2d-line and normalangle

[[La...]]

Hi!
I want to control a 2d line with PCM, but I'm struggling with understanding the "Normangle" value.

It is a single value but it does not seem to be relateable to the base system axes. How can you know where NormAngle = 0 is defined from? It's easy enough if you have lines along major axes, but as soon as you have lines that are not parallel with the base system I'm lost.

Does anyone know how NormAngle = 0 is defined?

best regards/
Lars Rönne
AB Sandvik Coromant

[[No...]]

I think this value relates not to the base alignment, but to the local feature coordinate system:

[[La...]]

Thank you Norbert.You are of course correct, it is related to the local feature alignment.

But I still do not see any connection to the definition of the line? (A1/A2 or if you do start/end points etc.)
All that define is the direction of the line. Somewhere (perpendicularly to the line direction) some other axis must exist where normangle = 0 is defined. That is what I'm looking for.


best regards/
Lars

[[Ma...]]

A1/A2 are angles in BA. And depends on 2d-line main axis ( ie X+ X- Y+ ... )

Normalangle is used in correction of probe with measured points.

[[To...]]

In a 2d Line, the space axis is the direction of the line relative to the closest primary axis, designated by the X leg of the trihedron. The A1 and A2 are the angles from the space axis. The origin is created by the first point and the depth (or length) is developed by the last point. The normal angle determines the vector or probing direction though I do not know the rules for how it is determined. In other words, why is it 90° on one side of the part and 270° on the opposite side. If creating a line from the features menu, pay attention to what direction the Z leg of the trihedron is pointing. It should always be perpendicular to the surface. The Y leg of the trihedron serves no useful purpose that I am aware of.

[[La...]]

Hi!
Thank you for the reply Martin and Tom.
I understand A1 A2 and normal angle are for, and I understand the x and z axes.

It is that last crucial information I need, what the normangle is determined from. I almost wish I was doing this in the terrible DMIS standard where a line is easily defined. 😃

FEAT/LINE,CART, x, y, z, i, j, k, ni ,nj, nk

In Calypso the ni, nj, nk (probe compensation direction) is defined by this elusive normangle.

Perhaps this is not something that anyone knows outside of the Calypso developers? It is probably an clear and easy algorithm somewhere inside Calypso but hidden away from the end users. And it is very hard to figure out how it works by testing and observing Calypso behaviour.

Does anyone know who I might adress this question to? Perhaps someone in the Calypso development team?



best regards/
Lars

[[Ma...]]

I just tried creating blank 2d line and i see no logic why is zero set like it is.
For X+ is zero in Y+
For X- is zero in Z+
Fox Y+ is zero in Z+
For Y- is zero in X+
Fox Z+ is zero in X+
Fox Z- is zero in Y+

[[La...]]

I agree Martin, I too have a hard time seing any logic.

Best regards/
Lars

[[No...]]

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The Calypso documentation doesn't even tell you exactly how the local feature system is established for each feature type. For some types it's obvious, but for others you can only try and guess.

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If you as John Doe, the Calypso customer, should ever get the chance to communicate with a developer directly, consider yourself highly honored. The most you can hope for is that an ordinary support guy is granted an audience and may be allowed to share with you the glimpse of wisdom he received from the ivory tower. 🤣 🤣 🤣 🤣

[[La...]]

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Well, it was worth a try. 😉

Best regards/
Lars

[[No...]]

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It was not meant to discourage you 😉 Maybe you succeed where others failed.
If you do, give me notice and I'll send you a long list of questions for the devs 🤠

When I I have some spare time, I'll take a look at the angle problem. Ther must be some logic behind it.

[[La...]]

Thank you Norbert.
I'm going to try to run the question through proper channels to see if I can get an answer. When the manager for the Calypso development team gives me a call, I'll give him your name and let him know you have some questions! 😉

Anyway, I did some digging around:
I defined a line using a CAD model and looked around in the inspection files on disk with a text editor. The face normal (since I used a CAD model I knew what to look for) is stored exactly the way I want it in the inspection file: i, j, k vector components. The normangle value however is nowhere to be found in the inspection files. So I guess they re-calculate the normal vector back to this obscure normangle value when the element is presented in the user interface.

After looking about in our CAD system I failed to find any logical connection between the normangle value, base system, the element system and the normal vector. I did intersections between all planes of these two systems and compared all angles I could think of but nothing gives me a clue as to how the normangle is calculated. Nothing matches up. 🙁

If you find something, please let me know. I'll report back as well if I manage to figure something useful out.

best regards/
Lars

[[No...]]

I don't know if this answers your question or just maximizes confusion 😃

Much of the following is commonplace, but I mention it for completeness.
If you look at the LEK of a 2D line, the X axis always defines the direction of the line, while the Z axis always defines the vector of the probing points. So the Y axis is automatically defined too.
Now, the direction of the LEK's X axis depends on which axis is selected as the space axis of the line (+X, +Y, +Z, -X, -Y, -Z) and the values of the angles A1 and A2.
A1 and A2 both refer to the space axis (the axis that is part of both projection planes: If A1 = Y/X and A2 = Z/X, space axis is +X).

Now comes the interesting part:
Assume that the line's space axis is +X, and A1 and A2 are zero. Then a norm angle of 0 degrees makes the LEK's Z axis point in the direction of the base alignment's Y axis. Why?
Because Y is the axis that "follows" X. What does that mean? Imagine you write down a repeating sequence of axes: XYZXYZ...... Then Y follows X, Z follows Y and X follows Z.
I don't know if this is a general rule in 3D calculations everywhere, but Zeiss does it like this since the early UMESS days.
The positive sense of rotation of the norm angle is counterclockwise as usual (when viewed against the direction of the space axis).

But how does it work if A1 and A2 are not zero? That's simple: exactly the same! Because Calypso first sets the norm angle as described above (assuming A1 and A2 are zero) and only then applies the A1/A2 transformations!

For negative space axes a similar logic applies, but it's a bit confusing as it isn't an exact continuation of the rule for the positive axes. Maybe, for whatever reason, point vectors must always point in a positive direction? I don't know.

Anyway, as already mentioned by Martin, the succession of axes is as follows:

Space axis:     -Z -Y -X (-Z -Y -X)  |  +X +Y +Z (+X +Y +Z)
Point vectors:  +Y +X +Z (+Y +X +Z)  |  +Y +Z +X (+Y +Z +X)

If you extend/repeat the sequence (brackets), you see that it all follows a rule.


Maybe someone can dumb all this down to a simple equation. I didn't have the time to do this. But at least there is some logic behind it. Now it's up to you to implement it in PCM 🤠

[[Er...]]

When I looked at this it looked like it was based on the A1 "plane." If the plane was normal(perpendicular) to the surface the normangle was zero. If the A1 plane was parallel to the surface then Calypso would calculate a normangle to approach from a normal vector.

Maybe this makes sense?

[[La...]]

Thank you Norbert and Erik.
I'm going to have to look into this, but it does sound plausible that normangle vector perhaps is transformed along with A1/A2 angles. Right now I'm swamped with other stuff at work, but I will try to look at it and signal back if I manage to boil it down to an equation or algorithm I can use.

best regards/
Lars

[[He...]]

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Is the first direction the space axis of the line? In that case I see the following logic:
+X is the YZ-plane. Therefor +Y will be 0° and +Z will be 90°
-X is the ZY-plane. Therefor +Z will be 0° and +Y will be 90°
+Y is the ZX-plane. Therefor +Z will be 0° and +X will be 90°
-Y is the XZ-plane. Therefor +X will be 0° and +Z will be 90°
+Z is the XY-plane. Therefor +X will be 0° and +Y will be 90°
-Z is the YX-plane. Therefor +Y will be 0° and +X will be 90°

Could this be the answer? Does this make sense? I can't check it out myself right now. To much that needs to be done.