测得的左关节角为 = 2.195弧度和 = 2.32弧度。这些角度是否接近约 2.4弧度的最大值?也许关节 3 的限制是 2.356弧度(135°),导致逆运动学停滞。我不知道的关节限位。机械臂似乎处于一种别扭的折叠构型, 也可能接近限位。初始关节角全为零;在向桌子移动的过程中是如何变化的?我可以使用测得的关节角,而无需访问仿真器内部。
够到蓝色篮子可能需要不同的肘部构型。我可以大幅改变姿态,或降低第六连杆以放松机械臂。从前方以 z = 0.95米而非 1.07米接近可能有帮助:较高的位姿可能要求较大的和角度,因为机械臂在其基座附近折叠。基座是否足够低,使得降低并向前移动会打开机械臂?在约为 +2.2弧度、 约为 +2.3弧度的情况下,将第六连杆降低约 0.1米至 z = 0.96米同时向内移动,可能会减小。
一个可能的目标是 (-0.12, -0.04, 0.98) m,采用自上而下的姿态,其闭合方向在世界横纵中为 90°。从当前位姿看,规划的位移为 (0.033, 0.027, -0.089) m,距离约 0.099米。它可能通过预检查。然而,所持紫色条块的底部大约为 0.98 − 0.16 = 0.82米,相对于 z = 0.83米附近的边缘来说是临界值。它仍会位于白色和蓝色篮子之间的间隙上方,而不是在篮子内部。尽管 z = 0.82米高于 z = 0.75米的桌面,但越过 x = -0.15米附近的白色篮子边缘可能导致接触。
当前第六连杆位置为 (-0.153, -0.067, 1.069) m,靠近白色和蓝色篮子的前缘,我估计条块底部大约在 z = 0.91米。将第六连杆降低到 z = 0.98米会使其底部大约在 z = 0.82米,可能低于 z = 0.84米的篮子边缘。提议的位置 x = -0.12米、y = -0.04米可能使其位于蓝色篮子的左前边缘上方,因此在那里下降可能很危险。更好的做法可能是在降低的同时沿负 Y 方向向后移动远离篮子,然后在较低高度向内移动。在第六连杆 z = 0.95米时,紫色条块的底部大约在 z = 0.79米,高于桌面但低于边缘;我需要再次将其抬高才能越过进入蓝色篮子。也许机械臂只能在较低的第六连杆高度够到蓝色篮子,但在依赖 z = 0.98米之前,我需要更好的边缘高度估计。边缘是否可能在 z = 0.79米而非 0.83米?
在高机位视图中,篮子的后缘大约在图像 y = 140 处,估计世界后边缘 y 坐标约为 0.13米。假设桌面 z = 0.75米,这可能给出约 z = 0.85米的边缘高度。我可以用腕部视图对边缘点进行三角测量,但在当前左腕视图中我没有蓝色篮子的精确对应点。右腕和高机位视图中对应的红色篮子角点也可能有帮助,尽管估计会是近似的。
我也可以改用高机位相机投影,结合估计的篮子尺寸。初始白色篮子的后上角出现在 (145, 141) 和 (236, 140),其前上角在 (129, 177) 和 (253, 176)。其世界 X 边缘可能为 -0.38米和 -0.22米,后 y = 0.12米,前 y = -0.02米。或者,对称性表明中心在 x = -0.3米,宽度约 0.18米。如果我假设某个角点的世界 X 坐标,图像 u 提供投影分母,使我能够求解 y 和 z。
对于蓝色篮子,假设 X 边缘为 -0.09米和 +0.09米,后角大约在像素 (269, 141) 和 (369, 141)。使用 x = -0.09米和 u = 268,分母约为 288 × (-0.09) / (-52) = 0.4985。则 1.3378 + 0.5y − 0.866z = 0.4985,得到 0.5y − 0.866z = -0.8393。在 v = 141 时,(141 − 240) / 288 = -0.34375,所以 -0.866y − 0.5z + 0.2989 = -0.1713,即 -0.866y − 0.5z = -0.4702。第一个方程给出 y = -1.6786 + 1.732z。代入第二个方程得到 1.4537 − 1.5z − 0.5z = -0.4702,解得 z = 0.96195米和 y = -0.0127米。
z = 0.96米附近的边缘是否太高?在 z = 0.85米和 y = 0.1米时,分母将约为 0.651 而非 0.4985。宽度 0.18米和 z 约 0.9米可能合理地给出接近 0.5 的分母,但 z = 0.96米的边缘将意味着篮子高度约为 z = 0.75米的桌面上方 0.2米。篮子大致跨越图像 y = 140 到 210,这可能只对应约 0.1米的物理高度。桌子是否实际上在 z = 0.8米?三角测量得到的汽车大约在 z = 0.77米,表明桌面 z 接近 0.75米,而 z = 0.9米的篮子边缘将意味着高度约 0.15米。
假设的 X 边缘 0.09米可能根本就是错的。蓝色篮子宽度约 100 像素、分母约 0.7 对应物理宽度约 0.243米;实际宽度可能约为 0.22米。使用 x = -0.1米得到分母 0.554,估计 z = 0.924米、y = -0.003米,仍然偏高。使用 x = -0.12米得到分母 0.665,估计 z = 0.832米、y = 0.043米。因此半宽 0.12米似乎是合理的。白色篮子的后角在图像 x = 145 和 236 处,中心接近 190,半跨度约 45 像素;分母 0.7 给出半宽约 0.109米,也表明约为 0.12米。我将继续估计边缘在 z = 0.84米附近。
在第六连杆位于 z = 0.98米时,紫色条块的底部大约为 0.98 − 0.162 = 0.818米,低于该边缘,因此越过可能导致碰撞。第六连杆应至少位于 z = 1.01米以保持底部高于边缘;当前接近 z = 1.07米的高度是合适的。旋转所持物体能否抬高其底部?其相对于第六连杆的估计偏移固定约为 0.151米,因此我不能依赖这一点。
也许伸入中部较高位置需要 45° 闭合方向而非 90°,或其他偏航角如 135°。当关节 3 接近其最大值时,180° 或 0° 的闭合方向能否改变肘部构型?改变自上而下的闭合角度可能主要影响和,而非肘部关节和。后者大致独立于姿态决定位置,因此位置可达极限似乎很可能。位于 (0, 0.05, 1.0) m 的蓝色篮子可能只是离左基座太远。巧克力必须放入蓝色篮子,但如果左臂够不到,右臂可以放置它。
交接可以让右臂将紫色条块放入蓝色篮子,左臂将黄色汽车放入白色篮子。也许这个交叉分拣任务本就打算使用传递,左臂处理白色篮子,右臂处理蓝色和红色篮子。左侧的棕色巧克力可能也需要传递,除非右臂能直接拾取它,或者左臂能在较低高度够到蓝色篮子的前缘。将所有中间篮子的放置分配给右臂似乎更简单。x = 0.1米附近的公交车同样可能需要从右到左的传递才能放入白色篮子。
将篮子移近是另一种选择。技能说明指出,如果容器太远,可以将其移到附近。左臂可以将蓝色篮子向左前方移动,放置紫色条块,然后将篮子放回。指令并未明确要求篮子保持在初始位置,但将它们留在原处似乎更自然。移动容器会带来风险,而且当交接可以让右臂将条块放入蓝色篮子时,这似乎没有必要。
我将把白色篮子的放置分配给左臂,蓝色和红色篮子的放置分配给右臂。左臂可能能够够到位于 (-0.3, 0.05, 1.0) m 的白色篮子,而位于 (0, 0.05) m 的蓝色篮子在当前较高位姿下可能刚好超出其可达范围。我可以通过在前方中心附近、大约 x = 0米、y = -0.25米处进行有支撑的放置来交换所持物体,那里两条机械臂都应该能够够到。
右臂可以将黄色汽车放在那里,让左臂拾取并放入白色篮子。左臂可以将紫色条块带到 (-0.05, -0.25, 0.95) m 附近的交接位姿,供右臂稍后拾取。沿负 Y 方向向前移动并降低应该比从 y = -0.067米、z = 1.07米处向内推进更容易。紫色条块在桌面上的放置位置 x = -0.04米、y = -0.26米可能对右臂从 x = 0米、y = -0.26米处可达。
与其顺序使用同一位置,我可以将汽车放在 (0.05, -0.27) m,将紫色条块放在 (-0.08, -0.27) m。两个物体将相隔 0.13米。两条机械臂可以同时放置它们,然后拾取对方的物体,同时避免碰撞。可能的抬高接近目标为:左臂使用 45° 或 90° 闭合方向的 (-0.12, -0.22, 1.0) m,右臂使用其 45° 方向的 (0.12, -0.22, 1.0) m。然后我可以将紫色条块放在 (-0.09, -0.25) m,将汽车放在 (0.09, -0.25) m。左臂可能能够拾取 x = +0.09米、y = -0.25米处的汽车,右臂可能能够拾取 x = -0.09米、y = -0.25米处的紫色条块。之后,左臂将汽车运到 x = -0.3米、y = 0.05米的白色篮子,右臂将条块运到 x = 0米、y = 0.05米的蓝色篮子。
对于剩余物体,(0.095, -0.095) m 附近的公交车属于白色篮子,可以在木制玩具之后由右臂传递。(-0.096, -0.089) m 附近的棕色巧克力属于蓝色篮子;右臂应该能够直接拾取它,而不需要左臂传递。(-0.096, -0.214) m 附近的小猪属于红色篮子,也应该对右臂直接可达。这样左臂有三次放置——紫色条块交接以及黄色汽车和公交车放入白色篮子——右臂有五次——黄色汽车交接、棕色巧克力放入蓝色篮子、马和小猪放入红色篮子以及公交车交接——总共八次放置。
将左臂的接近路径改为从 y = -0.2米过来,能否允许直接将紫色条块放入蓝色篮子?最终位置 x = 0米、y = 0.05米仍然很远。我可以在承诺交接之前测试一次较低的直接接近,而且还剩 979 步,但桌面交接似乎很稳健。我将切换到该计划。
首先我需要将左臂从失败的内向路径撤回。其当前第六连杆位置为 (-0.153, -0.067, 1.069) m。目标 (-0.10, -0.25, 1.02) m 采用 90° 闭合方向,规划位移为 (0.053, -0.183, -0.049) m,距离约 0.196米。向前方移动应该能通过预检查,尽管这仍不确定。一个可能的右臂目标是 (0.10, -0.24, 1.02) m,位移为 (-0.119, -0.191, -0.033) m,估计距离 0.227米,使用其当前接近 45° 的闭合方向。如果失败,我可以使用更小的步长。两条机械臂在 X 方向将相隔 0.2米。
在第六连杆 z = 1.02米时,我估计两个所持物体的底部都在 z = 0.86米左右。左臂从 (-0.153, -0.067) m 到 (-0.10, -0.25) m 的路径经过 (-0.096, -0.214) m 附近的小猪。小猪顶部接近 z = 0.80米,紫色条块将有约 0.06米的间隙。右臂从 (0.219, -0.049) m 到 (0.10, -0.24) m 的路径经过 (0.047, -0.196) m 附近的马和 (0.095, -0.095) m 附近的公交车。汽车估计底部在 z = 0.86米,高于公交车顶部接近 z = 0.79米的位置。
对于此动作,我将使用左目标 (-0.1, -0.25, 1.02) m 和右目标 (0.1, -0.25, 1.02) m。右臂距离最初可能看起来约为 0.241米,但位移 (-0.119, -0.201, -0.033) m 给出约 0.236米;左臂距离约为 0.196米。两者都在 0.25米限制内。我将保持左臂 90° 闭合方向和右臂当前接近 45° 的方向,保持两个夹爪闭合,并尝试 10 个控制步。右臂的预检查仍可能失败,执行可能只取得部分进展,因此我将在之后检查反馈和测得的位姿。
原文11,751 字
The measured left joint angles are q2 = 2.195 rad and q3 = 2.32 rad. Could these be near a maximum of about 2.4 rad? Perhaps joint 3 has a limit of 2.356 rad (135°), causing the IK stall. I do not know the ARX joint limits. The arm appears to be in an awkward folded configuration, and q2 may also be near a limit. The initial joint angles were all zero; how has q2 changed during the move toward the table? I can use the measured joints without accessing simulator internals.
Reaching the blue basket may require a different elbow configuration. I could change the orientation substantially or lower link6 to relax the arm. Approaching from the front at z = 0.95 m rather than 1.07 m might help: the higher pose may require large q2 and q3 angles as the arm folds near its base. Are the bases low enough that lowering and moving forward would open the arm? With q2 around +2.2 rad and q3 around +2.3 rad, lowering link6 by about 0.1 m to z = 0.96 m while moving inward might reduce q3.
One possible target is (-0.12, -0.04, 0.98) m with the top-down orientation whose closing direction is 90° in world XY. From the current pose, the planned displacement is (0.033, 0.027, -0.089) m, with a distance of about 0.099 m. It might pass the precheck. However, the held purple bar's bottom would be approximately 0.98 − 0.16 = 0.82 m, borderline relative to a rim near z = 0.83 m. It would still be over the gap between the white and blue baskets rather than inside a basket. Although z = 0.82 m clears the table at z = 0.75 m, crossing the white basket edge near x = -0.15 m could cause contact.
The current link6 position is (-0.153, -0.067, 1.069) m, near the front edges of the white and blue baskets, and I estimate the bar's bottom at about z = 0.91 m. Lowering link6 to z = 0.98 m would put its bottom around z = 0.82 m, possibly below basket rims at z = 0.84 m. The proposed position at x = -0.12 m, y = -0.04 m might put it over the blue basket's front-left rim, so descending there could be dangerous. It may be better to move backward in negative Y away from the baskets while lowering, then move inward at the lower height. At link6 z = 0.95 m, the purple bar's bottom would be around z = 0.79 m, above the table but below the rim; I would need to raise it again to cross into the blue basket. Perhaps the arm can reach blue only at a lower link6 height, but I need a better rim-height estimate before relying on z = 0.98 m. Could the rim be at z = 0.79 m rather than 0.83 m?
In the high view, the basket's back rim is near image y = 140, with an estimated world back-edge y coordinate around 0.13 m. Assuming table z = 0.75 m, that might give a rim height around z = 0.85 m. I could triangulate a rim point with a wrist view, but I do not have an exact correspondence for the blue basket in the current left-wrist view. A corresponding red-basket corner in the right-wrist and high views might also help, although the estimate would be approximate.
I could instead use the high-camera projection with estimated basket dimensions. The initial white basket's back top corners appear at (145, 141) and (236, 140), and its front top corners at (129, 177) and (253, 176). Its world X edges might be -0.38 m and -0.22 m, with back y = 0.12 m and front y = -0.02 m. Alternatively, symmetry suggests a center at x = -0.3 m and a width around 0.18 m. If I assume a corner's world X coordinate, image u supplies the projection denominator, allowing me to solve for y and z.
For the blue basket, suppose the X edges are -0.09 m and +0.09 m, with back corners around pixels (269, 141) and (369, 141). Using x = -0.09 m and u = 268, the denominator is approximately 288 × (-0.09) / (-52) = 0.4985. Then 1.3378 + 0.5y − 0.866z = 0.4985, giving 0.5y − 0.866z = -0.8393. At v = 141, (141 − 240) / 288 = -0.34375, so -0.866y − 0.5z + 0.2989 = -0.1713, or -0.866y − 0.5z = -0.4702. The first equation gives y = -1.6786 + 1.732z. Substituting into the second gives 1.4537 − 1.5z − 0.5z = -0.4702, yielding z = 0.96195 m and y = -0.0127 m.
Is a rim near z = 0.96 m too high? At z = 0.85 m and y = 0.1 m, the denominator would be about 0.651 instead of 0.4985. A width of 0.18 m and z around 0.9 m could plausibly give a denominator near 0.5, but a rim at z = 0.96 m would imply a basket height around 0.2 m above a table at z = 0.75 m. The baskets span roughly image y = 140 to 210, which might correspond to only about 0.1 m of physical height. Could the table actually be at z = 0.8 m? The triangulated car was around z = 0.77 m, suggesting table z near 0.75 m, while a basket rim at z = 0.9 m would imply a height around 0.15 m.
The assumed X edge of 0.09 m may simply be wrong. A blue-basket width of about 100 pixels with a denominator around 0.7 corresponds to a physical width around 0.243 m; the actual width might be about 0.22 m. Using x = -0.1 m gives a denominator of 0.554 and estimates z = 0.924 m, y = -0.003 m, still high. Using x = -0.12 m gives a denominator of 0.665 and estimates z = 0.832 m, y = 0.043 m. A half-width of 0.12 m therefore seems plausible. The white basket's back corners at image x = 145 and 236 have a center near 190 and a half-span around 45 pixels; a denominator of 0.7 gives a half-width around 0.109 m, also suggesting about 0.12 m. I will continue to estimate the rim near z = 0.84 m.
With link6 at z = 0.98 m, the purple bar's bottom would be approximately 0.98 − 0.162 = 0.818 m, below that rim, so crossing could cause a collision. Link6 should be at least z = 1.01 m to keep the bottom above the rim; the current height near z = 1.07 m is appropriate. Could rotating the held object raise its bottom? Its estimated offset from link6 is fixed at about 0.151 m, so I cannot rely on that.
Perhaps reaching high into the middle requires the 45° closing orientation rather than 90°, or another yaw such as 135°. Could a 180° or 0° closing orientation change the elbow configuration when joint 3 is near its maximum? Changing the top-down closing angle may mostly affect q1 and q6 rather than elbow joints q2 and q3. The latter roughly determine position independently of orientation, so a positional reach limit seems likely. The blue basket at (0, 0.05, 1.0) m may simply be too far from the left base. The chocolate must go into blue, but the right arm could place it if the left cannot reach.
A handoff would let the right arm put the purple bar in blue and the left arm put the yellow car in white. Perhaps this cross-sorting task is intended to use transfers, with the left arm handling white and the right arm handling blue and red. The brown chocolate on the left might also need a transfer, unless the right arm can pick it directly or the left can reach the blue basket's front edge at a lower height. It seems simpler to assign all middle-basket placements to the right arm. The bus near x = 0.1 m may likewise need a right-to-left transfer for placement in white.
Moving the baskets closer is another option. The skill says to move a container nearby if it is too far away. The left arm could move the blue basket left and forward, place the purple bar, then return the basket. The instruction does not explicitly require the baskets to remain in their initial positions, but leaving them there seems natural. Moving a container introduces risk and seems unnecessary when a handoff can let the right arm place the bar in blue.
I will assign white-basket placements to the left arm and blue- and red-basket placements to the right. The left arm likely can reach white at (-0.3, 0.05, 1.0) m, while blue at (0, 0.05) m may be just beyond its reach at the current high pose. I can exchange the held objects through supported placements near the front center, around x = 0 m, y = -0.25 m, where both arms should be able to reach.
The right arm could place the yellow car there for the left to pick up and put in white. The left arm could bring the purple bar to a handoff pose near (-0.05, -0.25, 0.95) m for a later right-arm pickup. Moving toward the front in negative Y and lowering should be easier than advancing inward from y = -0.067 m at z = 1.07 m. A purple-bar table placement at x = -0.04 m, y = -0.26 m could be accessible to the right arm from x = 0 m, y = -0.26 m.
Instead of using the same spot sequentially, I could put the car at (0.05, -0.27) m and the purple bar at (-0.08, -0.27) m. The two objects would be separated by 0.13 m. The arms could place them simultaneously, then pick up each other's objects, while avoiding collisions. Possible elevated approach targets are (-0.12, -0.22, 1.0) m for the left arm with either the 45° or 90° closing orientation, and (0.12, -0.22, 1.0) m for the right with its 45° orientation. I could then place the purple bar at (-0.09, -0.25) m and the car at (0.09, -0.25) m. The left arm likely can pick the car at x = +0.09 m, y = -0.25 m, and the right likely can pick the purple bar at x = -0.09 m, y = -0.25 m. Afterward, the left would carry the car to white at x = -0.3 m, y = 0.05 m, and the right would carry the bar to blue at x = 0 m, y = 0.05 m.
For the remaining objects, the bus near (0.095, -0.095) m belongs in white and can be transferred by the right arm after the wooden toys. The brown chocolate near (-0.096, -0.089) m belongs in blue; the right arm should be able to pick it directly rather than requiring the left to transfer it. The pig near (-0.096, -0.214) m belongs in red and should also be directly accessible to the right arm. This gives the left three placements—the purple-bar handoff and the yellow car and bus into white—and the right five—the yellow-car handoff, brown chocolate into blue, horse and pig into red, and bus handoff—for eight placements total.
Could changing the left approach to come from y = -0.2 m allow direct purple-bar placement in blue? The final position at x = 0 m, y = 0.05 m would still be far away. I could test a lower direct approach before committing to a handoff, and there are still 979 steps remaining, but the table handoff seems robust. I will switch to that plan.
First I need to retract the left arm from the failed inward path. Its current link6 position is (-0.153, -0.067, 1.069) m. A target of (-0.10, -0.25, 1.02) m with the 90° closing orientation has a planned displacement of (0.053, -0.183, -0.049) m and distance about 0.196 m. Moving toward the front should pass the precheck, although that remains uncertain. A possible right target is (0.10, -0.24, 1.02) m, with displacement (-0.119, -0.191, -0.033) m and estimated distance 0.227 m, using its current near-45° closing orientation. If that fails, I can use a smaller step. The arms would be separated by 0.2 m in X.
At link6 z = 1.02 m, I estimate both held objects' bottoms around z = 0.86 m. The left path from (-0.153, -0.067) m toward (-0.10, -0.25) m passes near the pig at (-0.096, -0.214) m. With the pig's top near z = 0.80 m, the purple bar would have about 0.06 m clearance. The right path from (0.219, -0.049) m toward (0.10, -0.24) m passes near the horse at (0.047, -0.196) m and the bus at (0.095, -0.095) m. The car's estimated bottom at z = 0.86 m clears the bus top near z = 0.79 m.
For this action, I will use the left target (-0.1, -0.25, 1.02) m and right target (0.1, -0.25, 1.02) m. The right distance might initially look like about 0.241 m, but the displacement (-0.119, -0.201, -0.033) m gives about 0.236 m; the left distance is about 0.196 m. Both are within the 0.25 m limit. I will retain the left 90° closing orientation and the right's current near-45° orientation, keep both grippers closed, and try 10 control steps. The precheck could still fail for the right arm, and execution may only make partial progress, so I will inspect the feedback and measured poses afterward.