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Coordinate System Transforms

Question asked by John Howell on Aug 6, 2019
Latest reply on Aug 7, 2019 by John Howell

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I know there are other topics - one of which I posted - however I need specific help. My apologies.

I'm on 2017 and using a VB.NET AddIn.

I have a point data extractor that re-curses the Feature Tree of a Part and returns X, Y, Z data for several DATUMPOINT's ("Point" Features). No sketch involved, thus no transform from Sketch to Model. All is well.

A user now wants this data relative to a user defined Coordinate System. No problem - I've added a ComboBox populated with any Coordinate Systems found in the model.

What I don't get - and I'm sure it's a fundamental block - is specifically how to do the Transform to a User Defined Coordinate System, which is selected in the ComboBox by name, let's say "Mikes Coordinate System";

So, I have an oPoint as a MathPoint populated with default coordinates.

How do I transform this Math Point's coordinate array from the default coordinate system to the user defined coordinate system "Mikes Coordinate System"? An example would be great - I'm stumbling on specifics, like how to get the 'transform' for the user selected coordinate system in the ComboBox.

So close yet so far... Any help much appreciated!

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Josh Brady Aug 6, 2019 12:09 PM

So... You have the coordinate system feature, right? So get it's feature data. The transform is a member of the feature data.

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Jacob Corder Aug 6, 2019 1:49 PM


public Function GetCoordinateSystemTransform(byval SwModel as modeldoc2, Byval CoordSystemName as string) as mathtransform
   Dim PDoc as partdoc = Swmodel
   Dim SwFeat as feature = PDoc.FeatureByName(CoordSystemName)
   if swfeat is nothing then return nothing
   if Swfeat.GetTypeName2 = "CoordSys" then
      Dim Cdata as coordinateSystemFeatureData = SwFeat.GetDefinition
      return cdata.Transform
   End If
   Return Nothing
End Function

If it is in the feature tree, it most likely can be retrieved by FeatureByName.

then as Josh brady stated Return The transform of the data definition.

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John Howell Aug 6, 2019 4:31 PM

Thank you for your reply, it got me on the right track. My apologies again - I'm not overly familiar with coordinate system transforms.

Yes, I was able to get a 'Transform' object from the User Defined Coordinate System.

I then took the MathPoint with the original model coordinates and did a .MultiplyTransform on it, passing in the .Inverse of the Transform object retrieved. Not sure why .Inverse, but without it the data was incorrect. With it all is GOOD!

This was not easy to figure out - not knowing how it was supposed to work. The Transform object of the 'alternate' CS is key, as is the .MultiplyTransform to do the actual transform. Noting to future self... and anyone else struggling with this one.

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Josh Brady @ John Howell on Aug 6, 2019 10:39 PM

So... The transform is like a set of directions to get from A to B. If you need to get from B to A, you have to follow the directions backwards. That's what the inverse is.

Have you seen

I need your help to improve this matrix transformations how-to

and the two blog posts referenced therein? Good stuff.

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Jacob Corder @ Josh Brady on Aug 7, 2019 7:37 AM

I love that transformation post you shared. It's very informative.

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John Howell @ Josh Brady on Aug 7, 2019 8:45 AM

I have now. Good stuff for sure!

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6 Replies

Josh Brady Aug 6, 2019 12:09 PM
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So... You have the coordinate system feature, right? So get it's feature data. The transform is a member of the feature data.
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John Howell Aug 6, 2019 4:31 PM
Mark Correct
Correct Answer
Thank you for your reply, it got me on the right track. My apologies again - I'm not overly familiar with coordinate system transforms.

Yes, I was able to get a 'Transform' object from the User Defined Coordinate System.

I then took the MathPoint with the original model coordinates and did a .MultiplyTransform on it, passing in the .Inverse of the Transform object retrieved. Not sure why .Inverse, but without it the data was incorrect. With it all is GOOD!

This was not easy to figure out - not knowing how it was supposed to work. The Transform object of the 'alternate' CS is key, as is the .MultiplyTransform to do the actual transform. Noting to future self... and anyone else struggling with this one.
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- Josh Brady @ John Howell on Aug 6, 2019 10:39 PM
  Mark Correct
  Correct Answer
  So... The transform is like a set of directions to get from A to B. If you need to get from B to A, you have to follow the directions backwards. That's what the inverse is.
  
  Have you seen
  
  I need your help to improve this matrix transformations how-to
  
  and the two blog posts referenced therein? Good stuff.
  Like • Show 1 Like1
  Actions
  - Jacob Corder @ Josh Brady on Aug 7, 2019 7:37 AM
    Mark Correct
    Correct Answer
    I love that transformation post you shared. It's very informative.
    Like • Show 0 Likes0
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  - John Howell @ Josh Brady on Aug 7, 2019 8:45 AM
    Mark Correct
    Correct Answer
    I have now. Good stuff for sure!
    Like • Show 0 Likes0
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