convention:
n = (b2 - b1)/dist, points from b1 to b2.
v_rel = v1 - v2 (velocity of b1 relative to b2).
v_rel_n = dot(v_rel, n).
If they're approaching, distance decreases => d(dist)/dt<0 => -v_rel_n <0 => v_rel_n>0. So approaching when v_rel_n>0.
Apply impulse when v_rel_n>0.
j_impulse magnitude along n: we add impulse J such that b2 gets +J*n (pushed away from b1 in +n, separating) and b1 gets -J*n.
Change in v_rel_n: v_rel_n = v1n - v2n. b1 gets -J*n/m1 => v1n -= J/m1. b2 gets +J*n/m2 => v2n += J/m2. So v_rel_n changes by -J/m1 - J/m2 = -J(1/m1+1/m2). New v_rel_n = v_rel_n - J(1/m1+1/m2). Want new = -e*v_rel_n. So -J(1/m1+1/m2) = -e*v_rel_n - v_rel_n = -(1+e)v_rel_n. Thus J = (1+e)v_rel_n/(1/m1+1/m2). With v_rel_n>0, J>0. ✓. b1.v -= J*n/m1, b2.v += J*n/m2. ✓ (matches my draft after correction).

Friction: t = perpendicular to n (tangential). v_rel_t = dot(v_rel, t). The friction opposes tangential relative motion. Impulse jt along t: b2 gets +jt*t, b1 gets -jt*t (so friction reduces v_rel_t). Change in v_rel_t = -jt*(1/m1+1/m2 + inertia terms). Want to reduce v_rel_t toward 0 (Coulomb). Max |jt| <= μ*J. jt = -v_rel_t / denom_t, clamped to [-μ*J, μ*J]. sign: if v_rel_t>0 (b1 moving along +t relative b2), friction should push b1 along -t => b1 gets -jt*t with jt>0 => b1.v -= jt*t/m1. With jt = -v_rel_t/denom_t <0 if v_rel_t>0... that gives b1.v -= (negative)*t = +|..|*t, pushing +t, wrong. Hmm.

Let me redo. We want friction impulse that reduces v_rel_t = dot(v1-v2,t). Friction on b1 should be opposite to b1's tangential relative motion. If v_rel_t>0, b1 moves relative along +t; friction on b1 = -t direction (reduces). So b1 gets impulse -|jt|*t, b2 gets +|jt|*t. Using jt as signed magnitude where b1 gets -jt*t, b2 gets +jt*t (so b1 along -jt*t). If jt>0, b1 gets -t. We want b1 to get -t when v_rel_t>0 (approaching tangential +t). So jt should be positive when v_rel_t>0. But standard: jt reduces relative velocity: new v_rel_t = v_rel_t - jt*denom_t. Want new = (1-f)*v_rel_t or 0 (clamped). For max reduction (kinetic), new=0 => jt = v_rel_t/denom_t. So jt = v_rel_t/denom_t (positive when v_rel_t>0). Clamp |jt| <= μ*J. Then b1.v -= jt*t/m1, b2.v += jt*t/m2. 

Wait but that gives friction impulse direction reducing v_rel_t: b1 gets -jt*t (opposite t when jt>0), reducing b1's +t motion. ✓. Good. So jt = clamp(v_rel_t/denom_t, -μJ, μJ). 

I had sign flipped before. Let me use this corrected version. denom_t includes inertia: denom_t = 1/m1+1/m2 + (r1×t)²/I1 + (r2×t)²/I2. We computed (r1×t)=r1 (scalar) and (r2×t) = -r2. Squared both r². So denom_t = 1/m1+1/m2 + r1²/I1 + r2²/I2.

Angular: b1 av change from friction impulse -jt*t applied at offset (contact point) = n*r1. Torque τ1 = offset1 × (-jt*t) = -jt*(offset1×t) = -jt*r1 (computed offset1×t = r1). av1 += τ1/I1 = -jt*r1/I1. Hmm wait earlier for ball-ball I derived av1 += -jt*r1/I1? Let me recompute: τ = offset × F. offset1 = (nx*r1, ny*r1). F1 = -jt*t = -jt*(-ny,nx) = (jt*ny, -jt*nx). τ1 = offset1.x*F1.y - offset1.y*F1.x = (nx*r1)*(-jt*nx) - (ny*r1)*(jt*ny) = -jt*r1*(nx²+ny²) = -jt*r1. So av1 += -jt*r1/I1 = av1 -= jt*r1/I1.

For b2: offset2 = (-nx*r2,-ny*r2). F2 = +jt*t = jt*(-ny,nx). τ2 = offset2.x*F2.y - offset2.y*F2.x = (-nx*r2)*(jt*nx) - (-ny*r2)*(jt*(-ny)) = -jt*r2*nx² - (-ny*r2)*(-jt*ny) = -jt*r2*nx² - jt*r2*ny² = -jt*r2. av2 += -jt*r2/I2 = av2 -= jt*r2/I2.

Hmm, so both av1 and av2 decrease by jt*r/I. That seems odd—both spin same direction? Let me sanity check: two balls, b1 left, b2 right, n points right (+x). t = (-ny,nx) = (0,1) (since n=(1,0), t=(0,1), +y/tangent up). They slide: b1 moving up relative (+t), v_rel_t>0. Friction opposes b1's up motion => b1 gets -t (down force) => F1 = -jt*t = (0,-jt), down. torque on b1: offset1=(r1,0) (right of center), force down at right side => clockwise (in y-down, clockwise visually; in math y-up that's... let me think in y-down: force down at right side of circle => rotates so right goes down => clockwise rotation, av (I'll define positive av as CCW visually). Let me define av positive CCW in canvas (y-down)? 

Ugh, the sign of av and rotation direction. Let me just define av such that angle += av*dt, and visually the ball rotates; the direction (CW or CCW) depends on sign. Let me define av positive => CCW in standard math (y-up). But canvas y is down, so to rotate a drawn object CCW visually, ctx.rotate(av*dt) with av>0 rotates CCW in canvas = clockwise visually? Confusing.

Let me just define: visual rotation of balls by ctx.rotate(ball.angle); ball.angle += ball.av * dt. The sign of av determines visual direction. It doesn't matter physically which is CW. The torque sign will follow from physics if I'm consistent with the 2D cross product convention.

Let me define 2D cross scalar cr(a,b)=a.x*b.y-a.y*b.x (z-component). And "av × offset" for velocity = av * perp(offset) where for av as a z-scalar and y-down... Standard: in y-up, ω×r = (-ω*ry, ω*rx). The velocity at offset due to rotation ω (positive CCW in y-up) is v = ω*(-ry, rx). In y-down, positive ω visually-CCW corresponds to... let me just use the formula v = ω * (-ry, rx) consistently, treating ω as the math CCW. Then visually (y-down) it'll appear clockwise. Whatever. The physics: torque τ = cr(offset, F) (z-component). av += τ/I * dt. With τ=cr(offset,F). Let me verify ball-ball friction gives consistent spin.

Let me recompute with cross cr(a,b)=ax*by-ay*bx:
τ1 = cr(offset1, F1) where offset1=(nx*r1,ny*r1), F1=-jt*t, t=(-ny,nx), so -jt*t=(jt*ny, -jt*nx). 
τ1 = cr((nx*r1,ny*r1),(jt*ny,-jt*nx)) = nx*r1*(-jt*nx) - ny*r1*(jt*ny) = -jt*r1*(nx²+ny²) = -jt*r1. av1 += τ1/I1*dt_sub = -jt*r1/I1. So av1 -= jt*r1/I1. ✓ same.

τ2 = cr(offset2, F2), offset2=(-nx*r2,-ny*r2), F2=jt*t=(jt*(-ny),jt*nx)=(-jt*ny, jt*nx).
τ2 = (-nx*r2)*(jt*nx) - (-ny*r2)*(-jt*ny) = -jt*r2*nx² - jt*r2*ny² = -jt*r2. av2 += -jt*r2/I2. So av2 -= jt*r2/I2. Both decrease.

Physically: b1 left, b2 right sliding up (b1 up relative). Friction makes b1 want to rotate so its contact point slows. b1 moving up relative to b2; friction down on b1. Force down at right side (offset +x) → in y-down, down-right force → torque? Force at (r1,0) downward (0,-jt): cr((r1,0),(0,-jt)) = r1*(-jt) - 0 = -jt*r1. av1 decreases. The visual rotation: angle += av*dt, av negative → rotates... with my draw (canvas y-down), a negative av via rotate(angle) where angle decreases → rotates clockwise visually. The contact point was moving up; friction down; the bottom... hmm. Anyway, consistent. Both balls spin same sign because the tangential relative motion is symmetric in that frame. Actually physically two balls sliding tangentially: friction acts to spin them in opposite directions usually? Let me think: b1 left moving up relative to b2; relative motion at contact: b1's contact point moves up. Friction on b1 is down (opposing), on b2 is up. Force down on b1 at its right side → b1 spins clockwise (visually). Force up on b2 at its left side → b2 spins... up force on left side → counterclockwise? In y-down, up = -y. Force at offset left (b2 contact at left side, offset (-r2,0)) upward (-y): cr((-r2,0),(0, -jt_up)). F2=jt*t with jt>0, t=(0,1) up? t=(-ny,nx)=(0,1) = +y = down in canvas. Hmm t=+y is downward visually. So F2 = jt*(0,1) = down on b2. Wait I said friction up on b2. Let me recheck sign.

I think I'm overcomplicating. The relative tangential velocity sign handles direction. The clamp and magnitude are what matter; the signs are self-consistent if I derived from cr. Let me trust the math: jt = clamp(v_rel_t/denom_t, -μJ, μJ), b1.v -= jt*t/m1, b2.v += jt*t/m2, av1 -= jt*r1/I1, av2 -= jt*r2/I2. 

Hmm but the av signs both being "-jt" — let me re-examine τ2 sign. Actually I might have an error. Let me recompute F2. The friction impulse on b2 = +jt*t (b2 gets +jt*t). For b1 moving up (+t, t=+y down) relative, friction opposes: b1 should get -jt*t (down... wait t=+y is down; -t is up). Let me re-express: if v_rel_t>0 means b1 moves along +t relative to b2. Friction on b1 opposite to its tangential relative motion => b1 gets impulse -jt*t (jt>0). b2 gets +jt*t (Newton's 3rd law). Good, that's what I have: b1 -= jt*t, b2 += jt*t.

Now torque signs. Let me re-derive τ2 for b2 with F2=+jt*t, t=(-ny,nx). In the example n=(1,0),t=(0,1)(+y, down). v_rel_t>0 => b1 moves down (+y) relative b2's contact. Friction: b1 up (-y) => F1 = -jt*t = (0,-jt) (up). Wait -jt*t with t=(0,1): F1=(0,-jt) = -y = up. Good b1 pushed up, opposing its down motion. b2 gets F2=+jt*t=(0,jt)=down. So b2 pushed down. 

b1 contact offset = n*r1 = (r1,0) (right side). F1=(0,-jt) up at right side. In y-down canvas, up=-y. cr((r1,0),(0,-jt)) = r1*(-jt) - 0*0 = -jt*r1. av1 -= jt*r1/I1. Negative av → angle decreases. Visual rotation angle-= => rotate... with ctx.rotate(angle) and angle decreasing, in y-down that's... a point at angle α (from center) at (cosα,sinα); decreasing α rotates clockwise? Let me not worry.

b2 contact offset = -n*r2 = (-r2,0) (left side). F2=(0,jt) down at left side. cr((-r2,0),(0,jt)) = (-r2)*jt - 0*0 = -jt*r2. av2 -= jt*r2/I2. Same sign as av1. Both decrease. Physically: b1 moving down relative, friction up on b1 (at right side → spins b1 one way), down on b2 (at left side → spins b2 same way?). For rolling without slip at a contact between two sliding bodies, they'd spin in the same direction (both such that contact points match). Yes! Two balls in contact sliding tangentially: friction spins them in the same sense (both clockwise or both counterclockwise) to reduce relative slip. So same sign is correct! Good, my derivation is consistent.

Great. So ball-ball friction correct.

Now wall friction: wall at bottom (for example). For ball hitting bottom wall: n_in points up (toward center). t = (-ny,nx) = (-1, 0)? n_in=(nx,ny)=(0,-1)? Wait bottom wall n_in points up → n_in=(0,1)? Center above the bottom wall, so inward (toward center) is up = +y? In canvas +y is down. Bottom wall: midpoint below center, inward toward center = upward = -y direction. So n_in=(0,-1)? dot(ball.x-mid, n_in)... ball is above wall (inside), ball - mid points up (-y), n_in=(0,-1), dot = (Δy)*(-1); if ball above, Δy<0 (up is negative y), dot>0. Yes d>0 inside. Good n_in=(0,-1) for bottom. t=(-ny,nx) = -(-1,0)=(1,0)? t = (-ny, nx) = (-(-1), 0) = (1,0) = +x (right). 

Ball moving right relative to wall (v_rel_t>0). Friction on ball along -t = -x (left). b1 gets -jt*t (jt>0) => -x. F_ball = (jt*ny, -jt*nx)? Wait b1.v -= jt*t/m1. In wall case, "b2" is wall (infinite). Impulse on ball = -jt*t (like b1 in ball-ball). So ball.v -= jt*t/m. 

torque τ = cr(offset, F) with offset = contact point relative to ball center. Contact point on ball = ball.x - n_in*r (toward wall). n_in=(0,-1) (up-ish? n_in=(0,-1) means inward is -y, i.e., up). offset = -n_in*r = -(0,-1)*r = (0, r) = +y (down), meaning contact point is below center (toward bottom wall). Yes ball touches bottom wall below center, offset (0,r). F = -jt*t = -jt*(1,0)=(-jt,0) left. cr((0,r),(-jt,0)) = 0*0 - r*(-jt) = r*jt. av += τ/I = jt*r/I >0 if jt>0. 

So ball moving right along bottom wall, friction left, spins av positive (which via my convention = ... visual). Anyway consistent. Good.

So for wall: ball.v -= jt*t/m + normal impulse; av += τ/I where τ = cr(offset, F_total_impulse). Let me just compute total impulse (normal + friction) on ball and torque = cr(offset, impulse_vec). Simpler and less error-prone:
P_total = normal_impulse_vec + friction_impulse_vec.
ball.v += P_total/m.
av += cr(offset, P_total)/I.

Where normal_impulse on ball = (J along n_in) = J*n_in (J as derived; recall for wall, impulse on ball = (j/m)*n_in... let me recompute. Earlier: j (I'll call Jn) = -(1+e)*v_rel_n (per mass) → impulse on ball = m*acceleration... let me just define impulse vector directly.

For wall (object2 infinite, n = n_in from wall to ball = inward):
J (impulse magnitude along n) such that b1.v += J/m * n, with J = (1+e)*v_rel_n... wait earlier for ball-ball J = (1+e)v_rel_n/(1/m1+1/m2) and b1 gets -J*n. For wall, the analog with n pointing wall->ball (inward), and ball is object 1, wall object 2: b1 gets +J*n (since n from 2 to 1). Using formula b1.v += J/m * n where J = (1+e)*v_rel_n*(...)? Let me recompute for wall: relative v_rel_n = dot(v1 - v2, n) = dot(ball.v - v_wall, n_in). Approaching (toward wall) => v_rel_n<0 (since n_in inward, ball moving outward/toward wall is -n_in). Apply when v_rel_n<0. J = (1+e)*v_rel_n / (1/m + 0) ... but careful: formula J = -(1+e)v_rel_n/... sign depends on n convention. Let me derive for wall directly: n=n_in (wall->ball, inward). b1 gets +J*n/m, b2(wall) gets -J*n/m2 → 0. v_rel_n = v1n - v2n. b1.v += J*n/m → v1n += J/m. v_rel_n new = v1n - v2n + J/m = v_rel_n + J/m. Want new = -e*v_rel_n (reflect, since approaching v_rel_n<0, after should be positive moving inward). So J/m = -e*v_rel_n - v_rel_n = -(1+e)v_rel_n. J = -(1+e)v_rel_n*m. v_rel_n<0 → J>0. b1.v += (J/m)*n_in = -(1+e)v_rel_n*n_in (positive along n_in, inward — correct, bounces inward). ✓.

So wall: if v_rel_n<0: Jn = -(1+e)*v_rel_n*m; impulse_n = Jn*n_in. Then friction jt = clamp(v_rel_t/denom_t, -μ*Jn/|?|,...). Clamp |jt| <= μ*Jn. denom_t = 1/m + r²/I. jt = clamp(v_rel_t/denom_t, -μ*Jn, μ*Jn). impulse_t = jt*t. total impulse = impulse_n + impulse_t. ball.v += impulse/m. av += cr(offset, impulse)/I. offset = -n_in*r (contact toward wall). Wait the contact point offset for angular: the force is applied at the contact point = ball.x - n_in*r (the point touching wall). relative offset = (ball.x - n_in*r) - ball.x = -n_in*r. So offset = -n_in*r. yes.

But careful: n_in for wall points inward; contact point on ball is at ball.x - n_in*r? If n_in points inward (toward center) and wall is outward from ball, the contact point is in the outward direction = -n_in from center. offset = -n_in*r. ✓.

For ball-ball, similarly offset1 = n*r1 (contact toward b2, n from b1 to b2), offset2 = -n*r2.

Let me now also reconsider normal collision condition for wall: penetration when d < r, where d = dot(ball.x - mid, n_in) (signed distance inward, positive inside). d<r means ball crosses wall. But also need relative velocity for impulse; I can resolve velocity even if slightly penetrating but only if approaching. I'll do positional correction always when d<r, and velocity impulse when v_rel_n<0 (and d<r, i.e., touching). Apply friction when sliding (v_rel_t nonzero) and in contact (d<r). Actually friction should apply during contact; if d slightly < r apply friction impulse. Let me apply both impulse and friction whenever d<r (clamped to not create separation issues). To avoid adding energy when separating and not in contact, only apply when d<r. Fine.

Let me now write code. I'll include multiple collision iterations.

Let me write it all out now.

I also want to make sure the wall velocity v_wall is computed at foot = projection of ball center onto wall line: foot = ball.x - d*n_in (since d = dot(ball.x-mid, n_in), foot = ball.x - d*n_in). Hmm check: ball.x - d*n_in; is foot on the wall line? The wall line passes through mid with direction along wall. foot = ball.x - d*n_in projects ball onto line (removing inward component). Yes foot on line. v_wall = ω*( -(foot.y-cy), foot.x-cx ).

Now coding.

Let me write the HTML file fully.

I'll also add a small HUD with pause/resume/reset.

Let me write.

Let me think about initial angular velocity to ensure visible spin early. During the initial fall, before hitting walls, no spin. The first impact spins them. Numbers will rotate after first bounce. That satisfies "spin is visible" once collisions occur. But to be safe for "sample several moments of initial fall"—spin visible after first wall contact. Should be fine. But let me also give a tiny random initial av (like ±0.5 rad/s) so numbers show rotation even during free fall, more clearly demonstrating "numbers rotate with balls." Hmm, but that might look unphysical. The requirement: "Give each ball angular motion and rotate its number with the ball so spin is visible." I think physics-driven spin is the intent. But to robustly show spin visible from the start, a small initial spin helps demonstration. I'll add a small initial av (random sign, ~0.8 rad/s) — it's angular motion, and reset restores it. Actually reset restores initial arrangement; restoring initial av too is consistent. Let me add small initial spin to make the rotation visible immediately and clearly. That's a reasonable design choice. I'll keep it small so it doesn't dominate.

Hmm, but then "angular motion" from physics is also nice. Let me add small initial av AND physics friction. Good.

Let me finalize constants after a test.

Let me write the file.

Let me write the code carefully.

I'll now produce index.html.

After writing, I'll open in browser, inspect, tune.

Let me also add that the heptagon should be "fully visible"—R=340 within 800 canvas is fine (vertex at 60..740). Good.

Let me write.

Code:

```html
<!DOCTYPE html>
<html>
<head>
<meta charset="utf-8">
<style>
html,body{margin:0;height:100%;background:#111;display:flex;align-items:center;justify-content:center;}
canvas{background:#000;display:block;border:2px solid #555;border-radius:6px;}
.ui{position:absolute;top:12px;right:12px;display:flex;gap:8px;}
button{background:#2a2a3a;color:#fff;border:1px solid #666;border-radius:5px;padding:8px 14px;font-size:15px;cursor:pointer;}
button:hover{background:#3a3a5a;}
</style>
</head>
<body>
<canvas id="c" width="800" height="800"></canvas>
<div class="ui"><button id="pause">Pause</button><button id="reset">Reset</button></div>
<script>
const canvas=document.getElementById('c');
const ctx=canvas.getContext('2d');
const W=800,H=800;
const cx=W/2,cy=H/2;
const PI=window.PI= Math.PI;
const R=340; // heptagon vertex radius
const N=7;
const omega=2*PI/5; // rad/s, one rev per 5s
const G=900; // gravity px/s^2
const ballR=13;
const NB=20;
const eWall=0.6;
const eBall=0.5;
const mu=0.35; // friction coeff
const linDamp=4; // linear damping coeff (a=-c*v)
const angDamp=3;
const Vmax=1800;
const sub=1/180;
const iter=3;

let balls=[];
let init=[];
let theta=0;
let paused=false;
let last=0;

function makeBall(x,y){
  const r=ballR;
  const m=r*r;
  const I=0.5*m*r*r;
  return {x,y,vx:0,vy:0,angle:0,av:(Math.random()-0.5)*1.6,r,m,I};
}
function initSim(){
  balls=[];
  const sx=2.3*ballR, sy=2.3*ballR;
  const cols=5,rows=4;
  const startX=cx - (cols-1)*sx/2;
  const startY=cy - 40;
  for(let i=0;i<NB;i++){
    const col=i%cols,row=(i/cols)|0;
    const x=startX+col*sx + (row%2?sx/2:0);
    const y=startY+row*sy;
    const b=makeBall(x,y);
    balls.push(b);
    init.push({x,y,angle:0,av:b.av});
  }
}
function resetSim(){
  for(let i=0;i<NB;i++){
    balls[i].x=init[i].x; balls[i].y=init[i].y;
    balls[i].vx=0; balls[i].vy=0;
    balls[i].angle=init[i].angle; balls[i].av=init[i].av;
  }
  theta=0;
}
function walls(th){
  const V=[];
  for(let k=0;k<N;k++){
    const a=th+k*2*PI/N;
    V.push({x:cx+R*Math.cos(a), y:cy+R*Math.sin(a)});
  }
  const Ws=[];
  for(let i=0;i<N;i++){
    const v1=V[i], v2=V[(i+1)%N];
    const mid={x:(v1.x+v2.x)/2, y:(v1.y+v2.y)/2};
    let nx=cx-mid.x, ny=cy-mid.y;
    let ln=Math.hypot(nx,ny)||1; nx/=ln; ny/=ln; // inward normal
    Ws.push({mid,nx,ny,v1,v2});
  }
  return Ws;
}
function cross(ax,ay,bx,by){return ax*by-ay*bx;}
function clamp(v,lo,hi){return v<lo?lo:v>hi?hi:v;}
function resolveBallWall(b,w){
  const {x,y,vx,vy,r,m,I,av}=b;
  const {mid,nx,ny}=w;
  const dx=x-mid.x, dy=y-mid.y;
  let d=dx*nx+dy*ny; // signed inward dist
  if(d<r){
    // positional correction
    const pen=r-d;
    b.x+=nx*pen; b.y+=ny*pen;
    // wall velocity at foot
    const fx=b.x - d*nx... 
```

Wait foot should use the original d before correction ideally. Let me compute foot from d (signed dist). foot = b.x - (d)*... Actually foot = mid + projection onto line. foot = b - d*n_in (removing inward component): foot.x = x - d*nx, foot.y=y-d*ny. Let me compute foot with the d before correction. Then v_wall = ω*(-(foot.y-cy), foot.x-cx).

But after positional correction b.x changes; for velocity impulse use original contact. Let me compute foot and v_wall before correction.

Let me restructure: compute d, foot, v_wall; if d<r resolve.

```
let d=dx*nx+dy*ny;
if(d<r){
  const fx=x-d*nx, fy=y-d*ny;
  const vwx=omega*(-(fy-cy)), vwy=omega*(fx-cx);
  // relative velocity at contact
  const offsetx=-nx*r, offsety=-ny*r; // contact point offset (toward wall)
  // ball velocity at contact including spin: v_contact = v + av×offset
  const cvx=vx + av*(-offsety), cvy=vy + av*offsetx;
  let rvx=vx-vwx, rvy=vy-vwy; // relative velocity of center? 
```

Hmm, for collision impulse with rotating wall, the relevant relative velocity is at the contact point. v_rel_contact = v_ball_contact - v_wall_contact. v_ball_contact = ball.v + av×offset. v_wall_contact = v_wall. 

n_in = (nx,ny). v_rel_n = dot(v_ball_contact - v_wall, n_in) = dot((vx+av*(-offsety) - vwx, vy+av*offsetx - vwy), (nx,ny)).

Let me compute:
```
const vbcvx = vx + av*(-offsety); // = vx + av*(-(-ny*r)) = vx + av*ny*r? offsety=-ny*r, -offsety=ny*r. so vbcx = vx + av*ny*r.
const vbcy = vy + av*offsetx; // offsetx=-nx*r, so vbcy = vy + av*(-nx*r)=vy - av*nx*r.
let rvx = vbcx - vwx;
let rvy = vbcy - vwy;
let vreln = rvx*nx + rvy*ny;
if(vreln<0){
  const Jn = -(1+eWall)*vreln*m; // impulse magnitude along n_in
  // friction
  const tx=-ny, ty=nx; // tangent
  let vrelt = rvx*tx + rvy*ty;
  const denom = 1/m + r*r/I;
  let jt = vrelt/denom;
  jt = clamp(jt, -mu*Jn, mu*Jn);
  // total impulse
  const ix = Jn*nx + jt*tx;
  const iy = Jn*ny + jt*ty;
  b.vx += ix/m; b.vy += iy/m;
  b.av += cross(offsetx,offsety, ix,iy)/I;
}
```
Wait sign: b1.v += J/m * n for wall (n from wall to ball = inward). We derived b.v += (Jn/m)*n_in with Jn = -(1+e)vreln*m (vreln<0 → Jn>0). impulse vector = Jn*n_in. Plus friction impulse jt*t (where b gets -jt*t? In ball-ball b1 got -jt*t). Here ball is "b1", wall "b2". So ball gets -jt*t, wall +jt*t (ignored). Let me recompute friction sign for wall: the friction impulse on the ball. Earlier ball-ball: b1.v -= jt*t/m1 with jt = clamp(v_rel_t/denom,...). The relative tangential velocity v_rel_t = dot(v1-v2, t). Friction opposes: b1 gets impulse -jt*t (jt>0 reduces v_rel_t). So for wall (ball is b1), ball.v -= jt*t/m? and the angular torque = cr(offset, (-jt*t)).

Hmm I'm now mixing "total impulse" signs. Let me be systematic.

For wall: ball = body 1, wall = body 2 (infinite). n = n_in (from wall 2 to ball 1). Approaching when v_rel_n = dot(v1_contact - v2_contact, n) < 0? v1_contact - v2_contact: if ball moves toward wall (outward, -n_in direction)... v_rel_n = dot(ball_contact_v - wall_v, n_in). Ball moving outward (toward wall) means ball_contact_v has negative n_in component, so dot <0. Yes approaching => v_rel_n<0. ✓.

Impulse magnitude J (along n, on body 1): body1 gets +J*n/m, body2 gets -J*n/m2(=0). J = -(1+e)*v_rel_n / (1/m). = -(1+e)v_rel_n*m. (v_rel_n<0 → J>0). body1.v += J*n_in/m. ✓ (b1.v += Jn*n_in/m, Jn=J). Good, impulse_n on ball = Jn*n_in (with Jn = -(1+e)v_rel_n*m). ball.v += impulse_n/m = Jn*n_in/m. 

Friction: coefficient, impulse jt along t. body1 gets -jt*t/m? Standard: the total impulse on body1 = J*n + jt*t (where jt chosen to reduce tangential rel velocity). We defined in ball-ball that b1 gets -jt*t. So here ball.v += (-jt*t)/m, with jt = clamp(v_rel_t/denom, -μJ, μJ), denom=1/m+r²/I. Let me recompute: we want to reduce v_rel_t (tangential relative). new v_rel_t = v_rel_t - jt*(1/m+r²/I) = v_rel_t - jt*denom. Set to 0 => jt = v_rel_t/denom (positive if v_rel_t>0). So ball gets -jt*t (jt>0) reducing its t motion. ✓. So friction impulse on ball = -jt*t.

Total impulse on ball = Jn*n_in + (-jt*t) = (Jn*nx - jt*tx, Jn*ny - jt*ty). 

torque = cr(offset, impulse) where offset = contact offset = -n_in*r = (-nx*r,-ny*r). 
av += cr(offset, impulse)/I.

Let me recompute cr for the example to ensure spin direction reasonable; trust math.

Let me re-derive denom and clamp using Jn (impulse magnitude, positive). μ*J is max friction impulse magnitude. jt = clamp(v_rel_t/denom, -μ*Jn, μ*Jn). Good.

So in code, total impulse vector on ball:
ix = Jn*nx - jt*tx
iy = Jn*ny - jt*ty
with Jn = -(1+eWall)*vreln*m, jt = clamp(vrelt/denom, -mu*Jn, mu*Jn), vreln<0, tx=-ny,ty=nx, denom=1/m+r²/I.

Ball-ball friction: earlier I had b1 gets -jt*t too, and jt = clamp(v_rel_t/denom), denom=1/m1+1/m2 + r1²/I1+r²/I2. And Jn = (1+e)v_rel_n/(1/m1+1/m2). impulse on b1 = -Jn*n - jt*t (n from b1 to b2, b1 gets -Jn*n). b1.v += (-Jn*n - jt*t)/... wait b1.v += impulse/m1 where impulse = -Jn*n + (-jt*t) = (-Jn*nx - jt*tx, -Jn*ny-jt*ty). Hmm but earlier for ball-ball I wrote b1.v -= J*n/m1 (J along n from 1 to2, b1 gets -J*n). J = (1+e)v_rel_n/(1/m1+1/m2). So impulse on b1 = -J*n, b1.v += (-J*n)/m1. So impulse vector = -J*n - jt*t. b1.v += impulse/m1; av1 += cr(offset1, impulse)/I1. offset1 = n*r1. 

Let me unify: define for each body the impulse it receives, then v += impulse/m, av += cr(offset,impulse)/I. For ball-ball: body1 receives impulse1 = -J*n - jt*t; body2 receives impulse2 = +J*n + jt*t (opposite). 

For wall: body(ball) receives impulse = +Jn*n_in - jt*t (since n from wall->ball is inward = +n_in for the ball; Jn>0 along n_in). impulse = Jn*n_in - jt*t. 

So I can write a helper applyImpulse(b, offset, impulse) { b.v += impulse/m; b.av += cross(offset,impulse)/I; }.

For ball-ball: 
impulse1 = (-Jn*nx - jt*tx, -Jn*ny - jt*ty); offset1 = (n*r1) wait offset1 = contact offset from b1 center = n*r1 = (nx*r1,ny*r1). 
impulse2 = (+Jn*nx + jt*tx, ...). offset2 = -n*r2 = (-nx*r2,-ny*r2).

For wall:
impulse = (Jn*nx - jt*tx, Jn*ny - jt*ty); offset = -n_in*r = (-nx*r,-ny*r). (here nx,ny = n_in).

Good, unified. Let me just code each separately to avoid confusion.

Let me write ball-ball:

```
function resolveBallBall(i,j){
  const a=balls[i], b=balls[j];
  let dx=b.x-a.x, dy=b.y-a.y;
  let dist=Math.hypot(dx,dy);
  if(dist===0){dx=1;dy=0;dist=1;}
  const rsum=a.r+b.r;
  if(dist<rsum){
    const nx=dx/dist, ny=dy/dist;
    const pen=rsum-dist;
    const invA=1/a.m, invB=1/b.m;
    const tot=invA+invB;
    a.x-=nx*pen*(invB/tot); a.y-=ny*pen*(invB/tot);
    b.x+=nx*pen*(invA/tot); b.y+=ny*pen*(invA/tot);
    // relative velocity at contact
    const offAx=nx*a.r, offAy=ny*a.r; // contact on a toward b
    const offBx=-nx*b.r, offBy=-ny*b.r;
    const vax=a.vx+a.av*(-offAy), vay=a.vy+a.av*offAx;
    const vbx=b.vx+b.av*(-offBy), vby=b.vy+b.av*offBx;
    let rvx=vax-vbx, rvy=vay-vby;
    const nxn = nx*nx+ny*ny; //1
    let vreln=rvx*nx+rvy*ny;
    if(vreln>0){
      let J=(1+eBall)*vreln/(1/a.m+1/b.m);
      const tx=-ny, ty=nx;
      let vrelt=rvx*tx+rvy*ty;
      const denom=1/a.m+1/b.m + a.r*a.r/a.I + b.r*b.r/b.I;
      let jt=vrelt/denom;
      jt=clamp(jt,-mu*J,mu*J);
      // impulse on a = -J*n - jt*t
      a.vx += (-J*nx - jt*tx)/a.m;
      a.vy += (-J*ny - jt*ty)/a.m;
      a.av += cross(offAx,offAy, -J*nx - jt*tx, -J*ny - jt*ty)/a.I;
      b.vx += ( J*nx + jt*tx)/b.m;
      b.vy += ( J*ny + jt*ty)/b.m;
      b.av += cross(offBx,offBy, J*nx + jt*tx, J*ny + jt*ty)/b.I;
    }
  }
}
```

Wait vreln>0 for approaching (computed earlier). ✓. J>0. impulse on a = -J*n - jt*t. Good.

Angular inertia term in denom: a.r²/a.I. With a.I=0.5*a.m*a.r² → a.r²/a.I = 2/a.m. denom = 3/a.m + 3/b.m basically. Fine.

Now ball-wall:

```
function resolveBallWall(b,w){
  const {mid,nx,ny}=w;
  let dx=b.x-mid.x, dy=b.y-mid.y;
  const d=dx*nx+dy*ny;
  if(d<b.r){
    const pen=b.r-d;
    b.x+=nx*pen; b.y+=ny*pen;
    const fx=b.x-d*nx... 
```
Wait foot computation: foot = b.x - d*n_in? The signed distance d = dot(b.x-mid, n_in). foot on line = b.x - d*n_in. But I just moved b.x by pen. Let me compute foot and v_wall BEFORE moving, then move. Let me reorder: compute foot & v_wall from current b (before correction). 

```
function resolveBallWall(b,w, th){
  const {mid,nx,ny}=w;
  let dx=b.x-mid.x, dy=b.y-mid.y;
  const d=dx*nx+dy*ny; // before correction
  if(d<b.r){
    const fx=b.x-d*nx, fy=b.y-d*ny;
    const vwx=omega*(-(fy-cy)), vwy=omega*(fx-cx);
    const offx=-nx*b.r, offy=-ny*b.r;
    // velocity at contact including spin
    const vcx=b.vx+b.av*(-offy), vcy=b.vy+b.av*offx;
    let rvx=vcx-vwx, rvy=vcy-vwy;
    let vreln=rvx*nx+rvy*ny;
    if(vreln<0){
      const Jn=-(1+eWall)*vreln*b.m; // along n_in (into polygon)
      const tx=-ny, ty=nx;
      let vrelt=rvx*tx+rvy*ty;
      const denom=1/b.m + b.r*b.r/b.I;
      let jt=vrelt/denom;
      jt=clamp(jt,-mu*Jn,mu*Jn);
      // impulse on ball = Jn*n_in - jt*t
      const ix=Jn*nx - jt*tx;
      const iy=Jn*ny - jt*ty;
      b.vx+=ix/b.m; b.vy+=iy/b.m;
      b.av+=cross(offx,offy,ix,iy)/b.I;
    }
    // positional correction
    b.x+=nx*pen; b.y+=ny*pen;
  }
}
```

Wait foot: foot = b.x - d*n_in? foot.x = b.x - d*nx, foot.y=b.y-d*ny. Yes fx=b.x-d*nx. Good. And d computed before correction (I compute d, then inside if). But I moved b after computing foot—good, foot from original. Then positional correction at end. Good.

But there's a subtlety: after positional correction, the ball might still be moving toward the wall if vreln condition didn't trigger (vreln>=0 means moving away, fine). Good.

Let me double check wall velocity sign and that it accounts for rotation properly. foot is on wall line; v_wall = omega * (-(fy-cy), fx-cx). Good (derived dp/dt).

Now the main step:

```
function step(dt){
  const nsub=Math.ceil(dt/sub);
  let s=dt/nsub;
  for(let k=0;k<nsub;k++){
    // forces
    for(const b of balls){
      b.vy+=G*s;
      // damping
      b.vx*=(1-linDamp*s); b.vy*=(1-linDamp*s); // hmm damping as multiplier
      b.av*=(1-angDamp*s);
    }
    // collisions
    const ws=walls(theta);
    for(let it=0;it<iter;it++){
      for(let i=0;i<NB;i++) resolveBallWall(balls[i], ... ws?);
      for(let i=0;i<NB;i++) for(let j=i+1;j<NB;j++) resolveBallBall(i,j);
    }
    // integrate
    for(const b of balls){
      b.x+=b.vx*s; b.y+=b.vy*s;
      b.angle+=b.av*s;
      // speed cap
      const sp=Math.hypot(b.vx,b.vy);
      if(sp>Vmax){const f=Vmax/sp; b.vx*=f;b.vy*=f;}
    }
    // containment safety: reflect by walls again? skip.
  }
  theta+=omega*dt; // advance heptagon
}
```

Hold on: walls() depends on theta; during step theta is constant (I advance after). Good, use walls(theta) once. But I call walls inside sub-step loop—wasteful but fine; compute once outside: const ws=walls(theta). Move outside loop.

Damping: using multiplier (1-c*s) each substep. For small s fine. Could use a=-c*v. Multiplier is OK.

Wait, applying damping as b.vx *= (1-linDamp*s): linDamp=4, s=1/180→ factor 1-4/180=0.9778. Over a second (180 substeps) → 0.9778^180 ≈ e^{-7.2}≈0.0007, strong damping. That might over-damp. Let me reduce linDamp to ~2. Then factor 1-2/180=0.9889, ^180≈e^{-3.6}=0.027, velocity dies in ~1s. Hmm gravity-driven; steady state v~g/linDamp = 900/2=450 px/s. That seems OK. Actually for damping force -c*v, terminal v=g/c. c=2 → vt=450. c=4→vt=225. Let me use c=2.5. Eh, but collisions inject energy; need enough damping to counter rotation pumping. Let me pick c=3 (vt=300) and test, adjust.

Actually the energy injection from rotating walls: when ball at wall moves inward, wall pushes it. The ball also gets lifted (centripetal). Let me just test and tune damping. Let me start c=2.5.

Hmm, also the angular damping angDamp: spin decays; set to ~2 so spins settle.

Let me reduce mu maybe 0.2 to not over-spin. Let me test.

Let me also reconsider: positional correction may push a ball into another ball; that's why we iterate. iterations=3 with substeps should keep stable.

Let me also add: after all, to be safe, clamp ball inside (re-run wall constraint once more). The iterations handle.

Let me set Vmax=1600.

Now draw function:

```
function draw(){
  ctx.clearRect(0,0,W,H);
  // heptagon
  const V=[];
  for(let k=0;k<N;k++){const a=theta+k*2*PI/N; V.push([cx+R*cos(a),cy+R*sin(a)]);}
  ctx.beginPath();
  for(const p of V) ctx.lineTo(p[0],p[1]);
  ctx.closePath();
  ctx.fillStyle='#0b0d10'; ctx.fill();
  ctx.strokeStyle='#9bc1ff'; ctx.lineWidth=3; ctx.stroke();
  // balls
  for(let i=0;i<NB;i++){
    const b=balls[i];
    const hue= (i/NB)*360;
    // ball circle
    ctx.beginPath(); ctx.arc(b.x,b.y,b.r,0,2*PI);
    const grd=ctx.createRadialGradient(b.x,b.y-b.r*0.4,b.r*0.2,b.x,b.y,b.r);
    grd.addColorStop(0,hsl(hue,70%,65%));
    grd.addColorStop(1,hsl(hue,70%,40%));
    ctx.fillStyle=grd; ctx.fill();
    ctx.strokeStyle='#111'; ctx.lineWidth=1.5; ctx.stroke();
    // number
    ctx.save();
    ctx.translate(b.x,b.y);
    ctx.rotate(b.angle);
    ctx.font=`bold ${b.r*1.2}px sans-serif`;
    ctx.fillStyle='#fff'; ctx.textAlign='center'; ctx.textBaseline='middle';
    ctx.strokeStyle='#000'; ctx.lineWidth=3;
    ctx.strokeText(i+1,0,0); ctx.fillText(i+1,0,0);
    ctx.restore();
  }
}
```

Use Math.cos etc. Let me alias. I'll just use Math.

UI buttons: pause toggles running. reset calls resetSim and set paused false (resume). Let me have reset restore and continue running.

Main loop:

```
function loop(timestamp){
  if(!last) last=timestamp;
  let dt=(timestamp-last)/1000; last=timestamp;
  if(dt>0.05)dt=0.05;
  if(!paused){
    step(dt);
    draw();
  }
  requestAnimationFrame(loop);
}
initSim(); resetSim();
requestAnimationFrame(loop);
```

Wait if paused we still might want to draw? Keep drawing. Actually if paused skip step but draw stays. Fine. But the requirement Pause/Resume—when paused, simulation stops. Drawing frozen. OK.

Reset button: resetSim(); paused=false. The "Pause" text toggles; let me update label.

Let me also start unpaused (running). 

Let me make sure theta resets and continues.

Also I want the first moment: balls placed near center, let them fall. With start at y=cy-40=360, they fall down. Good.

Let me also ensure balls start non-overlapping: grid spacing 2.3r=29.9, diam 26, gap ~4. Non-overlap ✓.

Now, the heptagon at rest orientation theta=0: vertex 0 at angle 0 → (cx+R, cy) = (740,400) rightmost. Top vertex at angle π/2? Vertices at 2πk/7. k=0 at 0° (right), k=1 at ~51° (upper right). There's no vertex straight up; top edge exists. Fine, fully visible (R=340, top edge midpoint at angle ~90-180/7... ). The topmost point is a vertex near 90°: k where 2πk/7 closest 90°=π/2 → k=7/4=1.75, k=2 at 102.85°, vertex at (cx+R cos102.85, cy+R sin102.85). cos negative small, sin~0.98. y=cy+340*0.98=cy+333=733 (near bottom). Hmm that's bottom. The top is k=... let me see which vertex has min y: sin most negative? For k=2, sin(102.8°)=0.97>0 → y below center (down). Negative y offset (up) corresponds sin<0 → k=4: 2π*4/7=205.7°, sin=-0.43? Hmm. Actually vertices symmetric: one at 0°, the "top-ish" vertices at ±90°ish. k=2 at 102.8° (down-left region, sin positive → lower). k=5 at 257° (sin negative → upper). So top vertex ~ at 257° → y=cy+R*sin257=cy+340*sin(257). sin(257)=sin(257-180)=sin77? 257° in third/quarter... 257° is in third quadrant (180-270), sin negative. sin(257)= -sin(77)? sin(180+77)= -sin77≈-0.97. So y=cy-333=67. So top vertex at y=67, x=cx+R*cos257=cos(257)=cos(180+77)=-cos77≈-0.22→ x=cx-75=325. So top vertex at (325,67). Min y=67. Good fully visible (margin 67). Good.

Now let me write final code and test.

Let me also double check the number "rotate its number with the ball": ctx.rotate(b.angle). The number drawn at (0,0) centered. Since text aligns center, rotating the number. But note text rotation: text is drawn upright normally; after rotate, it rotates. Good, "numbers rotate with the ball." ✓.

Let me write the file now and test.

Let me also reconsider the gravity sign and y-down: vy += G*s (G>0) → ball accelerates downward (+y). ✓. Falling direction down. Good.

Now let me also double-check resolveBallWall pen correction direction: pen=b.r-d (d<r so pen>0). b.x += nx*pen (inward, toward center). ✓.

Let me also handle the case where ball is far outside (shouldn't happen): d could be very negative; pen large; fine.

Let me finalize and write file.

I'll tune constants after testing. Let me write with initial guess and then run.

Let me write the HTML.

Let me reduce mu to 0.3, linDamp 2.5, angDamp 2.5, eWall 0.6, eBall 0.5. Initial av ±0.8.

Let me write.

Actually, I worry damping as multiplier (1-c*s) per substep with c=2.5,s=1/180 → factor 0.9986 each substep, over 180 substeps/sec → 0.9986^180=e^{-0.252}=0.777, so velocity retains 77% per second—that's weak damping. Eh, terminal v=g/c for force -c*v: at steady, G = c*v → v=G/c=900/2.5=360. The multiplier approx: v_{n+1}=v*(1-cs)+G*s; steady v=G/c=360. So terminal ~360 px/s. Reasonable. Damping time constant 1/c=0.4s. Fine. Let me keep c=2.5. If energy pumps up, increase.

Hmm but the rotating wall can pump significantly. Let me test.

Actually, wait: the wall at bottom rotates; bottom wall is near horizontal. Its velocity at contact (foot on bottom wall near x) = omega*(-(fy-cy), fx-cx). For bottom wall, foot y ~ cy+R_apoth (large y), so -(fy-cy) is large negative → vwx large negative (leftward) for positive omega? omega>0 → vwx = omega*(-(fy-cy)) = omega*(-(positive)) = negative (leftward), vwy = omega*(fx-cx). At bottom wall midpoint (fx=cx) vwy=0. So bottom wall moves horizontally (left) at speed omega*(fy-cy)=omega*apothem≈1.256*306=384 px/s. So bottom wall sweeps left fast. A ball on the bottom gets pushed left and also the wall friction spins it. Energy injection could be large. Damping must counter. Let me increase linDamp to 4 and test; also Vmax cap helps.

Actually 384 px/s wall speed, gravity-induced terminal 360 — comparable. The ball could get accelerated along wall. With damping c=2.5, steady-state velocity along a driven wall ~ v_wall*(friction coupling)... could exceed. Let me just test and raise damping as needed. Set initial c=4, mu=0.2.

Let me test empirically.

Let me write file and open browser.

I'll write now.

Let me write the JS fully and carefully.

Also need to make sure requestAnimationFrame handles pause (don't step). I'll still draw to keep frozen frame.

Let me write.

Let me also make reset also reset balls' angle/av to init values.

Let me write file.

After writing, open browser, screenshot, check console, observe for ~5-8 seconds simulation, check rotation, collisions, reset.

Let me write the code.

Potential issue: I compute walls(theta) once per step outside substep loop. But theta changes within step? No—I advance theta after step using dt. During step theta constant. Good. But for accurate collision with moving wall, within a substep theta is constant (fine, substep small).

Wait, but I advance theta = theta + omega*dt AFTER the whole step. The substeps use the same theta. Since omega*dt small (1.256*0.016=0.02 rad per frame), fine. Actually dt is frame dt (~0.016), theta advance 0.02 rad/frame; substeps don't re-evaluate theta. The walls() computed once. Good enough. Slightly less accurate but fine.

Hmm, but for moving-wall velocity I use global omega (constant). Good. The wall position uses theta (constant within step). Fine.

Let me write.

Let me also ensure the collision iterations re-evaluate after positional corrections move balls. They do (within the it loop).

Write file now.

Let me check if there's a browser available via playwright.

Playwright is available. I can use the `browser` tool provided by the harness. Let me write the index.html first.

Let me write the full file.

Wait, I have a bug in the positional correction for ball-ball: I wrote push a in -n. a.x -= nx*pen*(invA/tot). n points a->b, so -n pushes a away from b (correct, toward a's side) and +n pushes b away from a (correct). But the mass weighting should push more the lighter body. a moves by pen*(invA/tot) in -n direction; b moves pen*(invB/tot) in +n. Lighter body (smaller m, larger inv) moves more. ✓. Good.

Let me also double check: the initial av stored in init for reset is b.av (random). Good, reset restores it.

Let me now open the browser and test.

There's an error: "Cannot read properties of undefined (reading '0')". Likely in resolveBallWall: `mid` is an array `[(v1[0]+v2[0])/2,...]`, and I use `mid[0]`, `mid[1]`. That should be fine. The error reading '0' of undefined... maybe `V[i]` is undefined? V has N=7 entries indexed 0..6, v2=V[(i+1)%N]. Fine.

Wait, the error could be `mid` being used as `mid[0]` but I destructure `const {mid,nx,ny}=w;` and mid is an array. Fine.

"Cannot read properties of undefined (reading '0')" — something[0] where something is undefined. Could be in walls(): V[i] undefined? If theta fine. Or `b.r` ... no.

Hmm, maybe it's the `loop` runs and step called with dt, `walls(theta)` fine. The error appears in console. Let me get more detail. The browser tool only gave summary. Let me run with playwright directly to capture full console/error.

Actually "images: model does not support images" — the tool doesn't return images here. And the console error is terse. Let me run a playwright script to capture full error stack.

Let me write a small node script using playwright-core to load the file and capture console + pageerror, and take a screenshot.

Actually the harness has a `browser` tool. But the console output is terse. Let me use playwright directly via node to get full pageerror and console messages with stack traces.

Let me check if `@playwright/test` or `playwright` is available with chromium.

Top-level await not supported in CommonJS. Let me wrap in a function.

The error at resolveBallWall line 114, `mid[0]` reading '0' of undefined. So `mid` is undefined. In walls(), I push {mid, nx, ny} where mid is an array. But in resolveBallWall I do `const {mid,nx,ny}=w;` then `mid[0]`. That should work. Wait, the Ws array: I return Ws with objects having mid array. 

But the error says mid undefined. Let me check line numbers. The file lines: resolveBallWall `const {mid,nx,ny}=w;` then `const dx=b.x-mid[0]...`. If w is undefined... but I iterate `for(let i=0;i<NB;i++) resolveBallWall(balls[i],Ws);` passing Ws element. Ws should have 7 entries. Hmm unless walls() returned something wrong.

Wait, maybe the issue: in `walls()`, variable name `Ws` — but I also have outer... no. Let me check: `function walls(th){ ... return Ws; }`. Returns array of 7. In step `const Ws=walls(theta);` shadows? There's a parameter/local. Actually inside step I declared `const Ws=walls(theta);`. Then `resolveBallWall(balls[i],Ws)` passes the whole array, not an element! Bug: should pass `Ws[i]`. 

Yes — `resolveBallWall(balls[i],Ws)` passes the array. Then in resolveBallWall `const {mid,nx,ny}=w;` destructures array — arrays have no .mid/.nx/.ny properties, so mid is undefined. That's the bug.

Fix: `resolveBallWall(balls[i], Ws[i%Ws.length])` — but 7 walls, ball could be near any; should check all walls per ball, not just one. My loop `for(let i=0;i<NB;i++) resolveBallWall(balls[i],Ws)` only checks ONE wall per ball. That's wrong — a ball can collide with any of the 7 walls. Need to check all walls for each ball.

Let me fix: nested loop over walls too. For each ball, for each of 7 walls, resolve. Let me restructure.

So:
```
for(let it=0;it<iter;it++){
  for(let i=0;i<NB;i++) for(let k=0;k<N;k++) resolveBallWall(balls[i],Ws[k]);
  for(let i=0;i<NB;i++) for(let j=i+1;j<NB;j++) resolveBallBall(i,j);
}
```

That's the fix. Let me edit.

Also `window.theta` and `window.balls` not exposed (they're in module scope, not global). The eval returned "no balls global" because balls isn't on window. That's fine, just my probe. The real issue was the wall loop bug.

Let me fix the two issues:
1. Pass each wall: loop k over N.
2. Make sure Ws has N elements (it does).

Let me edit the collision loop.

I need to format edits as an array of objects.

No errors now. But "STATE: no balls global" because balls isn't a window global. Let me expose some state for probing, or just inspect via evaluating within the page. Actually I can't access module-scope vars. Let me add temporary debugging by exposing state to window, or just take a screenshot and also evaluate positions via reading canvas? Better: temporarily expose balls/theta to window for inspection. But I don't want to leave debug code. Let me add a quick temporary exposure for the probe, then remove. Actually, let me just take screenshots over time and visually verify via screenshot + DOM. But the tool said "model does not support images" — screenshots aren't returned as viewable. Hmm. The browser tool earlier returned images: "images: model does not support images". So I can't see screenshots visually.

Wait, the harness `browser` tool normally shows screenshots. The message "model does not support images" suggests I can't view images directly. But I can use probe.js to extract numeric state by exposing globals temporarily.

Let me temporarily add `window._dbg = {balls, theta, get theta(){...}}` — but balls is an array; I can expose a getter returning JSON. Let me add a debug hook that exposes ball states, take measurements at different times, then remove the debug hook before final.

Actually, simpler: I can inject an extra script at runtime via page.addInitScript? No. Let me just temporarily edit the HTML to expose state to window for probing, run probe capturing state at several timestamps, then revert the debug line.

Let me add `window._probe=()=>balls.map(b=>({x:+b.x.toFixed(1),y:+b.y.toFixed(1),vx:vx,vy,angle:+b.angle.toFixed(2),av:+b.av.toFixed(2)})).slice()` and theta exposure. Actually I'll just expose the arrays reference: `window._balls=balls; window._theta=...` but theta is a local let; I can expose via window._getTheta=()=>theta. Let me add temporary debug.

Let me add to the script end (temporarily):
```
window.__dbg=function(){return {theta:theta, balls:balls.map(b=>({x:b.x,y:b.y,vx:b.vx,vy:b.vy,angle:b.angle,av:b.av,r:b.r}))};};
```
Then probe calls window.__dbg() at intervals.

Let me add it temporarily.

Wait, I accidentally added an extra `;` — the oldText had `requestAnimationFrame(loop);` but I replaced with `...});};` — I put `};` after the function plus the existing `;`. Let me check: newText ends with `}});};` then I had `requestAnimationFrame(loop);`. Wait I replaced "initSim(); resetSim(); draw(); last=0;\nrequestAnimationFrame(loop);" with text that includes the function AND `requestAnimationFrame(loop);`? Let me re-read my newText:

"initSim(); resetSim(); draw(); last=0;\nwindow.__dbg=function(){...};\nrequestAnimationFrame(loop);"

Hmm my newText in the edit was: `initSim(); resetSim(); draw(); last=0;\nwindow.__dbg=function(){return {...};\nrequestAnimationFrame(loop);`

Wait I need to look. The newText I provided:
"initSim(); resetSim(); draw(); last=0;\nwindow.__dbg=function(){return {theta:+theta.toFixed(3), balls:balls.map(b=>({x:+b.x.toFixed(1),y:+b.y.toFixed(1),vx:+b.vx.toFixed(1),vy:+b.vy.toFixed(1),angle:+b.angle.toFixed(2),av:+b.av.toFixed(2),r:b.r}))};};\nrequestAnimationFrame(loop);"

So it ends with `};};` — I added an extra closing. Let me check: `...b.r}))}` closes the map return object and the function body `}`. Then `;` ends statement. Then `}`? No wait: `window.__dbg=function(){return {...}};`. The `return {theta, balls:[...]}`. So `return {...};` then `}` closes function. I wrote `...b.r}))};};` — let me count braces. Original inner: `{x,y,...}` is the map element object. `.map(b=>({x:...}))` arrow returns object. Then `balls: balls.map(...)`. So `{theta:..., balls: [...]}` — the outer return object. Structure:
`window.__dbg=function(){ return { theta:..., balls: balls.map(b=>({x:..., ...})) }; }`

So: `({x:...,r:b.r}))` — `({x...})` closes element obj and `)`, then `]`? No array. `.map(...)` then `}` closes return obj, then `}` closes function. Let me carefully reconstruct to avoid syntax errors. Let me just read the file around that area and fix.

Actually, let me just read the end of the file.

The debug line: `window.__dbg=function(){return {theta:..., balls:balls.map(b=>({...}))};};` — looks balanced: function(){ return { ... }; } — wait after `b.r}))` we have `}` closes return object, `;` then `}` closes function, `;`. So `}));};` then `;` → `}));};;`? Let me count:

`return {theta:..., balls:balls.map(b=>({x:...,r:b.r}))};`
- `b=>({...})` arrow returning object: `({...})`. `{x,y,...,r:b.r}` then `})` closes object and call.
- So `.map(b=>({...}))` ends with `}))`.
- Then `};` closes return object `{` → wait the return object is `{theta, balls}`. After `.map(...)` we need `}` to close return object, then `;`. So: `balls.map(...))` then `}` (close return obj) then `;`. So `...}))}` then `;`. Hmm but I wrote `b.r}))};`. Let me count in the actual text: `b.r}))};};`

Breaking: `b.r` `}` (close element obj) `)` (close map call) `}` (close arrow? no arrow is `=>({...})`). Ugh. Let me just parse: `b=>({x:...})` is `arrow(params)=> ( expr )`. The `(` after `=>` groups. `(` then `{...}` object then `)`. So `({x:...})`. Element: `({x:+b.x.toFixed(1),y:...r:b.r})`. That's `(` `{` ... `}` `)`. So `({...})`. Then `.map(b=>({...}))` has `.map(` `b=>` `({...})` `)`. So ends `})` `)` = `}))`? No: `({...})` then `)` closes map = `({...}))`? Wait `.map( ARG )` where ARG = `b=>({...})`. So `.map(b=>({...}))` = `.map(` + `b=>({...})` + `)`. The ARG ends with `)` (the grouping paren of arrow body). So `.map(b=>({x:...}))` ends with `)))? Let me write chars: `.map(b=>({x:...,r:b.r}))`. 
Characters: `. map ( b = > ( { x : ... r : b . r } ) )`. So after `b.r`: `}` closes object, `)` closes grouping paren `(` ... `)`, then `)` closes `.map(`. So `}))` then nothing else for the map. Then `balls:` value done. Close return object: `}`. So `balls.map(...)` `}` → `...}))` `}`. Hmm `}))` then `}` = `}))}`. Then `;`. 

So full: `balls:balls.map(b=>({x:...r:b.r}))}` then `;` → `b.r}))};`. Then close function `}` → `};`. Then `;`. Total ending: `b.r}))};};`? 

My text ends: `b.r}))};};` — that's `}` `)` `)` `}` `;` `}` `;`. Count opens vs closes in the function body return statement:
`return {` (1 open)
  `theta:...`
  `balls:balls.map(b=>({...})` 
    arrow body group `(` (2), object `{` (3) ... `}` (3 closes obj) `)` (2 closes group) 
  `)` ... wait map call opens `(` (2) and we closed it. Let me recount with depth.

This is getting confusing. The probe will tell me if syntax is broken (pageerror). Let me just run the probe; if syntax error, fix.

The simulation runs with no errors. Some observations:

1. theta is advancing correctly (0.628, 1.885, 3.791, 6.932, 11.33, 17.006). 2π/5 * t: at t=0.5, 1.2566*0.5=0.628 ✓. At t=4.5, 1.2566*4.5=5.655... but reported theta=17.006? That's because theta is not modulo 2π and accumulates; but 1.2566*4.5 = 5.655, not 17. Hmm wait the sampling is cumulative timing: t=0.5 means 0.5s into the run, theta=0.628 ✓. t=1.0 → 1.2566*1=1.256, but reported 1.885?? That doesn't match.

Wait, the sampling loop: I do `await p.waitForTimeout(t*1000)` but t runs over [0.5,1.0,1.5,2.5,3.5,4.5]. But waitForTimeout waits ADDITIONAL time, and I'm measuring from start of loop? Actually the first waitForTimeout(500) waits 500ms then evals. So theta ~0.628 ✓ (0.5s). Then waitForTimeout(1000) waits another 1000ms (total 1500ms = 1.5s), but theta=1.885 ≈ 1.5s? 1.2566*1.5=1.885 ✓. Good, cumulative. Then next waits 1500ms (total 3s), theta=3.791≈1.2566*3=3.77 ✓. Then 2500ms total, theta=6.932≈1.2566*5.5=6.91 ✓. Then 6500ms total? no: cumulative: 0.5+1+1.5+2.5+3.5+4.5 = 13.5s total but theta at last=17.0 ≈1.2566*13.5=16.96 ✓. Great, theta is correct and monotonic, no modulo but that's fine.

2. vmax ~150-185, bounded. Good, no unbounded speed.

3. Angular velocities develop (avmax up to ~17 at t=1.5, ~16.99 at t=4.5). Balls spin. Good. angle reaches 59.97 rad (spinning).

4. After Reset: theta=0.754??! That's wrong — Reset should restore theta=0. But reported theta=0.754. Also the positions after reset: balls at y=431.8, 458.4, etc. — those match init startY=cy-45=355? Wait startY=cy-45=355. But reset shows y=431.8. That's NOT the initial positions!

Wait, init positions: startX=cx-(cols-1)*sx/2. cols=5, sx=2.32*13=30.16. (5-1)*30.16/2=60.32. startX=400-60.32=339.68. startY=cy-45=355. Row 0 y = 355 + 0*... =355. But after reset, row0 y=431.8. 

Hmm, that's off by ~77. Also vy=176.6 nonzero after reset! That means reset didn't work properly, OR the probe clicked reset while simulation still running with the click handler `resetSim();setPause(false);draw();` — wait resetSim sets vx=vy=0. But probe shows vy=176.6. 

Oh wait — I called reset via `document.getElementById('reset').click()`. The reset handler does `resetSim();setPause(false);draw();`. resetSim sets velocities to 0. But then `setPause(false)` sets running=true. Then immediately requestAnimationFrame continues. The probe evaluates `__dbg()` 600ms later (waitForTimeout(600)) — by then the simulation has been running for 600ms again, so balls have fallen and gained velocity and moved! So the "AFTER RESET" reading is actually 600ms after reset, not the instantaneous reset state. That explains vy=176.6 (fallen for 0.6s) and positions dropped.

And theta=0.754 at 0.6s ≈ 1.2566*0.6=0.754 ✓. So theta WAS reset to 0, then advanced 0.6s. 

So reset works! The probe just measured 600ms later. Let me verify the instantaneous reset state by evaluating immediately (no wait) after clicking reset. Also verify positions match initial.

Also I notice the initial fall: at t=0.5, b0 x=339.7 (matches startX=339.68 ✓), y=414.4. Initial y row0 = 355. After 0.5s falling under g=950: y = 355 + 0.5*950*0.25 = 355+118.75=473.75? But reported y=414.4. Hmm, less than expected. Wait b0 is the first ball, index 0, col 0 row 0, x=339.7 ✓ startY=355. After 0.5s free fall y should be 355+118.75=473.75 if no collision. But reported 414.4. So it hit the bottom wall and bounced? Bottom wall apothem = R cos(π/7)=340*0.9009=306.3. Bottom wall line y = cy+(R*cos(π/7)?) Actually bottom edge: the heptagon bottom is a wall (an edge) not a vertex, since no vertex points straight down. The bottom wall's inward normal points up (toward center), midpoint at bottom. The apothem (distance from center to edge) = 306.3. Bottom edge is at y = cy + 306.3 = 706.3. The ball radius 13, so ball center reaches y=706.3-13=693.3 at resting on bottom. 

From 355 to 693.3 is 338px, free fall time sqrt(2*338/950)=0.846s. So at t=0.5 the ball is still falling (no bounce yet), y=355+950*0.25=355+237=592? But reported 414.4. That's much less. 

Wait, that's weird. b0 vy=170.8 at t=0.5. Free fall vy=950*0.5=475. But reported vy=170.8. So velocity is much less than free-fall. That means damping (linDamp=5) is reducing it strongly, OR something else.

With damping a=-c*v, terminal velocity = G/c = 950/5=190. So vy approaches 190, and at t=0.5 it's 170.8 (close to terminal 190). Position: with damping, y(t)= y0 + (v_term)*(t + (v0-v_term)(1-e^{-ct})/c). Starting v0=0: y=y0+v_term*(t - (1-e^{-ct})/c). v_term=190, c=5, t=0.5: (1-e^{-2.5})/5=(1-0.082)/5=0.1836. t - that = 0.5-0.1836=0.316. y=355+190*0.316=355+60=415. ✓ matches 414.4! Great, so damping works, terminal ~190.

So speeds are bounded ~190 terminal, vmax 150-185. Good, stable. Gravity + damping balanced. The rotating walls inject some energy but damping (c=5) keeps it bounded. Good.

Now is the motion interesting enough? Speeds ~180, with rotation ω giving wall speed ~384 at apothem. Balls bounce. av develops up to ~17 rad/s. angle accumulates. Spin visible. 

Let me verify:
- Enclosure rotates ✓ (theta advances, theta mod 2π shows rotation).
- Ball contacts produce bounce responses ✓ (vmax shows velocities changing, collisions occur).
- Numbers rotate with balls ✓ (angle grows to 59 rad).
- Reset restores arrangement & wall orientation: need to check instantaneous reset state.

Let me also double check: do balls stay contained (no escapes)? vmax bounded ~185, and they're bouncing near bottom. Let me verify no ball escapes the heptagon — check that all d>=r after simulation. Let me add a containment check in probe: compute min distance to any wall for each ball; if any < 1 (penetrating), report. Actually after collisions resolved, d should be ~r (just touching) or >r. Let me check min(d-r) across balls.

Also verify the initial arrangement is non-overlapping: check min pairwise distance >= 2r - epsilon.

Let me also confirm reset instantaneous restores theta=0 and exact initial positions and zero velocities.

Let me run a refined probe. Also remove the debug hook after.

Let me also reconsider: the reset click handler calls `resetSim();setPause(false);draw();`. But `running` might already be true (simulation running). setPause(false) sets running=true (redundant) and pause button text "Pause". Fine. But if user paused then reset, resetSim + setPause(false) resumes. Good. Reset also resets theta=0. Good.

One concern: the reset button handler calls draw() once but the loop will keep drawing. Fine.

Let me run refined probe: 
1. Capture initial positions (before any step) — expose via __dbg at time 0 (but loop already stepped during load? The page loads, requestAnimationFrame starts, first frame dt small. At time 0 the draw() called in init draws static. The loop starts. Hard to capture initial. But I can capture initial by reading init? init isn't exposed. Let me expose init too. Actually I can capture the reset state (which equals init positions with zero velocity) to verify arrangement, and theta=0.

Let me refine: after clicking reset, immediately evaluate __dbg (no wait) to get instantaneous reset state. Compare to init positions stored. Let me expose a function returning initial positions too. Let me modify debug to expose both current balls and init.

Actually simpler: I'll capture __dbg immediately after reset click (synchronous via evaluate, but the click triggers resetSim synchronously in the handler, so immediately after click returns, state is reset). Let me evaluate right after click with no waitForTimeout. But evaluate is async after click; the click handler runs synchronously? The handler `resetSim();setPause(false);draw();` runs synchronously during the click dispatch. evaluate runs after. So state is reset. Good. But setPause(false) doesn't change positions. Good.

Let me also verify initial non-overlap and containment. Let me add to debug: pairwise min dist, and min wall distance.

Let me update the debug hook to also return init positions and stats.

Let me also check that numbers "rotate with the ball" — angle values confirm rotation.

Let me refine and measure.

The probe3.js has a syntax error in my node script (the `const dd=(ball.x-w.mid[0])*w.nx...` line is fine, the issue is the line with `const pen=ball.r-dd;if(pen>worstPen)worstPen=pen;` — actually the error is "missing ) after argument list" at the destructure `const {cx:cx0,cy:cy0,R,N}= {cx:400,cy:400,R:340,N:7};` inside evaluate — that's a string of JS. Wait no, the evaluate callback is a function. The line numbering says probe3.js:30 which is inside the evaluate string? No—evaluate takes a function. The function body is the JS. Node parses the whole file including the function body as normal JS (since it's a function literal, not a string!). 

I passed `p.evaluate(()=>{const d=window.__dbg(); ...})` — a real function, so Node parses it as JS at file compile time. So syntax errors in it show up as file syntax errors. The error is at line 30 of probe3.js which corresponds to my function body line. Let me look: the issue might be `let worstPen=0;` then `const PI=Math.PI; const th=d.theta; const V=[];` ... `for(const w of Ws){const dd=(ball.x-w.mid[0])*w.nx+(ball.y-w.mid[1])*w.ny; const pen=ball.r-dd; if(pen>worstPen)worstPen=pen;}` — that looks fine.

Actually the error points at a long line ending. The "missing ) after argument list" — maybe `Math.cos(a),cy0+R*Math.sin(a)];}` inside the for with `for(let k=0;k<N;k++){const a=th+k*2*PI/N;V.push([cx0+R*Math.cos(a),cy0+R*Math.sin(a)]);}` fine.

Hmm the caret points mid-line at `}}}`. Let me just rewrite probe3 more carefully and avoid passing inline functions with complex logic; instead pass a string and use evaluate with a string expression, or simplify. Let me write the containment computation as a string passed to evaluate, and keep it simple. Actually easier: expose the containment helper from the page itself. But I don't want it in final. 

Alternative: compute containment in the page via evaluate with a string. Let me just rewrite probe3 cleanly with a string argument to evaluate containing a clean function.

Let me simplify: do two evaluate calls (strings) — one for t=2.5, one for reset instant — each returning JSON-serializable simple objects, with the containment logic inline but simpler. I'll write the JS as a string and be careful with syntax.

Actually, the cleanest: add the containment check to the debug hook temporarily too. Let me extend __dbg to compute worstPen itself (it has access to theta and the wall geometry). That avoids passing big strings. Let me update __dbg to include worstPen and a pairwise-overlap check using the page's own closures (it can reference cx, cy, R, N, etc., since it's in the same scope). The __dbg function is defined in the page script scope, so it can use cx, cy, R, omega, etc. Let me rewrite __dbg to compute containment and overlap.

Let me replace the debug hook.

I made a mistake with the path argument placement. Let me redo the edit with proper structure.

Let me analyze:

INIT (t~0): theta=0.0628 (advanced 0.05s). balls at y=355-434.6 (matching init, positions correct, non-overlapping minOverlap=4.16>0 ✓). worstPen=0 (all contained ✓). vy=41.5 (they've started falling). Good initial arrangement is compact and non-overlapping. ✓

t~2.5: theta=3.1625 (≈1.2566*2.5=3.14 ✓, advancing). Balls have spread to bottom region y~630-715 (bounced on bottom wall, contained). worstPen=3.29 — there's some penetration (3.29px), meaning after collisions a ball is slightly overlapping a wall. That's because positional correction in walls uses iter but only pushes one ball; with multiple balls near a wall, 3.29px residual overlap is acceptable but a bit high. minOverlap=-0.05 (slight overlap between two balls, negligible). Speeds: vmax ~ sqrt(108²+220²)≈245 for one ball; bounded by Vmax=1700. Good, no unbounded speed.

Wait worstPen=3.29 means a ball center is 3.29px beyond the wall line (penetrating). With r=13, the ball visually slightly overlaps the wall line. Not persistent/visible maybe. But let me see if it's a problem. It's small. The requirement: "do not persistently overlap." 3px transient is fine. But let me increase collision iterations or substeps to reduce. Actually worstPen is the max penetration at that instant (peak during a collision). It's transient. Fine.

Hmm, but one ball b4 (index 4) at y=714.7 with vy=166.3, near bottom. Bottom wall apothem: bottom wall midpoint y = cy + 306.3 = 706.3. Ball center y=714.7 is below that by 8.4, but r=13 so pen = 13 - (706.3 - 714.7)... wait d = signed inward dist = (ball.y - mid.y)*nx... for bottom wall n_in=(0,-1) (pointing up toward center). mid.y≈706.3. d = (ball.y-706.3)*... n_in=(0,-1): d = (ball.y-706.3)*(−1)*... actually d = dot(ball-mid, n_in) = (0, ball.y-706.3)·(0,-1) = -(ball.y-706.3) = 706.3-ball.y. For ball.y=714.7, d=706.3-714.7=-8.4. pen = r-d = 13-(-8.4)=21.4?? But reported worstPen=3.29. The max pen across all balls/walls is 3.29, so index 4 must be on a different wall or my computation in probe differs. Anyway worstPen=3.29 is the max; that's fine.

Actually wait, index 4 at y=714.7, bottom wall d=706.3-714.7=-8.4, pen=21.4. That would make worstPen≥21.4, but reported 3.29. So either index 4 isn't penetrating the bottom wall. Let me reconsider bottom wall. The bottom of heptagon — is it an edge or vertex? With vertex 0 at angle 0 (rightmost (740,400)). The bottom-most point: which vertex has max y? a=θ+k*2π/7. At θ=3.1625. The bottom point ~ where sin(a) max ≈1. Hmm, vertices rotate. The bottom wall changes as it rotates. So at θ=3.16, the "bottom" is a wall at some angle. The ball at y=714.7 might be near a bottom vertex region, not the bottom wall edge. The heptagon's bottom extent = cy + R = 740 (a vertex can be at bottom). Ball at y=714.7 with r=13 → bottom of ball at 727.7 <740, so still inside (vertex at 740). Good, contained. worstPen=3.29 is the worst wall penetration; fine.

AFTER RESET: theta=0 ✓. positions: balls at y=355,381.5,408.1,434.6 (exactly init) ✓. vx=vy=0 ✓. angle=0 ✓. av restored to init values ✓. minOverlap=4.16 (non-overlapping ✓), worstPen=0 (contained ✓).

But the probe said "positions restored? false" and "vel zero? false". Let me check why false. The probe compared rsj.balls[k] x,y to inj.balls[k]. The init in __dbg is recomputed each call (init array), and inj was captured at t~0. At t~0, init = the initial positions array. After reset, balls should equal init. The probe's `samePos` checks |b.x-inj.balls[k].x|<1 and y. inj.balls[k].x: inj is JSON.parse(init) where init was 'JSON.stringify(window.__dbg())' at t~0 — but that includes 'init' field (the init array) AND 'balls' field (current balls at t~0, which have fallen slightly, y=356.2 not 355). So inj.balls[k] is the t~0 current balls (vy=41.5, y=356.2), NOT the init positions! That's why samePos=false — I compared reset balls to the t~0 current balls (which had moved), not to init. 

Let me compare reset balls to rsj.init (the init array) instead. rsj.init at reset time = [{x:339.7,y:355,av:0.13},...] which matches balls after reset exactly. So reset DOES restore positions. The probe's comparison was just wrong (used inj.balls instead of inj.init).

Let me verify: rsj.balls[0] = {x:339.7,y:355,...} and rsj.init[0]={x:339.7,y:355,av:0.13}. They match exactly. So positions restored ✓. theta=0 ✓. velocities 0 ✓.

Vel zero check: rsj.vmax — I computed vmax from... in probe4 I didn't compute vmax in the after reset; I computed `rsj.vmax<1` but rsj doesn't have vmax field (since I JSON-stringified __dbg which includes balls,init,theta,worstPen,minOverlap but not vmax). So rsj.vmax is undefined, undefined<1 is false. That's why "vel zero? false". My probe bug, not a sim bug. The actual balls after reset have vx=vy=0 (visible in output). ✓.

So everything checks out:
- 20 balls numbered 1-20 ✓ (rendered with i+1).
- Compact non-overlapping initial arrangement near center ✓ (minOverlap=4.16>0, grid near center).
- Fall under gravity ✓ (vy becomes positive, positions drop).
- Heptagon rotates at 1 rev/5s constant ✓ (theta advances linearly: 0.0628 at 0.05s, 3.1625 at 2.5s → rate 1.256 rad/s = 2π/5 ✓).
- Ball-ball and ball-wall collision detection implemented manually ✓.
- Moving-wall velocity in bounce response ✓ (v_wall computed from rotation at contact foot).
- Damping/friction ✓ (linDamp, angDamp, friction mu).
- Each ball has angular motion, number rotates ✓ (angle grows, av up to ~15 rad/s).
- Stable: contained (worstPen small 3.29 transient), bounded speed (no runaway; gravity+damping balance). ✓
- Pause/Resume and Reset ✓. Reset restores arrangement & wall orientation ✓.

Now, a few things to double-check visually since I can't see images:
- Numbers rotating with balls: angle values change, and render uses ctx.rotate(b.angle). ✓ in code.
- The heptagon is fully visible: R=340, vertex at cy+R=740 (within 800). ✓.

Let me also verify Pause works. And that the simulation doesn't gain unbounded speed over a longer period (say 10s). Let me run a longer probe checking vmax over time and that it stays bounded, and check containment holds (no ball escapes = worstPen stays reasonable).

Also, I should double check: does worstPen grow over time (energy injection causing balls to be squeezed into walls)? Let me run 10s and sample worstPen and vmax periodically. If worstPen stays small (< a few px) and vmax bounded, we're good. If it grows, increase damping.

Let me run a 12-second probe sampling every 2s.

Also I want to verify Pause actually stops stepping (theta stops advancing). Let me test pause click then check theta doesn't advance.

Let me also reconsider the visual: the requirement "Keep it fully visible." R=340 with 800 canvas → 60px margin. Fine.

Let me also double check the number readability and that numbers 1-20 all render (monospace font, white with black stroke). Fine.

Let me run the longer stability + pause probe.

The simulation is stable and all behaviors work:
- theta advances linearly (2π/5): 2.5341 at t=0 (start) — wait t=0 sample is after 2s actually because my loop does waitForTimeout(2000) first then samples at "t=0"? No: loop t=0 does waitForTimeout(2000) (2s), samples → theta=2.534 ≈ 1.2566*2=2.513 ✓ (≈2s). Then t=2 → another 2s = 4s total, theta=5.068≈1.2566*4 ✓. Good constant rotation confirmed over 12s.
- vmax bounded 125-396, no runaway, no escapes.
- worstPen stays small (<2px transient), minOverlap near 0 (no persistent overlaps).
- Pause stops stepping (theta & positions unchanged). Resume advances (dtheta 1.885 over 1.5s ≈ 1.2566*1.5 ✓).

Everything works. Now I must remove the temporary debug hook (__dbg) and finalize index.html. Let me remove the debug line and keep the rest.

Let me edit out the debug window.__dbg line.

Current line:
```
initSim(); resetSim(); draw(); last=0;
window.__dbg=function(){const Ws=walls(theta); let worstPen=0,minOverlap=999;
for(const ball of balls){for(const w of Ws){const dd=(ball.x-w.mid[0])*w.nx+(ball.y-w.mid[1])*w.ny;const pen=ball.r-dd;if(pen>worstPen)worstPen=pen;}}
for(let i=0;i<NB;i++)for(let j=i+1;j<NB;j++){const d=Math.hypot(balls[i].x-balls[j].x,balls[i].y-balls[j].y)-balls[i].r-balls[j].r;if(d<minOverlap)minOverlap=d;}
return {theta:+theta.toFixed(4), balls:balls.map(b=>({x:+b.x.toFixed(1),y:+b.y.toFixed(1),vx:+b.vx.toFixed(1),vy:+b.vy.toFixed(1),angle:+b.angle.toFixed(2),av:+b.av.toFixed(2),r:b.r})), init:init.map(o=>({x:+o.x.toFixed(1),y:+o.y.toFixed(1),av:+o.av.toFixed(2)})), worstPen:+worstPen.toFixed(2), minOverlap:+minOverlap.toFixed(2)};};
requestAnimationFrame(loop);
```

Let me replace this whole block back to the clean version. I'll match the multi-line debug block and remove it.

Let me use edit to replace from `window.__dbg=...` through the end of that function with just `requestAnimationFrame(loop);`.

Now the debug hook is removed. Let me do a final verification: reload in browser, check no errors, and confirm the file is clean. Let me also delete the probe files (not necessary, but keep clean — the cwd should have index.html). Let me run a final quick check that the page loads without errors and the buttons exist.

Let me re-run the browser tool reload to confirm no page errors and the sim runs.

The "dbg hook removed? false" is because `window.__dbg` is now undefined, so JSON.stringify({theta:undefined}) = "{}" but my test compared to JSON.stringify({theta:"undefined"}) = '{"theta":"undefined"}'. Mismatch in the test logic, not a real issue. The hook is removed (window.__dbg is undefined). Let me quickly confirm __dbg is undefined — actually window.__dbg = undefined means accessing returns undefined. JSON.stringify({theta: undefined}) => "{}". My test string was '{"theta":"undefined"}' which is wrong. So "false" just means the comparison failed, not that the hook exists. It's fine.

Actually let me make sure the __dbg is truly gone — the eval `window.__dbg` returns undefined now (since removed). Good.

Everything is clean:
- index.html present (9824 bytes).
- No page errors.
- Canvas, Pause, Reset buttons present.
- No console errors.
- Probe scripts cleaned up.

The file is finalized. Let me give a brief report.

Let me do one final read of the key parts of index.html to ensure it's intact and clean (no leftover debug).