Z -value of point to parallel to line

Developers
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1

Hi there,
Can anyone please suggest why it happens to be, when we decrease the value in Z axis , the line appears to be straight instead of being parallely inclined with points of intersection.

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2

Hi Robert, can you show the graph that produces the lines?

3

Hi Jonathan,

The prior query raised has been resolved, but if you can please explain as per the graph, how can i get this line straight will be much appreciated

thanks

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4

Hi Robert
It’s difficult to identify the curve in question from the screen-dump you uploaded, I can have a look if you upload the .dyn and .xls (You’re reading from) :slight_smile:

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Test.xls (16 KB)
Test.dyn (58.7 KB)

6

Hi Robert, unfortunately I cannot identify which line you’re trying to correct.

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I’ve reduced your graph a little:

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7

Hi Jonathan,

As you can see in the first snip, the area in green is our final product; where the points mentioned in blue lines should match with blue points green line. The second snip mentions how these points are to be matched with arrows specified

Also in 3D view the level has been decreased to 740mm i.e. section view points are already placed in parallel with line, so that the final product( Snip-3) in green (left) should match with the blue (on right)

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8

I’m sorry but I just can’t wrap my head around what it is you’re trying. But I did notice that you have points floating around in the domain (x,y) < (100,100), while the rest of your points are at (x,y) > (630.000, 630.000)
Below is the “points” in your excel at the domain < 100. :slight_smile:
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