新的三角测量估计胡椒2在 z = 0.7976米处,胡椒1在 z = 0.7927米处。这些可能代表它们的中心,尽管像素匹配可能有偏差。我仍然估计桌子在 z = 0.75米处。 的估计值 z ≈ 0.776米与假设的扁平条高度 0.758米相比似乎太高了。头部和腕部像素可能不完全对应,也许是因为我匹配了标签的不同部分。红胡椒的估计值 z ≈ 0.778米看起来合理。
篮子前底部的估计值大约在 z = 0.79米和 y = -0.02米处,角点大约在中心两侧各 0.14米处,中心位于 x = ±0.28米。这是否意味着篮子高度仅比假设的桌子高 0.04米?我不确定我匹配了哪些特征。图像 v = 198 处的白色篮子特征可能是上前边缘的一部分,而不是底部。篮子可能高约 0.1米。其顶部边缘出现在图像 v = 139 附近,前底部在 v = 211 附近。头部相机角点 (126, 198) 和 (255, 198) 匹配到腕部像素 (174, 294) 和 (523, 294),实际上可能是前边缘特征。对于这些匹配,z = 0.79米附近的高度似乎是合理的,但它们的身份仍不确定。
我应该检查候选篮子中心在 (-0.28, 0.07, 0.80) m 处的投影。其头部图像 u 坐标会在 185 像素左右吗?使用光学深度 0.035 - 0.6928 + 1.337 ≈ 0.68米得到水平偏移 -0.28 × 288 / 0.68 ≈ -118.6 像素,即 u ≈ 201.4 像素。可见白色开口中心大约在 u = 194 处,所以其 x 坐标可能是 -0.3米。然而,三角测量的角点 x 坐标 -0.418 和 -0.142米给出的中心为 -0.280米。我将使用白色篮子的 x = -0.28米,蓝色篮子的 x = 0,红色篮子的 x = 0.28米。它们的宽度似乎约为 0.28米,前部 y 坐标约为 -0.02米,后部可能为 y = 0.18米,中心大约在 y = 0.08米。白色前侧底部可能在 z = 0.75米处,其上边缘在 z = 0.80米处。篮子会不会反而是 0.12米高?这仍不确定。
对于放置,我可以在篮子的中心 x 坐标和 y = 0.07米处释放,也许第六连杆在 z = 0.95米处。夹爪向下时,所持物体在第六连杆下方延伸约 0.13米。在第六连杆 z = 1.02米处,物体大约在 z = 0.87–0.90米处,高于估计的 z = 0.84米的边缘。然后它可能在释放后下落约 0.05米。这似乎是安全的。桌子很可能正好在 z = 0.75米处,尽管这仍然是一个假设。
一个可能的第一对是上部紫色条大约在 (-0.142, -0.122, 0.758) m 处,木质玩具2在 (0.129, -0.176, 0.785) m 处。机械臂起始于 x ≈ -0.3 和 0.3米处,因此它们可以在预抓取第六连杆高度 z = 1.03米处同时接近两个物体。然而,木质玩具形状不规则,紫色条需要不同的抓取方向。
上部紫色条似乎主要与图像 v 对齐,可能从垂直方向倾斜约 -15°。其顶部可能在图像 x = 251 附近,而中心大约在 (253, 252) 处。其在环境中的长轴大致沿 y 方向,带有一些 x 分量。使用第一个向下四元数时,局部 Y 是全局 -Y,因此闭合将跨越条的长维度。第二个向下四元数,按顺序为 [0.5, -0.5, 0.5, 0.5],沿全局 +X 闭合,应能横跨其短宽度抓取。我估计条的 y 长度约为 0.09米,x 宽度为 0.035米。夹爪最大开口约为 0.08米吗?沿 x 方向闭合看起来仍然更可取。
木质玩具2可能是一只松鼠,长约 0.08米,宽约 0.05米。我可以根据接触几何沿 x 或 y 方向闭合来抓取其身体。它大致跨越图像 x = 368–400 和 v = 255–286,呈对角线。较厚的身体可能在 (388, 268) 附近,头部在 (377, 275) 附近,但这些识别是暂定的。更近的裁剪可能有所帮助。红胡椒是更容易的第一选择,所以我将从左臂的上部紫色条和右臂的红胡椒开始。它们分别去往白色和蓝色篮子,其中心相距约 0.28米。
对于后续的配对,我可以使用下部紫色条与木质玩具2,然后是左侧胡椒与木质玩具1。我最初考虑将木质玩具1交给左臂并放入右篮,同时右臂留在原位或在右侧工作,但这会在右篮附近产生潜在冲突。更好的分配是左臂处理两个左侧胡椒和两个紫色条,右臂处理、红胡椒和两个木质玩具。这样每只臂处理四个物体。
暂定计划是:上部紫色条到白色,红胡椒到蓝色;下部紫色条到白色,木质玩具2到红色;然后一个左侧胡椒到蓝色,木质玩具1到红色。如果我在第三对中使用最左侧的胡椒,左臂从 x ≈ -0.39米移动到 x = 0,而右臂携带木质玩具1从 x ≈ -0.065米移动到 x = 0.28米。它们的抓取位置是分开的,但在运输过程中它们可能同时接近 x = 0。我能否让它们保持在高处但 y 坐标不同,或者在右臂持物时错开它们的移动?改为配对胡椒2可能更好:其起始 x ≈ -0.276米和木质玩具1的 x ≈ -0.065米相距约 0.2米,它们朝向 x = 0 和 0.28米的路径会发散。
这样就剩下最左侧的胡椒和。同时运输到蓝色和白色篮子可能会使机械臂在 x = -0.1米附近交叉,所以我应该避免这种情况。右臂可以转移,同时左臂保持在其左侧初始位姿附近,大约在 x 和 y 的 (-0.3, -0.35) m 处。右夹爪的目的地接近 (-0.28, 0.08) m,与该位姿在 y 方向上有约 0.4米的间隔。另一个可能的左臂保持位姿是 (-0.45, -0.2, 1.0) m。让左臂在 x ≈ 0.39米处拾取需要约 0.7米的伸展距离,可能不可行。如果左臂避开放置区域,右臂转移到白色应该更简单。
我仍然可以一起接近并提起最后两个物体:最左侧的胡椒在 x ≈ -0.39米处, 在 x ≈ 0.39米处。然后我将左侧胡椒移动到蓝色篮子,第六连杆在 (0, 0.08, 1.05) m 处,同时右臂保持在 (0.39, -0.13, 1.05) m 附近。释放胡椒后,左臂可以向左和向上收回,也许朝向 x = -0.4米和 y = -0.25米。然后右臂可以将携带到白色篮子,x = -0.28米和 y = 0.08米,同时左臂保持在最左侧,大约在 y = -0.3米处。这似乎是安全的。或者,我可以先完成,将最后一个胡椒留作单独的转移。
有 1100 个控制步骤可用,所以我不需要最大化同时运动。一次处理一个物体大约需要两个接近动作、一个下降、一个闭合、一个测试提升、一个或两个运输动作、一个释放和一个撤回:每个物体大约八或九个动作,总共大约 70 次调用。以独立配对工作可以将此减少到大约 40 次调用。我将在安全的地方使用配对动作,并验证每个抓取和放置。第一对将是上部紫色条和红胡椒。
在从初始位姿移动到向下抓取之前,我需要中间方向。对于初始四元数和 [0.5, 0.5, -0.5, 0.5],点积为 0.3535 + 0.3535 = 0.7071,给出方向距离 1.5708弧度。这超过了 1.0弧度的目标限制,所以我不能直接命令完整旋转。 四元数将给出约 0.785弧度的旋转。
左臂从大约 (-0.3, -0.352, 0.922) m 移动到 (-0.142, -0.122, 1.03) m 约为 0.295米,超过了 0.25米的位置限制。右臂移动到 (0.169, -0.098, 1.03) m 约为 0.296米。我可以先将每只臂移动到其物体的大致 x 坐标,y = -0.25米和 z = 1.03米,使用中间方向。左侧位移为平方根(0.158² + 0.102² + 0.108²) ≈ 0.217米,右侧位移为平方根((-0.131)² + 0.102² + 0.108²) ≈ 0.198米。每个中间旋转约为 0.785弧度。随后移动到最终物体上方位姿约为 0.13米,另加 0.785弧度的旋转。鉴于每步旋转增量为 0.1弧度,每次移动八个控制步骤可能就足够了。中间位姿是否通过预检查尚不清楚。我将使用科学计算库计算两个向下方向的四元数。
原文8,544 字
The new triangulation estimates pepper 2 at z = 0.7976 m and pepper 1 at z = 0.7927 m. These may represent their centers, although the pixel matches could be off. I still estimate the table at z = 0.75 m. The Snickers estimate of z ≈ 0.776 m seems too high compared with the assumed flat-bar height of 0.758 m. The head and wrist pixels may not correspond exactly, perhaps because I matched different parts of the label. The red pepper estimate of z ≈ 0.778 m looks reasonable.
The basket front-bottom estimates are around z = 0.79 m and y = -0.02 m, with corners approximately 0.14 m to either side of centers at x = ±0.28 m. Does that imply a basket height of only 0.04 m above the assumed table? I am not sure which features I matched. The white-basket feature at image v = 198 may be part of the upper front edge rather than the bottom. The basket might be about 0.1 m tall. Its top rim appears near image v = 139 and its front bottom near v = 211. The head-camera corners at (126, 198) and (255, 198), matched to wrist pixels (174, 294) and (523, 294), may actually be front-rim features. A height near z = 0.79 m seems plausible for those matches, but their identity remains uncertain.
I should check the projection of a candidate basket center at (-0.28, 0.07, 0.80) m. Would its head-image u coordinate be around 185 pixels? Using an optical depth of 0.035 - 0.6928 + 1.337 ≈ 0.68 m gives a horizontal offset of -0.28 × 288 / 0.68 ≈ -118.6 pixels, or u ≈ 201.4 pixels. The visible white opening center is around u = 194, so perhaps its x coordinate is -0.3 m. However, the triangulated corner x coordinates of -0.418 and -0.142 m give a center of -0.280 m. I will use x = -0.28 m for the white basket, x = 0 for the blue basket, and x = 0.28 m for the red basket. Their width appears to be about 0.28 m, their front y coordinate about -0.02 m, their back perhaps y = 0.18 m, and their center around y = 0.08 m. The white front-side bottom may be at z = 0.75 m and its upper edge at z = 0.80 m. Could the baskets instead be 0.12 m tall? That remains uncertain.
For placement, I could release at the basket’s center x coordinate and y = 0.07 m, perhaps with link6 at z = 0.95 m. With a downward gripper, the held object extends about 0.13 m below link6. At link6 z = 1.02 m, the object would be around z = 0.87–0.90 m, above an estimated rim at z = 0.84 m. It could then fall about 0.05 m after release. That seems safe. The table is likely exactly at z = 0.75 m, though this is still an assumption.
One possible first pair is the upper purple bar at approximately (-0.142, -0.122, 0.758) m and wooden toy 2 at (0.129, -0.176, 0.785) m. The arms start at x ≈ -0.3 and 0.3 m, so they could approach both objects simultaneously at a pregrasp link6 height of z = 1.03 m. However, the wooden toy is irregular, and the purple bar needs a different grasp orientation.
The upper purple bar appears mostly aligned with image v, perhaps tilted about -15° from vertical. Its top may be around image x = 251, while its center is around (253, 252). Its long axis in the environment is roughly along y with some x component. With the first downward quaternion, local Y is global -Y, so closing would span the bar’s long dimension. The second downward quaternion, [0.5, -0.5, 0.5, 0.5] in wxyz order, closes along global +X and should grasp across its short width. I estimate the bar’s y length at about 0.09 m and its x width at 0.035 m. Is the maximum gripper opening about 0.08 m? Closing along x still looks preferable.
Wooden toy 2 may be a squirrel about 0.08 m long and 0.05 m wide. I could grasp its body with closure along either x or y, depending on the contact geometry. It spans approximately image x = 368–400 and v = 255–286, diagonally. The thicker body may be around (388, 268), and the head around (377, 275), but those identifications are tentative. A closer crop could help. The red pepper is an easier first choice, so I will start with the upper purple bar in the left arm and the red pepper in the right arm. They go to the white and blue baskets, whose centers are separated by about 0.28 m.
For later pairs, I could use the lower purple bar with wooden toy 2, followed by a left pepper with wooden toy 1. I initially considered giving wooden toy 1 to the left arm and placing it in the right basket while the right arm stayed home or worked on the right, but that would create a possible conflict near the right basket. A better assignment is for the left arm to handle both left peppers and both purple bars, and for the right arm to handle the Snickers, red pepper, and both wooden toys. That gives each arm four objects.
The tentative schedule is: upper purple bar to white with red pepper to blue; lower purple bar to white with wooden toy 2 to red; then a left pepper to blue with wooden toy 1 to red. If I use the leftmost pepper in the third pair, the left arm travels from x ≈ -0.39 m to x = 0, while the right arm carries wooden toy 1 from x ≈ -0.065 m to x = 0.28 m. Their grasp positions are separated, but they might approach x = 0 at the same time during transport. Could I keep them high with different y coordinates, or stagger their moves while the right arm holds? Pairing pepper 2 instead may be better: its starting x ≈ -0.276 m and wooden toy 1’s x ≈ -0.065 m are about 0.2 m apart, and their paths toward x = 0 and 0.28 m would diverge.
That leaves the leftmost pepper and Snickers. Simultaneous transport to the blue and white baskets could make the arms cross near x = -0.1 m, so I should avoid that. The right arm could transfer the Snickers while the left arm stays near its left home pose at approximately (-0.3, -0.35) m in x and y. The right gripper’s destination near (-0.28, 0.08) m would have about 0.4 m of y separation from that pose. Another possible left-arm holding pose is (-0.45, -0.2, 1.0) m. Having the left arm pick the Snickers at x ≈ 0.39 m would require a reach of about 0.7 m and may not be feasible. The right-arm transfer to white should be simpler if the left arm is clear of the placement area.
I could still approach and lift the final two objects together: the leftmost pepper at x ≈ -0.39 m and Snickers at x ≈ 0.39 m. Then I would move the left pepper to the blue basket at (0, 0.08, 1.05) m for link6 while the right arm holds near (0.39, -0.13, 1.05) m. After releasing the pepper, the left arm could retract left and upward, perhaps toward x = -0.4 m and y = -0.25 m. The right arm could then carry the Snickers to the white basket at x = -0.28 m and y = 0.08 m while the left arm stays far left, around y = -0.3 m. That seems safe. Alternatively, I could complete the Snickers first and leave the last pepper for a separate transfer.
There are 1100 control steps available, so I do not need to maximize simultaneous motion. Handling one object at a time would require roughly two approach actions, one descent, one closure, one test lift, one or two transport actions, one release, and one withdrawal: about eight or nine actions per object, or roughly 70 calls overall. Working in independent pairs could reduce this to about 40 calls. I will use paired actions where safe and verify each grasp and placement. The first pair will be the upper purple bar and red pepper.
I need intermediate orientations before moving from the initial pose to a downward grasp. For the initial quaternion and [0.5, 0.5, -0.5, 0.5], the dot product is 0.3535 + 0.3535 = 0.7071, giving an orientation distance of 1.5708 rad. That exceeds the 1.0 rad target limit, so I cannot command the full rotation directly. A halfway SLERP quaternion would give a rotation of about 0.785 rad.
The left-arm move from approximately (-0.3, -0.352, 0.922) m to (-0.142, -0.122, 1.03) m is about 0.295 m, exceeding the 0.25 m position limit. The right-arm move to (0.169, -0.098, 1.03) m is about 0.296 m. I can first move each arm to its object’s approximate x coordinate, with y = -0.25 m and z = 1.03 m, using an intermediate orientation. The left displacement is sqrt(0.158² + 0.102² + 0.108²) ≈ 0.217 m, and the right displacement is sqrt((-0.131)² + 0.102² + 0.108²) ≈ 0.198 m. Each intermediate rotation would be about 0.785 rad. The following move to the final above-object pose would be about 0.13 m with another 0.785 rad rotation. Perhaps eight control steps per move would suffice, given a rotation increment of 0.1 rad per step. Whether the intermediate poses pass the precheck is not yet known. I will use SciPy to calculate the halfway quaternions for both downward orientations.