桌子很可能位于 z = 0.75米处,不过这仍然只是一个估计。手表表盘三角测量得到 z = 0.769米,它可能因表带倾斜而比桌面高出约 0.018米。辣椒主体点位于 z = 0.787米,茎顶部大约在 z = 0.81米。该主体点可能位于可见的前表面上,而不是物体的实际中心。如果它代表中心,那么辣椒总高度约为 0.075米,其中心高出桌面 0.037米,这似乎是合理的。
三角测量给出辣椒主体点大约为 (−0.396, −0.085, 0.787) m。将同一个高机位相机像素 (134, 230) 投影到 z = 0.75米上,则得到 x 和 y 为 (−0.424, −0.061) m。这说明了为什么我应该使用三角测量,而不是假设可见物体的中心位于桌面高度。同样,左侧手表表盘大约为 (−0.251, −0.119, 0.769) m,而不是先前在桌面平面上的估计值 (−0.262, −0.096) m。
篮子的几何形状仍需细化。它们的边缘可能位于 z = 0.84米,底部位于 z = 0.75米。它们的中心似乎大致位于 x = −0.31、0 和 0.31米处,y 可能为 0.08米。在高机位视图中,篮子顶部大约在像素 y = 140 处,前部底部大约在 y = 211 处。我可以用腕部视图对左侧白色篮子的特征进行三角测量。其内部底部可能位于 z = 0.76米,暂定的放置目标大约在 x = −0.30米、y = 0.09米处,但我应该稍后再细化这一点。
我当前的工作物体估计为:辣椒大约在 (−0.396, −0.085, 0.79) m;左侧手表表盘在 (−0.251, −0.119, 0.769) m;黄色小车大约在 (−0.31, −0.16, 0.79) m;右侧手表大约在 (0.12, −0.22, 0.77) m;白色小车大约在 (−0.075, −0.1, 0.79) m。左臂起始位置大约在 (−0.3, −0.352, 0.921) m,应该能够从上方接近辣椒,同时避开黄色小车。
对于向下抓取,局部 +X 指向下方,手指在第六连杆下方延伸 0.1576米。因此,接触高度接近 z = 0.79米时,使用大约 0.15米的偏移量,意味着第六连杆的高度大约在 z = 0.94米。在初始第六连杆高度 0.9215米处向下旋转,可能会使指尖到达 z = 0.764米,并有接触桌面的风险。我应该先抬起手臂,也许抬到 z = 1.05米;这处于距初始高度约 0.921米的 0.25米位置目标限制之内。
先前计算出的向下四元数 [−0.5, 0.5, −0.5, −0.5] 等价于 [0.5, −0.5, 0.5, 0.5],因为反转四元数的所有符号会保持旋转不变。我将使用后者。它将局部 X 映射到 [0, 0, −1],局部 Y 映射到 [−1, 0, 0],局部 Z 映射到 [0, 1, 0],因此夹爪沿全局 X 方向闭合。辣椒在 X 方向上的宽度约为 0.065米,在 Y 方向上的深度约为 0.065米,因此任一水平闭合方向都可能可行。手表表带在横纵平面内跨度约为 0.06米,但最佳闭合方向将取决于其形状和附近的物体。
对于小车,我应该沿其宽度方向闭合,并根据后续视图细化偏航角。白色小车在高机位图像中看起来是斜的,相对于图像 Y 方向可能约为 55°,尽管投影混合了桌面坐标轴。其端点似乎接近 (277, 225) 和 (296, 256),给出图像向量 (19, 31)。我可以在 z = 0.79米处反投影这些端点,或者在获得另一个视图后对它们进行三角测量。其在桌面平面上的长轴可能大约为 [0.4, 0.9],偏航角相对于 X 约为 65°,或相对于 Y 为 −25°。相应的宽度方向将是 [0.9, −0.4],主要沿 X 方向,因此向下四元数可能适用,只需将偏航角略微调整 −25°。黄色小车的可见前部接近 (178, 272),其后部车顶接近 (161, 250),给出纵向图像向量 (−17, −22)。这可能对应于环境 X 负方向和 Y 正方向,但我应该在移动后进行检查。
左侧手表表带可能呈环状立起,表盘平放朝向前方,表带在其后方拱起。右侧手表看起来类似。跨表盘外壳进行自上而下的抓取可能比抓取表带中心更可靠,因为在表带中心,手指可能会闭合到孔中或仅仅挤压表带。表盘直径可能约为 0.035米,即大约 3厘米,因此准确的位置和方向很重要。左侧表盘已经过三角测量。对于高机位相机像素 (379, 298) 处的右侧表盘,我可以假设 z = 0.769米并计算其位置;暂定估计为 x = 0.126米,y = −0.219米。对白色小车车身使用 z = 0.79米,高机位相机像素 (287, 243) 表明 x = −0.07米,y = −0.104米。黄色小车车顶可能位于 z = 0.80米,其中心大约在 x = −0.29米,y = −0.167米。这些估计可以稍后细化。
我需要选择一个避免手臂交叉的分拣顺序。辣椒起始于左侧,但属于右侧红色篮子。它对于右臂来说太靠左,因此左臂可以抓取它,抬升到大约 z = 1.05米,然后将其越过白色和蓝色篮子运送到大约 x = 0.32米,y = 0.1米处。右侧手表有相反的交叉需求:右臂会将其运送到左侧白色篮子。同时移动两者可能会导致碰撞,即使我尝试分离它们的高度或 Y 位置。更安全的做法是分阶段进行这些运输,让一只手臂保持静止,而另一只手臂交叉。
一种可能的分配是左臂处理辣椒、左侧手表和黄色小车,而右臂处理右侧手表和白色小车。我可以在相对两侧同时抓取辣椒和右侧手表,然后在右臂将其手表安全保持在右前方时放置辣椒。一旦左臂返回以腾出空间,右臂就可以放置其手表。通过垂直分离进行并发交叉是另一种可能性,但由于碰撞风险,它不太有吸引力。
1100 步的预算应该允许顺序运输。每个目标最多可以移动 0.25米,求解器将任务增量限制为每步约 0.02米。五步可能移动约 0.1米;在 15 步的动作中,穿越 0.25米可能需要大约 13 步,尽管必须检查实际到达情况。对于五个物体,大约两到三个接近动作、闭合、抬升、三个运输动作、放置和后退,每个物体总共大约十个动作。五十个动作平均八步将使用约 400 步,为仔细操作和返回留出空间。
另一个自然的顺序是:左臂将左侧手表放入白色篮子,同时右臂将白色小车放入蓝色篮子,然后处理黄色小车和右侧手表,最后处理辣椒。第一对具有分离的路径:左臂从 x ≈ −0.25 移动到 −0.32米,右臂从 x ≈ −0.07 移动到 0米。目的地的夹爪将相距约 0.32米。然而,将黄色小车从 x ≈ −0.3 移动到 0米,同时右侧手表从 x ≈ 0.12 移动到 −0.32米,可能会在 x = −0.1米附近交叉手臂,或者根据时机可能在 −0.2米附近。这些运输应该顺序进行或错开进行。在右臂放置白色小车时开始处理辣椒也可能发生冲突,因为辣椒会经过中间篮子上方。
我也可以改用左臂处理两辆小车和左侧手表,让右臂只在左臂安全离开时放置其手表。将辣椒分配给右臂需要向左长距离伸展,而且似乎没有必要进行交接。在这个预算下,没有必要积极优化并发性。
尽管左侧手表和白色小车这一对看起来很方便,但黄色小车可能会阻碍左侧手表的接近。手表位于 (−0.251, −0.119) m,辣椒大约在 (−0.396, −0.085) m,黄色小车大约在 (−0.316, −0.15) m。我最初估计辣椒距离接近 0.16米;根据这些坐标,它大约为 0.148米。黄色小车仅相距约 0.073米,大约 0.07米。手表宽度约为 0.06米,张开的夹爪沿 X 方向跨度约为 0.12米,每根手指距中心约 0.06米。左侧手指接近 (−0.311, −0.119) m 时,可能会接触到小车后部大约 (−0.316, −0.145) m 处,Y 方向间隔仅约 0.026米。
我可以先移开黄色小车,或者调整手表抓取方向,使手指沿 Y 方向展开。在这种方向下,手指将跨越 y = −0.119 ± 0.06米,在 x = −0.251米处达到 −0.179米。此时小车的 X 方向间隔将约为 0.065米,可能足够;辣椒的 X 方向间隔约为 0.145米。对于这种替代方案,局部 X 将指向 −Z,局部 Y 沿 +Y,它们的叉积将给出局部 Z 沿 +X。矩阵列将是 [[0, 0, 1], [0, 1, 0], [−1, 0, 0]]。我将其解释为绕 Y 旋转 −90°,四元数为 [0.707, 0, −0.707, 0]。从初始四元数 [0.707, 0, 0, 0.707] 出发,点积为 0.5,给出整体方向距离 120°,即 2.094弧度,在 1弧度的限制下需要三个动作。X 方向闭合的向下四元数与初始四元数的点积约为 0.3536 + 0.3536 = 0.7071,给出 90°,即 1.571弧度,可以分成两个动作。
先移开黄色小车将使 X 方向闭合的手表接近更容易。两辆小车可以一起抓取,但它们放入同一个中间篮子的放置应该顺序进行。右臂可以安全地握住白色小车,同时左臂放置黄色小车,然后放置白色小车。它们的抓取位置将相距约 0.24米,每个夹爪跨度约 0.12米,因此除非仔细选择它们的偏航角和接近路径,否则它们可能几乎接触。更安全的替代方案是单独处理黄色小车,同时右臂观察或准备处理右侧手表。我也可以先移开较大的辣椒,同时右臂抓取白色小车,然后安排它们的放置顺序。如果这样更可靠,我可以一次处理一个物体,让空闲手臂保持在其起始或抬升姿态。
在确定物体顺序之前,我将抬起两个夹爪并使其向下旋转,以获得更清晰的腕部视图。这可能需要两到三个动作。一个初步的可能性是将它们移动到 (−0.3, −0.30, 1.05) m 和 (0.3, −0.30, 1.05) m,并采用中间方向。预期的向下四元数是 [0.5, −0.5, 0.5, 0.5],与初始方向相差 90°,因此我可以将其分成两次旋转。
我需要正确归一化四元数和以得到中间方向。我的第一个猜测是 [0.6036, −0.3535, 0.3535, 0.6036],然后是 [0.60355, −0.24845, 0.24845, 0.60355]。将初始四元数 [0.707, 0, 0, 0.707] 与向下四元数相加得到 [1.207, −0.5, 0.5, 1.207]。其范数约为 1.706,我得到大约 [0.707, −0.293, 0.293, 0.707]。它与初始四元数的点积约为 0.924,对应于 45° 的中间旋转,即 0.785弧度。我可以使用中间目标 [0.7071, −0.293, 0.293, 0.7071],或者使用科学计算库计算中间四元数。大约 0.14米的位置变化和 0.785弧度的方向变化将在命令限制之内。然后我可以命令完整的向下方向,也许抬升到 z = 1.08米的高度。
手指的扫掠路径看起来是畅通的,并且在旋转时抬升应该会增加间隙。左侧初始指尖接近 y = −0.194米,z = 0.921米,而辣椒接近 y = −0.085米,z = 0.81米。黄色小车大约在 y = −0.15米,z = 0.80米,比初始指尖高度约 0.92米低约 0.12米。如果不抬升,将大约 0.158米的手指延伸部分向下旋转,会使指尖在同一 X 处从 (y, z) ≈ (−0.194, 0.921) m 扫向 (−0.352, 0.764) m。中间指尖位置大约在 y = −0.24米,z = 0.85米,仍在小车后方,最终指尖也在其后方。右侧夹爪的指尖下方没有附近的物体;右侧手表接近 x = 0.12米,y = −0.22米,而夹爪位于 x = 0.3米,其张开的手指跨度大约为 x = 0.24–0.36米。
我曾考虑将两只手臂都移动到 z = 1.06米和 y = −0.32米,但提议的 0.06米 Y 方向偏移是不必要的。我将保持它们的 X 和 Y 位置大致不变,目标 z = 1.07米,使用位置 (−0.3, −0.352, 1.07) m 和 (0.3, −0.352, 1.07) m。目标距离约为 0.149米,低于 0.25米,中间方向变化约为 0.785弧度。我将保持两个夹爪张开,并使用五步进行这次清空空间的抬升和旋转。在检查测量结果后,我可以继续朝向完整的向下方向,暂定再进行一个五步动作。
原文12,397 字
The table is likely at z = 0.75 m, though this remains an estimate. The watch dial triangulates to z = 0.769 m, perhaps tilted on its band about 0.018 m above the table. The pepper body point is at z = 0.787 m and the stem top around z = 0.81 m. The body point may be on the visible front surface rather than at the actual object center. If it represents the center, a total pepper height of roughly 0.075 m, with its center 0.037 m above the table, seems plausible.
Triangulation gives the pepper body point approximately (−0.396, −0.085, 0.787) m. Projecting the same high-camera pixel, (134, 230), onto z = 0.75 m instead gives (−0.424, −0.061) m in x and y. This illustrates why I should use triangulation rather than assume that a visible object center is at table height. Similarly, the left watch dial is approximately (−0.251, −0.119, 0.769) m, rather than the earlier table-plane estimate of (−0.262, −0.096) m.
The basket geometry still needs refinement. Their rims may be at z = 0.84 m and their bottoms at z = 0.75 m. Their centers appear roughly at x = −0.31, 0, and 0.31 m, perhaps with y = 0.08 m. In the high view, the basket tops are around pixel y = 140 and the front bottoms around y = 211. I could triangulate features of the left white basket with a wrist view. Its inner base may be at z = 0.76 m, with a tentative placement target around x = −0.30 m, y = 0.09 m, but I should refine this later.
My working object estimates are: pepper at about (−0.396, −0.085, 0.79) m; left watch dial at (−0.251, −0.119, 0.769) m; yellow car near (−0.31, −0.16, 0.79) m; right watch near (0.12, −0.22, 0.77) m; and white car near (−0.075, −0.1, 0.79) m. The left arm starts near (−0.3, −0.352, 0.921) m and should be able to approach the pepper from above, while avoiding the yellow car.
For a downward grasp, local +X points down and the fingers extend 0.1576 m below link6. A contact height near z = 0.79 m would therefore suggest a link6 height around z = 0.94 m, using an approximate 0.15 m offset. Rotating downward at the initial link6 height of 0.9215 m could put the tips at z = 0.764 m and risk table contact. I should first raise the arms, perhaps to z = 1.05 m; that is within the 0.25 m position-target limit from the initial height near 0.921 m.
The previously calculated downward quaternion [−0.5, 0.5, −0.5, −0.5] is equivalent to [0.5, −0.5, 0.5, 0.5], since reversing all quaternion signs preserves the rotation. I will use the latter. It maps local X to [0, 0, −1], local Y to [−1, 0, 0], and local Z to [0, 1, 0], so the gripper closes along global X. The pepper is roughly 0.065 m wide in X and 0.065 m deep in Y, so either horizontal closing direction may work. A watch band spans about 0.06 m in the XY plane, but the best closing direction will depend on its shape and nearby objects.
For the cars, I should close across their width and refine the yaw from later views. The white car appears diagonal in the high image, perhaps about 55° relative to the image Y direction, although projection mixes the table axes. Its endpoints appear near (277, 225) and (296, 256), giving an image vector of (19, 31). I could back-project those endpoints at z = 0.79 m or triangulate them once another view is available. Its table-plane long axis may be approximately [0.4, 0.9], with yaw around 65° from X or −25° from Y. A corresponding width direction would be [0.9, −0.4], mostly along X, so the downward quaternion may work with a slight yaw adjustment of −25°. The yellow car’s visible front is near (178, 272) and its rear roof near (161, 250), giving a longitudinal image vector of (−17, −22). That may correspond to negative environment X and positive Y, but I should check after moving.
The left watch band may be standing in a loop, with the dial lying flat toward the front and the band arching behind it. The right watch looks similar. A top-down grasp across the dial’s outer casing may be more reliable than grasping the center of the band, where the fingers might close into the hole or merely squeeze the strap. The dial may be about 0.035 m in diameter, or roughly 3 cm, so accurate position and orientation matter. The left dial is already triangulated. For the right dial at high-camera pixel (379, 298), I could assume z = 0.769 m and calculate its position; a tentative estimate is x = 0.126 m, y = −0.219 m. Using z = 0.79 m for the white car body and high-camera pixel (287, 243) suggests x = −0.07 m, y = −0.104 m. The yellow car roof may be at z = 0.80 m, with its center around x = −0.29 m, y = −0.167 m. These estimates can be refined later.
I need to choose a sorting order that avoids the arms crossing. The pepper starts on the left but belongs in the right red basket. It is far to the left for the right arm, so the left arm could grasp it, lift to around z = 1.05 m, and carry it over the white and blue baskets toward x = 0.32 m, y = 0.1 m. The right watch has the opposite crossing requirement: the right arm would carry it to the left white basket. Moving both across simultaneously could cause a collision, even if I tried separating their heights or Y positions. It would be safer to stage those transports, with one arm stationary while the other crosses.
One possible assignment is for the left arm to handle the pepper, left watch, and yellow car, while the right handles the right watch and white car. I could grasp the pepper and right watch simultaneously on opposite sides, then place the pepper while the right arm holds its watch safely at the right front. Once the left arm returns to clear space, the right could place its watch. Concurrent crossing with vertical separation is another possibility, but it is less attractive because of the collision risk.
The 1100-step budget should allow sequential transport. Each target can move at most 0.25 m, and the solver bounds task increments to about 0.02 m per step. Five steps might move about 0.1 m; traversing 0.25 m might take around 13 steps within a 15-step action, although actual arrival must be checked. For five objects, roughly two or three approach actions, closure, lift, three transport actions, placement, and retreat would total about ten actions per object. Fifty actions averaging eight steps would use about 400 steps, leaving room for careful work and return.
Another natural sequence would be to place the left watch in the white basket while the right arm places the white car in the blue basket, then handle the yellow car and right watch, and leave the pepper for last. The first pair has separate paths: the left arm moves from x ≈ −0.25 to −0.32 m, and the right from x ≈ −0.07 to 0 m. The destination grippers would be about 0.32 m apart. However, moving the yellow car from x ≈ −0.3 to 0 m while the right watch moves from x ≈ 0.12 to −0.32 m could cross the arms around x = −0.1 m, or perhaps near −0.2 m depending on timing. Those transports should be sequential or staggered. Starting the pepper while the right arm places the white car could also conflict as the pepper passes over the middle basket.
I could instead use the left arm for both cars and the left watch, leaving the right arm to place only its watch when the left arm is safely clear. Assigning the pepper to the right arm would require a long reach to the left, and a handoff seems unnecessary. There is no need to optimize concurrency aggressively with this budget.
Although the left-watch and white-car pair looks convenient, the yellow car may obstruct the left watch’s approach. The watch is at (−0.251, −0.119) m, the pepper around (−0.396, −0.085) m, and the yellow car around (−0.316, −0.15) m. I initially estimated the pepper distance near 0.16 m; it is roughly 0.148 m from these coordinates. The yellow car is only about 0.073 m away, roughly 0.07 m. With a watch width around 0.06 m and an open gripper spanning about 0.12 m along X, each finger is about 0.06 m from the center. A left finger near (−0.311, −0.119) m could contact the car rear around (−0.316, −0.145) m, with only about 0.026 m of Y separation.
I could clear the yellow car first or orient the watch grasp to spread the fingers along Y. In that orientation, the fingers would span y = −0.119 ± 0.06 m, reaching −0.179 m at x = −0.251 m. The car’s X separation would then be about 0.065 m, perhaps enough; the pepper’s X separation is about 0.145 m. For this alternative, local X would point along −Z, local Y along +Y, and their cross product would give local Z along +X. The matrix columns would be [[0, 0, 1], [0, 1, 0], [−1, 0, 0]]. I interpret this as a −90° Y rotation with quaternion [0.707, 0, −0.707, 0]. From the initial quaternion [0.707, 0, 0, 0.707], the dot product is 0.5, giving a whole-orientation distance of 120°, or 2.094 rad, requiring three actions under the 1 rad limit. The X-closing downward quaternion has a dot product of approximately 0.3536 + 0.3536 = 0.7071 with the initial quaternion, giving 90°, or 1.571 rad, which can be split into two actions.
Clearing the yellow car first would make the X-closing watch approach easier. Both cars could be grasped together, but their placements into the same middle basket should be sequential. The right could hold the white car safely while the left places the yellow car, then place the white car. Their grasp positions would be about 0.24 m apart, with each gripper spanning about 0.12 m, so they might nearly touch unless their yaw and approach paths are chosen carefully. A safer alternative is to handle the yellow car alone while the right arm observes or prepares for the right watch. I could also clear the larger pepper first while the right grasps the white car, then sequence their placements. I can handle one object at a time with the idle arm at its starting or raised pose if that proves more reliable.
Before committing to an object order, I will raise and turn both grippers downward to obtain clearer wrist views. This may take two or three actions. An initial possibility is to move them to (−0.3, −0.30, 1.05) m and (0.3, −0.30, 1.05) m with an intermediate orientation. The intended downward quaternion is [0.5, −0.5, 0.5, 0.5], a 90° change from the initial orientation, so I can split it into two rotations.
I need to normalize the quaternion sum correctly for the halfway orientation. My first guesses were [0.6036, −0.3535, 0.3535, 0.6036] and then [0.60355, −0.24845, 0.24845, 0.60355]. Adding the initial quaternion [0.707, 0, 0, 0.707] to the downward quaternion gives [1.207, −0.5, 0.5, 1.207]. With a norm of about 1.706, I obtain approximately [0.707, −0.293, 0.293, 0.707]. Its dot product with the initial quaternion is about 0.924, corresponding to a halfway rotation of 45°, or 0.785 rad. I can use the intermediate target [0.7071, −0.293, 0.293, 0.7071], or calculate the halfway quaternion with SciPy Slerp. A position change around 0.14 m and an orientation change of 0.785 rad would be within the command limits. I could then command the full downward orientation, perhaps raising as high as z = 1.08 m.
The swept finger paths appear clear, and raising while rotating should increase clearance. The left initial tip is near y = −0.194 m, z = 0.921 m, whereas the pepper is near y = −0.085 m, z = 0.81 m. The yellow car is around y = −0.15 m, z = 0.80 m, about 0.12 m below the initial tip height near 0.92 m. Without raising, rotating the roughly 0.158 m finger extent downward would sweep the tip from (y, z) ≈ (−0.194, 0.921) m toward (−0.352, 0.764) m at the same X. An intermediate tip position around y = −0.24 m, z = 0.85 m remains behind the car, and the final tip is also behind it. The right gripper has no nearby object beneath its tip; the right watch is near x = 0.12 m, y = −0.22 m, while the gripper is at x = 0.3 m and its open fingers span approximately x = 0.24–0.36 m.
I considered moving both arms to z = 1.06 m and y = −0.32 m, but the proposed 0.06 m Y shift is unnecessary. I will keep their X and Y positions approximately unchanged and target z = 1.07 m, using positions (−0.3, −0.352, 1.07) m and (0.3, −0.352, 1.07) m. The target distance is about 0.149 m, below 0.25 m, and the halfway orientation change is about 0.785 rad. I will keep both grippers open and use five steps for this clear-space raise and rotation. After inspecting the measured result, I can continue toward the full downward orientation, provisionally with another five-step action.