高机位相机点 (445, 207) 在假设高度 z = 0.765米处投影,得到金条的位置约为 x = 0.291米,y = -0.0077米。篮子的前底部大约在 y = 0.003米,因此金条中心在 y ≈ -0.008米处刚好在前部外侧,居中位于红色横条下方。它正对着篮子前方,而不是我之前估计的 x = 0.25米。在腕部视图中它看起来在篮子左侧,是因为相机的 X 和 Y 方向发生了旋转。
我能把金条推入篮子吗?我应该把它抬过边缘。我最初的重新抓取计划是张开空的右夹爪,移动到金条上方 (0.291, -0.008, 0.99) m 处,以 70 度偏航恢复朝下的姿态,然后将第六连杆下降到 z = 0.925米。对应的四元数约为 [0.1228, -0.6964, 0.1228, 0.6964]。当前倾斜的四元数是 [0.223, -0.671, 0.02, 0.707],相差约 0.4弧度。金条看起来从像素 (430, 202) 延伸到 (459, 216),具有正的世界 X 分量和负的世界 Y 分量,表明同样的长轴角度约为 -20 度。因此 70 度偏航看起来仍然适合横跨其宽度闭合。
然而,篮子壁紧挨着金条后方,沿正 Y 方向。在 z = 0.767米附近向下的手指在闭合过程中可能会碰到那面壁。前壁的顶部大约在 y = -0.027米,而底部大约在 y = 0.003米。在 z = 0.765米处,考虑到其坡度和厚度,其内部可能从 y = 0.005米左右开始。金条中心在 y = -0.008米处,略在壁底部前方,位于桌面上。金条长轴为 -20 度时,闭合方向约为 (0.34, 0.94),具有相当大的 Y 分量。一个手指在中心加上 0.05 × (0.34, 0.94) 会到达 y ≈ 0.039米,越过壁面。这能否解释为什么金条在边缘处卡住或掉落?我需要一种能避开壁面的抓取方式。
我可以在抓取前轻轻将金条向前移动,沿负 Y 方向,用闭合的夹爪推动它或用抓取来拖动它。另一种可能性是沿世界 X 方向以朝下姿态闭合,将手指放在 x ± 0.05米处而不越过壁面。但金条的长轴主要沿 X 方向,因此这需要横跨其长度张开。我估计的 0.13米金条长度超过了估计的最大开口 0.1米。那 -20 度偏航呢?我初步考虑一个主要沿 X 的闭合方向 [-0.94, -0.342],与金条方向 [0.94, -0.342] 相对,点积可能接近 -0.999。垂直的选择仍然是 70 度偏航。将金条向前移动约 0.04米可能更安全。
我能否在手指从上方跨过壁面的同时抓住前边缘?在第六连杆 z = 0.94米时,指尖将接近 z = 0.782米,低于 z = 0.84米的边缘。后手指在 y ≈ 0.039米处将位于篮子内部、其底部上方,但手指主体可能在 y = 0 附近越过壁面。这需要谨慎。
从篮子内部以向前或向后倾斜以及主要沿 X 的闭合方向接近是另一种选择。在零偏航时,闭合方向为 -X,几乎平行于金条的长轴,因此会沿长度方向夹持,可能放不下。在 20 度偏航时,闭合方向约为 [-0.94, -0.342],与金条长方向的点积约为 -0.766。这大约偏离平行方向 40 度。其投影长度可能约为 0.1米,因此如果金条实际长度为 0.11米,0.1米的开口可能放得下。不过,靠近中部的对角抓取会减少接触面积。
45 度偏航给出闭合方向 [-0.707, -0.707],与金条长轴的点积约为 -0.422。我初步认为这偏离垂直方向 65 度。期望的垂直闭合方向是 [0.342, 0.94] 或其相反方向;70 度偏航给出 [-0.342, -0.94],点积为零。45 度偏航可能仍然合理,但其 Y 分量会使后手指位于抓取中心后方约 0.035米处,仍然靠近壁面。在 20 度偏航时,Y 偏移仅约 0.017米。对于 0.12米的金条,沿该闭合方向的投影长度将为 0.12 × 0.766 = 0.092米,适合估计的 0.1米间隙。垫片力不会与形状法线共线,但抓取可能保持住。先向前拉金条仍然是一个替代方案。
也许最简单稳健的方法是用闭合的指尖在 z = 0.79米附近从后方推动金条向前约 0.06米。前方有开阔空间,但推动可能使其不可预测地滑动,正如技能所警告的那样。我能否用空的张开夹爪从前方接近并使用一个手指?以当前 70 度偏航在第六连杆 z = 0.96米处抓取会使指尖接近 z = 0.802米,后手指可能在 z = 0.80米附近越过前边缘,低于 z = 0.84米的壁顶。在第六连杆 z = 1.0米时,指尖将接近 z = 0.842米并避开边缘,但它们会太高而无法抓取桌面上 z = 0.765米处的金条。将其移离壁面似乎更可取。
左臂在 (-0.299, -0.248, 1.02) m 处是空的,但金条大约在 (0.291, -0.008) m 处,对它来说太远而无法触及。
右夹爪能否从篮子一侧接触金条的后端来将其向前推?一个提议的闭合夹爪位置在 x = 0.29米,y = 0.04米,z = 0.82米,将位于篮子内部、金条上方,然后沿负 Y 方向运动到 y = -0.08米。但这可能会扰动估计在 x = 0.30米,y = 0.06米的红色横条。夹爪在 y = 0.04米处会靠近它。我还考虑了一个位置 (0.24, 0.015, 0.84) m,在红色横条左侧,但向下的手指会使闭合指尖在 z = 0.68米左右,低于桌面。在第六连杆位于 z = 0.96米时,指尖接近 z = 0.802米;在 z = 0.765米处推动金条需要将第六连杆降低到约 z = 0.92米,使指尖位于 z = 0.762米并有接触前壁的风险。倾斜的夹爪可以从前方越过壁面,但以这种方式安排受控的推动很复杂。
金条是否已经足够位于篮子足迹内以计入?它看起来在前壁外侧,且原生判定成功为假。指令要求它在内部,所以我应该完成放置。
我可以再次尝试相同的 70 度偏航,也许一个手指在薄壁内侧,一个在外侧。篮子是否固定,因此如果被接触可能会保持不动?这不确定。更根本的是,壁面会位于后手指和金条之间,因此闭合可能会接触壁面而不是夹住金条。我应该改变抓取位置或先移动物体。
我能否沿金条的长轴靠近一端抓取,使两个手指都保持在壁面前方?我最初考虑向其左端移动,但向前突出的端在右侧:长轴方向约为 [+0.94, -0.34]。在 X 方向偏移 +0.05米、Y 方向偏移 -0.018米会使该端位于 y = -0.026米左右。在 (0.34, -0.026) m 附近以相同的后手指偏移 0.039米抓取仍会使该手指位于 y = 0.013米左右,在壁面内侧。向 y = -0.036米的端部进一步移动会使后手指位于 y = 0.003米左右,仍然在壁面处。45 度偏航的 Y 偏移为 0.035米,有类似的问题。在 20 度偏航且 Y 偏移为 0.017米时,以 y = -0.036米为中心的抓取会使后手指位于 y = -0.019米,在壁面前方。两个手指然后都可以保持在外侧。包装纸可能在端部附近变窄,使投影开口更小。相对的垫片是否与表面法线充分对齐?对于短距离抬入篮子来说可能足够稳定。一个候选抓取点大约在 (0.32, -0.025) m,带有修正的闭合角度;我还没有选定一个。
另一种可能性是通过推动在桌面上旋转物体,然后以更方便的方向抓取它。将其长轴与 X 对齐并沿 Y 闭合会使其接近现有的 -20 度角,所以我仍然需要先将其移开。以 90 度偏航抓取前包装纸端且中心在 y = -0.04米会使后手指位于 y = 0.01米,仍然越过壁面。70 度偏航有类似的估计偏移 0.047米。
我应该仔细检查金条相对于壁面的位置。在观测_000556 的右腕视图中,红色篮子内部在右侧,金条在桌面上、可见红色壁面的左侧。金条大约在像素 (130, 230) 处。壁面可能从大约 (0, 0) 斜向延伸到 (300, 480),而闭合的手指中心在 (320, 400) 附近。我能否从左侧或前方接近,使闭合线平行于该壁面,保持两个手指都在外侧?我最初想知道这是否意味着沿世界 Y 闭合,但篮子的前壁横跨世界 X。沿 X 闭合会使两个手指都在壁面外侧的同一 Y 坐标处。
那将意味着一个朝下、零偏航、沿世界 X 闭合的抓取。由于金条的长轴主要沿 X,它会横跨长度闭合。我之前的估计是辣椒长度 0.08米,金条长度 0.13米,最大物理夹爪间隙可能在 0.085米左右;我初步将开口命令 1 与约 0.08米关联。根据这些估计,它无法夹住金条的全长。端部抓取能否使手指抵住包装纸的侧面而不是超出两端,也许一个手指在顶部?那看起来不稳定。
我可以改为通过推动一端将金条的长轴转向世界 Y,然后沿 X 平行于壁面闭合,两个手指都在外侧。将右/前端沿正 Y 向后推,或将左/后端向前推,可以使其旋转。向篮子方向推是否会意外地将其抬起?我可以尝试在旋转的同时将其向前扫离壁面。这增加了操作,但可能可行。
在选择之前,我应该检查是否高估了金条的长度。如果放得下,在中心附近沿 X 的朝下抓取可能是最简单的。我考虑过长度 0.12–0.13米和实际开口间隙约 0.1米,但包装纸在 0.6米深度处跨越 50 像素表明约 0.104米而不是 0.13米。我将通过在假设高度 z = 0.765米处反投影端点来估计世界长度。最初的高机位相机端点大约为 (110, 226) 和 (160, 241)。当前端点看起来接近 (430, 201) 和 (462, 216),具有相似的视长度。对于立即计算,我将使用 (430, 202) 和 (462, 216),以及中心点 (445, 207),来估计端点位置、金条长度及其全向量角度。我将在选择下一次抓取之前检查这些结果。
原文10,887 字
The high-camera point (445, 207), projected at an assumed height of z = 0.765 m, gives the gold bar’s position as approximately x = 0.291 m, y = -0.0077 m. The basket’s front bottom is around y = 0.003 m, so the bar’s center at y ≈ -0.008 m is just outside the front, centered beneath the red bar. It is directly in front of the basket rather than at my earlier estimate of x = 0.25 m. It appears to the left of the basket in the wrist view because the camera’s X and Y directions are rotated.
Could I push the gold bar into the basket? I should lift it over the rim. My initial regrasp plan is to open the empty right gripper, move above the bar at (0.291, -0.008, 0.99) m, restore the downward orientation with a 70-degree yaw, and then lower link6 to z = 0.925 m. The corresponding wxyz quaternion is approximately [0.1228, -0.6964, 0.1228, 0.6964]. The current tilted quaternion is [0.223, -0.671, 0.02, 0.707], about 0.4 rad away. The gold bar appears to run from pixel (430, 202) to (459, 216), with positive world-X and negative world-Y components, suggesting the same long-axis angle of about -20 degrees. A 70-degree yaw therefore still looks appropriate for closing across its width.
However, the basket wall is immediately behind the bar, along positive Y. Downward fingers near z = 0.767 m might contact that wall during closure. The front wall’s top is around y = -0.027 m, while its bottom is around y = 0.003 m. At z = 0.765 m, allowing for its slope and thickness, its interior might begin around y = 0.005 m. The bar’s center at y = -0.008 m is slightly in front of the wall bottom, on the table. With the bar’s long axis at -20 degrees, a closing direction of approximately (0.34, 0.94) has a substantial Y component. One finger at the center plus 0.05 × (0.34, 0.94) would reach y ≈ 0.039 m, crossing the wall. Could this explain why the bar became stuck or dropped at the rim? I need a grasp that clears the wall.
I could gently move the bar forward, along negative Y, before grasping it, using a closed gripper to push it or a grasp to drag it. Another possibility is to close along world X with the downward orientation, placing the fingers at x ± 0.05 m without crossing the wall. But the bar’s long axis is mostly along X, so that would require opening across its length. My estimate of a 0.13 m bar exceeds the estimated maximum opening of 0.1 m. What about a -20-degree yaw? I tentatively consider a mostly-X closing direction of [-0.94, -0.342] against the bar direction [0.94, -0.342], with a dot product perhaps near -0.999. The perpendicular choice is still the 70-degree yaw. Moving the bar forward by about 0.04 m may be safer.
Could I grasp the front edge while the fingers straddle the wall from above? At link6 z = 0.94 m, the fingertip would be near z = 0.782 m, below the rim at z = 0.84 m. The rear finger at y ≈ 0.039 m would be inside the basket above its bottom, but the finger body could cross the wall near y = 0. This needs caution.
An approach from inside the basket with a forward or backward tilt and a mostly-X closing direction is another option. With zero yaw, the closing direction is -X, nearly parallel to the bar’s long axis, so it would grip along the length and might not fit. At a 20-degree yaw, the closing direction is approximately [-0.94, -0.342], with a dot product of about -0.766 against the bar’s long direction. That is about 40 degrees from parallel. Its projected length might be around 0.1 m, so an opening of 0.1 m could fit if the bar is actually 0.11 m long. A diagonal grasp near the middle would reduce contact area, though.
A 45-degree yaw gives a closing direction of [-0.707, -0.707], with a dot product of about -0.422 against the bar’s long axis. I tentatively think of that as an angle of 65 degrees from perpendicular. The desired perpendicular closing direction is [0.342, 0.94] or its opposite; the 70-degree yaw gives [-0.342, -0.94], with a zero dot product. The 45-degree yaw could still be reasonable, but its Y component would put the rear finger about 0.035 m behind the grasp center, still near the wall. At a 20-degree yaw, the Y offset would be only about 0.017 m. For a 0.12 m bar, the projected length along that closing direction would be 0.12 × 0.766 = 0.092 m, fitting an estimated 0.1 m gap. The pad forces would not be collinear with the shape normals, but the grasp might hold. Pulling the bar forward first remains an alternative.
Perhaps the simplest robust approach is to push the bar about 0.06 m forward with closed fingertips near z = 0.79 m, positioned behind it. There is open space in front, but pushing could make it slide unpredictably, as the skill warns. Could I approach from the front with an empty open gripper and use one finger instead? Grasping with the current 70-degree yaw at link6 z = 0.96 m would put the tips near z = 0.802 m, where the rear finger could cross the front rim around z = 0.80 m, below the wall top at z = 0.84 m. At link6 z = 1.0 m, the tips would be near z = 0.842 m and clear the rim, but they would be too high to grasp the bar on the table at z = 0.765 m. Moving it away from the wall seems preferable.
The left arm is clear at (-0.299, -0.248, 1.02) m, but the gold bar at approximately (0.291, -0.008) m is too far for it to reach.
Could the right gripper push the bar forward by contacting its back end from the basket side? A proposed closed-gripper position at x = 0.29 m, y = 0.04 m, z = 0.82 m would be inside the basket above the bar, followed by a negative-Y motion toward y = -0.08 m. But this could disturb the red bar, estimated at x = 0.30 m, y = 0.06 m. A gripper at y = 0.04 m would be close to it. I also consider a position at (0.24, 0.015, 0.84) m, left of the red bar, but downward fingers would put the closed tip around z = 0.68 m, below the table. With link6 at z = 0.96 m, the tips are near z = 0.802 m; pushing the bar at z = 0.765 m would require lowering link6 to about z = 0.92 m, putting the tips at z = 0.762 m and risking front-wall contact. A tilted gripper could reach over the wall from the front, but arranging a controlled push this way is complicated.
Could the bar already lie sufficiently within the basket footprint to count? It looks outside the front wall, and native_success is false. The instruction requires it inside, so I should complete the placement.
I could try the same 70-degree yaw again, perhaps with one finger inside and one outside the thin wall. Is the basket fixed, so it might remain still if contacted? That is uncertain. More fundamentally, the wall would lie between the rear finger and the bar, so closure could contact the wall rather than pinch the bar. I should change the grasp location or move the object first.
Could I grasp along the bar’s long axis nearer an end so both fingers remain in front of the wall? I initially consider shifting toward its left end, but the front-protruding end is on the right: the long-axis direction is approximately [+0.94, -0.34]. An offset of +0.05 m in X and -0.018 m in Y would put that end around y = -0.026 m. A grasp near (0.34, -0.026) m with the same rear-finger offset of 0.039 m would still put that finger around y = 0.013 m, inside the wall. Moving farther toward an end at y = -0.036 m would leave the rear finger around y = 0.003 m, still at the wall. A 45-degree yaw, with a Y offset of 0.035 m, has a similar problem. With a 20-degree yaw and a Y offset of 0.017 m, a grasp centered at y = -0.036 m would put the rear finger at y = -0.019 m, in front of the wall. Both fingers could then remain outside. The wrapper may taper near the end, making the projected opening smaller. Would the opposing pads be sufficiently aligned with the surface normals? It might be stable enough for a short lift into the basket. A candidate grasp point is around (0.32, -0.025) m, with a revised closing angle; I have not selected one.
Another possibility is to rotate the object on the table by pushing it, then grasp it in a more convenient orientation. Aligning its long axis with X and closing across Y would leave it close to its existing -20-degree angle, so I would still need to move it away first. Grasping the front wrapper end with a 90-degree yaw and a center at y = -0.04 m would put the rear finger at y = 0.01 m, still beyond the wall. The 70-degree yaw has a similar estimated offset of 0.047 m.
I should inspect the bar’s position relative to the wall carefully. In the right wrist view from obs_000556, the red basket interior is on the right and the gold bar is on the table to the left of a visible red wall. The bar is around pixel (130, 230). The wall may run diagonally from approximately (0, 0) to (300, 480), while the closed fingers are centered near (320, 400). Could I approach from the left or front with the closing line parallel to that wall, keeping both fingers outside? I initially wonder whether this means closing along world Y, but the basket’s front wall spans world X. Closing along X would keep both fingers at the same Y coordinate outside the wall.
That would mean a downward, zero-yaw grasp closing along world X. Since the bar’s long axis is mostly X, it would close across the length. My earlier estimates were a pepper length of 0.08 m and a bar length of 0.13 m, with a maximum physical gripper gap perhaps around 0.085 m; I am tentatively associating an opening command of 1 with about 0.08 m. On those estimates, it cannot grip the full bar length. Could an end grasp put the fingers against the wrapper’s sides rather than beyond both ends, perhaps with one finger on top? That does not look stable.
I could instead rotate the bar’s long axis toward world Y by pushing one end, then close along X parallel to the wall with both fingers outside. Pushing the right/front end backward along positive Y, or the left/back end forward, could rotate it. Would pushing toward the basket lift it unintentionally? I could try sweeping it forward away from the wall while rotating it. This adds manipulation, but it may be feasible.
Before choosing, I should check whether I have overestimated the bar’s length. A downward grasp across X near the center might be simplest if it fits. I have considered a length of 0.12–0.13 m and an actual open gap around 0.1 m, but a wrapper spanning 50 pixels at a depth of 0.6 m suggests about 0.104 m rather than 0.13 m. I will estimate the world length by inverse-projecting endpoints at the assumed height of z = 0.765 m. The initial high-camera endpoints were approximately (110, 226) and (160, 241). The current endpoints appear near (430, 201) and (462, 216), with a similar apparent length. For the immediate calculation, I will use (430, 202) and (462, 216), along with the center point (445, 207), to estimate the endpoint positions, the bar’s length, and its whole-vector angle. I will inspect those results before choosing the next grasp.