Beginner Questions about Querying Geometry

[Jason Ross]

Hello,

I'm interested in using this library to construct geometry and then query it later. Here's an example to get started:

-----

import cadquery as cq
import numpy as np


wp = cq.Workplane("front")
thickness = 1.0
height = 1.0
width = 1.0

pline = (
    wp
    .threePointArc((thickness/2, -thickness/2), (thickness, 0))
    .line(0, height)
    .tangentArcPoint((-thickness, thickness))
    .line(-width, 0)
    .tangentArcPoint((0, -thickness))
    .line(width, 0)
    .close()
)

points = 10 * (np.random.rand(20, 2) - 0.5)

------

I'd like to be able to do the following:

1. For each point in `points`, find if it is inside of `pline`
2. For each point in `points`, find the nearest point on `pline`
3. For each point in `points`, find the nearest segment of `pline`
4. For each point in `points`, find the parametric position of the nearest point on `pline`
5. Given a segment of `pline`, say the first arc, find its geometric properties (center, direction, etc.). I do know how to get the radius, but I don't know how to find the center of a circular arc.
   

Can someone help me with these?

Thanks

[Roger Maitland]

Here is something to get you started:

To work with and see the points, create Vector and Vertex versions of them:

# Convert the numpy array into cadquery vectors so they can be used as locations

random_vectors = [cq.Vector(*tuple(p), 0) for p in points]

# Convert the numpy array into cadquery vertices so they can be seen in cq-editor

random_vertices = [cq.Vertex.makeVertex(*tuple(p), 0) for p in points]

There are likely more efficient ways to find if a point is inside a wire but this might be enough for you:

test_sphere_radius = 0.1
test_sphere = cq.Workplane("XY").sphere(test_sphere_radius)
test_sphere_volume = test_sphere.val().Volume()
pline_offset_solid = (
    pline.toPending()
    .offset2D(test_sphere_radius)
    .extrude(test_sphere_radius, both=True)
    .val()
)
within = []
for p in random_vectors:
    within_shape = test_sphere.translate(p).intersect(pline_offset_solid)
    within.append(within_shape.val().Volume() == test_sphere_volume)

If the random point is within pline, the intersection of the test sphere at the location of each random point with a solid version of pline should result in a complete sphere.

There are cadquery selectors that can be helpful:

# Use a selector to find the closest vertex in pline to each point in points
nearest_pline_vertices = [
    pline.vertices(cq.NearestToPointSelector(tuple(p))) for p in points
]
This will find the nearest vertices. Custom selectors can be created if there isn't one that already fits.

The pline Workplane object can be converted back to Edge objects where a bunch of useful methods are available, such as:

pline_edges = pline.edges().vals()
arc_centers = []
for e in pline_edges:
    print(e.geomType(), e.positionAt(0), e.positionAt(1))
    if e.geomType() == "CIRCLE":
        arc_radius = e.radius()
        chord_center = (e.positionAt(1) + e.positionAt(0)) * 0.5
        arc_midpoint = e.positionAt(0.5)
        radial_direction = (chord_center - arc_midpoint).normalized()
        arc_center = arc_midpoint + radial_direction * arc_radius
        arc_centers.append(cq.Vertex.makeVertex(*arc_center.toTuple()))

`arc_centers` is a vertex at the center of the circle for each of the arcs in pline. By knowing if the edge is a CIRCLE (with end points, center and radius) or LINE (with end points), the remainder of your questions can be answered with some calculations.

Cheers,

Roger

[Jason Ross]

Hi Roger,

Thank you very much for the detailed explanation. I'm still not sure how to get positions along the curve the way I described.

This code

    # Use a selector to find the closest vertex in pline to each point in points
    nearest_pline_vertices = [
        pline.vertices(cq.NearestToPointSelector(tuple(p))) for p in points
    ]

returns vertices of `pline`, which are points that are already defined. What I'm looking for is the parametric position of the closest point on `pline` to each `p` in `points`, so if `p` is closest to the midpoint of `pline`, I'd like to get 0.5 back, rather than some vertex of `pline`.

As a side note, I found that this code

    test_sphere_radius = 0.1
    test_sphere = cq.Workplane("XY").sphere(test_sphere_radius)
    test_sphere_volume = test_sphere.val().Volume()
    pline_offset_solid = (
        pline.toPending()
        .offset2D(test_sphere_radius)
        .extrude(test_sphere_radius, both=True)
        .val()
    )
    within = []
    for p in random_vectors:
        within_shape = test_sphere.translate(p).intersect(pline_offset_solid)
        within.append(within_shape.val().Volume() == test_sphere_volume)

can be replaced by this code:

    pline_offset_solid = (
        pline.toPending()
        .extrude(0.01, both=True)
        .val()
    )
    within = [pline_offset_solid.isInside(v) for v in random_vectors]
and runs a lot faster. Is there any reason not to do it this way?

Thanks!

Jason

[Roger Maitland]

As I said Jason, "There are likely more efficient ways to find if a point is inside a wire" and you found it - excellent, I didn't know about the `isInside()` method.

I had a similar requirement to convert a Vector along a line back to a floating point position along the line between 0 and 1 and used a recursive binary search as follows:

def positiontOnEdge(

edge: cq.Edge,

target: cq.Vector,

window_start: float = 0.0,

window_end: float = 1.0,

tol: float = 0.00001,

) -> float:

"""

Search an edge for the position along it closest to target

"""

window_mid = (window_start + window_end) / 2

distance_to_start = (edge.positionAt(window_start) - target).Length

distance_to_mid = (edge.positionAt(window_mid) - target).Length

distance_to_end = (edge.positionAt(window_end) - target).Length

# Close enough to quit?

if distance_to_mid <= tol:

return window_mid

# Prepare for another try

if distance_to_start > distance_to_end:

window_start = window_mid

else:

window_end = window_mid

return positiontOnEdge(edge, target, window_start, window_end, tol)

which seems to work:

print(positiontOnEdge(pline_edges[0], pline_edges[0].positionAt(0.35)))

0.350006103515625

Again, there may be a more efficient way to do this but I don't know of one. I've found the binary search didn't need many loops to find a sufficiently accurate value for what I was doing.

Give the following a try to find the closest edge:

nearest_pline_edges = [pline.edges(cq.NearestToPointSelector(tuple(p))) for p in points]

If the closest edge is a CIRCLE, the same chord method used to find the center of the arcs for the circular edges, something like:

arc_closest = (

arc_center_vectors[0]

+ (random_vectors[0] - arc_center_vectors[0]).normalized() * arc_radii[0]

)

For LINE edges, you might try to use the Vector.getAngle() method to calculate the angles at each end and then solving the triangles to find the closest point.

I hope this helps.

Cheers,

Roger

--
cadquery home: https://github.com/CadQuery/cadquery
post issues at https://github.com/CadQuery/cadquery/issues
run it at home at : https://github.com/CadQuery/CQ-editor
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[Jason Ross]

Thanks, Roger. I think I have enough to work with.

Since I have your ear can I ask how you get syntax highlighting in google groups posts?

Thanks,

Jason

[Roger Maitland]

When replying to Gmail I just copy/paste from Visual Studio Code and the HTML sticks around. I couldn't get this to work directly from Google Groups though.

Cheers,

Roger

To view this discussion on the web visit https://groups.google.com/d/msgid/cadquery/4a6a2f29-8331-49f8-bdba-ef79317daa3dn%40googlegroups.com.

[Jeremy Wright]

I'm testing the use of an HTML pre tag for basic code separation to see if it works. https://www.w3schools.com/tags/tag_pre.asp

<pre>

import cadquery as cq

result = cq.Workplane().box(10, 10, 10)

show_object(result);

</pre>

[Jeremy Wright]

I'm not sure. Before they redisgned Google Groups there was an option for code under text formatting, now embedding HTML directly through the Groups UI doesn't even seem to work.