Load Data

You can copy actual loaddata value to load data of actual tool number. So you can use only one tool, with variable load data. You must activate the actual tool again, after copying load data, to also activate the new load data.

if you are only programming using inline forms, you can define two tools, one with and one without part. the TCP data would be identical, only the configured load value would be different.

Hello guys,

I think this is the thread where i should ask my question.

I pick a part from a random place with my robot, then i place it into a reposition fixture. After the re-positioning i place the part into the station. There are some calculation again and placing it. The problem is that I had to define many types of positioning in the robot, because the length of the part can be from 800 to 2500 mm and its width can also vary from 800 to 2000 mm. the thickness of the part also changes. If they want to produce a new type of product (which is included in the size range), the robot will calculate it and insert it into the machine exactly. My problem is that I don't know what load data to define when there are so many different types of packaging. I know I can enter zero values, but I don't want to overload the robot.

I have already calculated the tool (the empty gripper) in CAD and fed it to the robot, so there is no problem with this, but what should I do with this many types? I was thinking of a Switch-Case branch, but if one part is already 2 cm bigger than the other, then another load definition would be necessary.

Thanks for your help in advance!

Quote from Mate_271

I think this is the thread where i should ask my question.

In my opinion, not really. You should have opened a new one.

Quote from Mate_271

because the length of the part can be from 800 to 2500 mm and its width can also vary from 800 to 2000 mm. the thickness of the part also changes

Size doesn't matter that much .

You didn't say any word about the weight .

It's a big difference between handling a cardboard or steel.

May be you can calculate the weight from the dimensions and at least a rough calculation of the center of gravity to correct the load data.

Thanks for your answer hermann!:)

Quote from hermann

In my opinion, not really. You should have opened a new one.

Ohh ok, sorry.

Quote from hermann

Size doesn't matter that much .

You didn't say any word about the weight .

It's a big difference between handling a cardboard or steel.

May be you can calculate the weight from the dimensions and at least a rough calculation of the center of gravity to correct the load data.

The material of the missing part is wood, which can be wet or dry, but I really don't know how to filter this, so that the operator always sets it exactly. I wrote down the inclusive dimensions only because it means different weights, and not only the weight will be different, but also the center of mass will change as a result

I thought that I should create some kind of robot program, where I add the following to the load data of the already existing empty tool: inertia, center of mass, mass. Also, I don't know if this is possible or workable.

what do you mean by "it is wood"? there are many types of wood and there is a huge difference between species like balsa and some quebracho (black ironwood etc.). density varies by an order of magnitude.

also of you want to use robot to automate some task, you must have some idea of shape, size, mass, pick point... that is plenty of info to calculate load on the fly.

if you are picking piece of plywood that is 4'x8' and the pick point is off by 2-3 inches variation in load is insignificant. think about it, 4' is 48inches and that is the short side. using pick point that is 2 inches off is only 2/48 which is about 4%. this is even less significant if you consider the long side. and inertia is heavily influenced by the distribution of mass (parts far from center of rotation are much more significant for overall inertia) .

Thanks for your answer too Panic mode!:)

I simply wrote wood because we are talking about pallets (more precisely, the top of the pallet). Of course, there is a distance between the boards and it also matters how many boards are in the pallet and how thick the "top" of the pallet itself is. I would simply like to see the whole thing as a rectangle. They are mostly made of pine and poplar. It can be wet or dry, as I wrote before.

Of course, before positioning, they can be in the robot's hands completely randomly (shifted by 20-30-40 cm or maybe turned a little), but here too I would use the after positioning load data, because actually positioning is only necessary for precise insertion.

The problem, as you wrote, is that compared to the TCP of the robot Tool0, for example, the pallet is 2300mm away in world +x, so the center of mass changes quite a bit. It is possible that this 2500mm long pallet roof is 1 meter wide or 1.5m or a maximum of 2 meters wide.

As I took it from your words and 10-20-30 centimeter does not have such a big influencing factor, would it be good if I created the 60-70 pieces of load data in CAD and assigned it? Does this seem like the most reasonable and simple solution? Or create a robot program where can it be calculated?

i do not know. still having hard time visualizing your process, products and problem.

i mean i know what a wooden pallet is, but what i have in mind is not 2.5m long. and the whole idea of a pallet is to be uniform shape and size, not sure how come you have have large variety and different thicknesses, how pick point can be off by so much etc.

also no idea what your tool looks like, where the TCP is, where the pick points are etc. couple of pictures and sketches could go long way.

example tool with retractable hooks could pick wooden pallet like this and TCP would be always pretty close to center of the pallet. are you picking pallets or products that are on pallets? whatever you have is far from clearly described.

The manufacturer deals with the production of pallets of special sizes. The picture you sent is a good starting point. Due to the narrowness of the space, the size of the pallet, and the dimensions of the robot (KRC2 KR180), we cannot always grasp the center of the pallet. Sometimes the robot grabs the top of the pallet about a third of the way or even lower, towards the end of the pallet.

The gripper works with vacuum. The positions are ready and tested. The only problem is the load data.

Due to the reasons described above (small space, size of the pallet, reach of the robot's arm), the robot had to pick up the top of the pallet from such positions that it would fit. Because of this, sometimes the center of gravity really shifts, especially in the case of the largest size, which can weigh 30-40 kg.

The picture I sent shows the two parallel vacuum grippers and the TCP. The red part I circled is the top of the pallet. This is what the robot packs. Essentially, a pallet has 3 legs and a top. The robot always picks up the stacked pallet tops from one point, so it is conceivable that the load data will always change if the width, length, height and thus the weight of the pallet tops increases. These pallet top stacks arrive at the robot on a conveyor belt. The robot finds it and places it on the positioning table, where it is positioned. So it already has a 0 point (one of its corners) from which we can calculate. Here, the robot finds the center of X and, if necessary, even moves a bit in Y. It depended on how I fit in the area. If it fit, then you only had to move in x and grab it at the bottom, if it didn't fit because the top of the pallet is big, then we also move a bit in y. I hope I wrote clearly. Thanks!

i see no problem in this. You probably know the mass and inertia of the gripper, you should know what the pallet weighs or atleast should be able to calculate something plausible from the size, you can estimate where the center of mass of the pallet is. You can make a simplification and assume the pallet is a flat isotropic rectangular piece and calculate the inertias. Use steiners theorem get the combined inertias of pallet and gripper. Calculate the center of mass by "weighted averaging" (cant remember the correct translation). Mass is just to add up. Set the correct values in correct load_data[] and you are good to go.

agreed...

a day ago in private chat i offered this:

Quote

in KRL, load is structure of form {M 2.0, CM {X 0.0,Y 0.0,Z 0.0,A 0.0,B 0.0,C 0.0}, J {X 0.0,Y 0.0,Z 0.0}}

where:

M is total mass of everything attached to robot flange (gripper + whatever is held by gripper)

CM is center of gravity (XYZ) as well as orientation of axes of inertia (with respect to flange). When axes of inertia are parallel to flange axes, rotation angles are zero.

J is inertia of the payload (gripper + whatever is held by it)

Determining mass is easiest since weight is easily measured and values simply get added (this quantity is a scalar).

M = m1+m2+m3...

determining center of mas requires solving vector equation for all masses.

breaking it down into three equations we get

Xcenter = (m1*x1 + m2*x2+m3*x3...)/M

Ycenter = (m1*y1 + m2*y2+m3*y3...)/M

Zcenter = (m1*z1 + m2*z2+m3*z3...)/M

in your case there are only two of them:

m1 = gripper (always present)

m2 = product (top layer of the pallet)

so CM equations simply become:

Xcenter = (m1*x1 + m2*x2)/(m1+m2)

Ycenter = (m1*y1 + m2*y2)/(m1+m2)

Zcenter = (m1*z1 + m2*z2)/(m1+m2)

the next part is orientation but as mentioned, if you keep things parallel, ABC values are zero.

and the last part is the inertia. one can compute it from product shape and density.

pallet top is not very consistent - there are gaps between the planks (And this changes from pallet to pallet).

the quick and easy solution is to treat it like one solid object ("slab" or "board"). it will not give you perfect result but - it will be close and more than good enough. this means much simpler computation using popular inertia equations for simple geometric shapes.

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so this is the most user-friendly and realistic solution. I also thought that I would have to use the robot to calculate each mass, center of gravity, inertia, but I was wondering how you would solve it. Thanks for the help guys, I'm on it!:)

Have a nice day!