绿色小车已经完全掉到右侧篮子外面,倒扣在高机位相机像素 (453, 226) 附近的桌面上。我需要重新抓取它。
马最初看起来在下半身周围被稳稳夹住,但我不确定。腕部视图显示它在手指之间显得很大,尽管手指可能靠近它的头部。左侧第六连杆上升了 0.033米,但马看起来仍在高机位相机像素 (179, 218) 附近,几乎和之前一样。对比另一对估计,它在观测 133 中位于 (181, 223),在观测 136 中位于 (179, 219),变化仅约 4 像素。对于 0.033米的物体抬升,我原本预期大约 10 像素的变化。
腕部图像证据也不明确。在观测 133 中,白色篮子遮挡了马的大部分,使其身体的一部分位于图像 y = 350 附近。在观测 136 中,马的更多部分可见,大致跨越 y = 170–450。它的底部似乎从 (400, 480) 移动到了 (400, 330)。如果马留在桌面上,抬高相机可能使它变小并向上移动。如果它随夹爪刚性移动,其相对图像尺寸应保持不变。张开命令稳定在 0.577,但手指是否可能被黄色小车或马挡住,而实际上并没有夹住马?
另一次高机位相机对比显示,马现在位于 (180, 218) 附近,而在观测 90 中位于 (182, 222),这表明只有轻微移动,而不是明确的 0.033米抬升。在观测 133 中,一根内侧手指位于图像 x = 180 附近;我不确定另一根手指的位置,而马的身体跨越 x = 280–400。在观测 136 中,手指似乎位于 x = 150 和 460 附近,马跨越 x = 180–470。它可能被夹住并朝相机方向向上旋转,这可能改变其腕部投影,同时使其高机位相机位置几乎不变。或者,抓取可能失败了,马可能仍在桌面上。我应该进一步抬升到第六连杆 z = 1.05米并检查高机位视图。如果没有被夹住,手指应该空着上升。
绿色小车现在在红色篮子前部附近可以够到。在像素 (453, 226) 处,假设 z = 0.775米,它的世界位置可能在 x = 0.29米和 y = −0.04米附近;我需要计算一下。它倒扣的位置可能因其平坦的底面而使稳定重抓更容易。我还需要系统地解决放置可达性问题。
我可以根据当前手臂姿态和目标预检查来推理可达性。较低的放置高度可能有帮助。我考虑过近似正运动学或从机器人模型文件推断 – 参数,但访问模拟器内部或另一个控制器是被禁止的,所以我将改为根据可用的观测进行推理。
初始末端执行器方向是水平的。手臂看起来是拟人化的,其基座可能在世界 y = −0.55米附近;一个基座似乎在高机位相机像素 (120, 430) 附近。我估计篮子在世界 y = 0.08米附近,桌面在 z = 0.75米。如果夹爪水平向前指向,可达性可能足够。例如,第六连杆姿态在 y = −0.08米和 z = 0.85米附近,手指沿世界 +Y 方向指向,将使指尖位于 y = 0.08米和 z = 0.85米附近,以便释放到篮子中。因此,将夹持的物体从向下抓取旋转到倾斜或水平方向,可以减少所需的第六连杆 Y 位置和高度。
向下抓取便于拾取物体。对于放置,我可以将局部 X 从世界 −Z 旋转到世界 +Y,或旋转到 +Y 和 −Z 的组合,同时保持闭合轴水平。初始 (90°) 方向具有局部 X 沿世界 +Y、局部 Y 沿世界 −X、局部 Z 沿世界 +Z。它沿世界 X 方向闭合,同时手指沿 +Y 方向向前延伸。
我当前的向下四元数按顺序为 [0.5, −0.5, 0.5, 0.5]。我最初将其轴解释为局部 X 沿 −Z、局部 Y 沿 −X、局部 Z 沿 −Y。按照这种解释,绕世界 +X 旋转 +90° 会将 −Z 映射到 +Y,但会将局部 Z 从 −Y 映射到 −Z,而不是期望的 +Z。绕世界 X 旋转 −90° 则会将 −Z 映射到 −Y,这同样是错误的。等价地,绕局部 Y(指向世界 −X)旋转 −90° 得到世界 +X 旋转,但仍似乎产生错误的局部 Z 符号。
这种不一致表明我的轴映射有误。局部 X 沿 +Y、局部 Y 沿 −X 的正当旋转必须具有局部 Z 沿 +Z,因为 Y × (−X) = +Z。同样,在当前向下姿态中,(−Z) × (−X) = +Y,所以局部 Z 必须是 +Y,而不是 −Y。我之前的矩阵 [[0, −1, 0], [0, 0, −1], [−1, 0, 0]] 符号错误,将是不正当的。
我定义的一般向下旋转矩阵是 [[0, 余弦 θ, 正弦 θ], [0, 正弦 θ, −余弦 θ], [−1, 0, 0]]。其列是局部 X = (0, 0, −1)、局部 Y = (余弦 θ, 正弦 θ, 0) 和局部 Z = (正弦 θ, −余弦 θ, 0),与局部 X × 局部 Y = 局部 Z 一致。在 θ = π 时,余弦 θ = −1 且正弦 θ = 0,所以局部 Y = (−1, 0, 0) 且局部 Z = (0, +1, 0)。正确的矩阵是 [[0, −1, 0], [0, 0, +1], [−1, 0, 0]]。我在其第二行中错误地写成了 −1 而不是 +1。
使用修正后的映射,绕世界 +X 旋转 +90°,等价于绕沿世界 −X 的局部 Y 旋转 −90°,将局部 X 从 −Z 映射到 +Y,并将局部 Z 从 +Y 映射到 +Z。这给出了初始向前方向, 四元数为 [0.707, 0, 0, 0.707]。它与当前四元数 [0.5, −0.5, 0.5, 0.5] 的点积为 0.3535 + 0 + 0 + 0.3535 = 0.707,给出方向距离 π/2。我可以通过两步达到该向前方向。在夹持物体时,我可以向前旋转并将第六连杆移动到 y = −0.08米和 z = 0.9米附近,使手指和夹持的物体伸入篮子。我应该对马和绿色小车都使用这种方法。
在重新抓取绿色小车之前,我将根据高机位相机像素 (453, 226) 计算其世界位置,使用估计高度 z = 0.775米和桌面高度 z = 0.75米。其 x 坐标可能约为 0.285米,但我需要计算。马可能仍需要另一次抬升检查。我将使用投影辅助函数来估计其余物体在假设高度处以及篮子前缘的位置。
原文6,276 字
The green car has fallen completely outside the right basket and is upside down on the table near high-camera pixel (453, 226). I need to regrasp it.
The horse initially looks securely held around its lower body, but I am not certain. The wrist view shows it large between the fingers, although the fingers may be near its head. The left link6 rose by 0.033 m, yet the horse appears near high-camera pixel (179, 218), almost where it was before. Comparing another pair of estimates, it was at (181, 223) in observation 133 and (179, 219) in observation 136, a change of only about 4 pixels. I would have expected roughly 10 pixels for a 0.033 m object lift.
The wrist-image evidence is also ambiguous. In observation 133, the white basket occluded much of the horse, leaving part of its body near image y = 350. In observation 136, more of the horse is visible, spanning approximately y = 170–450. Its base appears to have moved from (400, 480) to (400, 330). If the horse stayed on the table, raising the camera could make it smaller and shift it upward. If it moved rigidly with the gripper, its relative image size should remain constant. The opening command is stable at 0.577, but could the fingers be blocked by the yellow car or the horse without actually holding the horse?
Another high-camera comparison puts the horse near (180, 218) now versus (182, 222) in observation 90, suggesting only slight motion rather than a clear 0.033 m lift. In observation 133, one inner finger was near image x = 180; I am uncertain about the other finger’s position, while the horse body spanned x = 280–400. In observation 136, the fingers appear near x = 150 and 460, and the horse spans x = 180–470. It might be held and rotating upward toward the camera, which could change its wrist projection while leaving its high-camera position nearly unchanged. Alternatively, the grasp may have failed and the horse may still be on the table. I should make a further lift to link6 z = 1.05 m and inspect the high view. If it is not held, the fingers should rise empty.
The green car is now accessible near the front of the red basket. At pixel (453, 226), assuming z = 0.775 m, its world position might be around x = 0.29 m and y = −0.04 m; I need to calculate that. Its upside-down position may make a stable regrasp easier because of its flat underside. I also need to address the placement reach problem systematically.
I can reason about reach from the current arm poses and the target prechecks. A lower placement height may help. I considered approximating ARX X5 forward kinematics or inferring Denavit–Hartenberg parameters from robot model files, but accessing simulator internals or another controller is prohibited, so I will reason from the available observations instead.
The initial end-effector orientation was horizontal. The arms appear anthropomorphic, perhaps with their bases near world y = −0.55 m; one base appears around high-camera pixel (120, 430). I estimate the baskets near world y = 0.08 m and the table surface at z = 0.75 m. Reach may be sufficient if the gripper points horizontally forward. For example, a link6 pose near y = −0.08 m and z = 0.85 m, with the fingers pointing along world +Y, would put the fingertips near y = 0.08 m and z = 0.85 m for release into the basket. Rotating a held object from a downward grasp to a tilted or horizontal orientation could therefore reduce the required link6 Y position and height.
A downward grasp is convenient for picking objects up. For placement, I could rotate local X from world −Z toward world +Y, or toward a combination of +Y and −Z, while keeping the closing axis horizontal. The initial Rz(90°) orientation has local X along world +Y, local Y along world −X, and local Z along world +Z. It closes across world X while extending the fingers forward along +Y.
My current downward quaternion is [0.5, −0.5, 0.5, 0.5] in wxyz order. I initially interpreted its axes as local X along −Z, local Y along −X, and local Z along −Y. With that interpretation, rotating about world +X by +90° would map −Z to +Y but map local Z from −Y to −Z, rather than the desired +Z. Rotating about world X by −90° would instead map −Z to −Y, which is also wrong. Equivalently, rotation about local Y, which points along world −X, by −90° gives the world +X rotation, but still seemed to produce the wrong local Z sign.
That inconsistency points to an error in my axis mapping. A proper rotation with local X along +Y and local Y along −X must have local Z along +Z, since Y × (−X) = +Z. Likewise, in the current downward pose, (−Z) × (−X) = +Y, so local Z must be +Y, not −Y. My earlier matrix, [[0, −1, 0], [0, 0, −1], [−1, 0, 0]], had the wrong sign and would be improper.
The general downward rotation matrix I defined is [[0, cos θ, sin θ], [0, sin θ, −cos θ], [−1, 0, 0]]. Its columns are local X = (0, 0, −1), local Y = (cos θ, sin θ, 0), and local Z = (sin θ, −cos θ, 0), consistent with local X × local Y = local Z. At θ = π, cos θ = −1 and sin θ = 0, so local Y = (−1, 0, 0) and local Z = (0, +1, 0). The correct matrix is [[0, −1, 0], [0, 0, +1], [−1, 0, 0]]. I mistakenly wrote −1 instead of +1 in its second row.
With the corrected mapping, rotating about world +X by +90°, equivalently about local Y along world −X by −90°, maps local X from −Z to +Y and local Z from +Y to +Z. That gives the initial forward orientation, with wxyz quaternion [0.707, 0, 0, 0.707]. Its dot product with the current quaternion [0.5, −0.5, 0.5, 0.5] is 0.3535 + 0 + 0 + 0.3535 = 0.707, giving an orientation distance of π/2. I can reach that forward orientation in two steps. While holding an object, I could rotate forward and move link6 near y = −0.08 m and z = 0.9 m, allowing the fingers and held object to extend into the basket. I should use this approach for both the horse and green car.
Before regrasping the green car, I will calculate its world position from high-camera pixel (453, 226), using the estimated height z = 0.775 m and table height z = 0.75 m. Its x coordinate may be about 0.285 m, but I need the calculation. The horse may still need another lift check. I will use the projection helper to estimate positions for the remaining objects at their assumed heights and for the basket’s front edge.