The user wants a single-page web app in one self-contained index.html file. Let me analyze the requirements:

1. 20 numbered balls (1 to 20) dropping from the center of a heptagon
2. The heptagon spins 360 degrees every 5 seconds
3. Balls are affected by gravity and friction
4. Balls collide with each other
5. Balls bounce realistically off the rotating walls
6. Each ball's spin is shown via rotation of its number
7. Implement all collision detection myself

Let me think about the physics here.

A heptagon is a 7-sided polygon. The balls drop from the center. They need gravity, friction, collision with each other, and collision with the rotating walls.

Let me think about the physics implementation:

**Heptagon geometry:**
- Center at (cx, cy), radius R
- 7 vertices
- Rotating at angular velocity ω = 2π/5 rad/s

For a rotating wall (edge of heptagon), I need to:
- Find the closest point on the edge to a ball
- If the ball penetrates the edge, push it out
- Reflect the velocity considering the wall's motion (the wall is moving because of rotation)

The proper way to handle collision with a moving wall: transform into the wall's reference frame, compute the collision, transform back.

Actually, for a rotating wall, the tangential velocity at a point on the wall is v = ω × r, where r is the position relative to center.

Let me think about the approach:

**Ball-wall collision:**
For each ball, for each edge of the heptagon:
- Find the closest point on the edge segment to the ball center
- Compute distance d
- If d < ball radius, there's a collision
- The edge has a velocity at that point due to rotation: v_edge = ω × (point - center)
- Reflect the relative velocity

Let me work in the rotating frame or the lab frame?

Actually, let me work in the lab frame (world coordinates). The heptagon rotates, so the edges move.

**Approach for wall collision:**
1. Compute the edge's normal (pointing inward) and the point on the edge closest to the ball.
2. Check if ball penetrates: distance < radius.
3. If penetration, compute the edge's velocity at the contact point: v = ω × r (where r is position relative to center). In 2D, for angular velocity ω (scalar, positive = counterclockwise), the velocity at point p is v = (-ω*py, ω*px) relative to center... let me be careful.

Angular velocity ω (scalar). The linear velocity at point r relative to center is:
v = ω × r (cross product). In 2D with ω = (0,0,ω):
v = (-ω*ry, ω*rx)

4. Relative velocity: v_rel = v_ball - v_edge
5. Normal component: v_rel_n = (v_rel · n) where n is the inward normal (toward center side... actually n points inward toward the ball's side).

Let me define the edge normal pointing inward (into the heptagon interior, where balls are). 

6. If v_rel_n < 0 (approaching the wall), apply collision:
   v_rel_n' = -e * v_rel_n (e = restitution)
   v_ball' = v_edge + v_rel_n' * n + (v_rel - v_rel_n * n) (tangential component preserved, possibly with friction)

Actually for friction on the wall, we could reduce the tangential component, but let me add wall friction too for realism. The balls sliding on the wall.

Hmm, but the main friction is on the floor. Let me think about where gravity points. Gravity is downward. The heptagon rotates, so the "floor" changes.

Let me just set gravity downward (positive y), and the balls bounce around inside the rotating heptagon.

Let me think about the implementation more carefully.

**Ball-ball collision:**
For each pair of balls:
- Distance between centers d
- If d < r1 + r2, collision
- Compute normal n = (b2 - b1) / d
- Relative velocity v_rel = v1 - v2
- Normal component v_rel_n = v_rel · n
- If approaching (v_rel_n > 0... depends on convention), apply impulse
- Separate positions

Let me set conventions clearly.

Let me define:
- Positions in world coordinates (canvas), y down (standard canvas).
- Gravity g pointing down (y positive).
- ω for heptagon rotation, say positive = clockwise or counter-clockwise, doesn't matter, pick one. Let's say the heptagon rotates such that it spins. 360 degrees every 5 seconds. Let's make it rotate counter-clockwise when viewed normally... actually in canvas y is down so let me just pick ω and see.

Actually, let me use a coordinate system where I track things and just render. Let me use canvas with y-down, and gravity = (0, g) with g > 0.

**Heptagon:**
Vertices at angle θ + 2π k/7 for k=0..6, radius R.
θ(t) = ω t.

Each vertex position: v_k = center + R * (cos(θ + 2πk/7), sin(θ + 2πk/7)).

Edges connect v_k to v_{k+1}.

For each edge, the inward normal (pointing toward center) is the average of the two vertex directions normalized:
n_in = normalize(v_k + v_{k+1} - 2*center) ... actually the direction from midpoint of edge toward center.
midpoint = (v_k + v_{k+1})/2
n_in = (center - midpoint) normalized.

The edge direction: e = v_{k+1} - v_k, and a normal to edge pointing inward is n_in as computed.

**Closest point on edge to ball:**
Project ball position onto the edge line segment, clamp to endpoints.
point P = closest point on segment.
distance = |ball_pos - P|
if distance < ball_radius: penetration

**Wall velocity at P:**
The heptagon rotates about center with angular velocity ω. Velocity at point P: v_wall = ω × (P - center).
In 2D with ω scalar: v_wall = (-ω*(Py - cy), ω*(Px - cx)).

**Collision response:**
Relative velocity v_r = v_ball - v_wall.
Normal component (using inward normal n, which points toward ball from wall): 
v_rn = v_r · n
If v_rn < 0 (ball moving toward wall, i.e., relative velocity pointing outward... wait).

Hmm let me think. n points inward (from wall toward center). If the ball is inside and near the wall, the wall is outside relative to the ball. The ball moving toward the wall means moving in the -n direction (outward). v_rn = v_r · n. If ball moves toward wall, v_rn < 0. So if v_rn < 0, approaching.

Collision: 
v_rn_new = -e * v_rn (reverse, with restitution)
Add friction: tangential component v_rt = v_r - (v_rn) n. Reduce by friction factor: v_rt_new = v_rt * (1 - friction_wall) or use Coulomb.

v_ball_new = v_wall + v_rn_new * n + v_rt_new

Actually, the wall friction would reduce tangential velocity. But careful with direction. Let me just do:
v_r_new = v_wall + v_rn_new * n + v_rt * friction_wall_coeff

where friction reduces the tangential relative velocity. Let's use a coefficient like 0.9 per collision (damping).

Actually let me implement Coulomb-like friction for the wall: 
max_friction = μ * |v_rn_new|
v_rt_mag = |v_rt|
v_rt_new = v_rt * max(0, 1 - μ * |v_rn_new|/v_rt_mag) ... this is getting complex. Let me keep it simpler with a damping factor on tangential: v_rt_new = v_rt * 0.95. That's fine for a demo.

**Position correction:** push ball out along n by penetration depth.

**Ball spin:** Each ball has an angular velocity (spin ω_ball). The number drawn on the ball rotates. When the ball hits a wall or another ball, spin changes due to friction at contact.

For the number rotation, I'll render the ball with a number, and rotate the canvas by the ball's angle. The spin angular velocity updates based on friction during collisions and rolling.

Actually, to show "each ball's spin via the rotation of its number," I'll give each ball an orientation angle. When the ball collides (with wall friction / ball friction), the spin gets an impulse. Also, when the ball slides on the ground, rolling without slipping would set spin = v/r. 

Let me add: when a ball is on the floor/wall, friction creates spin. Actually a simpler approach: during wall collision, if there's tangential relative velocity, transfer some of it to spin and damp it (rolling).

Let me add rolling friction: each frame, the ball's spin tends toward matching the velocity at contact for rolling.

Hmm, this is getting complex. Let me think about a clean implementation.

Let me give each ball:
- position (x, y)
- velocity (vx, vy)
- radius r
- orientation angle (for number)
- spin (angular velocity of the ball)

Gravity, ball-ball collision with spin transfer, wall collision with spin transfer.

For simplicity and to "show each ball's spin via rotation of its number," I'll:
- Update orientation += spin * dt each frame.
- During collisions, update spin based on tangential impulse (friction at contact).

Let me write the code.

Actually, let me reconsider the whole thing and make it robust. Let me use a fixed timestep for physics to avoid instability, and a rendering loop (requestAnimationFrame).

Let me think about parameters:
- Canvas: say 800x800, center (400, 400)
- Heptagon radius R = 300
- Ball radius: 20 balls, need them small enough. Let's say r = 12. Actually balls drop from center and spread out. Let me use r = 10.
- Gravity g = 800 (pixels/s²) - need to tune
- Restitution e = 0.6
- Friction: 0.99 per frame (air resistance) and ground friction
- Heptagon ω = 2π/5 ≈ 1.2566 rad/s

Let me think about stability. With small balls and collisions, I need a small timestep. Let me use dt = 1/120 or substep.

Let me use a fixed timestep of 1/120 (≈8.3ms) and run multiple substeps if needed. With requestAnimationFrame (~60fps), I can do substepping: physics dt = 1/240, 2 substeps per frame. Or just use 1/120 with clamping.

Let me write the engine.

Let me structure:

```html
<!DOCTYPE html>
<html>
<head>
<style>
  body { margin: 0; overflow: hidden; background: #111; }
  canvas { display: block; }
</style>
</head>
<body>
<canvas id="c"></canvas>
<script>
// physics + rendering
</script>
</body>
</html>
```

Let me write the JS.

```js
const canvas = document.getElementById('c');
const ctx = canvas.getContext('2d');
let W, H, dpr;

function resize() {
  dpr = window.devicePixelRatio || 1;
  W = canvas.offsetWidth; H = canvas.offsetHeight;
  // Actually let's set explicit size
}
```

Let me set canvas to full window.

```js
function resize() {
  W = window.innerWidth;
  H = window.innerHeight;
  canvas.width = W * dpr;
  canvas.height = H * dpr;
  ctx.scale(dpr, dpr);
}
```

Hmm, let me keep it simpler with W, H as CSS pixels and not worry about dpr too much, but I'll handle dpr.

Let me define the world. Center = (W/2, H/2).

```js
const CW = W, CH = H;
const center = { x: CW/2, y: CH/2 };
const R = Math.min(CW, CH) * 0.38; // heptagon radius
const omega = 2*Math.PI / 5; // rad/s, counterclockwise in canvas? 
```

Wait, canvas y is down. Positive ω (standard cross product ω×r gives v=(-ω*ry, ω*rx)). Let me just define rotation angle θ = omega * t and increment. Direction: let's make it spin one way; I'll pick θ increasing (counterclockwise in standard math, which is clockwise rotation visually because y is down... actually in canvas, positive angle rotates clockwise because y-down. It doesn't matter for the demo).

Let me just have theta = omega * t.

Balls start at center with zero velocity, then they "drop" - but from center, gravity pulls them down. They'll spread out as they fall due to the walls. Actually starting all at center with zero velocity means they overlap. Let me give them a slight random spread from center so they don't all stack, OR they'll push apart naturally through collision resolution. Let me give a tiny random initial position offset.

Actually, to make them "drop from the center," I'll spawn them in a small cluster at center with a tiny velocity downward or slight spread. The collision resolution will separate them.

Let me spawn them with small random offsets.

Let me now write the full physics.

```js
class Ball {
  constructor(i) {
    this.r = 11;
    this.x = center.x + (Math.random()-0.5)*20;
    this.y = center.y + (Math.random()-0.5)*20;
    this.vx = 0;
    this.vy = 0;
    this.sp = 0; // spin angular velocity
    this.ang = 0; // orientation
    this.id = i+1;
  }
}
```

Hmm, but they need to "drop from center." With gravity they'll fall and bounce. Good.

**Edge list:** precompute the 7 edges as pairs of (k, k+1 mod 7). At render, compute vertex positions from theta.

Let me compute vertices each frame:
```js
function vertices(theta) {
  const v = [];
  for (let i=0;i<7;i++){
    const a = theta + 2*Math.PI*i/7;
    v.push({x: center.x + R*Math.cos(a), y: center.y + R*Math.sin(a)});
  }
  return v;
}
```

**Wall collision for one ball against all edges:**
```js
function resolveWalls(ball, theta) {
  const vs = vertices(theta);
  for (let i=0;i<7;i++){
    const a = vs[i], b = vs[(i+1)%7];
    // edge vector
    const ex = b.x - a.x, ey = b.y - a.y;
    // closest point on segment
    const len2 = ex*ex + ey*ey;
    if (len2 < 1e-9) continue;
    // vector from a to ball
    const ax = ball.x - a.x, ay = ball.y - a.y;
    const t = (ax*ex + ay*ey)/len2;
    const tc = Math.max(0, Math.min(1, t));
    const px = a.x + tc*ex, py = a.y + tc*ey;
    const dx = ball.x - px, dy = ball.y - py;
    const dist = Math.sqrt(dx*dx + dy*dy);
    if (dist < ball.r) {
      // inward normal
      let nx = center.x - px, ny = center.y - py;
      const nl = Math.sqrt(nx*nx+ny*ny);
      nx /= nl; ny /= nl;
      // wall velocity at contact point
      const wx = -omega*(py - center.y);
      const wy = omega*(px - center.x);
      // relative velocity
      const rvx = ball.vx - wx, rvy = ball.vy - wy;
      const vn = rvx*nx + rvy*ny;
      if (vn < 0) { // approaching
        const e = 0.5;
        // normal impulse
        // for spin: tangential
        // contact point relative to ball center
        ...
      }
      // position correction
      const pen = ball.r - dist;
      ball.x += nx*pen; ball.y += ny*pen;
      // ... velocity update
    }
  }
}
```

Wait, I need to be careful with the inward normal direction and the cross product for spin.

Let me reconsider. The wall velocity due to rotation. I defined omega > 0. v_wall at point p = (-omega*(py-cy), omega*(px-cx)). Let me double check: angular velocity vector ω = (0,0,ω) in a right-handed system where x right, y up, z out. v = ω × r = (-ω*ry, ω*rx, 0) where r = (rx, ry) in the xy-plane with y up. But canvas has y down. So in canvas coordinates with theta increasing... ugh, let me not worry about the sign and just ensure the rotation is consistent: I'll compute wall velocity from the derivative of the vertex positions, OR just define it consistently with the visual rotation.

Actually, the cleanest: define the rotation by angle θ = omega*t, vertex at center + R*(cos(θ+offset), sin(θ+offset)). The velocity of a point fixed on the rotating body is d/dt of that = R*(-sin, cos)*omega for the direction... For a point at (cx + R cos φ, cy + R sin φ), velocity = (-R omega sin φ, R omega cos φ) = omega*(-R sin φ, R cos φ). And the position relative to center is r = (R cos φ, R sin φ). So v = omega * (-sin φ, cos φ)*R = omega * perp(r) where perp(r) = (-ry, rx)? Let's check: r = (rx, ry) = (R cos φ, R sin φ). (-ry, rx) = (-R sin φ, R cos φ) = v/omega. Yes! So v_wall = omega * (-ry, rx) where (rx,ry) = (px-cx, py-cy). So v_wall = ( -omega*(py-cy), omega*(px-cx) ). Good, matches.

But this is the velocity in the frame where theta increases with time. If theta = omega*t, then this is correct. Good.

Now, the collision with a rotating wall is like a moving wall. The standard approach: transform to wall frame, collide, transform back. But for friction and spin, let me handle it more directly.

Let me reconsider the whole approach to make spin meaningful.

For wall collision with spin:
- Contact point on ball is at ball center minus r*n (n is inward normal, so contact point is at ball_pos - r*n... wait n points from wall toward center, ball is on center side. The ball's contact point with wall is at ball_pos - r*n? The ball center is inside at distance dist from wall along -n direction (since n points inward toward center, the wall is in +n... no).

Let me recompute. n = inward normal = direction from wall (contact point P) toward center. The ball is at position ball_pos. The vector from P to ball_pos is (dx, dy) with magnitude dist. dist < r means penetration. The direction from P to ball center is (dx,dy)/dist. Is this the same as -n? 

n points from P toward center. (dx, dy) = ball_pos - P points from P toward ball center. These should be roughly the same direction (both toward center side) if ball is inside. So (dx,dy)/dist ≈ n (when not deeply penetrated). Actually yes, for a ball near the wall inside the polygon, the closest point P is on the wall, and ball center is toward the interior, so (dx,dy) points inward ≈ n direction. Good, so n = (dx,dy)/dist essentially (normalized). Let me just use n = (dx, dy)/dist (pointing from contact toward ball center = inward). That's cleaner.

So n = (dx/dist, dy/dist), pointing from wall contact point toward ball center (inward). Good.

Now, the ball's velocity. Relative velocity v_r = v_ball - v_wall. 
vn = v_r · n (component along n, which is inward). If ball moving toward wall = moving in -n direction = vn < 0. So approaching when vn < 0.

After collision, vn_new = -e * vn (now positive, moving away).

For spin: The angular velocity of the ball ω_b. The velocity at contact point due to spin is v_spin = ω_b × ... in 2D, velocity at contact point = ω_b * perp(contact - center) = ω_b * (-(cp-cy), (cp-cx)) where (cp-cx, cp-cy) is contact point relative to ball center.

Contact point relative to ball center = -r * n (the contact is at distance r in the -n direction from center). So contact - center = -r*n.
v_spin = ω_b * perp(-r*n) = ω_b * (-(-r*ny), -r*nx) = ω_b * (r*ny, -r*nx).

Hmm, let me define perp(v) = (-vy, vx). Then v_spin at contact = ω_b * perp(contact - center) = ω_b * perp(-r*n) = ω_b * r * perp(-n) = ω_b * r * (ny, -nx).

The tangential direction (along the wall) is t = perp(n) = (-ny, nx).

The no-slip condition relates spin to tangential velocity at contact.

This is getting complicated. Let me simplify: I'll model spin with a simpler approach. I'll apply the normal collision (with wall motion) and a friction that creates/opposes spin, plus air resistance on spin.

Actually, let me simplify the spin model to make it visually clear without overcomplicating:

For each wall collision:
1. Compute n (inward), v_wall at contact.
2. v_rel = v_ball - v_wall.
3. vn = v_rel · n. If vn < 0 (approaching):
   v_rel_n = vn
   v_rel_n' = -e * v_rel_n  (bounce)
   Apply to v_ball: v_ball = v_wall + v_rel_n' * n + (v_rel - v_rel_n * n) ... but we also want to dampen tangential and create spin.

Let me also apply tangential friction:
   v_rel_t = v_rel - (v_rel·n)*n  (tangential part of relative velocity)
   friction reduces tangential by factor fr (e.g., 0.8): v_rel_t' = v_rel_t * fr
   The change in tangential velocity Δv_t = v_rel_t' - v_rel_t = (fr-1)*v_rel_t (opposite to motion).
   This tangential change creates spin. Impulse at contact with normal n: the friction impulse is along t. The torque = r × friction_impulse. 

Let me compute the spin impulse: The friction impulse J_t = ball.mass * (fr-1) * |v_rel_t|... in direction -t (opposite to tangential motion). Actually let me just transfer: 

The change in tangential velocity dvt = (fr - 1) * |v_rel_t| component... let me be vectorial.

v_rel_t_vec = v_rel_t (vector). 
magnitude reduction: the new tangential relative velocity = v_rel_t_vec * fr.
Δv_rel_t = (fr - 1) * v_rel_t_vec.  This is the change in the relative (ball) tangential velocity (since wall tangential is already in v_wall, and we computed v_rel = v_ball - v_wall, so changing v_rel_t by Δ means v_ball changes by Δ too, assuming wall velocity unchanged).

So v_ball_tangential_new = v_wall_t + v_rel_t_vec*fr + v_rel_n'*n. (v_wall_t = v_wall - (v_wall·n)*n)

Now spin: The tangential impulse magnitude. The impulse on ball = m * Δv = m*(fr-1)*|v_rel_t_vec| in the -t direction (if fr<1). This impulse acts at contact point (distance r from center), creating torque τ = r * |J_t| (sign depending). The angular acceleration: Δω_b = τ*dt/(I) but impulses are instantaneous: ΔL = r × J. J vector = m*(fr-1)*v_rel_t_vec (the impulse). r_vec = contact - center = -r*n. 

τ_impulse (angular momentum change) = r_vec × J (z-component) = (r_vec.x*J.y - r_vec.y*J.x).
r_vec = -r*n = (-r*nx, -r*ny).
J = m*(fr-1)*v_rel_t_vec.
v_rel_t_vec = v_rel_t (the tangential vector). Note v_rel_t_vec · n = 0, so v_rel_t_vec is along ±t where t = (-ny, nx). Let's say v_rel_t_vec = |v_rel_t| * sign * t.

r_vec × J = (-r*nx)*J.y - (-r*ny)*J.x = r*(-nx*J.y + ny*J.x) = r * (ny*J.x - nx*J.y).
If J is along t = (-ny, nx): J = |J|*t direction. ny*J.x - nx*J.y = ny*(-ny*|J|) - nx*(nx*|J|) = -(ny²+nx²)|J| = -|J|. So r_vec × J = r * (-|J|) = -r*|J|.

Hmm, the sign of spin. Let me just compute numerically: 
Δω_b = (r_vec × J) / I_ball where I_ball = m * r² (for a disk/hoop). For a solid disk I = m*r²/2; let's use I = m*r² (treating as ring, or just pick). Actually let me use I = m*r² (simpler, and gives reasonable spin). Then Δω_b = (r_vec × J)/(m*r²) = -|J|/(m*r) = -(fr-1)*|v_rel_t_vec|*m/(m*r) = -(fr-1)*|v_rel_t_vec|/r = (1-fr)*|v_rel_t_vec|/r.

Since fr<1, (1-fr)>0, so Δω_b = (1-fr)*|v_rel_t_vec|/r > 0 when... sign depends on direction of v_rel_t. Let me just compute J vector and use the cross product formula directly with sign:

Δω_b = (r_vec.x * J.y - r_vec.y * J.x) / (m*r²).

where J.x = m*(fr-1)*v_rel_t.x, J.y = m*(fr-1)*v_rel_t.y.

So Δω_b = (m*(fr-1) / (m*r²)) * (r_vec.x * v_rel_t.y - r_vec.y * v_rel_t.x)
= (fr-1)/r² * (r_vec × v_rel_t)  [z-cross]

with r_vec = -r*n = (-r*nx, -r*ny).

r_vec × v_rel_t = (-r*nx)*v_rel_t.y - (-r*ny)*v_rel_t.x = r*(ny*v_rel_t.x - nx*v_rel_t.y).

And t = (-ny, nx), so if v_rel_t = v_tmag * t = v_tmag*(-ny, nx), then ny*v_rel_t.x - nx*v_rel_t.y = ny*(-ny*v_tmag) - nx*(nx*v_tmag) = -v_tmag*(ny²+nx²) = -v_tmag.

So r_vec × v_rel_t = r*(-v_tmag) = -r * v_tmag where v_tmag = v_rel_t · t = (v_rel_t.x*(-ny) + v_rel_t.y*nx)... 

OK this is getting deep. Let me just compute numerically in code:

```js
const tx = -ny, ty = nx; // tangential (along wall)
let vt = rvx*tx + rvy*ty; // tangential relative speed (v_rel · t)
// friction impulse
const fr = 0.8;
const impulse = (1-fr)*vt; // magnitude of tangential velocity removed
// apply to ball velocity (reduce tangential relative velocity)
// v_ball_t_new = v_wall_t + (v_rel_t)*fr  ... 
// equivalently reduce v_ball by (1-fr)*v_rel_t_vec plus account for wall
// v_rel_t_vec = v_rel - (v_rel·n)*n = (rvx,rvy) - vn*(nx,ny)... but we only want tangential part
// Let me just: 
const Jx = -m*(1-fr)*vt*tx;  // impulse on ball (negative along t if vt>0... ) 
const Jy = -m*(1-fr)*vt*ty;
// Actually friction opposes tangential relative motion: J = -sign(vt)*... = -(1-fr)*vt*m*t
```

Hmm, let me think again. vt = v_rel · t. If vt > 0, ball moves in +t direction relative to wall. Friction opposes: impulse in -t direction: J = -(1-fr)*m*|vt|*t_sign... = -(1-fr)*m*vt*t (since if vt>0, -vt*t points -t). So J = -(1-fr)*m*vt*t.

Jx = -(1-fr)*m*vt*tx, Jy = -(1-fr)*m*vt*ty.

Apply: v_ball += J/m = -(1-fr)*vt*t. This reduces the tangential relative velocity: new vt_rel = vt - (1-fr)*vt = fr*vt. Good, that's the damping factor fr on tangential. 

Wait but v_rel = v_ball - v_wall, and applying J/m to v_ball changes v_rel by same amount (v_wall fixed). New v_rel_t = vt + (-(1-fr)*vt) = fr*vt. So tangential relative velocity becomes fr*vt. 

Now spin from this impulse: 
contact relative to ball center: r_vec = -r*n (from center to contact point is toward wall = -n direction, distance r). Wait, contact point = ball_center + r_vec where r_vec points from center to contact. Contact is on the wall side = inward-n direction... the contact point is at ball_center - r*n? n points inward (from wall to center). The contact point between ball and wall is on the side of the ball facing the wall = -n direction from center. So r_vec = contact - center = -r*n = (-r*nx, -r*ny). Yes.

Angular impulse: ΔL = r_vec × J (z): = r_vec.x*J.y - r_vec.y*J.x.
Δω_b = ΔL / I. I = m*r². 

Let me compute in code: 
```js
const rx_vec = -r*nx, ry_vec = -r*ny;  // r_vec
const deltaL = rx_vec*Jy - ry_vec*Jx;  // z cross
ball.sp += deltaL / I;  // I = m*r², m=1
```

With Jx,Jy = -(1-fr)*m*vt*tx, -(1-fr)*m*vt*ty, m=1:
deltaL = (-r*nx)*(-(1-fr)*vt*ty) - (-r*ny)*(-(1-fr)*vt*tx)
= r*nx*(1-fr)*vt*ty - r*ny*(1-fr)*vt*tx... 
= r*(1-fr)*vt*(nx*ty - ny*tx)
t = (-ny, nx) so tx=-ny, ty=nx.
nx*ty - ny*tx = nx*nx - ny*(-ny) = nx² + ny² = 1.
So deltaL = r*(1-fr)*vt.
Δω_b = r*(1-fr)*vt / (m*r²) = (1-fr)*vt/(m*r).

So ball.sp += (1-fr)*vt/(1*r) = (1-fr)*vt/r. 

So: when ball slides on wall with relative tangential speed vt (along wall), friction (1-fr) transfers to spin: sp += (1-fr)*vt/r. And reduces vt to fr*vt. That makes physical sense: it's like partial rolling. Good.

But wait, I need to double check sign convention so the number rotates appropriately. The sign will come out from vt. Let me just trust the formula.

Also I should make sure the tangential velocity reduction is applied to v_ball, and the normal bounce applied too. Let me combine.

Let me also reconsider: maybe simpler to not over-think and just implement and tune. The key deliverables:
- balls drop from center
- gravity + friction
- ball-ball collision
- bounce off rotating walls (with wall motion)
- number rotation shows spin

Let me write it.

**Ball-ball collision with spin:**
```js
function resolveBalls(balls) {
  for (let i=0;i<balls.length;i++){
    for (let j=i+1;j<balls.length;j++){
      const a = balls[i], b = balls[j];
      const dx = b.x - a.x, dy = b.y - a.y;
      const dist2 = dx*dx+dy*dy;
      const rsum = a.r + b.r;
      if (dist2 < rsum*rsum && dist2 > 1e-12) {
        const dist = Math.sqrt(dist2);
        const nx = dx/dist, ny = dy/dist; // from a to b
        // relative velocity
        const vrx = a.vx - b.vx, vry = a.vy - b.vy;
        const vn = vrx*nx + vry*ny;
        if (vn < 0) { // approaching (a moving toward b in +n direction... a.v - b.v, if a approaches b from behind)
          // impulse normal
          const e = 0.7;
          const m1=1, m2=1;
          const jimp = -(1+e)*vn/(1/m1+1/m2);
          const ix = jimp*nx, iy = jimp*ny;
          a.vx += ix/m1; a.vy += iy/m1;
          b.vx -= ix/m2; b.vy -= iy/m2;
          // friction -> spin
          const tx = -ny, ty = nx;
          const vt = vrx*tx + vry*ty;
          const fr = 0.6;
          const jf = -(1-fr)*vt*... 
          // transfer tangential to spin
          ...
        }
        // positional correction
        const pen = rsum - dist;
        const p = pen/ (1/m1+1/m2);
        a.x -= nx*p/m1; ... wait need center of mass weighted. a.x -= nx*pen*0.5; b.x += nx*pen*0.5;
        a.y -= ny*pen*0.5; b.y += ny*pen*0.5;
      } else if (dist2 < rsum*rsum) {
        // overlapping exactly, separate by random
      }
    }
  }
}
```

For ball-ball spin from friction, similar: contact point on ball a is at +r*n (toward b), on ball b at -r*n. The tangential velocity vt = v_rel · t. Friction impulse creates spin on both.

For ball a: r_vec_a = +r*n = (a.r*nx, a.r*ny). Δω_a = r_vec_a × J_t / I.
J_t = -(1-fr)*vt*t * (reduced mass factor). Let me apply J per ball.

Actually let me use a combined friction impulse approach similar to wall. Let me keep it but simpler: 

For ball-ball, after normal impulse, apply tangential damping and spin:
```js
// tangential (using pre-impulse relative velocity)
const tx=-ny, ty=nx;
const vt = vrx*tx + vry*ty;
const fr = 0.5;
const jt = -(1-fr)*vt; // per unit mass impulse along t on relative velocity...
// apply to velocities: reduce relative tangential
const ix_t = jt*tx*0.5, iy_t = jt*ty*0.5; // distribute equally (equal mass)
a.vx += ix_t; a.vy += iy_t;
b.vx -= ix_t; b.vy -= iy_t;
// spin
const ra = a.r;
const rav = {x: ra*nx, y: ra*ny}; // contact relative to a center (toward b)
const rbv = {x: -rb*nx, y: -rb*ny}; // contact relative to b center
// angular impulse on a: rav × J_t_a. J_t_a = m*(jt along t)/2... 
```

Hmm, the distribution. Let me think: the impulse J (vector) acts on a in direction ... For ball-ball friction, the impulse on a due to friction = -(1-fr)*vt*t applied to a? No.

Let me reconsider. The relative tangential velocity vt = (a.v - b.v)·t. To damp it by factor fr, we want new vt' = fr*vt. Change needed: Δvt = (fr-1)*vt. We apply equal and opposite impulses to a and b along t: J on a = -(1-fr)*vt*0.5*t? No—for equal masses, to change relative velocity by Δvt along t, apply impulse J along t to a and -J to b: relative velocity change = (J/m_a + J/m_b) along... = J*(1/m_a+1/m_b). For m=1: = 2J. We want 2J = (fr-1)*vt => J = (fr-1)*vt/2. So impulse on a along t = (fr-1)*vt/2 * t... 

Wait sign. If vt > 0 (a moving in +t relative to b), friction on a should be in -t direction (opposing). So J on a = -|...|*t. (fr-1) is negative, so (fr-1)*vt/2 is negative, and (negative)*t points -t. Good: J_on_a_vec = ((fr-1)*vt/2) * t. Applied to a.v: a.v += J_on_a_vec (since m=1). And b.v -= J_on_a_vec.

Let me verify: a.v becomes a.v + J, b.v becomes b.v - J. New v_rel = (a.v+J) - (b.v - J) = v_rel + 2J. v_rel along t: 2*(J·t) = 2*((fr-1)*vt/2) = (fr-1)*vt. New vt_rel = vt + (fr-1)*vt = fr*vt. 

Now spin on a: angular impulse = r_vec_a × J_on_a_vec / I_a. r_vec_a = +r*n (contact point on a, toward b). 
Δω_a = (r_vec_a × J_a)/I.
J_a = ((fr-1)*vt/2)*t.
r_vec_a = r*n.
r*n × ((fr-1)*vt/2)*t = r*(fr-1)*vt/2 * (n × t). n×t z = nx*ty - ny*tx = nx*nx - ny*(-ny) = 1 (since t=(-ny,nx)). So = r*(fr-1)*vt/2.
I_a = m*r² = r². Δω_a = r*(fr-1)*vt/2 / r² = (fr-1)*vt/(2r).
Since fr<1, negative if vt>0. Sign: if a moves in +t (=(-ny,nx)) relative to b, friction tries to spin a... The contact point velocity from spin should reduce relative slip. Anyway, formula gives consistent spin.

Similarly Δω_b: r_vec_b = -r*n (contact on b, toward a, = -n*r). J_b = -J_a. 
Δω_b = (r_vec_b × J_b)/I = (-r*n × (-J_a))/r² = (r*n × J_a)/r² = same as Δω_a = (fr-1)*vt/(2r). 

Interesting both get same sign spin (both rotate same way), which makes sense for rolling between them. OK.

Let me just implement these formulas. I'll keep I = m*r² (ball as ring). 

Actually, for the spin to be visually prominent, maybe I want the spin to also come from the ball "rolling" on walls. The wall friction already does that. Good.

Also, I want balls to eventually roll on the wall (the floor of the rotating heptagon). With wall friction transferring spin, they'll spin as they slide. Good. Let me make wall friction stronger (fr=0.7) so they quickly gain spin matching rolling.

Let me also add air resistance (linear damping 0.99) and spin damping.

Let me also reconsider: gravity g. With R≈300, g should make balls move reasonably. g = 600? Let me use g = 500.

Let me think about the number rendering on each ball. For ball i, draw circle, then draw the number (i+1) rotated by ball.ang. Set ball.ang += ball.sp * dt.

Let me also color the balls.

**Rendering order:** clear, draw heptagon (spinning), draw balls, draw numbers on balls (rotated), maybe draw center. 

Heptagon: draw 7 edges, with a moving appearance. Maybe shade based on rotation. Keep simple: stroked polygon with thick lines, maybe with some accent.

Actually, to make it look nice, I'll draw the heptagon with a gradient or just white stroke. And maybe show the rotation by a marker.

Let me make the balls bright colored, numbers white/black with a contrasting ring.

Let me also add: the balls spawn and there's an initial "drop." 

Let me also handle the case where balls need to fit. 20 balls r=11, total... the heptagon interior radius (inradius) = R*cos(π/7) ≈ 300*0.9009 ≈ 270. Area ≈ large. 20 balls of r=11 area = 20*π*121 ≈ 7600, polygon area ≈ (7/2)*R²*sin(2π/7) ≈ 3.5*90000*sin(0.8976)≈3.5*90000*0.7818≈246000. Plenty of room. Good. r=12 maybe. Let me use r=11.

Hmm, but if balls pile at the bottom (the rotating floor), they might not all fit in one layer and could jam. With r=11, a single row along the floor edge (length ≈ side * ... ) 7 sides. Bottom area. Let me not worry; collisions push them around. If too many pile, they might jitter. With friction they'll settle. 20 balls should be OK with r=11.

Actually the heptagon is rotating, so the "floor" rotates and balls will be flung. This creates a dynamic scene—good for showing spin and bounces.

Let me think about potential issue: balls getting stuck in walls due to position correction pushing them into another wall or each other. Iterative position correction (multiple passes) helps. Let me do a few iterations of wall + ball correction.

Let me write the code now.

Let me also double-check the wall normal orientation issue. n = (dx,dy)/dist points from contact point P toward ball center. For a ball inside the polygon near an edge, P is on the edge, ball center is inside, so n points inward (toward center). That's the inward normal. Good. We want to push the ball inward (into polygon) by penetration, correction: ball.x += nx*pen, ball.y += ny*pen (along n, inward). Good.

vn = v_rel · n. v_rel = v_ball - v_wall. If ball approaches wall (moving outward, away from center, in +n? no). Wait, n points inward (toward center). Approaching the wall means moving toward the wall = outward = opposite to n = -n direction. So v_rel · n < 0 means approaching. vn < 0 → bounce. Good (matches earlier).

Let me now also reconsider the wall velocity used for v_rel: at the contact point P. v_wall = omega*(-(Py-cy), (Px-cx)). Good.

Now let me write the full code.

```js
const canvas = document.getElementById('c');
const ctx = canvas.getContext('2d');
let W=window.innerWidth, H=window.innerHeight;
canvas.width=W;canvas.height=H;
const CX=W/2, CY=H/2;
const R = Math.min(W,H)*0.36;
const omega = 2*Math.PI/5; // rad/s
const g = 700;
const N = 20;
const balls = [];

for (let i=0;i<N;i++){
  balls.push({
    r:11,
    x: CX + (Math.random()-0.5)*30,
    y: CY + (Math.random()-0.5)*30,
    vx: (Math.random()-0.5)*40,
    vy: (Math.random()-0.5)*40,
    sp: 0,
    ang: Math.random()*Math.PI*2,
    c: `hsl(${i*18},80%,60%)`
  });
}

let theta = 0;
const DT = 1/120;
const SUB = 2;
const E = 1/60;

function vertices(t){
  const v=[];
  for(let i=0;i<7;i++){
    const a=t+2*Math.PI*i/7;
    v.push({x:CX+R*Math.cos(a), y:CY+R*Math.sin(a)});
  }
  return v;
}

function physics(dt){
  theta += omega*dt;
  const vs = vertices(theta);
  for(const b of balls){
    b.vy += g*dt;
    // air resistance
    b.vx *= 0.998; b.vy *= 0.998;
    b.sp *= 0.995;
    b.x += b.vx*dt;
    b.y += b.vy*dt;
    b.ang += b.sp*dt;
  }
  // collisions
  for(let iter=0;iter<4;iter++){
    // walls
    for(const b of balls){
      resolveWall(b, vs);
    }
    // balls
    resolveBalls();
  }
  // bounds safety: if somehow outside, clamp (shouldn't happen)
}

function resolveWall(b, vs){
  for(let i=0;i<7;i++){
    const a1=vs[i], a2=vs[(i+1)%7];
    const ex=a2.x-a1.x, ey=a2.y-a1.y;
    const len2=ex*ex+ey*ey;
    if(len2<1e-9) continue;
    const ax=b.x-a1.x, ay=b.y-a1.y;
    const t2=(ax*ex+ay*ey)/len2;
    const tc=Math.max(0,Math.min(1,t2));
    const px=a1.x+tc*ex, py=a1.y+tc*ey;
    const dx=b.x-px, dy=b.y-py;
    const dist=Math.sqrt(dx*dx+dy*dy);
    if(dist<b.r){
      let nx,ny;
      if(dist<1e-6){ nx=CX-px; ny=CY-py; }
      else { nx=dx/dist; ny=dy/dist; }
      const nl=Math.sqrt(nx*nx+ny*ny);
      nx/=nl; ny/=ny... 
```

Hmm let me be careful with normalization when dist tiny. Let me write a helper.

Let me just write the complete file now. I'll be careful.

Let me write it fully and cleanly.

For the number rendering, I'll draw the ball, then a ring, then the number rotated.

Let me also draw the heptagon nicely—maybe with an inner glow and tick marks at vertices to show rotation.

Let me also add a subtle rotation indicator. Keep it clean and impressive.

Let me consider performance: 20 balls, 4 iterations, walls 7, ball pairs 190. Fine.

One concern: the wall collision normal when ball is near a vertex (closest point is a vertex). The normal should point from vertex toward ball. Using n=(dx,dy)/dist from the vertex (px,py) to ball works. But then pushing along that normal might not be the true polygon inward—could push ball outside adjacent edge. With iterations and the fact that vertices are on the heptagon, it's usually fine. Let me keep it.

Let me also make sure balls don't escape: after all corrections, if a ball is outside (beyond edge), the wall resolver with position correction handles it.

Let me write the file.

For visual quality, let me use a slightly larger canvas with dpr, draw antialiased.

Let me also add: numbers should be readable. Ball radius 11, font size ~ about 0.7*r = ~8px? That's tiny. Let me scale: at r=11, draw number with radius ~8. Maybe increase ball size if window large. Let me set r relative to window: r = Math.min(W,H)*0.018 maybe. For 800px → 14. For 1920 → 34. Let me use r = Math.min(W,H)*0.02 → 16 at 800. Number font ~ 0.85*r.

Let me parameterize nicely.

Let me write it.

Actually, let me reconsider gravity and the "drop from center." When balls start at center and gravity pulls down, the heptagon is also rotating. The balls will fall and hit walls, bounce. Good, dynamic.

But there's a subtlety: with the heptagon rotating fast-ish (full turn per 5s) and balls bouncing around, it'll be lively. Good for showcasing.

Let me make sure restitution and friction feel "realistic" - not too bouncy to settle. Let me use e=0.6 for walls, friction reducing tangential. 

Let me write the final code.

Let me also think: should balls bounce "realistically off rotating walls." The wall motion effect: when the wall moves toward the ball, extra bounce; the reflection includes the wall's velocity. I've used v_rel = v_ball - v_wall and reflected. Good.

Let me finalize and write.

I'll write helper normalize, and the resolver.

Let me also double check the spin update sign for wall: I derived Δω_b = (1-fr)*vt/r where vt = v_rel·t, t = (-ny, nx). And n = (dx,dy)/dist (inward). Let me just compute via the cross product numerically to avoid sign mistakes:

```
// tangential relative velocity
const tx=-ny, ty=nx;
let vt = rvx*tx + rvy*ty;  // v_rel · t
// friction impulse magnitude (reduces |vt| by factor fr)
const fr = 0.8;
const jmag = (1-fr)*vt; // the amount of tangential relative velocity removed
// apply impulse to ball velocity: reduce tangential relative vel
// v_ball_t new = v_wall_t + (vt - jmag)*t = ... change v_ball by -jmag*t
b.vx -= jmag*tx; // wait sign
```

Hold on. v_rel = v_ball - v_wall. We want new v_rel_t = fr*vt (reduced). Change in v_rel_t = (fr-1)*vt. Since v_rel changes by same as v_ball change (v_wall fixed), change in v_ball_t = (fr-1)*vt. Direction t with sign vt. So Δv_ball_t_vec = (fr-1)*vt * t. Applied to v_ball: b.v += (fr-1)*vt*t. (fr-1) negative → reduces. Good.

```js
b.vx += (fr-1)*vt*tx;
b.vy += (fr-1)*vt*ty;
```

Then the velocity update for normal bounce must also be applied. But careful: I'm modifying b.v for friction; the normal impulse also modifies b.v. Let me combine: 

Compute v_rel = b.v - v_wall. vn = v_rel·n. vt = v_rel·t.
If vn<0 (approaching):
  Apply normal: b.v = v_wall + (-e*vn)*n + (v_rel_t)*1 ... 
Hmm, let me restructure: decompose v_rel into normal and tangential parts, then recombine.

```js
const rvx = b.vx - wx, rvy = b.vy - wy; // v_rel
const vn = rvx*nx + rvy*ny;
const tx=-ny, ty=nx;
const vt = rvx*tx + rvy*ty;
if (vn < 0){
  const e=0.7, fr=0.7;
  // new relative velocity
  const vn2 = -e*vn;
  const vt2 = fr*vt;
  // new v_rel = vn2*n + vt2*t
  // b.v = v_wall + new v_rel
  b.vx = wx + vn2*nx + vt2*tx;
  b.vy = wy + vn2*ny + vt2*ty;
  // spin from friction impulse (magnitude (1-fr)*|vt| removed)
  // impulse along friction: Δv_ball_t = (fr-1)*vt*t ; angular impulse
  const rx = -b.r*nx, ry = -b.r*ny; // contact rel to center
  const Jx = (fr-1)*vt*tx, Jy = (fr-1)*vt*ty; // impulse per unit mass? J with m=1: J_vec = m*Δv = Δv since m=1
  // angular impulse = rx*Jy - ry*Jx
  let dL = rx*Jy - ry*Jx;
  b.sp += dL/(b.r*b.r); // I = m r² = r² (m=1)
}
// position correction
const pen = b.r - dist;
b.x += nx*pen; b.y += ny*pen;
```

Wait, but I only apply spin/friction when vn<0 (approaching). But tangential friction should also apply when the ball is sliding on the wall even if not "approaching" in normal (e.g., resting). Hmm. For resting contact, vn≈0 and we'd want friction to damp tangential and create rolling. Let me apply tangential damping always (regardless of vn<0) but normal only when approaching. Actually, to avoid sticking when separating, let me apply tangential friction always but it's minor. Let me apply friction always (vt damping + spin), and normal bounce only when vn<0, and a small positional correction always.

Actually if I apply tangential reduction always, a ball sliding along wall constantly gets vt damped and sp increased (rolling). Good for realism. Let me do that.

But reducing vt always when there's contact (dist<b.r) even slightly—what if ball is just touching and moving along? It'll keep damping vt and adding spin. That's fine—it simulates rolling friction. But could over-damp. Let me apply with small factor per collision frame; since we do 4 iterations, vt gets damped 4x per frame. Let me use fr=0.95 to be gentle, and rely on iterations. Actually let me restructure: I'll apply the full contact logic once per (ball, edge) but with gentle damping, and do few iterations.

Hmm, doing 4 iterations each damping tangential by 0.7 would kill tangential velocity fast (0.7^4=0.24 per frame → very high friction, balls stop). Let me use 0.98 per iteration and 2 iterations, plus a separate linear air drag. Let me tune: fr=0.95, iterations=3. (0.95^3=0.855 per frame along wall contact). Plus air drag 0.998. Combined stopping reasonable.

Actually for "realistic bounce," let me reduce iterations to 3 and fr=0.92. I'll tune by testing mentally; can't test visually but let me make reasonable.

Let me reduce complexity: do collision resolution with 3 passes. For each wall contact, apply normal bounce (if approaching) + tangential damping+spin. 

Let me also reconsider: do I want friction to create spin when ball is at rest on wall (vn≈0 but ball sliding due to wall motion)? Yes that's the rolling. Fine.

Let me also handle ball-ball friction similarly (transfer spin).

Let me write the final code and tune constants. I'll go with:
- g = 800
- e_wall = 0.6, fr_wall = 0.9 (rolling-ish)
- e_ball = 0.7
- air drag 0.998
- spin drag 0.99

Let me write it.

I realize I should double check: with balls starting very close (cluster at center, 30px spread, r=16 say), they overlap initially. The ball collision resolver will separate them over iterations. Starting velocities tiny. Should be OK; they'll pop apart then fall. Could cause a brief jitter but fine. Maybe start them slightly spread by placing on a small circle to avoid heavy overlap. Let me place initial positions on a small circle radius 20, so they don't overlap: angle = i*2π/20, pos = center + 20*(cos,sin). With r=16, circumference/20 ~ 6.3, so spacing 20*2π/20=6.3 < 2r... they'd still overlap. Let me place with radius 80 initial: spacing = 80*2π/20 ≈ 25 > 2*16=32? No, 25 < 32. r=16 needs spacing≥32 → radius ≥ 32*20/(2π)=101.9. Let me place initial cluster radius ~120 to avoid overlap, or just place them at center with zero overlap by staggering... 

Simplest: place them along a small line/circle with enough radius. Let me place on circle radius = 3*r... no. Let me just place each ball at center + small offset and let physics separate; use the position correction which separates them. Initial heavy overlap → position correction in ball resolver pushes them apart; over 3 iterations they separate partially then fall. Acceptable "drop from center" effect. The user said "dropping from the center," so a cluster that disperses is fine.

Actually heavy initial overlap might cause issues (balls tunneling). Let me place initial positions on a circle of radius ~ 4*r around center, with 20 balls. 4*r = 64, circumference = 2π*64≈402, per ball arc = 20, chord ≈ 20 > 2r=32? 20<32, still overlap. Need radius≥102. Let me just do radius = 130, no overlap, looks like a ring dropping. But "from the center" suggests center origin. Hmm.

Compromise: place them in a small filled circle (radius ~ 30) with random positions, accept minor overlap, let resolver separate. Since resolver does positional correction (push apart), they'll separate cleanly. The initial velocity 0. This looks like they're dropping from the center and spreading as they fall. I'll go with cluster radius ~30, r=11 → overlap is mild (20 balls in radius 30 circle: total ball area 20*π*121≈7600, circle area π*900≈2827, so they definitely overlap). Heavy overlap. The resolver might be unstable with 20 heavily overlapping circles (pushing apart but constrained).

Let me instead place them on a circle of radius 90 (just above where they fit) so they're slightly overlapping or touching, then "drop." With radius 90: chord spacing 2*90*sin(π/20)=180*0.156=28, 2r=22, so 28>22, no overlap! radius 90 works for r=11. Wait r=11 → 2r=22, spacing 28 > 22, no overlap. 

But that's a ring of radius 90, not "center." The user wants them dropping from center. Let me start them at center but with a tiny outward velocity or just let resolver handle. Actually, let me start them all at the exact center with zero velocity and let the first physics step... they'd all be at the same point (total overlap, dist=0, can't compute normal). That breaks.

So I must start them non-overlapping. Let me start them in a tiny cluster but non-overlapping: use a small circle packing. The simplest robust start: place them on a circle of radius ~ 1.5*r*... Let me place them within a small radius using a grid/jitter but ensure separation by running a few separation steps at startup, OR just place on a small circle.

Let me place initial positions on a small circle of radius = 2.5*r = 27.5 for r=11. Spacing = 27.5*2π/20 = 8.6 < 2r=22 → overlap. So they'd overlap and resolver separates them over the first frame into a ring. The positional correction (push apart by pen/2 each) would expand the ring quickly. That actually looks like an explosion from center then settling under gravity—dramatic! Could be fun. But instability risk.

Let me instead just start them already separated on a modest circle and add a downward initial velocity to make them "drop." That's clearly "dropping from the center." Place on circle radius 80, r=11, no overlap, give each a slight downward velocity and let gravity take over. The heptagon is stationary-ish at start (theta=0) so they fall, hit walls, bounce, heptagon rotates. Nice.

Actually even simpler and matches "dropping from center": start them all near center but in a small non-overlapping arrangement (e.g., 4 rows of 5, or a small hexagonal packing within radius ~ 40), zero initial velocity. They fall and spread. Let me do a small grid cluster at center: 20 balls in ~5x4 grid within radius 45, r=11 (spacing 25 > 22). They start stationary, gravity pulls down, they "drop" and immediately interact. 

Let me do that: positions = center + offsets on a 5x4 grid with spacing 25, centered. No overlap (25>22). r=11. They drop from center. 

Let me code initial positions:
```js
const cols=5, rows=4, sp=25;
let idx=0;
for(let i=0;i<cols;i++)for(let j=0;j<rows;j++){
  if(idx>=N) break;
  balls.push({x: CX + (i-(cols-1)/2)*sp, y: CY + (j-(rows-1)/2)*sp, ...});
  idx++;
}
```
But then numbering: ball id should be 1..20. Let me assign id = idx+1.

Good. This gives a centered cluster dropping.

Wait, but "balls dropping from the center of a heptagon"—maybe they want balls to appear at center and fall outward? The grid cluster at center captures "from the center" well. Good.

Let me finalize code.

Let me also reconsider r: with grid spacing 25 and r=11, fits. But 5 wide = 5*25=125 span + 2*11 margin ~ within heptagon inradius 270. Fine.

Let me make r scale with window: r = Math.min(W,H)*0.016 → at 1080: ~17, at 800:~13. Grid spacing = 2.4*r... let me set sp = 2.3*r. 

Let me just compute r and sp from r.

Let me write the code now, then review.

Let me write index.html.

```html
<!DOCTYPE html>
<html lang="en">
<head>
<meta charset="utf-8">
<title>Heptagon Ball Drop</title>
<style>
html,body{margin:0;padding:0;background:#111;height:100%;overflow:hidden}
canvas{display:block}
</style>
</head>
<body>
<canvas id="c"></canvas>
<script>
(() => {
  const canvas = document.getElementById('c');
  const ctx = canvas.getContext('2d');
  let W,H;
  function resize(){
    W=canvas.width=window.innerWidth;
    H=canvas.height=window.innerHeight;
  }
  resize();
  window.addEventListener('resize',resize);
  const CX=W/2, CY=H/2;
  const RR=Math.min(W,H)*0.36; // heptagon radius
  const omega=2*Math.PI/5; // 1 revolution per 5s
  const g=900;
  const N=20;
  const br= Math.max(8, Math.min(W,H)*0.016);
  const balls=[];
  {
    const cols=5, rows=4;
    const sp=2.4*br;
    let k=0;
    for(let i=0;i<cols;i++)for(let j=0;j<rows;j++){
      if(k>=N) break;
      balls.push({
        x:CX+(i-(cols-1)/2)*sp,
        y:CY+(j-(rows-1)/2)*sp,
        vx:(Math.random()-0.5)*30,
        vy:(Math.random()-0.5)*30,
        r:br,
        sp_:0, // spin ang vel
        ang:Math.random()*Math.PI*2,
        id:k+1,
        c:`hsl(${k*19},80%,58%)`
      });
      k++;
    }
  }
  let theta=0;
  function verts(t){
    const v=[];
    for(let i=0;i<7;i++){
      const a=t+2*Math.PI*i/7;
      v.push({x:CX+RR*Math.cos(a), y:CY+RR*Math.sin(a)});
    }
    return v;
  }
  function resolveWall(b,vs){
    for(let i=0;i<7;i++){
      const p1=vs[i], p2=vs[(i+1)%7];
      const ex=p2.x-p1.x, ey=p2.y-p1.y;
      const L2=ex*ex+ey*ey;
      if(L2<1e-10) continue;
      const ax=b.x-p1.x, ay=b.y-p1.y;
      const t2=(ax*ex+ay*ey)/L2;
      const tc=Math.max(0,Math.min(1,t2));
      const px=p1.x+tc*ex, py=p1.y+tc*ey;
      let dx=b.x-px, dy=b.y-py;
      const dist=Math.hypot(dx,dy);
      if(dist<b.r){
        let nx,ny;
        if(dist<1e-6){ nx=CX-px; ny=CY-py; }
        else { nx=dx/dist; ny=dy/dist; }
        const nl=Math.hypot(nx,ny);
        if(nl<1e-9) continue;
        nx/=nl; ny/=nl;
        // wall velocity (rotation about center)
        const relx=px-CX, rely=py-CY;
        const wx=-omega*rely, wy=omega*relx;
        const rvx=b.vx-wx, rvy=b.vy-wy;
        const vn=rvx*nx+rvy*ny;
        const tx=-ny, ty=nx;
        const vt=rvx*tx+rvy*ty;
        if(vn<0){
          const e=0.6;
          const vn2=-e*vn;
          // new relative velocity: normal bounce + tangential damping
          const fr=0.85;
          const vt2=fr*vt;
          b.vx=wx+vn2*nx+vt2*tx;
          b.vy=wy+vn2*ny+vt2*ty;
        } else {
          // still damp tangential, allow rolling
          const fr=0.98;
          b.vx-= (1-fr)*vt*tx;
          b.vy-= (1-fr)*vt*ty;
        }
        // spin from tangential friction (always when in contact)
        const frSp=0.9; // ... 
        // impulse: reduce vt further -> spin
        ...
      }
    }
  }
```

Hmm, I'm double-dampening tangential. Let me simplify: when vn<0, do full bounce with tangential damping fr and spin transfer. When vn>=0 (resting), skip (or gentle). Let me just always: compute the tangential damping and spin transfer using the relative velocity, but only apply normal bounce if vn<0.

Let me define a single approach: 
```
if(vn<0){
  // approaching: bounce + friction
  const e=0.6;
  const fr=0.8;
  const vt2=fr*vt;
  b.vx = wx + (-e*vn)*nx + vt2*tx;
  b.vy = wy + (-e*vn)*ny + vt2*ty;
  // spin transfer from friction:
  transferSpinWall(b,nx,ny,tx,ty,vt,fr);
} 
// positional correction always
const pen=b.r-dist;
b.x+=nx*pen; b.y+=ny*pen;
```

transferSpinWall: angular impulse from tangential velocity reduction.
Δv_ball_t = (fr-1)*vt*t (this is the change). Angular impulse = r_vec × J where J = m*Δv = (fr-1)*vt*t (m=1). r_vec = -r*n (contact from center). 
dL = r_vec.x*J.y - r_vec.y*J.x = (-r*nx)*(((fr-1)*vt)*ty) - (-r*ny)*(((fr-1)*vt)*tx)
= (fr-1)*vt * [ -r*nx*ty + r*ny*tx ] = (fr-1)*vt*r*(ny*tx - nx*ty)
tx=-ny, ty=nx: ny*tx - nx*ty = ny*(-ny) - nx*nx = -(ny²+nx²) = -1.
So dL = (fr-1)*vt*r*(-1) = -(fr-1)*vt*r = (1-fr)*vt*r.
Δω = dL/I = (1-fr)*vt*r/(r²) = (1-fr)*vt/r.
With m=1, I=r². So b.sp_ += (1-fr)*vt/r.

So:
```
b.sp_ += (1-fr)*vt/b.r;
```
where vt = v_rel·t before damping. Good, and sign handles direction.

Let me verify sign with intuition: ball slides along wall in +t direction (vt>0), friction should spin ball so bottom (wall side) moves ... rolling. The spin sign: if ball rolls in +t direction without slipping, the contact point velocity from spin = ω_b * perp(rel). For rolling, ω_b relates to +t... It'll come out; trust the math. Let me just implement and the number rotation will reflect actual physics.

For ball-ball:
```
function resolveBalls(){
  for(i..j){
    dx,dy,dist,rsum
    if(dist<rsum && dist>1e-10){
      nx=dx/dist;ny=dy/dist;
      tx=-ny,ty=nx;
      vrx=a.vx-b.vx, vry=a.vy-b.vy;
      vn=vrx*nx+vry*ny;
      vt=vrx*tx+vry*ty;
      if(vn<0){ // approaching (a.v-b.v along n; if a moves toward b... a to b is +n; a.v-b.v .n <0 means moving apart actually!)
```

Wait, careful. n points from a to b. v_rel = a.v - b.v. vn = v_rel·n. If vn > 0, a moves toward b (in +n) faster → approaching (since a is behind, moving toward b). If vn<0, separating. So bounce when vn>0. Let me re-examine: a and b, a is at lower, n from a→b. After they approach, a.v > b.v in n direction means closing. So approaching when vn>0. So bounce when vn>0. Let me fix: bounce if vn>0, impulse reduces it.

Impulse j = -(1+e)*vn/(1/m1+1/m2). For m=1: j=-(1+e)*vn/2. Negative (since vn>0). Applied to a: a.v += j*n → a.v decreases in +n (slows/sends back). Good.

Let me rewrite ball-ball:
```
if (vn>0){
  const e=0.7;
  const j=-(1+e)*vn/2;
  ix=j*nx; iy=j*ny;
  a.vx+=ix; a.vy+=iy; b.vx-=ix; b.vy-=iy;
  // friction: damp vt, create spins
  const fr=0.4;
  const jt=(1-fr)*vt/2; // per ball? 
  a.vx += (1-fr)*vt*0... 
```

Hmm let me redo tangential. v_rel_t = vt (a.v-b.v along t). We want reduce to fr*vt. Change in v_rel_t = (fr-1)*vt. Apply equal/opposite impulses along t: a gets ((fr-1)*vt/2)*t, b gets -that (since m equal, change in v_rel = 2*(impulse along t) = (fr-1)*vt → impulse = (fr-1)*vt/2 per ball). 
```
const fx=(fr-1)*vt/2*tx; const fy=(fr-1)*vt/2*ty;
a.vx+=fx; a.vy+=fy; b.vx-=fx; b.vy-=fy;
```
Check: v_rel_t new = (a.v+fx - (b.v-fx))·t = (v_rel + 2(fx,fy))·t = vt + 2*(fx*tx+fy*ty) = vt + 2*((fr-1)*vt/2) = fr*vt. Good.

Spin transfer:
a: r_vec_a = +r_a*n (contact on a toward b). J_a_t = (fx,fy) (the impulse on a along t is (fx,fy)? wait fx,fy already = impulse vector on a). Angular impulse on a = r_vec_a × J_a = (r_a*nx)*fy - (r_a*ny)*fx = r_a*(nx*fy - ny*fx).
nx*fy - ny*fx = nx*((fr-1)*vt/2*ty) - ny*((fr-1)*vt/2*tx) = (fr-1)*vt/2*(nx*ty - ny*tx) = (fr-1)*vt/2*(nx*nx - ny*(-ny)) = (fr-1)*vt/2*(1) = (fr-1)*vt/2.
So dL_a = r_a*(fr-1)*vt/2. Δω_a = dL_a/r_a² = (fr-1)*vt/(2*r_a).
b: r_vec_b = -r_b*n. J_b = -(fx,fy). dL_b = r_vec_b × J_b = (-r_b*nx)*(-fy) - (-r_b*ny)*(-fx) = r_b*nx*fy - r_b*ny*fx... wait: (-r_b*nx)*(-fy) - (-r_b*ny)*(-fx) = r_b*nx*fy - r_b*ny*fx = r_b*(nx*fy-ny*fx) = r_b*(fr-1)*vt/2. Δω_b = (fr-1)*vt/(2*r_b). Same sign as a. Both spin same way—correct for rolling between them.

```
a.sp_ += (fr-1)*vt/(2*r_a);
b.sp_ += (fr-1)*vt/(2*r_b);
```

With fr=0.4, (fr-1)=-0.6, so spin opposes... sign: if vt>0 (a moves +t relative to b), the impulse reduces it, and spin created. Fine.

Positional correction ball-ball: pen = rsum - dist. push a by -n*pen/2, b by +n*pen/2.
```
a.x -= nx*pen*0.5; a.y -= ny*pen*0.5;
b.x += nx*pen*0.5; b.y += ny*pen*0.5;
```

Good.

Now rendering. Each ball: fill circle gradient, then number rotated.

Heptagon: draw edges with thick stroke, maybe a glowing effect, and vertex markers + a tick to show rotation.

Let me render:
```
function render(){
  ctx.clearRect(0,0,W,H);
  // heptagon
  const vs=verts(theta);
  // background ring
  ctx.fillStyle='#181c2a';
  ctx.fillRect(0,0,W,H);
  ctx.save();
  ctx.translate(CX,CY);
  ctx.strokeStyle='#3a4260';
  ctx.fillStyle='rgba(60,70,110,0.12)';
  ctx.lineWidth=3;
  ctx.beginPath();
  for(let i=0;i<7;i++){ const p=vs[i]; if(i==0)ctx.moveTo(p.x-CX,p.y-CY); else ctx.lineTo(p.x-CX,p.y-CY);}
  ctx.closePath(); ctx.fill(); ctx.stroke();
  // rotation indicator - a line from center to first vertex
  ctx.strokeStyle='#ffd24f';
  ctx.lineWidth=2;
  ctx.beginPath(); ctx.moveTo(0,0); const v=vs[0]; ctx.lineTo(v.x-CX,v.y-CY); ctx.stroke();
  ctx.restore();
  // balls
  for(const b of balls){
    ctx.save();
    ctx.translate(b.x,b.y);
    ctx.rotate(b.ang);
    const grad=ctx.createRadialGradient(-b.r*0.3,-b.r*0.3,1,0,0,b.r);
    grad.addColorStop(0,b.c);
    grad.addColorStop(1,'#000');
    ctx.fillStyle=grad;
    ctx.beginPath(); ctx.arc(0,0,b.r,0,Math.PI*2); ctx.fill();
    ctx.strokeStyle='rgba(0,0,0,0.9)';
    ctx.lineWidth=1.5; ctx.stroke();
    // number
    ctx.textAlign='center'; ctx.textBaseline='middle';
    ctx.font=`bold ${b.r*1.1}px sans-serif`;
    ctx.fillStyle='#fff';
    ctx.fillText(b.id,0,0);
    ctx.restore();
  }
}
```

Wait, the number should rotate with the ball (b.ang), and we rotate canvas by b.ang so the number draws rotated—good, shows spin. But the ball circle itself also rotates (cosmetic, fine). Actually if I rotate the whole canvas including the circle, the gradient moves—minor. To keep circle upright and only number rotated, draw circle first then rotate for number only. Let me draw circle unrotated, then save again for number rotation. Cleaner. But it's fine to rotate both; balls rotating visually is OK and the gradient shift is subtle. Let me rotate only number for clarity:

```
ctx.save();
ctx.translate(b.x,b.y);
// circle
...fill circle...
// number rotated
ctx.save();
ctx.rotate(b.ang);
ctx.fillText(b.id,0,0);
ctx.restore();
ctx.restore();
```

Good.

Main loop:
```
function loop(){
  const now=...;
  for(let s=0;s<SUB;s++){
    physics(DT);
  }
  render();
  requestAnimationFrame(loop);
}
requestAnimationFrame(loop);
```

Fixed DT, ignore real elapsed (or use real elapsed but substep). Fixed is stable. Let me use real elapsed with clamp and substeps:
Actually fixed DT with SUB=2 → 2*120=240 physics steps/s. Good. Let me just use fixed.

But then animation speed independent of frame rate—fine.

Let me set SUB=2, DT=1/120 → 240 physics/sec. Each physics call does collision iter 3 + walls. 240 * (3*(20*7 + 190)) ≈ 240*3*330=237600 ops/sec, fine.

Let me set collision iterations=3.

Let me also cap velocities to prevent explosion: if speed huge, scale down. Add a safety clamp.

Let me also ensure balls stay inside: wall resolver with position correction should keep them in. 

Let me write final code and create the file.

Let me reconsider constants for good visuals:
- g=900, omega=2π/5≈1.257
- e_wall=0.6, fr_wall=0.85 (rolling), e_ball=0.7, fr_ball=0.4
- air 0.997, spinDrag 0.993

Spin: b.sp_ *= spinDrag each frame; transfers during collisions. Should see numbers spin as balls roll on walls. 

Let me make sure spin magnitude is visible: a ball rolling at v=200, r=16: rolling ω=v/r=12.5 rad/s. Over collisions, sp builds toward that. Number rotates 12.5 rad/s ~ 2 rev/s, visible. Good.

But my wall friction factor fr=0.85 per contact iteration, 3 iterations → vt*0.85³=0.61 per physics step, too aggressive (kills tangential fast, sp builds fast). Hmm that means balls sliding on wall lose tangential velocity quickly (high friction) → they'd stick/roll immediately. That's like high friction floor—realistic for "friction" but might stop motion too fast. Let me reduce fr_wall to 0.95 per iteration. 0.95³=0.857 per step, times 240 steps... that's per second huge damping. Wait DT fixed; physics step 1/120, 2 substeps→ 1/60s per frame. Per second: (0.95^3)^(60) ≈ 0.857^60 ≈ e^(60*ln.857)=e^(60*-0.154)=e^-9.2≈0.0001. That kills tangential velocity in <1 second even without gravity. That's too much friction!

The issue: applying friction every physics step while in contact. A ball sliding along wall contacts continuously, so per physics-step damping accumulates. With dt=1/120 and 0.95 factor, that's a coefficient of 0.95 per 1/120s → equivalent to continuous damping rate. 0.95 per (1/120)s ≈ exp(-0.0513 per 1/120) → per second exp(-6.16)=0.002. So velocity decays to ~0 in ~1s. Too sticky.

I should model friction as Coulomb with the normal force, not a per-step velocity multiplier. But that's more complex. Alternatively, apply friction only at the moment of impact (when vn<0), not every frame while resting. That way rolling only gets spin boost on bounces, and between bounces the ball keeps tangential velocity (sliding with air drag). But then a ball resting on the floor wouldn't spin.

Hmm. For a more physical model: friction while in contact. Let me use impulse-based friction with a coefficient tied to normal force and dt:
Friction impulse magnitude = μ * |normal impulse|, applied tangentially (clamp to vt). This is per-collision (impact) based.

For continuous contact (resting), use the same: each substep, the ball is penetrating wall → contact → apply Coulomb friction: impulse = μ * (mass*gravity-ish normal)... Actually for resting contact, normal impulse balances gravity → friction = μ*N. 

This is getting complex. Let me take a pragmatic approach: apply tangential damping only when vn<0 (on impact), with fr=0.7. And separately, for rolling, derive spin from the no-slip condition when sliding: if ball in contact and vt != 0, apply a friction impulse bounded by μ*|vn|... 

Simpler pragmatic: Each physics step, for wall contacts, apply Coulomb friction impulse: J_f = -μ * vn * m * sign(vt), but clamped to not exceed |vt*m| (so it doesn't reverse tangential velocity—prevents sticking unless truly rolling). Then spin transfer from J_f. This gives realistic limiting friction.

Let me use: For each wall contact:
- if dist < r (contact/collision):
  - normal: if vn<0, jn = -(1+e)*vn (bounce). Also we need a normal impulse to prevent penetration (for resting, to counter gravity). For resting (vn≈0), apply jn to push out: jn = something to prevent sinking. Let me add: a normal impulse to resolve penetration via velocity: if vn<0 apply bounce; if vn slightly negative or zero, apply just enough to prevent gravity pulling in: jn = -m*(gravity·n)... eh.

This is getting too deep. Let me step back and choose a robust-ish scheme that looks good:

Per physics step, per wall contact:
1. Position correction: push ball out (penetration). This prevents sticking via position.
2. Normal velocity: if vn<0 (approaching), bounce: vn→-e*vn. (For resting where gravity pulls ball into wall making vn slightly>0 after bounce... position correction handles penetration but gravity keeps pushing; without continuous normal force the ball sinks through? Position correction pushes out each step, gravity pushes in; with air drag it oscillates tiny. To prevent sinking, add a small normal velocity out: if vn>0 small... hmm.)

Actually, for a resting ball on a floor: gravity increases vy each step. The floor pushes back via position correction (teleport out) each step and via bounce when it starts moving into floor. The position-correction-out each step plus the velocity bounce creates a jittery but stable rest if I also zero tiny normal velocities and apply a floor friction. 

Let me use a simpler classic platformer-style floor contact per step:
- After gravity, if ball's bottom below floor line... but floor is rotating/angled.

OK I think the cleanest robust approach for this simulation: use penalty/contact forces per step:

For wall contact (penetrating):
- n (inward normal), dist<r.
- Project relative velocity: vn = (v_ball - v_wall)·n.
- If vn<0 (moving into wall): set normal impulse to reflect with restitution AND add continuous normal force to support weight. 
- Tangential: Coulomb friction with μ.

Let me implement with impulses:

```
const m=1, e=0.5, mu=0.9;
// normal
let vn = (b.vx-wx)*nx + (b.vy-wy)*ny;
if (vn < 0) {
  const jn = -(1+e)*vn; // full bounce
  b.vx += jn*nx; b.vy += jn*ny;
  // friction
  let vt = ((b.vx-wx)*tx + (b.vy-wy)*ty);
  const maxJt = mu * Math.abs(jn);
  let jt = -vt; // clamp
  const jtClamped = Math.max(-maxJt, Math.min(maxJt, -vt)); // reduce vt
  // apply to ball
  b.vx += jtClamped*tx; b.vy += jtClamped*ty;
  // spin
  const rvecX=-b.r*nx, rvecY=-b.r*ny;
  const dL = rvecX*(jtClamped*ty) - rvecY*(jtClamped*tx);
  b.sp_ += dL/(b.r*b.r);
} else {
  // resting contact: apply normal force to counter gravity + friction
  // support: cancel normal velocity component due to gravity
  ...
}
```

For resting, I can apply a normal impulse that cancels the gravity-induced normal velocity: 
```
// resting: make vn = 0 (no penetration velocity) by applying -m*vn
if (Math.abs(vn) < some small && penetration) {
  b.vx += (-vn)*nx; b.vy += (-vn)*ny; // zero normal vel
  // tangential friction (kinetic)
  const vt = (b.vx-wx)*tx + (b.vy-wy)*ty;
  const jt = Math.sign(vt)*Math.min(Math.abs(vt), mu*... );
  ...
}
```

This is the standard approach. Let me combine: 

```
if (penetrating){
  if (vn<0){
    // collision
    const jn=-(1+e)*vn;
    b.vx+=jn*nx; b.vy+=jn*ny;
  } else {
    // resting: clamp normal velocity to 0 to prevent sinking
    b.vx += (-vn)*nx; b.vy += (-vn)*ny; // jn = -vn (m=1)
  }
  // tangential friction (Coulomb, kinetic)
  const vt = (b.vx-wx)*tx + (b.vy-wy)*ty;
  if (Math.abs(vt)>0.001){
    // determine max friction = mu * |normal impulse|. Approximate normal impulse = |jn| or gravity term.
    const jnUsed = (vn<0)? (-(1+e)*vn) : (-vn);
    const maxJt = mu*Math.abs(jnUsed);
    let jt = -vt;
    if (Math.abs(vt) > maxJt) jt = -Math.sign(vt)*maxJt;
    b.vx += jt*tx; b.vy += jt*ty;
    // spin
    const rvecX=-b.r*nx, rvecY=-b.r*ny;
    const dL = rvecX*(jt*ty) - rvecY*(jt*tx);
    b.sp_ += dL/(b.r*b.r);
  }
  // position correction
  const pen=b.r-dist;
  b.x+=nx*pen; b.y+=ny*pen;
}
```

For resting, jnUsed = -vn (small, since vn≈0). maxJt small. That limits friction. To support weight, the normal impulse should be at least m*g*dt/n... Hmm. The gravity adds g*dt to vy each step; if ball rests on floor with normal n, gravity's normal component = g*dt*(n·yhat). We need to remove that. Setting vn=0 removes the normal component of (v after gravity). But then gravity re-adds next step. So effectively the floor supports: we zero the normal velocity (which includes the gravity contribution) → ball stays at floor without sinking. Good, position correction keeps it on surface. The tangential: gravity also has tangential component along floor → ball accelerates down the slope (rolling). Friction opposes up to μ. With μ=0.9>1 and slopes <90°, static friction holds → ball rolls down without sliding → spin builds via no-slip. 

But my code uses kinetic friction reduction of vt by jt up to maxJt. If vt small, jt can fully zero it (if |vt|<maxJt) → no slip → then to spin it correctly, when vt zeroed, spin should be set to rolling. Currently if I zero vt but add spin only from jt (=vt removed), and if jt removes all vt, then dL from that. But that spin would be the rolling spin. Let me check: if vt fully removed (sliding→rolling), the removed tangential velocity vt maps to spin. Δω = dL/r² where dL = rvec×J_t. With J_t removing vt: if J_t = -vt*t (full), then dL = r*(1)*... earlier with full reduction (|J_t|=|vt|) we got dL = (1-fr)... for fr=0 full reduction. Let me compute full-reduction spin for wall: dL = (1-0)*vt... wait with full removal jt=-vt: Δv_t = jt = -vt → vt becomes 0. dL = rvec × J_t. rvec=-r*n. J_t = -vt*t. dL = (-r*n) × (-vt*t) = r*vt*(n×t) = r*vt*(nx*ty-ny*tx)=r*vt*1=r*ty? n×t = nx*ty-ny*tx = nx*(nx) - ny*(-ny) = nx²+ny²=1. So dL=r*vt. Δω = r*vt/r² = vt/r. So b.sp += vt/r. 

Is that rolling? For rolling without slip on wall, spin ω_b such that contact velocity from spin = -vt (cancels). Contact velocity from spin = ω_b * perp(rvec). rvec from center to contact = -r*n. perp(-r*n) = rotate 90: (-(-r*ny), -r*nx) = (r*ny, -r*nx)... = r*(ny,-nx). Hmm perp as (-vy,vx): perp(-r*n)=(-(-r*ny), -r*nx)=(r*ny,-r*nx). Contact velocity = ω_b*(r*ny,-r*nx). We want this to cancel vt along t=(-ny,nx): contact vel along t = ω_b*(r*ny*(-ny) + (-r*nx)*nx) = ω_b*r*(-(ny²+nx²)) = -ω_b*r. We want = -vt (so total vt - ω_b*r ... wait relative tangential vel = vt (ball) - wall_t - spin_contact_t. Set 0: spin_contact_t = -vt. spin_contact_t = -ω_b*r. So -ω_b*r = -vt → ω_b = vt/r. Yes! So Δω = vt/r gives rolling. So when we fully remove vt via friction, setting sp += vt/r gives perfect rolling. 

But in my incremental code, if friction only partially removes vt (kinetic), spin gets partial. For resting static, I want full removal → full rolling spin. Let me, when in resting/rolling regime, enforce no-slip: set spin to vt/r (or adjust toward it). 

This is getting complicated but let me implement a "rolling enforcement" for wall contacts: if the ball is in contact and the tangential velocity (relative) is small enough to be static friction (|vt| <= μ*|normalForce|/m... ) treat as static: set tangential relative velocity to match rolling.

Ugh. Let me simplify to something that looks good and is stable:

I'll do per-step:
1. Apply gravity, air drag, update positions, update ang from spin.
2. Wall collisions: for each contact, do bounce (if vn<0) with e, then Coulomb friction clamp jt = clamp(-vt, -maxJt, maxJt) with maxJt = μ*|jn|, apply to velocity, and transfer to spin (b.sp += dL/r²). Then position correction (push out). Also clamp residual penetration velocity: ensure ball not moving into wall after (set vn<=0... actually after bounce vn=-e*old<0 becomes positive, good).

For resting, the bounce only triggers when vn<0 (sinking into wall from gravity). Gravity makes vn>0 (into wall) each step → triggers bounce → ball pushed up → bouncy rest. With position correction it stabilizes as a damped bounce on the floor. The spin accumulates from the tangential friction each impact. The number spins. Visually decent.

But the bounce-on-floor each step with e=0.5 and gravity means the ball jiggles (small bounce) and slowly settles due to air drag. Spin builds at each floor contact from vt. Good enough.

Actually the floor "bounce" each step: gravity adds g*dt to vy. Ball sinks into floor, bounce reflects with e=0.5 → moves up by e*g*dt, then gravity brings back. Oscillates ~ g*dt height. Tiny jiggle. Fine. Spin: each impact vt removed partially → spin. With μ and small jn (g*dt small), maxJt small, so vt barely reduced per step → spin builds slowly. Might be too slow.

Hmm. Let me increase μ and make floor friction transfer more spin. Or add direct rolling: when ball in contact with wall (penetrating) and not bouncing hard, set spin toward vt/r (rolling). 

Let me add a rolling enforcement: if dist < r (in contact) and |vn| small (resting or light contact), enforce rolling on the tangent: set b.sp toward vt/b.r gradually, and zero vt. Specifically:
```
if (penetrating){
  ... normal bounce ...
  const vt = rel_tangential;
  // rolling: transfer vt to spin
  // if static (small normal force available)... just do: 
  b.sp += vt/b.r * 0.5;  // partial roll
  b.vx -= vt*0.5*tx; b.vy -= vt*0.5*ty;  // reduce sliding
  // clamp residual
  ....
}
```

Let me use a "rolling friction" that gradually matches spin to vt/r and kills vt, coefficient per step. Since contact is per-step (dt=1/120), I can use a strong per-step factor because it only acts while in contact (short contact arcs). Let me use: 
```
// reduce tangential slip toward rolling
const slip = vt - b.sp*b.r; // difference between current vt and rolling vt (ω*r)
// apply correction
const corr = 0.5*slip; // 
b.vx -= corr*tx; b.vy -= corr*ty;
b.sp += corr/b.r;  // add to spin
```
Wait sign. Rolling: contact velocity = ω_b*r should cancel vt (relative). slip = vt - ω*r... if ω*r = vt then no slip, slip=0. We want to drive slip→0. The change: reduce vt by corr and increase ω by corr/r. Δ(vt) = -corr, Δ(ω*r)=+corr → Δslip = -corr - corr = -2corr. So corr=0.5*slip halves slip per step. With 0.5 factor and dt small, over contact it converges. Good and bounded (won't blow up). And it conserves angular momentum-ish (energy not conserved but friction dissipates—OK).

Let me use this rolling enforcement for both wall and ball contacts. It's clean.

So final wall contact logic:
```
if (pen){
  // normal bounce
  const vn = rel·n
  if (vn<0){
    b.vx += (-(1+e)*vn)*nx;  // wait need jn then split. Let me do directly:
    // v_ball normal rel component -> -e
    b.vx = wx + (-e*vn)*nx + (rel - vn*n)·t * t ... let me just:
  }
  // rolling friction
  const vt = rel·t (recomputed)
  const slip = vt - b.sp*b.r;
  const corr = 0.6*slip;
  b.vx -= corr*tx; b.vy -= corr*ty;
  b.sp += corr/b.r;
  // position correction
  b.x += nx*pen; b.y += ny*pen;
}
```

I need rel after normal bounce for vt. Let me compute carefully. Let me just write it step by step in code. I'll compute rel = b.v - v_wall. Handle normal. Then handle tangent+roll on updated rel. Then position correction.

OK let me just write the code now; I've spent enough planning. I'll make it stable and look good, tune friction factors to reasonable values (corr=0.5-0.7). 

Let me also add: after physics, clamp speeds to a max (e.g., 1500) to avoid NaNs.

Let me also make sure spin doesn't explode: b.sp *= 0.993 (spin drag).

Final code writing now. Let me also render motion blur? No, keep clean. Maybe add a subtle trail via semi-transparent clear? That'd look cool but with 20 balls maybe noisy. Let me use a fading clear: ctx.fillStyle='rgba(17,17,22,0.15)'; fillRect. That gives motion trails. Combined with the rotating heptagon, looks dynamic. But might make numbers blurry. Let me use clearRect (full clear) for readability. Actually motion trails are nice for "bouncing realistically." Let me do a mild trail: fillRect with rgba(15,19,28,0.25). I'll make heptagon drawn fully each frame (no trail on it? trails affect it too). Heptagon trails would be confusing. Let me do full clear + redraw heptagon each frame (clean), and maybe balls have slight trail. Hmm, simpler: full clear, no trails. Clean and readable. I'll go clean.

Actually, a motion trail on balls only: render balls to a separate step... too much. Full clear.

Let me add a nice touch: the heptagon edges glow faintly, and a bright marker at the rotating vertex. Numbers in white with black stroke for readability.

Let me write it. I'll make the heptagon fairly large and centered, balls inside.

Let me write the file now.

I need to be careful that the number rotation (b.ang) shows spin: b.ang += b.sp*dt per frame. The number drawn at angle b.ang. Good.

Let me write final HTML.

Let me reconsider initial spawn: grid 5x4 at center, r=16. They drop under gravity, bounce off each other and walls, while heptagon rotates. The rotation of heptagon will sweep walls; balls get flung. Should be a nice chaotic scene with spinning numbers. 

Let me make initial velocities a bit downward so they "drop" clearly. Add vy=50ish. Fine.

Let me write.

Let me set:
- g=1000
- e_wall bounce 0.5; rolling corr 0.7; 
- e_ball 0.65; rolling corr 0.5;
- air 0.996, spindrag 0.992
- mu style via corr.

Cap speed 1200.

Physics: SUB=2, DT=1/120, collision passes=3.

Let me write the code carefully and completely.

I'll also guard against NaN.

Let me write the file.

Let me write the index.html file.

Let me review the code for issues. The ball-ball friction line has a weird `slip` variable that's unused. Let me check that section.

```
const slip = vt - (a.sp*a.r + b.sp*b.r)/2 * (-1); // use average
// simpler: drive each ball's contact slip to rolling
const corrA = 0.25 * (vt - a.sp*a.r) * dt * 60;
const corrB = 0.25 * (vt + b.sp*b.r) * dt * 60;
```

The `slip` is unused. Let me remove it. Also the formulas `corrA = 0.25*(vt - a.sp*a.r)*dt*60` — dt*60 = (1/120)*60 = 0.5. So corrA = 0.25*0.5*(vt - a.sp*a.r) = 0.125*(vt - a.sp*a.r). Small correction per sub-step. Over an impact that may not build much spin. 

Actually the ball-ball friction spin transfer is minor; the main spin comes from wall rolling. Let me simplify and make sure it's stable. Let me remove the unused slip var and clean up.

Also there's a concern: in `resolveWall`, after normal bounce I recompute vrx2 and apply rolling correction `corr = 0.7*slip*dt`. With dt=1/120, corr = 0.7*(vt - sp*r)/120 ≈ 0.0058*(...). That's very small per substep. Over the contact (which lasts while ball is near wall, many substeps), it accumulates. But the factor dt makes it tiny. The wall friction should be stronger. The `*dt` is wrong dimensionally here because other parts already account per-step. Let me reconsider.

In the wall resolver, I'm not scaling by dt consistently. The position correction is per-collision (not dt-scaled). The normal bounce is instantaneous (not dt-scaled). The tangential corr `0.7*slip*dt` is dt-scaled (continuous). Mixing. Let me make the rolling correction instantaneous per contact (not dt-scaled) so it acts at impact, like the wall friction. Since the contact resolver runs ITER=3 times per physics step, an instantaneous factor of ~0.4-0.6 per iteration gives strong rolling. Let me set corr = 0.5*slip (no dt). Then per physics step (3 iterations) slip reduced by 0.5^3=0.125, converges fast. Good for rolling without slip.

But careful: this runs every physics step while ball is in contact (sliding along floor). 0.5 per iteration * 3 = strong. The ball on floor sliding will quickly reach rolling. That's like high friction floor. Acceptable for "realistic bounce + friction." Actually realistic rolling friction on a floor is high (μ~0.7-1). Good.

But this also damps the ball-ball tangential strongly. Let me set ball-ball corr to 0.3 (less). Fine.

Let me also reconsider: the rolling correction changes spin but I also have spin drag 0.993 per frame. Over a floor contact the spin builds. Good.

Let me fix the wall corr to remove dt (instantaneous per contact). And clean ball-ball.

Actually, wait: if corr is applied as a velocity change (b.vx -= corr*tx), and corr is large (0.5*slip), this changes velocity instantaneously which could make the ball move through the wall if it also changes normal? No, it's tangential only. Fine. But applying a large tangential impulse that overcorrects spin? The formula corr=0.5*slip zeroes half the slip per iteration → stable (Gauss-Seidel-ish), converges, no blowup. Good.

Let me also double check the wall normal bounce doesn't conflict with position correction: after bounce vn becomes positive (moving inward), position correction pushes inward—fine.

One concern: for a ball resting on floor, each step gravity adds gn to vn (into wall, vn>0? wait vn = rel·n where n inward. Into wall = outward = -n direction = vn<0). Gravity adds g*dt downward; the floor's normal n points up-ish (inward toward center). vn = (b.v - wv)·n. The gravity contribution to vn each step ≈ g*dt*(downward·n). For floor near bottom, n points up (toward center=up), downward·n ≈ -1, so vn gets -g*dt each step → vn<0 → triggers bounce. Bounce sets vn=+e*old. So ball bounces tiny each step. With position correction keeping it on surface and the tangential rolling, it'll have a high-frequency tiny bounce (jitter). With e=0.5 and g=1100, the tiny bounce velocity ~ e*g*dt. Over dt the ball moves e*g*dt²... tiny. Visually a slight jitter but with rolling spin it'll look like rolling. Acceptable.

To reduce jitter, I could clamp small normal velocities. Let me add: in the wall resolver, if vn is small negative (tiny penetration velocity), treat as resting: set vn=0 (no bounce), apply rolling friction. Let me add a threshold: if vn < 0 and vn > -g*dt*2 (small), set to 0. Else bounce. This removes the jitter for resting contact.

Let me implement: 
```
const vnThresh = -G*DT*3; // small
if (vn < 0) {
  if (vn > vnThresh) {
    // resting: zero normal velocity (no penetration)
    // set b.v such that normal rel component = 0
    ... 
  } else {
    // real bounce
  }
}
```
Resting: we want (b.v - wv)·n = 0. So b.v -= vn*n (remove the inward penetration velocity). Then apply rolling friction on tangent.

Let me restructure the wall resolver cleanly:

```
if (dist < b.r) {
  // normals etc
  const vrx=b.vx-wx, vry=b.vy-wy;
  const vn = vrx*nx+vry*ny;
  // tangent tangent already have tx,ty
  // tangential velocity
  let vt = vrx*tx+vry*ty;
  if (vn < 0) {  // moving into wall (penetrating velocity)
    if (vn > -G*DT*4) {
      // resting: cancel normal velocity (support weight)
      b.vx -= vn*nx; b.vy -= vn*ny;
    } else {
      const e=0.6;
      b.vx = wx + (-e*vn)*nx + vt*tx;
      b.vy = wy + (-e*vn)*ny + vt*ty;
    }
  }
  // rolling friction (always, when in contact)
  // recompute vt after normal handling
  const vrx2=b.vx-wx, vry2=b.vy-wy;
  vt = vrx2*tx+vry2*ty;
  const slip = vt - b.sp*b.r;
  const corr = 0.55*slip;   // converge to rolling
  b.vx -= corr*tx; b.vy -= corr*ty;
  b.sp += corr/b.r;
  // position correction
  b.x += nx*(b.r-dist); b.y += ny*(b.r-dist);
}
```

This is cleaner. Let me also make sure the resting threshold `G*DT*4` is reasonable: G=1100, DT=1/120 → 1100/120*4 ≈ 36.7. So if penetration normal velocity magnitude < 36 px/s, treat as resting. Balls rolling gently on floor will have small vn → resting → no jitter. Good.

But wait, when resting, I cancel vn each substep. Over ITER=3 per physics step, fine.

One more: when resting and the wall is the floor under gravity, the ball's downward velocity gets partly into the wall (canceled) and partly tangential (along floor) → ball accelerates down the slope, friction → spin. Good, that's rolling down an incline.

Also when the heptagon rotates, the floor moves; the ball's tangential relative velocity includes wall motion → spin. Good.

Let me rewrite the file with this cleaner version. Let me also remove the unused slip in ball-ball.

Let me also double-check ball-ball: vn>0 means approaching (a.v-b.v along n where n a→b). Good. Bounce. Then rolling corr.

Let me set ball-ball rolling corr = 0.3*slip where slip = vt - (a.sp*a.r - b.sp*b.r)? The contact slip for two balls: relative tangential velocity at contact = vt - (ω_a*r + ω_b*r) ... contact point on a moves at ω_a*r along... Let me keep it simple: corrA = 0.2*(vt - a.sp*a.r), corrB = 0.2*(vt + b.sp*b.r). Apply to damp vt and add spin. With small factor 0.2 per iteration it's gentle.

Actually, let me reconsider the ball-ball spin signs to ensure correct rolling direction. For two balls touching, contact point: on ball a, contact is at +r*n (toward b). Tangential velocity from a's spin at contact = ω_a * r along... perp direction. Hmm. Let me just use a heuristic: reduce vt between them and add equal-opposite spin. The exact rolling condition for two balls: (ω_a + ω_b)*r... 

Let me use: slip = vt - (a.sp*b.r + b.sp*b.r)? Actually the relative tangential velocity of the two contact surfaces = vt - (ω_a*r + ω_b*r) because both spins contribute to relative surface velocity if they have same sign... depends on geometry. For two balls, if both spin same direction, surfaces move same direction → relative contact slip = vt - (r*ω_a - r*ω_b)? 

This is fiddly. Ball-ball spin is a minor visual effect (the big spin is from walls). Let me keep a simple version: reduce vt by corr and add spin to each ball with opposite signs (so they don't artificially sync). Actually let me just damp the inter-ball tangential relative velocity and transfer half to each ball's spin:

```
const corr = 0.25 * vt;  // reduce relative tangential
a.vx -= corr*tx; a.vy -= corr*ty;
b.vx += corr*tx; b.vy += corr*ty;
a.sp += corr/a.r;
b.sp -= corr/b.r;  // opposite spin
```
This removes vt*0.25 per iteration (gentle stickiness) and adds opposite spins. Reasonable for friction. Let me use this simpler form.

Let me rewrite the file cleanly.

Let me also reconsider: the `corr = 0.55*slip` for walls—slip could be large on first impact (ball hits wall with high vt). corr=0.55*high → big velocity change + big spin kick. 0.55 factor converges (0.55³=0.166). The spin kick b.sp += corr/r could be large but spin drag limits. Should be OK, maybe dramatic bounces. Let me cap corr magnitude to avoid huge spin: corr = clamp(0.55*slip, -something, +something). Actually let me cap the spin rate: b.sp capped to ±40 rad/s. And corr capped. Let me just cap final sp.

Let me add: b.sp = clamp(b.sp, -50, 50) each frame.

Let me rewrite.

Also I realize in the rolling correction for walls, applying corr to velocity AND it's not impulse-based (units: corr has units of velocity since slip is velocity). b.vx -= corr*tx (velocity), b.sp += corr/r (1/s). Consistent. Good. The factor 0.55 per iteration (3 per step, 2 substeps = 6 per frame) → 0.55^6=0.028 per frame of slip remaining → very strong rolling (ball fully rolls ~ within one frame of contact). That might be too aggressive (walls act like sticky ice). Let me reduce to 0.3 per iteration: 0.3^6=0.0007 → still very strong. Hmm, 6 applications. The factor compounds across substeps+iterations.

Let me reconsider: per render frame we do SUB*ITER = 2*3 = 6 resolver passes. A factor f per pass → f^6 per frame. For moderate rolling (~0.5 per frame remains), f^6=0.5 → f=0.5^(1/6)=0.89. So f≈0.9 gives gentle rolling over a frame. f=0.7 → 0.7^6=0.12 (strong). f=0.8 → 0.26 (moderate). 

But the resolver only acts while dist<b.r (in contact). A ball sliding along floor is in contact each frame, so 6 passes/frame continuously. f=0.85 → 0.85^6=0.38 per frame → half-life ~1 frame → too fast (instant stick). We want rolling to develop over ~0.3-0.5s while allowing slide. 

Hmm, since it acts every frame on a ball in contact, the effective continuous damping is high. Let me use f=0.93: 0.93^6=0.65 per frame → half-life ~1.8 frames ~ 0.03s. Still fast. 

The issue is contact is continuous. To model realistic Coulomb, friction impulse magnitude = μ*N where N balances gravity → limited, not a fraction of slip. My slip-based approach is overly aggressive for continuous contact.

Let me switch to impulse-based Coulomb for the tangential, which naturally limits friction to μ*|normal force|:

For wall contact, the normal force supports the ball against gravity (in normal direction). The normal impulse per frame to support weight = m*g*dt / (n·down)... approximately m*G*DT per frame (the gravity added). So friction impulse max = μ * (normal support impulse). 

Per frame, gravity adds vy+=G*DT. The floor's normal cancels the normal component of that. The normal "impulse" magnitude ~ G*DT (for flat floor). Friction max = μ*G*DT ≈ μ*9 (at DT=1/120, G=1100 → 9.17). So tangential velocity removed per frame ≤ μ*9 ≈ 9 (for μ=0.5 → 4.5). A ball sliding at vt=200 loses 4.5/frame → stops in ~44 frames ~0.7s. That's reasonable sliding. And spin builds by ~4.5/r ≈ 0.4/frame → over 0.7s builds 40/... reaches rolling ω=v/r=200/11≈18. Good.

But this requires computing the normal support impulse. Let me estimate normal impulse as the amount of normal velocity corrected (the resting cancel or bounce). 

This is getting heavy. Given this is a demo and I can't iterate visually, let me pick a pragmatic middle ground that's stable and looks rolling-ish:

Use slip-based friction with a moderate per-iteration factor that only acts significantly on impacts, and accept continuous contact damping. To prevent total stickiness on the floor, I'll make the rolling factor smaller and rely on the fact that contact isn't always full-penetration (ball oscillates slightly so dist is just under r, contact, but the position correction lifts it each frame).

Honestly, let me just use a Coulomb-impulse approach properly; it's not that long. Let me define it per ball-ball and wall. I'll implement the standard sequential impulse with normal + tangent + spin.

Let me write a cleaner sequential-impulse contact solver. Actually, given time, let me go with the slip-based approach but tune to look good: factor 0.6 per pass but recognize it acts 6x/frame on contacts. For impacts (brief, 1-3 frames of contact) it's fine (strong bounce spin). For floor resting, it makes balls roll without slipping almost immediately (high friction) — which actually looks "realistic" (a ball on a tilted rotating floor rolls). Many bouncing-ball demos have high floor friction. I think strong rolling on floors looks good. Let me keep f=0.6 per pass but cap spin and velocity to avoid blowup, and use resting threshold to avoid jitter.

Actually, let me reduce passes' compounding by using DT-scaled continuous friction for the resting/sliding case and full impulse for impacts. Two regimes:

- If vn < 0 and vn > -G*DT*4 (resting): continuous Coulomb. Compute normal impulse = -vn (cancel). maxFriction = μ * |normal impulse|. Apply tangential impulse = clamp(-vt, -maxFriction, maxFriction) (full kinetic friction to slow vt, or zero if within static). Then position correction. This gives controlled friction.
  μ=0.6. normal impulse ~ G*DT*... for resting on floor, the cancel -vn ≈ G*DT*cos ≈ 9. maxFriction=0.6*9=5.4 per frame. vt loses up to 5.4/frame. Reasonable.

- If vn < 0 and vn <= -G*DT*4 (real bounce): bounce (e*vn), then apply tangent impulse = clamp to μ*|(1+e)vn| (impact friction), + spin.

For the resting tangent impulse to create spin: impulse jt (tangential velocity removed). spin: Δω = r_vec × J / r² = (r*... ) → for wall, Δω = jt/r (full removal). Let me compute: r_vec=-r*n, J=jt*t. dL = r_vec × J = (-r*n)×(jt*t) = -r*jt*(n×t) = -r*jt*1 = -r*jt. Δω=dL/r²=-jt/r. Hmm sign. Let me just compute numerically: dL = rvec.x*J.y - rvec.y*J.x with rvec=(-r*nx,-r*ny), J=(jt*tx,jt*ty). 

Let me just compute in code with the cross product to avoid sign errors, using jt the actual impulse vector applied to ball velocity (J/m, m=1). 

OK let me write it properly with impulse-based friction. This is the robust way. Let me write the full file.

Let me define for wall contact:
```
function resolveWall(b, vs){
  for each edge {
    ... closest point, dist, nx,ny,inward, px,py ...
    if (dist < b.r){
      const wx=-OMEGA*(py-CY), wy=OMEGA*(px-CX);
      const vrx=b.vx-wx, vry=b.vy-wy;
      const vn=vrx*nx+vry*ny;
      const tx=-ny,ty=nx;
      const vt=vrx*tx+vry*ty;
      const mu=0.6;
      if (vn<0){  // into wall
        if (vn > -G*DT*6){  // resting: support body
          // normal impulse to zero normal velocity (m=1)
          const jn = -vn;   // velocity removed = jn, applied as +jn*n to b
          b.vx += jn*nx; b.vy += jn*ny;
          // tangent: Coulomb kinetic friction (or static if small)
          let jt = Math.abs(vt) < mu*Math.abs(jn) ? -vt : -mu*Math.abs(jn)*Math.sign(vt);
          // the static branch zeros vt; kinetic reduces by mu*|jn|
          b.vx += jt*tx; b.vy += jt*ty;
          applySpinWall(b,nx,ny,tx,ty,jt);
        } else {  // dynamic bounce with friction
          const e=0.6;
          const jn = -(1+e)*vn;  // normal impulse
          b.vx += jn*nx; b.vy += jn*ny;
          // recompute vt after normal
          const vrx2=b.vx-wx, vry2=b.vy-wy;
          const vt2=vrx2*tx+vry2*ty;
          const jt = Math.sign(vt2)*Math.min(Math.abs(vt2), mu*Math.abs(jn));
          b.vx += -jt*tx; b.vy += -jt*ty;   // reduce tangential
          applySpinWall(b,nx,ny,tx,ty,-jt);
        }
      }
      // position correction
      const pen=b.r-dist;
      b.x+=nx*pen*0.8; b.y+=ny*pen*0.8;  // partial to avoid jitter, rely on iterations
      // also kill residual normal velocity into wall
      const vrx3=b.vx-wx, vry3=b.vy-wy;
      const vn3=vrx3*nx+vry3*ny;
      if (vn3<0){ b.vx -= vn3*nx; b.vy -= vn3*ny; }
    }
  }
}
```

Hmm the position correction 0.8*pen with iterations might still jitter. Let me just use full pen (1.0) and rely on ITER passes; for resting balls the position correction each frame keeps them at surface. The "kill residual normal velocity" at end prevents sinking. Good.

applySpinWall:
```
function applySpinWall(b,nx,ny,tx,ty, jtVel){
  // jtVel is the tangential velocity change applied to ball (J/m, m=1, along t)
  // impulse J = jtVel * t (m=1)
  const rvecX=-b.r*nx, rvecY=-b.r*ny; // contact relative to center (toward wall = -n)
  const Jx=jtVel*tx, Jy=jtVel*ty;
  const dL = rvecX*Jy - rvecY*Jx;
  b.sp += dL/(b.r*b.r);
}
```

Check: For wall, when we reduce vt (sliding) by jtVel along -sign... let me just pass the actual applied tangential velocity delta. In rest branch, applied delta = jt (could be -vt to zero, or -mu*|jn|*sign). So pass jt. applySpinWall(b,...,jt). dL = rvec × J. With rvec=-r*n, J=jt*t. = (-r*nx)*(jt*ty)-(-r*ny)*(jt*tx)= jt*r*(-nx*ty+ny*tx)= jt*r*(ny*tx - nx*ty). ny*tx-nx*ty = ny*(-ny)-nx*nx = -1. So dL = -jt*r. Δω = dL/r² = -jt/r = (1/r)*(-jt). Since jt is the applied tangential delta (negative of slide), spin increases. Good.

For dynamic bounce branch, I apply `-jt*tx` etc (reducing tangential by jt magnitude), so the applied delta = -jt*t. Pass that. Let me restructure to compute a single `tangentDelta` (velocity) and pass it.

Let me just write applySpinWall(b, nx, ny, tx, ty, deltaVx, deltaVy) where (deltaVx,deltaVy) is the tangential velocity added to ball; compute dL = rvec × deltaV (since m=1, J=deltaV). dL = rvecX*deltaVy - rvecY*deltaVx. b.sp += dL/r².

That's cleanest: pass the actual tangential velocity delta.

Let me rewrite with this.

OK let me also handle ball-ball similarly (impulse based with spin). 

Ball-ball contact a,b:
```
n from a to b (nx,ny), t=(-ny,nx).
vrel = a.v - b.v. vn=vrel·n. vt=vrel·t.
if (vn>0) approaching:
  e=0.6, m=1.
  jn = -(1+e)*vn/2;  // impulse (each gets half since equal mass; actually jn/(1/m+1/m)=jn/2 per ball)
  Actually standard: j = -(1+e)*vn / (1/m1+1/m2). For m=1: j=-(1+e)*vn/2. Apply +j*n to... which one? a gets pushed back, b forward: a.v -= (j/?) Let me define j as the impulse magnitude on b along n: jn (scalar) along n. 
  Let me: a.v += (jn*n)*? 
  
Let me use the common formulation: 
impulse scalar Jn (along n, positive pushes a away from b): 
J = -(1+e)*vn / (1/m1 + 1/m2), with vn = vrel·n (a-b along n), m1=m2=1 → J = -(1+e)*vn/2.
This J is the magnitude of impulse along n applied to b (and -J to a)? Sign: vn>0 approaching. J = -(1+e)*vn/2 <0. The impulse on a along n is -J (positive, pushes a in -n = away from b). Standard: a gets -J*n, b gets +J*n.
So: a.vx -= J*nx; a.vy -= J*ny; b.vx += J*nx; b.vy += J*ny. (J negative → a gets positive n... wait -J*nx with J<0 = +|J|*nx → a moves in +n (toward b)?? That's wrong.)

Let me be careful. Let me look at the standard. Let n = (b-a)/|b-a| (from a to b). Relative velocity vrel = va - vb. vn = vrel·n. If vn > 0, a is moving toward b along n (closing) → but wait if a moves toward b, va has +n component, vrel·n = (va-vb)·n >0 means va_n > vb_n, a catches up → closing. Yes vn>0 closing.

Impulse: Jn = -(1+e)*vn / (1/m1+1/m2). Jn <0 (since vn>0). This Jn is... the scalar such that va' = va + Jn*n/m1? Let me recall: the impulse on body 1 (a) is Jn*n, on body 2 (b) is -Jn*n. Where Jn is the scalar defined above (negative for closing). Check: a along n gets Jn (negative) → reduces a's +n velocity → correct (a slows in n). b gets -Jn (positive) → increases b's +n velocity → b pushed away (in +n, away from a) → correct (they separate). 

So: a.v += (Jn/m1)*n = Jn*n (m1=1). b.v += (-Jn/m2)*n = -Jn*n.
Since Jn<0: a.v += negative n (slows toward b), b.v += positive n (away). Good.

So: a.vx += Jn*nx; a.vy += Jn*ny; b.vx -= Jn*nx; b.vy -= Jn*ny.

Tangential friction (Coulomb): similar, Jt = -vt/(1/m1+...) = -vt/2, clamped to μ*|Jn|.
a.v += Jt*t; b.v -= Jt*t.
Spin: contact point on a is +r_a*n (toward b); on b is -r_b*n.
Δω_a = (rvec_a × J_impulse_a)/I, rvec_a = r_a*n, J_impulse_a = Jt*t (the tangential impulse on a) + Jn*n (normal, but normal through center → no spin from normal... actually normal impulse at contact point with center offset does create spin if not head-on, but n passes through... contact point = a + r_a*n, offset r_a*n from center, impulse Jn*n at that point → torque = (r_a*n)×(Jn*n)=0 since parallel. So normal gives no spin. Only tangential.)
So Δω_a = (rvec_a × Jt_vec)/I_a = (r_a*n) × (Jt*t)/r_a².
(r_a*n)×(Jt*t) z = r_a*(nx*(Jt*ty) - ny*(Jt*tx)) = r_a*Jt*(nx*ty-ny*tx) = r_a*Jt*1 = r_a*Jt.
Δω_a = r_a*Jt/r_a² = Jt/r_a.
Δω_b: rvec_b = -r_b*n, impulse on b = -Jt*t. dL = rvec_b × (-Jt*t) = (-r_b*n)×(-Jt*t)= r_b*Jt*(n×t)=r_b*Jt. Δω_b=J t/r_b. Same sign (both +Jt/r). 

Wait both same sign? For two balls, when a slides in +t relative to b, friction on a is -t, on b is +t. Jt = -|...|*t (negative if vt>0). Then Δω_a = Jt/r_a (negative), Δω_b=Jt/r_b (negative). Both spin negative (same direction). Is that right for rolling? If a and b are side by side and a moves +t faster, friction slows a (spins it... a's contact point should move -t to resist, spin ω_a such that contact vel = ω_a*r along... for a, contact toward b is +n, spin ω_a gives contact velocity perp = ω_a*r in direction perp(n) = t direction (since t=perp(n)). For rolling a on b: if a moves +t, spin ω_a negative → contact vel = ω_a*r*t negative → opposes motion → good. And b spun same direction (negative) → b contact vel negative too → b dragged in -t by a (a pulls b). Both spin same direction while vt reduced. Makes sense. OK.

So Δω_a = Jt/r_a, Δω_b = Jt/r_b, with Jt signed impulse (along t) on a.

Let me implement:
```
const Jn = -(1+e)*vn/2;
const Jtmag = Math.sign(vt)*Math.min(Math.abs(vt)/(2?... ), mu*Math.abs(Jn));
```
Wait Jt total impulse magnitude: Jt = -vt/(1/m1+1/m2) = -vt/2 (full stop vt), clamped to μ*|Jn|. The relative tangential velocity change = Jt*(1/m1+1/m2)=Jt*2. To zero vt, need Jt*2=vt → Jt=vt/2... 

Hmm I'm mixing. Let me define Jt as the impulse applied to a along t (like Jn applied to a along n). Then a.v += Jt*t, b.v -= Jt*t. Relative tangential vel change = (a.v - b.v)·t change = 2*Jt. To reduce vt to fr... For kinetic: Jt = clamp(-vt/2, -μ|Jn|, μ|Jn|)? To zero vt fully need 2*Jt = -vt → Jt=-vt/2. Clamped: |Jt| ≤ μ|Jn|. So Jt = clamp(-vt/2, -μ|Jn|, μ|Jn|).

Let me just compute:
```
let Jt = -vt/2;
const maxJt = mu*Math.abs(Jn);
Jt = Math.max(-maxJt, Math.min(maxJt, Jt));
```
Then a.v += Jt*t, b.v -= Jt*t.
Δω_a = Jt/r_a; Δω_b = Jt/r_b. (a.b.sp += Jt/r)

Wait check sign with rolling: if vt>0, Jt=-vt/2<0. a.v += Jt*t (reduces a's +t). b.v -= Jt*t = +|Jt|*t increases b +t. Δω_a=Jt/r_a<0. For a rolling (a moving +t, should spin negative): ω_a negative → contact vel on a = ω_a*r*t (t direction) negative → opposes +t motion. Good, rolling. 

But hold on, the normal gives no spin, and the standard also includes normal friction if off-center—but contact is along n through the line of centers, so normal impulse is collinear with offset → no torque. Correct, no spin from normal. Good.

Let me also do positional correction for balls.

For wall spin, let me recompute with impulse approach: Jn (normal impulse on b along n) = -(1+e)*vn (m=1, just one body) for bounce; for resting support, Jn = -vn. Then Jt = clamp(-vt/2? but single body: to reduce vt, impulse Jt along t on b: relative tangential change = Jt/m = Jt (single body, wall immovable). To zero vt: Jt=-vt. clamp |Jt|≤μ|Jn|.
Δω_b = Jt/r_b? Let me recompute single body. rvec_b = -r*n (contact), impulse tangential = Jt*t on b. dL = rvec_b × (Jt*t) = (-r*n)×(Jt*t) = -r*Jt*(n×t) = -r*Jt*1 = -r*Jt. Δω = -r*Jt/r² = -Jt/r. 

Whereas earlier with `deltaV` approach Δω = dL/r² = -Jt... let me just compute numerically in code to avoid sign errors:
```
function spinFromImpulse(b, rvecX, rvecY, impX, impY){
  const dL = rvecX*impY - rvecY*impX;
  b.sp += dL/(b.r*b.r);
}
```
rvec for wall contact = center→contact = -r*n = (-b.r*nx, -b.r*ny). imp = Jt*t = (Jt*tx, Jt*ty). Call spinFromImpulse(b, -b.r*nx, -b.r*ny, Jt*tx, Jt*ty). Numerically correct whatever signs. 

Let me use this helper everywhere. Cleaner.

Now let me also make sure the normal impulse for resting wall gives enough friction to support weight without sliding unrealistically. For a ball on floor: each frame gravity adds g*dt to downward vy. The floor cancels the normal component: Jn ≈ g*dt*|down·n|. vt on floor ~ 0 if not sliding down slope. If slope, gravity tangential accelerates. Friction μ*Jn resists. With μ=0.6 and slope angle < arctan(0.6)=31°, static holds → ball rolls without sliding. Heptagon edge angle: edges are nearly horizontal only near the bottom vertex; near bottom the floor is roughly horizontal. Good, balls roll at bottom.

Alright. Let me also add: to avoid balls piling and jittering, the position correction full and 3 iterations should separate them. Good.

Let me also handle the very first frame overlap (grid spacing 2.5*BR; for BR=13, sp=32.5, 2r=26, OK no overlap). Good.

Let me finalize the code with this impulse-based approach. Let me rewrite the whole file cleanly.

Let me write it.

I'll set mu_wall=0.8 (grippy floor so balls roll and don't slide forever), e_wall=0.6, mu_ball=0.5, e_ball=0.6.

Let me write applySpin helper and final code.

Also I should double check the wall normal sign for the bounce impulse Jn applied to b: Jn along n (inward). For bounce, vn = (b.v - wv)·n, if vn<0 (into wall), Jn = -(1+e)*vn >0 (along n, inward). b.v += Jn*n → pushes inward (away from wall). Correct.

For resting support: Jn = -vn (n>0). b.v += Jn*n → cancels normal velocity. Correct.

Now tangential Jt along t. b.v += Jt*t. To reduce vt: Jt = -vt (zero) or clamp. For single body, full stop vt needs Jt=-vt. clamp |Jt|≤μ|Jn|. 

Let me write.

After applying, recompute vt maybe not needed.

Let me also add the residual penetration velocity kill at the end (if still vn<0, zero it) to prevent sinking—optional. The resting branch already zeros vn. After position correction, velocities unchanged. Should be fine; I'll keep a small kill.

Let me write the final file.

For ball-ball, after impulse, positional correction.

Let me also ensure no division by zero when dist tiny in ball-ball.

Let me write it out fully and replace the file.

Let me also reconsider rendering performance and the spinning number. Good.

Let me write final.

Let me verify the file is valid by checking it can be parsed. Let me also do a quick sanity check of the logic and look for syntax errors by running it through node (won't render but can check syntax).

Let me extract and validate.

Syntax OK. Now let me do a logic review by actually simulating the physics in Node (headless) to check for blow-ups / NaN and see some sample behavior. I can't render but I can run the physics and log ball states over time. Let me extract the physics and run it.

Actually, let me just do a quick headless simulation to verify stability: spawn the balls, run physics for a few hundred frames, ensure no NaN and counts stay inside.

Let me write a quick test that replicates the core physics inline (or extracts). Simpler: I'll write a small node script that re-implements the same logic? That duplicates. Instead, let me eval the script in a fake canvas/raf environment. The script references document.getElementById, canvas.getContext, window. Let me stub those and run the file via a headless approach: load index.html content, inject stubs, run.

Let me create a stub environment with jsdom-like minimal. Actually I can use the 'canvas' package? Not installed. Let me just stub.

I'll read the file, extract the script, create a fake `document`, `window`, `requestAnimationFrame` that just steps, and a fake ctx that no-ops, run physics for N steps, and report.

But the script auto-starts via requestAnimationFrame. Let me instead make the test harness call physics directly. Simpler: I'll create a modified copy where I expose physics. Hmm.

Alternative: Run the actual file using a headless DOM. Let me check if there's a way. Let me check if `node` has undici or if puppeteer/playwright available. Probably not.

Let me just do a careful manual logic check and also a lightweight numerical test: I'll write a small node script replicating the physics functions exactly (copy them) and run stability checks. That validates the math.

Let me do that.

I made a syntax error in my test harness (the ctxStub line has a double comma `},};`). Let me fix the test harness. Actually it's the test harness, not the actual file. Let me rewrite the test more simply.

The closure captures `stubCanvasObj` from the outside scope, but it's not in the sandbox. I need to pass it into the sandbox or define it inside. Let me define the stub inside the wrapped context, or expose stubCanvasObj in the sandbox as a global. Let me add stubCanvasObj and stubCtx to the sandbox globals.