第一块手表在左侧篮子里,狮子在右侧篮子里。剩余物体包括:位于约 x = -0.3米、y = -0.135米处的黄色木制玩具;两块位于中央的手表,分别在 x ≈ 0.009米和 x = 0.108米处,均靠近 y = -0.13米;以及位于横纵平面中 [-0.091, -0.212] m 和 [0.377, -0.149] m 处的巧克力棒。玩具应放入右侧篮子,手表应放入左侧篮子,巧克力应放入中间篮子。
接下来,我可以让左臂在 [-0.3, -0.14] m 附近抓取黄色玩具,同时右臂抓取右侧的巧克力并将其放入中间篮子。然而,将玩具运送到右侧篮子可能会超出左臂的可达范围。另一种方案是让每只手臂各抓取一块中央手表,并将两块都放入左侧篮子,但右臂必须越过中心区域,可能与左臂产生干涉。
在分配剩余物体之前,我需要考虑可达范围。如果左侧基座位于 x = -0.3米,而右侧篮子靠近 x = 0.32米,则水平间距为 0.62米,超过了我估计的最大可达范围约 0.55米。让右臂直接抓取黄色玩具需要从 x = 0.3米伸展到 x = -0.3米,位移为 -0.6米,这也可能太远,尽管在较低高度下或许可行。我不确定基座和肩部的位置:我考虑过基座位于 x = ±0.3米,以及肩部或基座位于 x = ±0.45米的情况。如果基座位于 ±0.45米,0.75米的交叉可达距离似乎不可能实现。如果左肩位于 x = -0.3米,臂长约 0.65米,那么在较低高度下或许能够到达位于 x = 0.32米的右侧篮子。我尚未测试这些交叉可达情况。左臂在 [-0.405, -0.12, 0.925] m 处的预抓取是可达的,右臂在 x = 0.291米附近抓取狮子也是可达的,但这些并不能解决这个问题。
中央转移可以避免依赖那个不确定的可达范围。左臂可以在 x = 0米处直接将玩具递给右臂,但那样会很复杂。更稳健的替代方案是将玩具暂时放在桌子中央的空旷区域,也许在横纵平面中的 [0, -0.25] m 处,然后让右臂重新抓取它并放入右侧篮子。这会增加一次放置和第二次抓取,但我还剩余 976 个控制步。
我考虑过先将两块巧克力放入中间篮子,每只手臂各放一块,然后再处理两块中央手表。这仍然留下了右臂交叉到达左侧篮子的不确定性问题。更好的分配可能是让左臂处理两块中央手表,分别在 x = 0.009米和 x = 0.108米处,并将它们放入靠近 x = -0.325米的左侧篮子。右臂可以处理两块巧克力;到达位于 x = -0.091米的左侧巧克力需要约 0.39米的交叉可达距离,这似乎很可能可行,而位于 x = 0米的中间篮子应该很容易到达。如果需要,左臂也可以处理位于 x ≈ -0.09米处的左侧巧克力。
我偏好的顺序是:左臂抓取黄色玩具,右臂同时抓取右侧巧克力。左臂将玩具放在一个临时的中央位置,也许在靠近前方的 [-0.05, -0.28] m 处,而右臂将巧克力放入中间篮子。然后左臂可以抓取第一块中央手表,同时右臂从转移点抓取玩具,分别将它们放入左侧和右侧篮子。最后,左臂可以抓取第二块中央手表,同时右臂抓取左侧巧克力,分别将它们放入左侧和中间篮子。由于玩具被抓取两次,这将需要剩余六次抓取。我估计这需要五个循环加上临时放置。另一个可能的转移位置是 [0, -0.2] m。我不会依赖右臂直接到达 x = -0.3米处的玩具。
两个夹爪目前都在篮子处张开,俯仰角为 60 度。我需要缩回到空旷区域,并将它们恢复到向下的抓取姿态。我最初考虑过玩具的头顶位置 [-0.30, -0.135, 1.0] m 和巧克力的 [0.377, -0.149, 1.0] m。对于此动作,我将使用 z = 1.02米。从当前左侧位姿 [-0.354, -0.019, 0.987] m 到 [-0.30, -0.135, 1.02] m,我估计距离为 0.128米。从当前右侧位姿 [0.319, 0.001, 0.99] m 到 [0.377, -0.149, 1.02] m,我估计为 0.166米。从 60 度俯仰变为 90 度向下姿态需要约 0.524弧度,在 1.0弧度的姿态限制之内。向负 Y 方向并略微向上移动应能使夹爪安全地离开篮子。我可以为此撤退使用 10 个控制步。
我还需要为巧克力适当地调整右夹爪的对齐。金色矩形棒长约 0.09米,宽约 0.025米,呈对角线放置。其长度可能超过夹爪估计的最大开口 0.08米,因此最好沿其短宽度方向闭合,使夹垫位于长边上。在向下姿态中,闭合方向沿世界 X 轴。巧克力的长轴在高机位视图中似乎大致从像素 [480, 257] 延伸到 [522, 276],在图像中向右下方倾斜。这表明世界方向为 X 正、Y 负,尽管其确切角度仍不确定。
我可以在向下姿态之前施加一个世界 Z 轴偏航来调整闭合方向:q = ×。在零偏航时,局部 Y 轴指向世界 -X 方向。偏航后,其方向为 [-余弦, -正弦, 0],轴的正负号对闭合无关紧要。因此,闭合轴的角度为 + 180 度。要垂直于世界横纵平面中长轴角度 θ_ 闭合,我将使用 = θ_ + 90 度。如果棒的长轴在 -25 度,则 65 度的偏航是合适的。手指仍将指向下方,而腕部相机将随偏航旋转。大约 60 度的偏航可能可行,但将 65 度偏航与从当前 60 度俯仰变为向下姿态的变化相结合,可能会超过 1.0弧度的动作限制,并需要中间姿态。
如果棒有足够的对角线对齐,我也可以改为沿世界 X 轴以零偏航抓取。我对其投影宽度的估计为 0.09 × 正弦(25 度) + 0.025 × 余弦(25 度),大约 0.09 × 0.42 + 0.025 × 0.91 = 0.061米,在估计的 0.08米开口范围内。夹垫将接触对角线边缘,而不是精确垂直于长轴闭合,因此 45 或 60 度的偏航仍然更可取。一旦手臂更靠近棒,我将从右腕视图细化该选择。
黄色猪形玩具似乎侧躺呈水平状态。我估计其形状长约 0.07米,宽约 0.04米,尽管其可见宽度可能在 0.05米左右。它的头部在高机位图像中可能指向左上方,靠近像素 [180, 260]。以零偏航向下抓取,沿世界 X 轴横跨宽大的身体闭合,似乎是合理的。
目前,我将把两只手臂撤退到 z = 1.02米的头顶目标位置,手指张开,向下四元数为顺序的 [0.5, -0.5, 0.5, 0.5]。我考虑过在撤退期间给右臂添加 45 度偏航,但在此动作中我将保持其偏航为零,并在检查巧克力方向后于安全预抓取位置稍后调整。我将使用 10 个控制步。
原文7,282 字
The first watch is in the left basket and the lion is in the right basket. The remaining objects are the yellow wooden toy at approximately x = -0.3 m, y = -0.135 m; two center watches at x ≈ 0.009 m and x = 0.108 m, both near y = -0.13 m; and chocolate bars at [-0.091, -0.212] m and [0.377, -0.149] m in the XY plane. The toy belongs in the right basket, the watches in the left basket, and the chocolates in the middle basket.
I could next pick the yellow toy with the left arm near [-0.3, -0.14] m while the right arm picks the right chocolate and places it in the middle basket. However, transporting the toy across to the right basket may exceed the left arm’s reach. Another option is to pick one center watch with each arm and place both in the left basket, but the right arm would have to cross the center, creating possible interference with the left arm.
I need to consider reach before assigning the remaining objects. If the left base is at x = -0.3 m and the right basket is near x = 0.32 m, the horizontal separation is 0.62 m, beyond my estimated maximum reach of about 0.55 m. Having the right arm pick the yellow toy directly would require reaching from x = 0.3 m to x = -0.3 m, a displacement of -0.6 m, which may also be too far, although it might be possible at a lower height. I am uncertain about the base and shoulder locations: I have considered bases at x = ±0.3 m and shoulders or bases at x = ±0.45 m. If the bases are at ±0.45 m, a cross reach of 0.75 m seems impossible. If the left shoulder is at x = -0.3 m and the arm length is about 0.65 m, reaching the right basket at x = 0.32 m might be possible at a low height. I have not tested these cross reaches. The left pregrasp at [-0.405, -0.12, 0.925] m was reachable, as was the right lion grasp near x = 0.291 m, but those do not settle this question.
A central transfer would avoid relying on that uncertain reach. The left arm could hand the toy directly to the right arm at x = 0 m, but that would be complex. A more robust alternative is to place the toy temporarily on a clear central part of the table, perhaps at [0, -0.25] m in XY, and let the right arm regrasp it for placement in the right basket. This adds a placement and a second grasp, but I have 976 control steps remaining.
I considered placing both chocolates in the middle basket first, one with each arm, followed by the two center watches. That still leaves the right arm’s cross reach to the left basket uncertain. A better assignment may be for the left arm to handle both center watches, at x = 0.009 m and x = 0.108 m, and place them in the left basket near x = -0.325 m. The right arm can handle both chocolates; reaching the left chocolate at x = -0.091 m would require a cross reach of about 0.39 m, which seems likely to be possible, and the middle basket at x = 0 m should be easy to reach. The left arm could also handle the left chocolate at x ≈ -0.09 m if needed.
My preferred sequence is to pick the yellow toy with the left arm and the right chocolate with the right arm in parallel. The left arm will place the toy at a temporary central location, perhaps [-0.05, -0.28] m near the front, while the right places the chocolate in the middle basket. Then the left arm can pick the first center watch while the right picks the toy from the transfer point, placing them in the left and right baskets respectively. Finally, the left arm can pick the second center watch while the right picks the left chocolate, placing them in the left and middle baskets. That would require six remaining grasps because the toy is grasped twice. I estimate this as five cycles plus the temporary placement. Another possible transfer location is [0, -0.2] m. I will not rely on the right arm reaching the toy directly at x = -0.3 m.
Both grippers are currently open at the baskets with a 60-degree pitch. I need to retract into clear space and return them to downward grasp orientations. I initially considered overhead positions of [-0.30, -0.135, 1.0] m for the toy and [0.377, -0.149, 1.0] m for the chocolate. For this action, I will use z = 1.02 m. From the current left pose [-0.354, -0.019, 0.987] m to [-0.30, -0.135, 1.02] m, I estimate a distance of 0.128 m. From the current right pose [0.319, 0.001, 0.99] m to [0.377, -0.149, 1.02] m, I estimate 0.166 m. Changing from the 60-degree pitch to the 90-degree downward pose requires about 0.524 rad, within the 1.0 rad orientation limit. Moving toward negative Y and slightly upward should take the grippers away from the baskets safely. I can use 10 control steps for this retreat.
I also need to align the right gripper appropriately for the chocolate. The gold rectangular bar is approximately 0.09 m long and 0.025 m wide, lying diagonally. Its length may exceed the gripper’s estimated maximum opening of 0.08 m, so closing across its short width, with the pads on the long sides, would be preferable. In the downward pose, the closing direction is along world X. The chocolate’s long axis appears to run from roughly pixel [480, 257] to [522, 276] in the high view, down and right in the image. That suggests a world direction with positive X and negative Y, although its exact angle remains uncertain.
I can adjust the closing direction by applying a world-Z yaw before the downward orientation: q = Rz(yaw) × R_down. At zero yaw, local Y points along world -X. After yaw, its direction is [-cos(yaw), -sin(yaw), 0], and the axis sign does not matter for closure. Thus, the closing-axis angle is yaw + 180 degrees. To close perpendicular to a long-axis angle θ_long in the world XY plane, I would use yaw = θ_long + 90 degrees. If the bar’s long axis is at -25 degrees, a yaw of 65 degrees would be suitable. The fingers would still point downward, while the wrist camera would rotate with the yaw. A yaw around 60 degrees may work, but combining a 65-degree yaw with the change from the current 60-degree pitch to the downward pose might exceed the 1.0 rad action limit and require an intermediate orientation.
I could instead grasp with zero yaw across world X if the bar has enough diagonal alignment. My estimate for its projected width is 0.09 × sin(25 degrees) + 0.025 × cos(25 degrees), approximately 0.09 × 0.42 + 0.025 × 0.91 = 0.061 m, within the estimated 0.08 m opening. The pads would contact diagonal edges rather than close exactly perpendicular to the long axis, so a yaw of 45 or 60 degrees would still be preferable. I will refine that choice from the right wrist view once the arm is nearer the bar.
The yellow pig-like toy appears to lie horizontally on its side. I estimate its shape as about 0.07 m long and 0.04 m wide, though its visible width may be around 0.05 m. Its head may point left and upward in the high image near pixel [180, 260]. A downward grasp with zero yaw, closing along world X across the broad body, seems reasonable.
For now, I will retreat both arms to the overhead targets at z = 1.02 m with open fingers and the downward quaternion [0.5, -0.5, 0.5, 0.5] in wxyz order. I considered adding a 45-degree yaw to the right arm during the retreat, but I will keep its yaw at zero for this action and adjust later at the safe pregrasp position after inspecting the chocolate’s orientation. I will use 10 control steps.