Need a self-contained physics simulation: balls inside a rotating regular heptagon, gravity, ball-ball collisions, wall collisions with the line segments of the rotating polygon, friction, spin (angular velocity) displayed by rotating the number. Implementing collision detection myself.

Key physics: rotating walls — the wall has velocity at the contact point: v_wall = ω × r. When a ball collides with a wall, use the relative velocity and apply a coefficient of restitution + tangential friction, and additionally add spin from the tangential friction.

Let's try writing it.

Heptagon: 7 vertices, rotating at ω = 2π/5 rad/s. Center at the canvas center. Balls spawn from the center with small random velocities.

Wall collision: for each edge (A,B), compute the distance from the ball center to the line segment; if less than the radius, push it out along the normal (inward). Normal points inward (toward the center). Wall point velocity: v = ω × r, where r is the vector from the center to the contact point: v_wall = ω * (-ry, rx).

Relative velocity vr = v_ball - v_wall. Normal component: if vn < 0 then... apply reflection: vn' = -e*vn. Tangential: apply friction impulse that changes tangential velocity and spin. Model the ball's angular velocity by rolling friction: a surface-contact impulse jt = -mu * jn or a clutch-like thing. Simplification: compute the tangential relative velocity at the contact point, including spin: v_rel_t = vr_t - ω_ball * r (sign conventions). Apply a friction impulse proportional to the normal impulse, clamped.

Ball-ball collisions: standard equal-mass impulse with coefficient of restitution; friction transfers spin. Keep it reasonably simple but with spin visible.

Drawing: circle with a gradient color, number rotated by the ball's angle.

Write a canvas-based animation loop with a fixed dt substep (e.g. 4 substeps).

Let's write the code carefully.

Constants: canvas size ~ 700. Heptagon circumradius R ~ 300. Ball radius ~ 14. 20 balls spawning from center with random offsets/velocities.

Physics per substep dt:
- Integrate: v += g*dt; pos += v*dt; ball.angle += ball.angVel*dt.
- Apply light air resistance / rolling friction to angVel.
- Ball-ball: for each pair, dist < 2r → separate, impulse.
- Walls: for each edge, closest point on segment to ball center; if dist < r push it out.

Wall collision details:
Edge from V1 to V2 (rotated vertices). Closest point P on segment. n = (center - P)/dist — but the ball is inside, so the normal points from the wall to the ball (inward). Wall point velocity: ω_rot × (P - center0): v = ω*(-(P.y-cy), (P.x-cx)).

rel = v_ball - v_wall. vn = dot(rel, n). If vn < 0:
 jn = -(1+e)*vn (mass=1)
 Tangent t = perp(n) = (-n.y, n.x).
 Contact relative tangential velocity including spin: rel_t = dot(rel, t) + ball.angVel * r? Contact point on ball at position -n*r from ball center. Velocity at ball contact point = v_ball + ω × (-r n) = v_ball + ω*( r*n.y, -r*n.x )... cross product ω×r_vec in 2D = ω * (-ry, rx). r_vec = -n*r → ω*(n.y*r, -n.x*r). Dot with t = (-n.y, n.x): ω*r*(-n.y² - n.x²) = -ω r. So slip = dot(rel,t) - ω r.

Friction impulse jt = -slip * something; clamp with |jt| <= mu*jn. Apply to tangential velocity and angular velocity: torque = r × jt... change in angular velocity: Δω = (r_vec × J_t)/(I), I = (1/2) m r² (solid disk) or (2/5)mr² sphere; use 0.5*r². Cross product in 2D: r_vec × J = rx*Jy - ry*Jx. r_vec = -n*r, J = jt*t = jt*(-n.y, n.x). Cross product = (-n.x*r)*(jt*n.x) - (-n.y*r)*(jt*(-n.y)) = -jt*r*(n.x² + n.y²)... let's compute: rx*Jy - ry*Jx = (-n.x r)(jt n.x) - (-n.y r)(-jt n.y) = -jt r n.x² - jt r n.y² = -jt r. So Δω = -jt*r / I, where I = 0.5 r² (m=1) → Δω = -2*jt/r.

Change in tangential velocity: Δv_t = jt * t.

Use a jt that fully cancels slip, clamped: to stop slip, jt_full = -slip / (1 + r²*I⁻¹)? Effective mass: 1/m + r²/I = 1 + 2 = 3 (for a disk I=0.5r²). jt_stop = -slip/3. Clamp with mu*jn.

Same approach for ball-ball friction.

Also, position correction to prevent sinking.

Spawn: since the heptagon rotates and balls drop from center, spawn all 20 initially at the center with small random velocities — they overlap; separate them gradually by spawning one at a time every few frames, or with random offsets near the center. Spawning one ball roughly every 150ms from the center with a small random velocity looks good ("falling from center").

Drawing: subtle background, heptagon edges as lines with glow, balls as radial gradients colored by hue, numbers drawn in white bold with a rotation applied.

Handle devicePixelRatio and resize: keep it simple with fixed size and scaling.

Now let's write the complete file.

Wall restitution e ~ 0.55, mu ~ 0.4. Ball-ball e ~ 0.85? Maybe 0.8. Gravity ~ 1200 px/s². Angular velocity of the heptagon: 2π/5 rad/s.

Also, since the walls are moving, spin is induced.

To prevent tunneling, substeps: 6 substeps per frame with dt capped.

Code:

```html
<!DOCTYPE html>...
```

Let's write it all out.

Quick sanity check — is there an issue with the spawn while balls are overlapping at the center? No problem. Ball-ball slip sign: for ball b, the contact offset is +n*r, and ω×(nr)·t = ω*(-nr_y... let me verify: ω×r_vec = ω*(-ry, rx). If r_vec = n*r, then = ω*r*(-ny, nx) = ω*r*t. Dot product with t is ω*r. So the surface velocity contribution from b is +ωb*r... wait, I had written "- b.angVel*b.r - a.angVel*a.r".

For a: r_vec = -n*r → ω×r_vec = ω*r*(ny, -nx) → dot product with t=(-ny,nx) is ω*r*(-ny² - nx²) = -ω*r. So a's contribution: -ωa*r. ✓

For b: r_vec = +n*r → ω*r*(-ny, nx)... wait: ω×(x,y) = ω*(-y, x). For r_vec = (nx*r, ny*r): ω×r_vec = ω*(-ny*r, nx*r) = ω*r*t. Dot product with t is +ω*r. So b's contribution is +ωb*r, not -. In my slip formula I used -b.angVel*b.r. Sign error at b.

slip = (rv·t) + ωb*r + ωa*r... no wait: slip = v_contact_b - v_contact_a = (vb + ωb*r*t) - (va - ωa*r*t) = rv + (ωb + ωa)*r*t. So slip_t = rv·t + (ωb+ωa)*r. And then torque: impulse J = jt*t is applied to b at r_vec = n*r: torque = r_vec × J = (nx r)(jt nx)... cross product rx*Jy - ry*Jx = nx*r*jt*nx - ny*r*jt*(-ny) = jt*r. Therefore Δωb = jt*r/I = 2jt/r. For a, the impulse is -jt*t applied at -n*r: cross product = (-nx r)(-jt nx) - (-ny r)(-jt(-ny)) = jt r nx² + jt r ny²... let me try computing: rx*Jy - ry*Jx with r_vec=(-nx r, -ny r), J=(-jt*tx, -jt*ty) = jt*(ny, -nx). = (-nx r)(-jt nx) - (-ny r)(jt ny) = jt r nx² + jt r ny² = jt r. Δωa = 2jt/r as well? Hmm, that gives both positive 2jt/r.

Hmm, wait, that seems off. Let me check the effect of jt on slip: d(slip)/d(jt) should be negative (stabilizing). Δslip from b's velocity: jt*t·t/m ×2 bodies = 2jt. From ω: Δωb = 2jt/r contributes to slip through +ωb r → +2jt. Δωa = 2jt/r contributes through +ωa r → +2jt. Total = 6jt. So slip_new = slip + 6jt, to zero it out jt = -slip/6. ✓ Matches the code magnitude, and all signs are consistent with slip = rv·t + (ωa+ωb)r.

But in the code slip = (rv·t) - ωb*r - ωa*r, and Δωa += 2jt/r, Δωb -= 2jt/r. The signs are flipped relative to my derivation, but let me re-check the cross product convention. Actually, I need to be careful about the direction of t and the cross product.

t = (-ny, nx). For b with r_vec = n*r: velocity at the contact point from spin = ω × r_vec = ω*(-ry, rx) = ω*(-ny*r, nx*r) = ω*r*t. Dot product with t: ω*r. So b's spin adds +ωb*r to the tangential slip. In the code it's minus. Inconsistency.

For the torque on b: J = jt*t applied at r_vec = n*r. τ = r_vec × J = rx*Jy - ry*Jx = nx*r*(jt*nx) - ny*r*(jt*(-ny)) = jt r nx² + jt r ny² = jt*r. Δωb = τ/I = jt*r/(0.5r²) = 2jt/r. In the code Δωb -= 2jt/r. Also flipped.

Both signs flipped for b → they cancel? Let's see: code's slip' = slip_true - 2(ωa+ωb)r. The jt obtained by clamping from code's jt = -slip_code/6 will be different; and applying Δωb = -2jt/r to the correct physics.

Effective slip dynamics in the code's model: code assumes Δslip = 6jt with its own slip definition. True slip = code_slip + 2(ωa+ωb)r. After impulse: ωb changes by -2jt/r (code) but correct Δωb should be... hmm, the code's torque sign: the correct torque is +jt*r giving +2jt/r; code applies -2jt/r. And a: correct Δωa? a receives J_a = -jt*t at r_vec = -n*r. τ = rx*Jy - ry*Jx = (-nx r)(-jt*(-ny))... J = -jt*t = jt*(ny, -nx). τ = (-nx r)(-jt nx) - (-ny r)(jt ny) = jt r nx² + jt r ny² = jt r. So correctly Δωa = +2jt/r, and the code has Δωa += 2jt/r. Correct!

So the code is correct for a, wrong for b (both slip contribution and torque flipped). Let me reconsider: could my sign for b's contribution be wrong? Ball b is at +n direction from a. Contact point is between them. r_vec from b's center to contact = -n*r! Because the contact point is on b's side facing a, i.e., in the -n direction. I flipped it: for b, r_vec = -n*r, and for a, r_vec = +n*r (a's contact side faces toward b, +n).

Redo: for a with r_vec = +n*r: velocity from spin = ω*(-ny r, nx r) = ω r t → contribution +ωa*r. Hmm, so this means a's contribution is +, opposite of before.

I keep flipping; let me do this carefully. n points from a to b. a's contact point is a + n*r. So r_vec_a = +n*r. b's contact point is b - n*r, r_vec_b = -n*r.

Surface velocity at contact from a's spin: ω_a × r_vec_a = ωa*(-rvec_y, rvec_x) = ωa*(-ny r, nx r) = ωa r * t. Dot product with t: +ωa r.

From b's spin: ω_b × r_vec_b = ωb*(ny r, -nx r) = -ωb r t. Dot product: -ωb r.

slip = (v_b + v_spin_b) - (v_a + v_spin_a) all · t = rv·t - ωb r - ωa r. ✓ Matches the code!

Torque: on a, J_a = -jt*t applied at r_vec_a = +n*r: τ = rx Jy - ry Jx = nx r*(-jt*ty) - ny r*(-jt*tx), where t=(-ny,nx), ty=nx, tx=-ny. = nx r*(-jt nx) - ny r*(-jt*(-ny)) = -jt r nx² - jt r ny² = -jt r. Δωa = -2jt/r. But the code has a.angVel += 2jt/r. Flipped!

Hmm wait — friction impulse direction: if slip > 0, friction acts to reduce slip: J_b = -jt... I defined jt = -slip/6, and applied b.vx += jt*t, a.vx -= jt*t. If slip>0, jt<0, so b is pushed in -t direction ✓.

Torque on b: J_b = jt*t applied at r_vec_b = -n*r: τ = (-nx r)(jt nx) - (-ny r)(jt(-ny)) = -jt r nx² - jt r ny² = -jt r. Δωb = -2jt/r. Code: b.angVel -= 2jt/r ✓.

Torque on a: J_a = -jt*t applied at +n*r: τ = nx r * (-jt·ty) - ny r * (-jt·tx) = nx r(-jt nx) - ny r(jt ny)... let me redo: J_a = -jt*t = (jt ny, -jt nx). τ = rx·Jy - ry·Jx = (nx r)(-jt nx) - (ny r)(jt ny) = -jt r(nx²+ny²) = -jt r. So Δωa = -2jt/r, but the code has +2jt/r. Sign error on a.

Hmm, interesting — both should be -2jt/r? Physically: two equal balls, friction impulse jt*t on b and -jt*t on a. Both torques have the same sign? Check slip dynamics: Δslip from Δωa: slip includes -ωa r, Δωa = -2jt/r → Δslip = +2jt. From ωb: slip has -ωb r, Δωb = -2jt/r → +2jt. From linear: b gets jt t, a gets -jt t → Δ(rv·t) = 2jt. Total Δslip = 6jt ✓, jt = -slip/6 zeros it out. Good — so correct code: a.angVel -= 2jt/r; b.angVel -= 2jt/r. Wait, that's odd — both balls spin the same way? Yes! Friction at the contact surface: imagine slipping, the tangential force on b is -t (jt<0) and on a is +t. Torque on a: r_vec × J = (+nr) × (+jt'|t|)... with jt' = -jt > 0: J_a = jt' t applied at n r: τ = nx r (jt' nx) - ny r (jt'(-ny)) = jt' r. Hmm, now it's positive.

I keep making sign errors. Let me be very systematic. t = (-ny, nx). Cross product 2D: u × v = ux·vy − uy·vx.

For a: r_vec = (nx r, ny r). J_a = -jt · t = -jt(-ny, nx) = (jt·ny, -jt·nx).
τ_a = rvec_x·Jy − rvec_y·Jx = (nx r)(−jt nx) − (ny r)(jt ny) = −jt r (nx² + ny²) = −jt r.
Δωa = τ/I = −jt r / (0.5 r²) = −2jt/r.

For b: r_vec = (−nx r, −ny r). J_b = jt·t = (−jt ny, jt nx).
τ_b = (−nx r)(jt nx) − (−ny r)(−jt ny) = −jt r nx² − jt r ny² = −jt r.
Δωb = −2jt/r.

Both −2jt r? Let me verify with a simple physical example. Two balls arranged horizontally, n = (1,0) pointing right (b on right side). t = (0,1) pointing up. Suppose b is sliding upward relative to a: slip > 0. Friction pushes b downward (−t... wait, jt = −slip/6 < 0, J_b = jt·t = downward ✓, J_a = −jt·t = upward on a ✓ (Newton's third law).

Torque on a: force upward (0, +F) applied at contact point which is to the right of a's center (r_vec = (+r, 0)). τ = rx·Fy − ry·Fx = r·F > 0 → a spins counterclockwise. By the formula with J_a = (jt ny, −jt nx), jt<0, nx=1, ny=0: J_a = (0, −jt) = (0, +F) ✓. τ_a = −jt r = +F r > 0 ✓ consistent. So Δωa = −2jt/r, positive here. The code has `a.angVel += (2*jt)/a.r`, which gives negative — wrong sign.

Torque on b: force downward (0, −F) applied at the point to the left of b's center (r_vec = (−r, 0)). τ = (−r)(−F) − 0 = +rF > 0, b also spins CCW. Physically: b's bottom-left point is dragged downward... a downward force on the left side of b rotates b counterclockwise? Force at left point pushing down: rotation about center — point on left moving down = counterclockwise? Point on left at (−r,0), velocity downward (0,−v): ω × r_vec = ω(−ry, rx) = ω(0, −r). For this to equal (0, −v): ω = v/r > 0 CCW. ✓. So both spin CCW — makes sense, like two touching gears slipping, friction tries to spin them the same direction? Actually no — rolling contact spins gears in opposite directions, but friction from slipping... here both end up CCW. Fine, the math says both Δω = −2jt/r.

Wait, but that contradicts the gear intuition. Rolling without slipping between two balls requires ωa and ωb to be opposite (gears). Here friction opposes slip; if b is sliding up relative, friction is down on b and up on a — down on b's left surface spins b CCW, and up on a's right surface spins a also CCW? Up on right side: point (r, 0), force (0, +F): τ = r·F > 0 CCW. Hmm, both CCW, yes. Like two balls, if one rolls right on top of the other... whatever, the math is consistent, and the check of slip dynamics closed (Δslip = 6jt zeros it out).

So fix: a.angVel -= 2jt/r. Currently: a.angVel += 2jt/r. Need to flip.

Also check wall collision: the ball's r_vec from center to contact point = −n·r (contact is on wall side, direction −n). J = jt·t applied at r_vec = (−nx r, −ny r): τ = (−nx r)(jt nx) − (−ny r)(jt(−ny)) = −jt r nx² − jt r ny² = −jt r. Δω = −2jt/r. Code: ball.angVel −= 2jt/r ✓ correct.

Wall slip: velocity at ball's contact point from spin: ω × r_vec = ω(ny r, −nx r) = −ω r t. Dot product with t: −ω r. slip = (v_ball − v_wall)·t − ω r ✓ matches code.

Sanity of wall friction sign: ball rolling along the floor. Floor at bottom, n = (0,−1) pointing up? n is wall→ball, floor below ball: n = (0,−1). t = (−ny, nx) = (1, 0). Ball moving right: rv·t > 0, ball not spinning ω=0 → slip > 0 → jt < 0. J = jt t = leftward friction ✓ decelerates. Δω = −2jt/r > 0 CCW. Ball moving right on the floor should roll... clockwise when moving right! ω×r_vec at contact (below center, r_vec = (0, r)): ω×(0,r) = ω(−r, 0). Rolling: contact velocity is zero: v + ω(−r,0) = 0 → ω = v/r > 0 for v>0. Positive ω is CCW in math coords but with y-down on screen it appears CW — no matter, the sign convention is consistent; what matters is that the slip→0 condition holds. slip_new = slip + Δ(rv·t) + Δ(−ωr) = slip + jt + (2jt... Δω = −2jt/r → −Δω·r = 2jt. Hmm, plus linear jt → total 3jt, and jt = −slip/3 ✓. Code uses /3 ✓.

So there's just one fix: the ball-ball ωa term.

Also recheck the effective mass for ball-ball: Δslip from linear: b gets jt t (+jt), a gets −jt t → contribution to rv·t = jt − (−jt) = 2jt ✓. Spin terms: slip has −ωa r − ωb r; Δωa=Δωb=−2jt/r → contribution +2jt+2jt = 4jt. Total 6jt ✓ /6 correct.

Fix and done. Also a small check: the spawn position is at the center (0,0), that's fine.

Check that the file exists, and quickly verify there are no syntax errors with node? Let me check if node is available: `node -e` to parse the script? More simply: extract the script and run a syntax check.

I need to investigate this further. Let me check the details.

Done. Let me summarize.