估计的小车位置为 (-0.236, -0.116, 0.774) m,前方的辣椒位于 (-0.285, -0.145, 0.807) m。它们的中心相距约 0.058米,因此这些物体很可能相互接触。小车的长度是否与世界 Y 轴对齐?在头部图像中它看起来是竖直的,宽度沿 X 方向。辣椒在它的左侧。虽然我最初认为它会落在沿 X 方向的闭合路径之外,但左手指可能位于 x = -0.286米处并与其发生碰撞。
我可以旋转夹爪,使其沿 Y 方向闭合,从而避开左侧的辣椒,或许还能避开前方的篮子。沿 X 方向闭合对于小车看似较长的轴线来说很理想,但辣椒挡住了这条路径。沿 Y 方向抓取小车的前后两端可能不太稳定:其长度约为 0.09米,而夹爪的最大开口可能约为 0.1米。另一个选择是先把前方的辣椒移开,以清出通往小车的路径。
右臂目前位于篮子处,夹爪张开,可以在左臂抓取辣椒的同时处理另一个物体。然而,要把左侧的辣椒放入右侧篮子,可能需要在两臂之间进行交接。左基座大约在 x = -0.3米,右侧篮子在 x = +0.28米,跨桌面的可达距离似乎很长。我估计在 y = 0 和 z = 1.0米处的 X 跨度为约 0.56米,可能接近 0.6米的臂展。我可以暂时把辣椒放在中间的篮子里,让右臂来转运,虽然这不符合其最终目的地,而且会增加工作量。右臂也许可以直接伸到 (-0.285, -0.145) m 处的辣椒,但这也涉及大约 0.6米的跨桌面可达距离。
从头部视角看,基座位置不确定;另一个估计是 x = ±0.4米和 y = -0.55米。按这个估计,左臂伸到右侧篮子的距离可能约为 0.9米,很可能无法实现。也许每条臂都能到达所有目的地,或者也许在 y = 0.06米处的失败是由向下的朝向而非总体可达范围造成的。不同的朝向可能会延长可达距离。我还剩 998 个控制步,因此有空间来处理这个问题。推动也许也能把物体弄进篮子,但我会使用抓取。
可能需要在中心附近进行交接。左臂可以抓起一个辣椒,把它移到 x = 0、y = -0.1米处,让右臂抓住它,然后松开,由右臂放置。红色小车同样可以转交给左臂。该技能不鼓励在非必要情况下用两个夹爪同时作用于一个物体,但跨类别放置可能使其合理。我对的粗略可达距离估计为 600毫米,基座间距为 600毫米。从左基座 x = -0.3米到对侧篮子 x = +0.3米,0.6米的 X 差加上 0.4米的 Y 差,得到约 0.72米,超出了该估计可达范围。右侧篮子上更近的一点 x = 0.2米、y = -0.03米可能把各分量减小到约 0.5米,但这仍然给出约 0.707米。用右臂直接伸到左侧辣椒也有类似的估计 0.72米距离。因此,中心转运似乎很可能。
我可以把左侧物体——两个辣椒、两辆小车和一块巧克力棒——分配给左臂,把红色小车、一块巧克力棒和已经放好的辣椒分配给右臂。左侧的辣椒将转交给右臂,红色小车转交给左臂。在两个都夹着物体的夹爪之间同时交换辣椒和小车,需要复杂而精确的接触。顺序转运会更简单。在 x = 0、y = -0.15米处临时放置到桌面上,可以让一条臂先松开物体,再由另一条臂拿起,从而避免同时接触。直接交接仍然是一种替代方案,但桌面转运似乎更稳健。
转运区域必须保持畅通。红色小车最初在 (-0.01, -0.08) m 附近。右臂可以把它放到 (-0.05, -0.25) m 处给左臂,而左臂可以把一个辣椒放到 (+0.05, -0.25) m 处给右臂。黄色公交车在 (-0.139, -0.18) m 附近,绿色小车在 (-0.236, -0.116) m 附近,巧克力棒在 x = ±0.37米附近。中心前方区域 x = 0、y = -0.25米看起来是空的,并且应该两条臂都能够到。
我会首先考虑移动前方的辣椒,以清出绿色小车。在当前向下的朝向并沿世界 X 方向闭合的情况下,辣椒右侧的绿色小车可能会挡住右手指。沿 Y 方向闭合的俯视辣椒抓取可以避开这种相邻关系。辣椒大致是对称的,因此我可以让夹爪绕世界 Z 轴旋转 90°,把局部 Y 映射到世界 -Y,并使手指前后分开。篮子在 y = +0.06米附近的前方,巧克力棒在 y = -0.199米附近的后方。辣椒在 y = -0.145米,而小车在其右侧 y = -0.116米处。后方的辣椒在 (-0.385, -0.091) m,巧克力棒在 (-0.366, -0.199) m,都远在左侧;黄色公交车在 (-0.139, -0.18) m,远在右侧。沿 Y 方向的手指路径看起来是畅通的。
或者,我可以沿 Y 方向抓取绿色小车。手指围绕 y = -0.116米以约 ±0.05米的间距分开,会使前手指位于 y = -0.066米附近,可能避开 y = 0 附近的篮子。位于 (-0.285, -0.145) m 的辣椒会在侧面。小车约 0.09米的长度可能刚好适合估计的 0.1米开口,从而能够接触前保险杠和后保险杠。它倒置的底盘是外露且平坦的;另一个长度估计是 0.085米。更深的抓取可能可行,但沿其宽度方向闭合仍然更可取。我也可以把小车稍微向右移,或者抓取位置离辣椒更远。先用沿 Y 对齐的抓取移动辣椒更简单。
右臂可以离开篮子,接近红色小车或右侧巧克力棒。对于红色小车,估计位置是 (-0.01, -0.08, 0.77) m。从当前右第六连杆位姿 (0.28, 0.007, 1.038) m 出发,接近距离约为 0.31米,超过了 0.25米的目标限制。位于 (0.12, -0.08, 1.0) m 的路径点距离约为 0.19米,允许后续接近。红色小车沿世界 X 方向拉长,因此当前沿 X 方向的闭合会沿其长度方向抓取。旋转 90° 偏航以沿 Y 方向闭合会更好,并且可以在移动过程中完成。
对于该朝向,我可以把当前向下的旋转乘以一个 +90° 的世界 Z 旋转:(90°) 把局部 Y 从 -X 映射到 -Y,同时保持局部 X 向下。当前向下的四元数按顺序为 [0.5, -0.5, 0.5, 0.5]。我暂定的乘法对 Z 旋转使用 [0.707, 0, 0, 0.707]。我得到 w = 0.3535 - 0.3535 = 0,x = -0.3535 - 0.3535 = -0.7071,y = 0.3535 + 0.3535 = 0.7071,z = 0.3535 + 0.3535 = 0.7071,得到 [0, -0.7071, 0.7071, 0.7071]。我需要用科学计算库检查这一点。整个旋转是 90°,超过了 1弧度的朝向限制,因此我需要一个中间的 45° 旋转。我暂定的中间四元数是 [0.353553, -0.612372, 0.353553, 0.612372],也需要检查。
两条臂目前都是向下的朝向。我可以在移动左臂到前方辣椒 (-0.285, -0.145, 1.0) m 上方、右臂到红色小车路径点 (0.12, -0.08, 1.0) m 附近的同时,让两条臂依次旋转 45° 和 90°。左臂从 (-0.28, -0.033, 1.025) m 移动约 0.114米,右臂移动约 0.19米。在左第六连杆 z = 1.025米处,我估计指尖在 z = 0.867米,高于约 0.83米的物体顶部。绕 Z 轴旋转使手指保持向下,约 0.1米的张开间隙在物体上方扫过。右指尖在第六连杆 z = 1.038米时大约在 z = 0.88米,比估计的 0.835米篮子边缘高约 0.045米。这似乎足以在从篮子中撤出的同时进行旋转。
一个更简单的并行动作是让右臂去收集它的巧克力棒,而左臂处理辣椒,推迟红色小车。右侧巧克力棒估计在 (0.371, -0.13, 0.783) m。它的长度沿 Y 方向,因此当前向下朝向并沿 X 方向闭合是合适的。从右臂当前 (0.28, 0.007, 1.038) m 位姿移动到 (0.37, -0.13, 1.0) m 的接近位置约为 0.164米,且不需要旋转。然后它可以前往位于 x = 0、y = 0 的中间篮子。对于右臂在 z = 1.0米处、基座在 x = 0.3米附近,这能够到吗?似乎合理。这样可以在左臂旋转去抓辣椒的同时分拣右侧的巧克力棒;并行动作不需要处理匹配的物体类型。
我将选择该计划:让左夹爪绕世界 Z 轴旋转 45°,并把它移到前方辣椒 (-0.285, -0.145, 1.0) m 上方,同时让右夹爪以其当前向下朝向移到巧克力棒 (0.371, -0.13, 1.0) m 上方。我正在考虑为最终运动分配 12 个控制步。首先,我会用科学计算库计算并检查左夹爪 45° 和 90° 的四元数。
原文8,777 字
The estimated car position is (-0.236, -0.116, 0.774) m, and the front pepper is at (-0.285, -0.145, 0.807) m. Their centers are about 0.058 m apart, so the objects are probably touching. Is the car’s length aligned with world Y? It appears vertical in the head image, with its width across X. The pepper is to its left. Although I initially thought it would be outside an X-directed closing path, the left finger could be at x = -0.286 m and collide with it.
I could rotate the gripper to close along Y, avoiding the pepper on the left and perhaps the basket in front. Closing across X would be ideal for the car’s apparent long axis, but the pepper blocks that approach. Grasping the front and back of the car along Y might be less stable: its length is about 0.09 m, while the maximum gripper opening may be about 0.1 m. Another option is to move the front pepper first to clear access to the car.
The right arm is currently at the basket with its gripper open and could handle another object while the left arm picks the pepper. However, getting a left-side pepper into the right basket may require a transfer between arms. With the left base around x = -0.3 m and the right basket at x = +0.28 m, the cross-table reach seems long. I estimated the X span as about 0.56 m at y = 0 and z = 1.0 m, perhaps near an arm reach of 0.6 m. I could temporarily put the pepper in the middle basket and have the right arm transfer it, though that would not satisfy its final destination and would add work. The right arm might instead reach the pepper directly at (-0.285, -0.145) m, but that also involves roughly 0.6 m of cross-table reach.
The base positions are uncertain from the head view; another estimate is x = ±0.4 m and y = -0.55 m. On that estimate, the left arm’s reach to the right basket could be about 0.9 m and likely impossible. Perhaps each arm can reach all destinations, or perhaps the failures at y = 0.06 m were caused by the downward orientation rather than general reach. A different orientation might extend the reach. I have 998 control steps remaining, so there is room to work through this. Pushing might also get objects into baskets, but I’ll use grasps.
A handoff near the center may be necessary. The left arm could pick a pepper, move it to x = 0, y = -0.1 m, let the right arm grasp it, then release it for the right arm to place. The red car could similarly be transferred to the left arm. The skill discourages using both grippers on one object unless necessary, but cross-category placement may justify it. My rough reach estimate for an ARX X5 is 600 mm, with 600 mm base spacing. From a left base at x = -0.3 m to an opposite basket at x = +0.3 m, the 0.6 m X difference plus a 0.4 m Y difference gives about 0.72 m, beyond that estimated reach. A nearer right-basket point at x = 0.2 m, y = -0.03 m might reduce the components to about 0.5 m each, but that still gives about 0.707 m. Reaching the left peppers directly with the right arm has a similar estimated 0.72 m distance. A central transfer therefore seems likely.
I could assign the left arm the left-side objects—two peppers, two cars, and one chocolate bar—and the right arm the red car, one chocolate bar, and the pepper already placed. The left peppers would transfer to the right arm, and the red car to the left. Exchanging a pepper and car simultaneously between occupied grippers would require complicated, precise contact. Sequential transfers would be simpler. A temporary table placement at x = 0, y = -0.15 m would let one arm release the object before the other picks it up, avoiding simultaneous contact. Direct handoffs remain an alternative, but a table transfer seems more robust.
The transfer area must remain clear. The red car is initially near (-0.01, -0.08) m. The right arm could place it at (-0.05, -0.25) m for the left arm, while the left arm could place a pepper at (+0.05, -0.25) m for the right. The yellow bus is near (-0.139, -0.18) m, the green car near (-0.236, -0.116) m, and the chocolate bars near x = ±0.37 m. The center-front area at x = 0, y = -0.25 m appears clear and should be reachable by both arms.
I’ll first consider moving the front pepper to clear the green car. With the current downward orientation and closing along world X, the green car on the pepper’s right could obstruct the right finger. A top-down pepper grasp with closure along Y would avoid that adjacency. The pepper is roughly symmetric, so I can rotate the gripper 90° about world Z, mapping local Y to world -Y and spacing the fingers front to back. The basket is forward near y = +0.06 m, and the chocolate bar is behind near y = -0.199 m. The pepper is at y = -0.145 m, while the car is to its right at y = -0.116 m. The back pepper at (-0.385, -0.091) m and the chocolate bar at (-0.366, -0.199) m are well to the left; the yellow bus at (-0.139, -0.18) m is well to the right. The Y-directed finger path appears clear.
Alternatively, I could grasp the green car along Y. Fingers spaced by about ±0.05 m around y = -0.116 m would put the forward finger near y = -0.066 m, perhaps clear of the basket near y = 0. The pepper at (-0.285, -0.145) m would be to the side. The car’s approximately 0.09 m length might just fit the estimated 0.1 m opening, allowing contact on the front and rear bumpers. Its upside-down chassis is exposed and flat; another length estimate is 0.085 m. A deeper grasp could work, but closing across its width would still be preferable. I could also shift the car slightly right or grasp farther from the pepper. Moving the pepper first with a Y-aligned grasp is simpler.
The right arm can leave the basket and approach either the red car or the right chocolate bar. For the red car, the estimated position is (-0.01, -0.08, 0.77) m. From the current right link6 pose at (0.28, 0.007, 1.038) m, the approach distance would be about 0.31 m, exceeding the 0.25 m target limit. A waypoint at (0.12, -0.08, 1.0) m would be about 0.19 m away, allowing a subsequent approach. The red car is elongated along world X, so the current X-directed closure would grip along its length. A 90° yaw rotation to close along Y would be better and could happen during travel.
For that orientation, I can multiply the current downward rotation by a +90° world-Z rotation: Rz(90°) maps local Y from -X to -Y while leaving local X downward. The current downward quaternion is [0.5, -0.5, 0.5, 0.5] in wxyz order. My tentative multiplication uses [0.707, 0, 0, 0.707] for the Z rotation. I get w = 0.3535 - 0.3535 = 0, x = -0.3535 - 0.3535 = -0.7071, y = 0.3535 + 0.3535 = 0.7071, and z = 0.3535 + 0.3535 = 0.7071, giving [0, -0.7071, 0.7071, 0.7071]. I need to check this with SciPy. The whole rotation is 90°, exceeding the 1 rad orientation limit, so I need an intermediate 45° rotation. My tentative intermediate quaternion is [0.353553, -0.612372, 0.353553, 0.612372], also to be checked.
Both arms currently have the downward orientation. I could rotate both through 45° and then 90° while moving the left above the front pepper at (-0.285, -0.145, 1.0) m and the right toward the red-car waypoint at (0.12, -0.08, 1.0) m. The left move from (-0.28, -0.033, 1.025) m is about 0.114 m, and the right move is about 0.19 m. At left link6 z = 1.025 m, I estimate the fingertips at z = 0.867 m, above object tops around 0.83 m. Rotating about Z keeps the fingers downward, with the approximately 0.1 m open gap sweeping above the objects. The right fingertips are around z = 0.88 m at link6 z = 1.038 m, about 0.045 m above the estimated 0.835 m basket rim. That appears sufficient for rotating while withdrawing from the basket.
A simpler parallel action is for the right arm to collect its chocolate bar while the left handles the pepper, deferring the red car. The right chocolate bar is estimated at (0.371, -0.13, 0.783) m. Its length is along Y, so the current downward orientation with closure along X is suitable. Moving from the right arm’s current (0.28, 0.007, 1.038) m pose to an approach at (0.37, -0.13, 1.0) m is about 0.164 m and requires no rotation. It could then go to the middle basket at x = 0, y = 0. Is that reachable for the right arm at z = 1.0 m from a base near x = 0.3 m? It seems plausible. This would sort the right-side chocolate bar while the left rotates for the pepper; parallel actions do not need to handle matching object types.
I’ll choose that plan: rotate the left gripper 45° about world Z and move it above the front pepper at (-0.285, -0.145, 1.0) m, while moving the right gripper above the chocolate bar at (0.371, -0.13, 1.0) m with its current downward orientation. I’m considering 12 control steps for the eventual motion. First, I’ll compute and check the 45° and 90° left-gripper quaternions with SciPy.