我可以对两条手臂都使用四元数 [0.5, 0.5, 0.5, -0.5]。我最初从起始朝向估计其旋转为 120 度,即 2.09弧度。一个中间朝向可以让局部 X 轴斜向前下方,沿世界坐标 (0, 0.707, -0.707),同时保持局部 Y 轴沿世界 -X,这样手指就沿世界 X 方向闭合。那会得到什么四元数?用科学计算库生成中间四元数可能更容易。
在旋转之前,我应该考虑间隙。初始夹爪靠近左边的大象和右边的手表,手指可能已经接近接触。原地旋转会扫动手指,手指向前伸出约 0.158米,扫过的弧线可能撞到物体。我可以先把两条手臂抬高 0.12米并保持其朝向,然后在物体上方旋转。初始第六连杆高度约为 0.922米,手指处于同一高度,我估计桌面在 z = 0.75米。抬升到 z = 1.06米将在初始高度约 0.14米范围内。手指朝下伸出 0.158米,指尖将位于 z = 0.902米,或许能避开约 z = 0.86米的篮子边缘。
对于自上而下的抓取,我是否应该把第六连杆的 z 设为期望的物体接触高度加上约 0.13米?手指伸出最远达 0.1576米,但接触垫可能大约在局部 x = 0.11–0.15米处。对于 z = 0.77米的物体,第六连杆高度接近 0.91米会使指尖位于约 z = 0.752米,这可能低到足以碰到桌面。我希望指尖在 z = 0.755米左右,同时在 z = 0.77米处抓取,所以第六连杆 z = 0.91米似乎仍然合理。抓取中心可能大约在局部 x = 0.13米处,接触点接近 0.14米。因为局部 X 垂直向下,这不需要水平偏移:第六连杆的 x 和 y 可以与期望的抓取中心一致。我可以从第六连杆 z = 0.97米接近,指尖约在 z = 0.81米,然后下降到 z = 0.91米。
对于 z = 0.77米的手表,我可以沿其直径抓取表带顶部。第六连杆 z = 0.90米是否合适?圆环直径似乎约为 0.055米,张开的夹爪间隙可能约为 0.09米。在 z = 0.765米处抓取外侧表带两侧、指尖在 z = 0.752米,意味着第六连杆 z = 0.91米。对于车身宽度约 0.045米、高度约 0.025米的小车,我可以在第六连杆 z = 0.915米处沿其宽度闭合,使指尖位于约 z = 0.757米,侧面接触点位于约 z = 0.77米。大象的身体宽度可能为 0.06米,高度为 0.09米。在其身体中部约 z = 0.80米处抓取、指尖在 z = 0.78米,意味着第六连杆 z = 0.93 或 0.94米。白色玩具是否也大约 0.06米?我应该检查一下。使用初始朝向进行水平抓取是另一种选择,但自上而下似乎更好。
我可以先同时抓取左边和右边的小车。一种方法是把左臂向上移动约 0.1米并移向 (-0.15, -0.19, 1.04) m,把右臂移向 (0.27, -0.11, 1.04) m,同时保持初始四元数。左侧目标是否符合 0.25米的距离界限?其近似分量为 = 0.15米、 = 0.16米、 = 0.12米,总计接近 0.25米,也许是 0.249米。更安全的做法是直接抬升到 z = 1.06米,然后再部分旋转并平移。
在无障碍空间中,一个动作可以使用 15 步。由于每步旋转增量限制在 0.1弧度,2.09弧度的旋转至少需要 21 步,因此至少需要两次调用。每个目标还必须保持在测量朝向的 1弧度以内。我可以选择约 0.9弧度的中间旋转,并允许 10–12 步。朝向距离为 2.094弧度的直接自上而下目标会被拒绝。在比例 0.45 处进行会得到约 0.942弧度;我可以在物体上方移动时将其用作中间值。两条手臂可以同时旋转。我应该计算中间四元数并考虑抬升后的位置。另一个自上而下的候选 [0, 0.707, 0, -0.707] 与初始 [0.707, 0, 0, 0.707] 的点积似乎也是 -0.5,得到 2.094弧度。起初,这些候选似乎需要 120 度的旋转。
我应该检查这一解释。初始局部 X 沿世界 +Y,局部 Y 沿世界 -X,局部 Z 沿世界 +Z。绕世界 X 旋转 +90 度会使手指朝上,而 -90 度会使手指朝下。手指朝下且沿世界 X 闭合的期望轴为局部 X = -Z、局部 Y = -X、局部 Z = +Y,因为 (-Z, -X) = +Y。绕世界 X 的单个 -90 度旋转将三个初始轴都映射到这些期望轴,因此其相对角度应为 90 度,而不是 120 度。(-90) 乘以初始 (90) 可能得到 [0.5, -0.5, 0.5, 0.5]。我误读了之前的表格:[0, 0.707, 0, -0.707] 对应局部 X = -Z 和局部 Y = -Y,而不是局部 Y = -X。表中对应局部 X = -Z 和局部 Y = -X 的四元数是 [-0.5, 0.5, -0.5, -0.5],等价于 [0.5, -0.5, 0.5, 0.5]。它与初始四元数的点积约为 0.3535 + 0.3535 = 0.707,得到 90 度。对于沿世界 Y 闭合,局部 Y = +Y 得到 [0.707, 0, 0.707, 0],它与初始四元数的点积为 0.5,得到 120 度。
沿世界 X 闭合对小车是否可行?在高机位相机图像中,左边小车的长轴似乎大致水平,因此它大约沿世界 X。因此我应该沿世界 Y 跨其较短宽度闭合,使用 [0.707, 0, 0.707, 0] 和 120 度旋转。其轴线在世界 X–Y 中可能略有斜向。沿世界 X 闭合会跨越小车的长度,约 0.11米,这可能超过夹爪约 0.08米的最大跨度。
右边的小车在图像中似乎斜向约 45 度。其长轴大约从像素 (432, 260) 到 (461, 229),位移为 (+29, -31)。世界 x 水平投影,而正世界 y 向上并向深度投影,因此小车的长方向可能具有大致相等的正 x 和 y 分量。其宽度将垂直,沿 x = -y。我可以将局部 Y 设为 (0.707, -0.707, 0) 或 (-0.707, 0.707, 0),局部 X = -Z,并通过它们的叉积计算第三轴。该四元数是否会与初始朝向相差超过 1弧度?如果是,我可以使用一个中间值。
左边小车的长轴似乎从像素 (224, 286) 到 (264, 274),位移为 (+40, -12)。这表明世界方向具有正 x 和较小的正 y 分量。其宽度方向可能大致为 (-0.1, 1),主要沿 Y。局部 Y 大约为 (-0.1, 0.995) 或相反 (0.1, -0.995) 的自上而下朝向可能合适。
我可以先处理手表,但小车似乎仍然方便。黑色手表大约在 (0.16, -0.1) m,右边小车大约在 (0.27, -0.11) m,最右边的手表大约在 (0.42, -0.13) m。初始右侧第六连杆大约在 (0.3, -0.352) m,其尖端大约在 (0.3, -0.194) m,靠近右边小车和手表。左边小车靠近 (-0.15, -0.19) m,而初始左侧尖端大约在 (-0.3, -0.194) m,因此左臂需要向正 x 方向移动。两条手臂向内移动是否会导致碰撞?小车目标在 x = -0.15米和 x = +0.27米,相隔约 0.42米,这似乎安全。
我应该细化小车坐标。左边小车的图像中心可能是 (245, 280);假设其车身中心在 z = 0.77米,则大约为 (x, y) = (-0.149, -0.193) m。右边小车的中心在 (449, 244),z = 0.772米,则大约为 (0.273, -0.114) m。其右腕中心可能大约在 (261, 402),因此三角测量可以细化该估计。另一个可能的匹配对是高机位相机 (448, 245) 和右腕 (260, 405)。对于车顶中心,我可以使用高机位相机 (447, 243) 和腕部 (263, 395) 作为对应点。左边小车可能在左腕图像下方,该图像似乎只显示大象和手表。右腕提供了右边小车的有用视图。我将编写一个通用的三角测量计算,并计算自上而下的四元数和中间值。
原文8,142 字
I could use quaternion [0.5, 0.5, 0.5, -0.5] for both arms. I initially estimate its rotation from the starting orientation as 120 degrees, or 2.09 rad. An intermediate orientation could point local X diagonally forward and down, along world (0, 0.707, -0.707), while keeping local Y along world -X so the fingers close across world X. What quaternion would that give? It may be easier to generate an intermediate quaternion with SciPy Slerp.
Before rotating, I should consider clearance. The initial grippers are near the left elephant and right watch, and the fingers may already be close to contact. Rotating in place would sweep the fingers, which extend about 0.158 m forward, through an arc that could knock objects. I could first lift both arms by 0.12 m while maintaining their orientation, then rotate above the objects. The initial link6 height is about 0.922 m, with the fingers at the same height, and I estimate the table at z = 0.75 m. A lift to z = 1.06 m would be within about 0.14 m of the initial height. With downward-pointing fingers extending 0.158 m, the fingertips would then be at z = 0.902 m, perhaps clear of basket rims around z = 0.86 m.
For a top-down grasp, should I set link6 z to the desired object contact height plus about 0.13 m? The fingers extend as far as 0.1576 m, but the contact pads are likely around local x = 0.11–0.15 m. A link6 height near 0.91 m for an object at z = 0.77 m puts the fingertip at about z = 0.752 m, which may be low enough to touch the table. I want the tips around z = 0.755 m while gripping at z = 0.77 m, so link6 z = 0.91 m still seems reasonable. The grasp center may be around local x = 0.13 m, with contact near 0.14 m. Because local X points vertically downward, this does not require a horizontal offset: link6 x and y can match the desired grasp center. I could approach at link6 z = 0.97 m, with tips around z = 0.81 m, then descend to z = 0.91 m.
For a watch at z = 0.77 m, I could grasp the top of the band across its diameter. Would link6 z = 0.90 m be appropriate? The ring diameter appears to be about 0.055 m, and the open gripper gap may be about 0.09 m. Gripping the outer band sides at z = 0.765 m with fingertips at z = 0.752 m suggests link6 z = 0.91 m. For a car with body width about 0.045 m and height about 0.025 m, I could close across its width at link6 z = 0.915 m, putting the fingertips around z = 0.757 m and the side contacts around z = 0.77 m. The elephant may have a body width of 0.06 m and height of 0.09 m. A grasp around the middle of its body at z = 0.80 m, with fingertips at z = 0.78 m, suggests link6 z = 0.93 or 0.94 m. Is the white toy similarly about 0.06 m? I should examine it. A horizontal grasp using the initial orientation is another option, but top-down seems better.
I could start by picking the left and right cars simultaneously. One approach would move the left arm upward by about 0.1 m and toward (-0.15, -0.19, 1.04) m, and the right arm toward (0.27, -0.11, 1.04) m, while retaining the initial quaternion. Would the left target fit the 0.25 m distance bound? Its approximate components are dx = 0.15 m, dy = 0.16 m, and dz = 0.12 m, giving a total near 0.25 m, perhaps 0.249 m. It would be safer simply to lift to z = 1.06 m before rotating partway and translating.
In unobstructed space, an action could use 15 steps. With rotation increments bounded at 0.1 rad per step, a 2.09 rad rotation would need at least 21 steps, so at least two calls. Each target must also remain within 1 rad of the measured orientation. I could choose an intermediate rotation of about 0.9 rad and allow 10–12 steps. A direct top-down target at an orientation distance of 2.094 rad would be rejected. Slerp at fraction 0.45 would give about 0.942 rad; I could use that as an intermediate while moving above the objects. Both arms could rotate simultaneously. I should calculate the intermediate quaternion and consider the positions after lifting. Another top-down candidate, [0, 0.707, 0, -0.707], also appears to have a dot product of -0.5 with the initial [0.707, 0, 0, 0.707], giving 2.094 rad. At first, these candidates seem to require 120-degree rotations.
I should check that interpretation. The initial local X points along world +Y, local Y along world -X, and local Z along world +Z. A rotation of +90 degrees about world X would send the fingers upward, whereas -90 degrees would send them downward. The desired axes for downward fingers with closure along world X are local X = -Z, local Y = -X, and local Z = +Y, since cross(-Z, -X) = +Y. A single -90-degree rotation about world X maps all three initial axes to those desired axes, so its relative angle should be 90 degrees, not 120 degrees. Rx(-90) multiplied by the initial Rz(90) may give [0.5, -0.5, 0.5, 0.5]. I misread the earlier table: [0, 0.707, 0, -0.707] corresponds to local X = -Z and local Y = -Y, not local Y = -X. The table’s quaternion for local X = -Z and local Y = -X is [-0.5, 0.5, -0.5, -0.5], equivalent to [0.5, -0.5, 0.5, 0.5]. Its dot product with the initial quaternion is approximately 0.3535 + 0.3535 = 0.707, giving 90 degrees. For closure along world Y, local Y = +Y gives [0.707, 0, 0.707, 0], whose dot product with the initial quaternion is 0.5, giving 120 degrees.
Would closure along world X work for the cars? The left car’s long axis appears roughly horizontal in the high-camera image, so it is approximately along world X. I should therefore close across its shorter width along world Y, using [0.707, 0, 0.707, 0] and a 120-degree rotation. Its axis may be slightly diagonal in world X–Y. Closing along world X would span the car’s length, about 0.11 m, which may exceed a maximum gripper span of about 0.08 m.
The right car appears diagonal at about 45 degrees in the image. Its long axis runs approximately from pixel (432, 260) to (461, 229), a displacement of (+29, -31). World x projects horizontally, while positive world y projects upward and into depth, so the car’s long direction may have roughly equal positive x and y components. Its width would be perpendicular, along x = -y. I could set local Y to (0.707, -0.707, 0) or (-0.707, 0.707, 0), with local X = -Z, and calculate the third axis by their cross product. Would that quaternion be more than 1 rad from the initial orientation? If so, I can use an intermediate.
The left car’s long axis appears to run from pixel (224, 286) to (264, 274), a displacement of (+40, -12). That suggests a world direction with positive x and a smaller positive y component. Its width direction may be roughly (-0.1, 1), mostly along Y. A top-down orientation with local Y approximately (-0.1, 0.995), or the opposite (0.1, -0.995), could be suitable.
I could start with watches instead, but the cars still seem convenient. The black watch is around (0.16, -0.1) m, the right car around (0.27, -0.11) m, and the rightmost watch around (0.42, -0.13) m. The initial right link6 is around (0.3, -0.352) m, with its tip around (0.3, -0.194) m, near the right car and watch. The left car is near (-0.15, -0.19) m, while the initial left tip is around (-0.3, -0.194) m, so the left arm needs to move toward positive x. Could both arms moving inward cause a collision? The car targets at x = -0.15 m and x = +0.27 m are separated by about 0.42 m, which seems safe.
I should refine the car coordinates. The left car’s image center may be (245, 280); assuming its body center is at z = 0.77 m gives approximately (x, y) = (-0.149, -0.193) m. The right car’s center at (449, 244), with z = 0.772 m, gives approximately (0.273, -0.114) m. Its right-wrist center may be around (261, 402), so triangulation could refine that estimate. Another possible matching pair is high-camera (448, 245) and right-wrist (260, 405). For the roof center, I could use high-camera (447, 243) and wrist (263, 395) as corresponding points. The left car may be below the left wrist image, which seems to show only the elephant and watch. The right wrist provides a useful view of the right car. I’ll write a general triangulation calculation and calculate the top-down quaternion and Slerp intermediates.