大象的估计位置为 x = −0.115米,y = −0.057米,靠近蓝色篮子的左前角,而不是在白色篮子内部。白色篮子的右边缘大约在 x = −0.12米,它似乎横跨在间隙上。我的推动是否把它从 x = −0.13米向左移动了?它似乎没有移动多少。要把它推入白色篮子,我需要将它大幅向左移动,朝向 x = −0.22米,并向后朝向 y = 0.02米。当前左夹爪位于 x = −0.144米,y = −0.155米,在大象的前方偏左。合适的推动方式应该从它的前方偏右开始,即 x = −0.06米,y = −0.12米,然后向左和向后移动,朝向 [−0.22, −0.02] m。我之前的推动减小了 y,即向前移动,这对于进入篮子来说是错误的方向。我需要先抬起闭合的夹爪,并绕到右前方而不发生碰撞。
小车的估计位置为 x = 0.258米,y = −0.052米,在红色篮子的前缘。右夹爪位于 x = 0.236米,y = −0.04米,是否在它后面?其 y 坐标大于小车的 y 坐标,所以它在后面而不是前面。要将小车向后推入红色篮子,我需要将夹爪放在前方 y = −0.10米处,降低它,并沿正 Y 方向移动。
重新抓取小车似乎比执行所有这些推动操作更好。我最初考虑的是它较早的 x 坐标 0.232米,但其当前估计中心是 [0.258, −0.052] m。我可以打开右夹爪,在该中心处以第六连杆 z = 0.95米对齐,闭合,然后抬起。然后我可以用标准的 45 度前倾来运输它。最初的白色小车放置在该倾角和第六连杆位置 [0.253, −0.033, 0.953] m 下成功了。它的释放是垂直的吗?之后我沿对角线撤回,但白色小车留在了里面。对于黄色小车,相同的倾斜方向和放置位姿 [0.25, −0.035, 0.95] m 会将物体向前延伸,朝向红色篮子的中心。
在向下抓取之后,我可以在高仰角处分两步旋转到标准的 45 度倾斜方向,并放置小车。小车在旋转过程中可能会滑动,因为它被夹持在前部和后部,就像之前发生的那样,尽管之前 0.347 的右开口命令保持稳定,而且小车很轻。水平归位方向是另一种可能性,但会更复杂。我估计从对角线向下方向到标准 45 度倾角的旋转为 1.098弧度,所以我需要一个中间四元数。我只有该四元数大约 0.68 的暂定值,应该在使用前计算并归一化它。我认为这种恢复可能需要大约 60 个控制步。
对于大象,我仍然可以尝试直接将它推入白色篮子,而不进行完整抓取。从其估计中心 [−0.115, −0.057] m 到期望中心 [−0.25, 0.02] m,得到推动方向 [−0.135, +0.077] m。一个闭合的、朝下的夹爪可以从 [−0.06, −0.12, 0.94] m 开始,在 x 和 y 方向上相对大象偏移 [+0.055, −0.063] m,然后朝 [−0.24, −0.015, 0.94] m 移动。起点在蓝色墙壁前方,似乎安全。在第六连杆 z = 0.94米时,指尖大约在 z = 0.782米,可以推抵主体。
然而,继续进入白色篮子可能会使夹爪在 y 经过 −0.059米时撞到其前壁。我可以在 y = −0.07米附近停下,希望大象通过惯性滑过边缘。另一个候选终点是 [−0.23, −0.075, 0.94] m。从起始横纵位置 [−0.06, −0.12] m 出发,这要求 [−0.17, +0.045] m 的位移,主要是向左,带有较小的向后分量。将夹爪保持在 y = −0.075米会使其留在墙壁前方,但大象会绕过右前角并掉进去吗?白色内部在墙壁后方,即 y > −0.059米,因此停留在 y = −0.06米附近的物体可能会留在边缘上。在将它向左推之后,我可能需要第二次向后轻推。
如果白色篮子可移动,将其向前移动是另一种可能性,但这可能会扰乱已经分拣在里面的物体。我不会采用这个替代方案。
稳定的象抓取,然后在其腕部方向固定的情况下运输,似乎更可取。在观测_000785 处成功的测试抬升显示它直立且被牢固夹持;它在后来的腕部旋转过程中滑动了。重复从观测_000770 到观测_000785 的抓取并保持四元数固定,可以允许在白色篮子前部内部 y = −0.035米处向下放置。
在观测_000831 和观测_000843 处的最后一次重新抓取未能通过其测试抬升,开口命令达到 0。我使用了 x = −0.13米,而不是成功的第六连杆 x 坐标 −0.109米。成功的主体抓取在 x = −0.109米,y = −0.086米,z = 0.95米,四元数略有偏差。后来的尝试使用了 x = −0.13米,y = −0.079米,z = 0.952米,这太靠左,夹住了头部而不是主体。
在滑动之后,大象的实际位置是否保持相似?在观测_000843 处,我估计它接近高机位相机像素 (260, 220),而在观测_000778 处可能是 x = 280 像素,所以它似乎稍微向左移动了。其当前估计的环境 x 坐标为 −0.115米,而在成功抓取期间可能是 −0.104米,相差 −0.011米。我瞄准了 −0.13米,我认为这太靠左了 0.026米。在 x = −0.10米或 −0.105米,y = −0.086米,z = 0.95米处,以标准向下方向进行抓取目标,应该能重新获得主体接触。当前 x = −0.115米的估计可能指的是身体;我将选择抓取 x = −0.105米。失败的尝试在 12 个闭合步后开口命令为 0.281,然后在抬升期间为 0,这与我在 x = −0.13米处夹住了耳朵或头部一致。
我将放弃大象推动,并在 [−0.105, −0.086, 0.95] m 处重新抓取。我将打开,缓慢闭合直到接触,并进行测试抬升。如果牢固,我将保持四元数固定,并将其朝下运输到白色篮子的前部内部,位置为 x = −0.25米,y = −0.035米,z = 0.95米。我认为这很可能成功,并且我可以在右臂恢复小车的同时准备它。
对于立即动作,左夹爪在第六连杆位置 [−0.144, −0.155, 0.961] m 处闭合。我将打开它,并移动到高预抓取位置 [−0.105, −0.086, 1.0] m,使用向下四元数 [0.5, −0.5, 0.5, 0.5]。右第六连杆位于 [0.236, −0.04, 0.99] m,几乎处于对角线向下方向,开口命令为 0。我将打开它,并在小车上方的 [0.258, −0.052, 1.02] m 处对齐,使用对角线向下四元数。我将为双臂使用六个控制步。之后,我计划在闭合前进行短的三步下降,朝向左右第六连杆高度 0.95米。剩余 232 个控制步,应该足够。
原文6,837 字
The elephant’s estimated position is x = −0.115 m, y = −0.057 m, near the blue basket’s front-left corner rather than inside the white basket. With the white basket’s right edge around x = −0.12 m, it seems to straddle the gap. Did my push move it left from x = −0.13 m? It does not seem to have moved much. To push it into the white basket, I would need to move it substantially left, toward x = −0.22 m, and backward toward y = 0.02 m. The current left gripper at x = −0.144 m, y = −0.155 m is in front and to the left of the elephant. A suitable pushing approach would instead start in front and to its right, at x = −0.06 m, y = −0.12 m, then move left and backward toward [−0.22, −0.02] m. My previous push decreased y, moving forward, which was the wrong direction for entering the basket. I would need to raise the closed gripper first and move it around to the front-right without colliding.
The car is estimated at x = 0.258 m, y = −0.052 m, at the red basket’s front rim. Is the right gripper at x = 0.236 m, y = −0.04 m behind it? Its y coordinate is greater than the car’s, so it is behind rather than in front. To push the car backward into the red basket, I would need to position the gripper in front at y = −0.10 m, lower it, and move along positive Y.
Regrasping the car seems better than making all those pushing maneuvers. I initially considered its earlier x coordinate of 0.232 m, but its current estimated center is [0.258, −0.052] m. I could open the right gripper, align at that center with link6 z = 0.95 m, close, and lift. Then I could transport it with the standard 45-degree forward tilt. The initial white-car placement succeeded with that tilt and link6 position [0.253, −0.033, 0.953] m. Was its release vertical? I withdrew diagonally afterward, but the white car stayed inside. For the yellow car, the same tilted orientation and placement pose [0.25, −0.035, 0.95] m would extend the object forward toward the red basket’s center.
After the downward grasp, I could rotate at high elevation to the standard 45-degree tilted orientation in two steps and place the car. The car could slip during rotation while held at its front and rear, as happened before, although the earlier right opening command of 0.347 stayed stable and the car is lightweight. A horizontal home orientation is another possibility, but it would be more complicated. I estimate the rotation from the diagonal downward orientation to the standard 45-degree tilt at 1.098 rad, so I would need a halfway SLERP quaternion. I have only a tentative value around 0.68 for that quaternion and should calculate and normalize it before using it. I think this recovery could take about 60 control steps.
For the elephant, I could still try pushing it directly into the white basket without a full grasp. From its estimated center [−0.115, −0.057] m, the desired center [−0.25, 0.02] m gives a push direction of [−0.135, +0.077] m. A closed, downward-facing gripper could start at [−0.06, −0.12, 0.94] m, offset from the elephant by [+0.055, −0.063] m in x and y, then move toward [−0.24, −0.015, 0.94] m. The start is in front of the blue wall and seems safe. With link6 z = 0.94 m, the fingertips would be around z = 0.782 m and could push against the body.
However, continuing into the white basket could make the gripper hit its front wall as y passes −0.059 m. I could stop around y = −0.07 m and hope the elephant slides over the rim through inertia. Another candidate endpoint is [−0.23, −0.075, 0.94] m. From the starting xy position [−0.06, −0.12] m, that requests [−0.17, +0.045] m of displacement, mostly left with a smaller backward component. Keeping the gripper at y = −0.075 m would leave it in front of the wall, but would the elephant move around the front-right corner and drop inside? The white interior is behind the wall at y > −0.059 m, so an object remaining near y = −0.06 m could stay on the rim. I might need a second backward nudge after pushing it left.
Moving the white basket forward is another possibility if it is movable, but that could disturb the objects already sorted inside it. I will not pursue that alternative.
A stable elephant grasp followed by transport with its wrist orientation fixed seems preferable. The successful test lift at obs_000785 showed it upright and securely held; it slipped during the later wrist rotation. Repeating the grasp from obs_000770 through obs_000785 and keeping the quaternion fixed could allow a downward placement in the white basket’s front interior at y = −0.035 m.
The last regrasp at obs_000831 and obs_000843 failed its test lift, with the opening command reaching 0. I used x = −0.13 m rather than the successful link6 x coordinate of −0.109 m. The successful torso grasp was at x = −0.109 m, y = −0.086 m, z = 0.95 m, with a slightly deviated quaternion. The later attempt used x = −0.13 m, y = −0.079 m, z = 0.952 m, which was too far left and pinched the head rather than the torso.
Has the elephant’s actual position remained similar after the slips? At obs_000843, I estimated it near high-camera pixel (260, 220), compared with perhaps x = 280 pixels at obs_000778, so it seems to have moved slightly left. Its current estimated environment x coordinate is −0.115 m, compared with perhaps −0.104 m during the successful grasp, a difference of −0.011 m. I targeted −0.13 m, which I regard as 0.026 m too far left. A grasp target at x = −0.10 m or −0.105 m, y = −0.086 m, z = 0.95 m, with the standard downward orientation, should regain the torso contact. The current x = −0.115 m estimate may refer to the body; I’ll choose grasp x = −0.105 m. The failed attempt had an opening command of 0.281 after 12 closure steps and then 0 during the lift, consistent with my having pinched the ear or head at x = −0.13 m.
I will abandon the elephant push and regrasp at [−0.105, −0.086, 0.95] m. I’ll open, close slowly until contact, and test-lift. If secure, I’ll keep the quaternion fixed and transport it downward-facing to the white basket’s front interior at x = −0.25 m, y = −0.035 m, z = 0.95 m. I think this is likely to succeed, and I can prepare it while the right arm recovers the car.
For the immediate action, the left gripper is closed at link6 position [−0.144, −0.155, 0.961] m. I’ll open it and move to the high pre-grasp position [−0.105, −0.086, 1.0] m with downward wxyz quaternion [0.5, −0.5, 0.5, 0.5]. The right link6 is at [0.236, −0.04, 0.99] m, nearly in the diagonal downward orientation, with opening command 0. I’ll open it and align above the car at [0.258, −0.052, 1.02] m with the diagonal downward quaternion. I’ll use six control steps for both arms. Afterward, I plan short three-step descents toward left and right link6 heights of 0.95 m before closure. There are 232 control steps remaining, which should be enough.