exp+fix: 候选塌缩定案(steps=2 把K条候选抹平35倍)+z_scale反例；补 n_plans/brake_rate 核算§9.2刹停率；记录我在链式作业排队期间改库导致两臂未跑的事故

ml/sim/closed_loop.py

@@ -204,6 +204,7 @@ def rollout(model, kin, tasks, device, K=8, safety=False, seed=0, max_sec=None,
     vratio = torch.zeros(B, device=device); aratio = torch.zeros(B, device=device); prev_exec = torch.zeros(B, 6, device=device)
     rev = torch.zeros(B, device=device); nact = torch.zeros(B, device=device); prev_sgn = torch.zeros(B, 6, device=device)
     nmov = torch.zeros(B, device=device)
+    nplan = torch.zeros(B, device=device)   # 每题规划次数 = 推理周期数（§9.2 的"刹停 <2% 推理周期"要这个分母）
     traj = [q.clone()] if return_traj else None
     plans = {k: [] for k in ('it', 'r', 'qe', 'commit', 'points', 'midpoints', 'gaps', 'certified')} if return_traj else None
     n_start = int(round(0.5 / DT)); moved_start = None          # R2 起步失败：0.5 s 内关节总行程 < 1°
@@ -216,6 +217,7 @@ def rollout(model, kin, tasks, device, K=8, safety=False, seed=0, max_sec=None,
     for it in range(nmax):
         tnow = it * DT
         if (queue.shape[1] <= 3) if replan == 'r3' else (queue.shape[1] == 0):
+            nplan += 1.0        # 验收判据要"刹停 / 推理周期"的比值，分母就是这个（批内所有行同步规划）
             r = queue.shape[1]
             pre = torch.zeros(B, 3, 6, device=device)
             qe = q
@@ -513,6 +515,7 @@ def rollout(model, kin, tasks, device, K=8, safety=False, seed=0, max_sec=None,
     side = torch.where(j1_first != 0, j1_first, torch.where(j1_max >= -j1_min, torch.ones_like(j1_max), -torch.ones_like(j1_max)))
     per = {'done': done.cpu(), 'ok': ok.cpu(), 't_done': t_done.cpu(), 'limit': limit.cpu(), 'col': col.cpu(), 'travel': travel.cpu(), 'err': e_end.cpu(),
            'side': side.cpu(), 'start_fail': start_fail.cpu(), 'brakes': brakes.cpu(), 'backoff': bo.cpu(), 'first_hit': first_hit.cpu(), 'dls_it': dls_it.cpu(),
+           'n_plans': nplan.cpu(), 'brake_rate': (brakes / nplan.clamp_min(1.0)).cpu(),
            'goal_ratio_2s': ratio2.cpu(), 'speed_ratio': vratio.cpu(), 'accel_ratio': aratio.cpu(),
            # 口径与离线分解一致：换向的 (步,轴) 对数 / 活动的 (步,轴) 对数 ∈ [0,1]（除以 6 个轴）
            'reverse_rate': (rev / (6.0 * nact.clamp_min(1.0))).cpu(),