[ Solved ] Normal of a Sketch - Calculate Normal from Placement ?

[paullee]

Hi, it seems i am not familiar / confused by those geometrical terms.

Assume a Sketch map on XY plane (regardless of its content) has a Normal (0,0,1) correct?
(Map on XZ plane has a Normal (0,-1,0) )

Try to 'calculate' this Normal, (from the sketch's placement?), searching googling but find no previous discussion.

Any idea? Thanks.

[Chris_G]

I would do :

Code: Select all

normal = sketch.Placement.Rotation.multVec(FreeCAD.Vector(0,0,1))

[openBrain]

I would do :

Code: Select all

normal = sketch.Placement.Rotation.multVec(FreeCAD.Vector(0,0,1))

This is perfect to get the normal vector in the LCS.
If normal vector in the GCS is needed, just change to :

Code: Select all

normal = sketch.getGlobalPlacement().Rotation.multVec(FreeCAD.Vector(0,0,1))

[paullee]

Thanks both for the solutions!

If normal vector in the GCS is needed, just change to :

Code: Select all

normal = sketch.getGlobalPlacement().Rotation.multVec(FreeCAD.Vector(0,0,1))

Do some exercise, haven't figured out the difference between sketch.Placement and sketch.getGlobalPlacement() :-

1. Make 2 Sketches
   (1st rectangle, 2nd triangle)
2. Map the 2nd to 1st with ObjectXY mapmode
3. Attachment Offset 0,0,50
4. With 1st Sketch Placement at 0,0,0, 2nd's Placement / getGlobalPlacement are (0,0,50), so far so good.

Code: Select all

>>> s1.Placement
Placement [Pos=(0,0,0), Yaw-Pitch-Roll=(0,0,0)]
>>> s2.Placement
Placement [Pos=(0,0,50), Yaw-Pitch-Roll=(0,0,0)]
>>> s2.getGlobalPlacement()
Placement [Pos=(0,0,50), Yaw-Pitch-Roll=(0,0,0)]

1. Now set 1st Sketch placement to (10,10,10), 2nd sketch move altogether, ok
2. 2nd's Placement / getGlobalPlacement both return (10,10,60) ...

Code: Select all

>>> s2.Placement
Placement [Pos=(10,10,60), Yaw-Pitch-Roll=(0,0,0)]
>>> s2.getGlobalPlacement()
Placement [Pos=(10,10,60), Yaw-Pitch-Roll=(0,0,0)]

... I do not understand the correct way to use it ?

Screenshot from 2019-09-15 09-58-04.png (227.76 KiB) Viewed 680 times

Screenshot from 2019-09-15 10-09-29.png (195.41 KiB) Viewed 680 times

[openBrain]

... I do not understand the correct way to use it ?

Quite simple when you know.
'Attachment' properties in PartDesign are directly injected in 'Placement', thus changing them makes no difference as it doesn't rely in a LCS.

To demonstrate, you have to use a 'container' that has an LCS, such as Body in the below example (also works with Part container).
In the attached file, 'Sketch' is mapped on XY plane with no offset/rotation.
However I changed the 'Body' Placement to offset 10,10,10 and rotation of 90° around X.
And now we see a difference.

Code: Select all

>>> App.ActiveDocument.Sketch.Placement
Placement [Pos=(0,0,0), Yaw-Pitch-Roll=(0,0,0)]
>>> App.ActiveDocument.Sketch.getGlobalPlacement()
Placement [Pos=(10,10,10), Yaw-Pitch-Roll=(0,0,90)]

[paullee]

Quite simple when you know.
'Attachment' properties in PartDesign are directly injected in 'Placement', thus changing them makes no difference as it doesn't rely in a LCS.

To demonstrate, you have to use a 'container' that has an LCS, such as Body in the below example (also works with Part container).
In the attached file, 'Sketch' is mapped on XY plane with no offset/rotation.
However I changed the 'Body' Placement to offset 10,10,10 and rotation of 90° around X.
And now we see a difference.

Code: Select all

>>> App.ActiveDocument.Sketch.Placement
Placement [Pos=(0,0,0), Yaw-Pitch-Roll=(0,0,0)]
>>> App.ActiveDocument.Sketch.getGlobalPlacement()
Placement [Pos=(10,10,10), Yaw-Pitch-Roll=(0,0,90)]

Thanks, got it and it become quite simple now