Search origin with MFRAME

[Paolo]

Hello.

Knowing the position of the robot in points 1/2/3 in the example shown, from time to time as I can find the origin of the frame using MFRAME?
I accept all the suggestions of techniques possible with MFRAME.

Thank you very much  

[Paolo]

   

Does anyone have any technique?

[Paolo]

  

[Paolo]

No one has tips?

[motomichael]

Hi!

What do you mean by "position of the robot"? If "position" is coordinates in some frame, then the point (0,0,0) is the origin of that frame. Or I've got something wrong?

Regards,
Michael

[Paolo]

I need to dynamically set the FRAME.
The orientation of the figure on display changes every time.
Moving with the robot, I can figure out where are the points 1,2,3.
When these are known locations of the robot, I'd like to use to set each time the MFRAME.
What technique can I use to dynamically recalculate MFRAME?

[Paolo]

    

[smurrill73]

Paolo,

If you know the Origin, XX, and XY coordinates for the user frame, you can put these points into P-variables, then use the MFRAME instruction to set the user frame:

MFRAME UF#(1) PX010 PX011 PX012

You would probably want to take some sort of precautions to make sure those points are reachable and will not cause collisions.

Hope this helps,
Scott

[Paolo]

Yes, but I do not know ORG.
My problem is that I calculate ORG

[motomichael]

Well, it seems that I got the point.

Actually, if the part is flat and parallel to plane Z=0 in the base frame, then two points will be enough.

Suppose you know the positions of points in the base frame, let them be (x1,y1) and (x2,y2). Construct the frame such that both points lies on one of its X axis; then Y axis must be perpendicular to the first one.  Put the frame origin into the point 1; now you must find some point of Y axis, let it be (x3,y3) (x3 and y3 are unknown yet).

(1) such a point must satisfy the following equation:

(x3 - x1) * (x2 - x1) + (y3 - y1) * (y2 - y1) = 0

This express the fact that scalar product of orthogonal vectors is 0.

(2) Now you have 1 equation with 2 unknown quantities. Fix the distance between (x1,y1) and (x3,y3), let it be d:

(x3 - x1)^2 + (y3 - y1)^2 = d^2.

Denote x3 - x1 as dx, y3 - y1 as dy; then

dx * (x2 - x1) + dy * (y2 - y1) = 0;
dx^2 + dy^2 = d^2.

Then

(d^2 - dy^2) * (x2 - x1)^2 = dy^2 * (y2 - y1)^2;
dx^2 = d^2 - dy^2,

=>

dy = (- (x2 - x1)^2 +- sqrt((x2-x1)^4 + 4 * (y2 - y1)^2 * d^2 * (x2 - x1)^2)) / (2 * (y2 - y1)^2);
dx = sqrt(d^2 - dy^2).

(Recheck math, please)

Solutions with plus and minus correspond to points, which lie on the different sides w.r.t. X axis.

(3) Now you know (x1,y1), (x2,y2) and (x3,y3) = (x1 + dx, y1 + dy) (they all are in the base frame).

GETS PX000 $PX001
SUB PX000 PX000
SETE P000 (1) x1
SETE P000 (2) y1
' P001 is (x1,y1)
SET P001 P000
SETE P000 (1) x2
SETE P000 (2) y2
' P002 is (x2,y2)
SET P002 P000
SETE P000 (1) x3
SETE P000 (2) y3
' P003 is (x3,y3)
SET P003 P000
MFRAME UF#(1) P001 P002 P003

Now UF#(1) is linked with your part; you can define another frame with the origin in the point of part you like. You must know its coordinates in UF#(1). Be them (x0,y0), do the following:

GETS PX000 $PX000
' Work in UF#(1) rather than the base frame
CNVRT PX000 PX000 UF#1
SUB P000 P000
SETE P000 (1) x0
SETE P000 (2) y0
' origin at P001
SET P001 P000
SETE P000 (1) 100000
SETE P000 (2) y0
' XX at P002
SET P002 P000
SETE P000 (1) x0
SETE P000 (2) 100000
' XY at P003
SET P003 P000
MFRAME UF#(2) P001 P002 P003

UF#(2) has the origin in the desired point, and its X axis is parallel to line connecting point 1 with point 2.

I hope this will help.

Regards,
Michael

[Paolo]

Ciao, parli italiano?

[motomichael]

Ciao!

No, only Russian 

Michael

[Paolo]

I can leave my email?

[potis]

Hello,

Paolo we have already answered your question in previous post.
http://www.robot-forum.com/robotforum/yaskawa_motoman_robot_forum/search_org_mframe-t8101.0.html;msg32809#msg32809