What's new
What's new

Mcosmos Probe angle to vector

Larry Dickman

Titanium
Joined
Jan 30, 2014
Location
Temecula, Ca
Anyone using Mcosmos? I need to rotate my probe to a specific angle (PH20), but in the Change Probe by Angle function, I need to use the IJK vectors.
I can't seem to make the connection between the two.
The manuals are pathetic and I couldn't find anything online.
Seems like it shouldn't be that hard. I just want to use IJK to rotate to a specific A/B angle.
 

John Garner

Titanium
Joined
Sep 1, 2004
Location
south SF Bay area, California
Good Morning, Larry --

There isn't a general answer to your question(s); we need a full description of the (a = 0, b = 0) condition, if the object being rotated is a single-axis or a three-axis system, which A-or-B axis remains parallel to which X-or-Y-or-Z axis no matter how the other B-or-A axis is rotated.

And, unfortunately, I am swamped enough over the next few days that I probably won't be able to devote any significant time until next week.

I'll offer a few generalities, that may help you figure out your own answer:

The Direction (a mathematical construct that is often defined as "a Vector of undefined Radius", which is a pretty-poor definition since a Vector is defined as having "a Direction and Radius".

Vectors are quantified most often as a Cartesian Triple, with the Origin of the Vector at the Origin of the Cartesian Reference Frame, and the End Point of the Vector at the specified (x, y, z) location. The Magnitude of such a Vector is the straight-line distance between , the Origin and End Point which can be calculated as the Square Root of (x^2 + y^2 + z^2).

If the Magnitude of the Vector is exactly 1, the (x, y, z) triple is exactly equal to the (i, j, k) triple.

Directions can also be quantified as a Spherical Pair (a, b) where a = Rotation around A and b = Rotation around B. Since a and b are rotation angles, finding the equivalent Cartesian Triple (x, y, z) requires a bit of trigonometry. So let's assume that the (a = 0, b = 0) position is with A exactly parallel to X, and B exactly parallel to Y. Let us also assume that there is a C axis that is exactly parallel to Z . . . which means C is exactly perpendicular to A, B, X, and Y.

If we rotate the A, B, C system around A by some amount a, A does not change position with respect to the X, Y, Z Coordinate Frame, but B and C both move in the Y, Z Plane, "becoming" B' and C', which effectively creates a new Coordinate Reference Frame A' B' C'.

Subsequent rotation about B' rotates A' and C' in the plane that they define, creating another new Coordinate Reference Frame A", B", C".

If, as I suspect, you are asking how to determine the (i, j, k) values of C", and the geometry of the starting position is as I speculated . . . I THINK that the simple algorithms are these:

i of C" = Sine b
j of C" = -1 x Sine a x Cosine b
k of C" = Cosine a x Cosine b

By way of warning, I haven't done this type of arithmetic in a few years, and I'm pretty well oxidized. I strongly urge you to draw your own cartoons, and try to understand the geometry of the situation.

Finally, by way of a tip: When you draw your cartoons, set the A-or-B rotational axis that DOESN'T change position relative to the X,Y,Z Cartesian frame perpendicular to the paper.

Lunch hour's over, gotta run. Hope this helps.

John
 

John Garner

Titanium
Joined
Sep 1, 2004
Location
south SF Bay area, California
I should add that I used the word "parallel" to mean pointing in the same Direction; pointing in exactly-opposite Directions is what I call "Anti-parallel".

I assumed that both the A, B, C and X, Y, Z coordinate frames are right-hand orthogonal.
 

Larry Dickman

Titanium
Joined
Jan 30, 2014
Location
Temecula, Ca
Here's a view of what I'm doing. I had to block out some of it for NDA reasons
X is left to right
Y is forth and back
Z is up and down

DSCN2815.JPG
This angle is pretty close to what I wanted, but I found it through trial and error
The actual angle is A 61.00 deg B-39.15 deg
the vector is I -.54431, J -.68396, K -.48516
 

John Garner

Titanium
Joined
Sep 1, 2004
Location
south SF Bay area, California
Larry --

At a quick glance, I'm thinking that your Axis A is always perpendicular to Axes X and Y, while Axis B is always parallel to the plane defined by Axes X and Y.

If so, I'm wondering how the probe tip points at (a = 0, b = 0). If parallel to Axis X or Axis Y, it's reasonable to assume Axis B is perpendicular to both Axis A and the probe tip. But if the probe tip is parallel to Axis A, we need to know which (Axis X or Axis Y) axis Axis B is parallel.

As said earlier, I assumed that at (a = 0, b =0) Axis A would be parallel to Axis X while B would be parallel to Axis Y. That assumption is consistent with machine-tool axis conventions, but doesn't appear to be consistent with your hardware reality.

John
 

Larry Dickman

Titanium
Joined
Jan 30, 2014
Location
Temecula, Ca
I think this might be the wrong picture, I think B is off 180 deg. I'll check tomorrow.
With the probe pointing straight down is A0 /B0, which has the vector I0, J0, K-1
So, B is really C as it rotates around Z
With B at 0 deg, A is A as it rotates around X
 








 
Top