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Find tangents from point to cubic Bézier #288

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merged 3 commits into from
Aug 22, 2024
Merged

Find tangents from point to cubic Bézier #288

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ed63f43
Find tangents from point to cubic Bézier
raphlinus Jun 2, 2023
5cdb803
Naming and docstring changes.
raphlinus Jun 2, 2023
294c243
Merge branch 'main' into tangent
raphlinus Aug 14, 2024
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22 changes: 21 additions & 1 deletion src/cubicbez.rs
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Original file line number Diff line number Diff line change
Expand Up @@ -12,7 +12,7 @@ use crate::{Line, QuadSpline, Vec2};
use arrayvec::ArrayVec;

use crate::common::{
solve_quadratic, GAUSS_LEGENDRE_COEFFS_16_HALF, GAUSS_LEGENDRE_COEFFS_24_HALF,
solve_quadratic, solve_quartic, GAUSS_LEGENDRE_COEFFS_16_HALF, GAUSS_LEGENDRE_COEFFS_24_HALF,
GAUSS_LEGENDRE_COEFFS_8, GAUSS_LEGENDRE_COEFFS_8_HALF,
};
use crate::{
Expand Down Expand Up @@ -348,6 +348,26 @@ impl CubicBez {
.filter(|t| *t >= 0.0 && *t <= 1.0)
.collect()
}

/// Find lines from the point to the curve tangent to the curve.
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@Keavon Keavon Jun 2, 2023 •
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On my initial few read-throughs of this sentence I had a hard time groking it. I can also take a shot at wording it, which might also be flawed, but maybe we can meet in the middle with something that's clearer to unbiased readers:

Find points on the curve where each one's tangent line passes through the given point p in space.

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I made it a little shorter (it should be one line), but agree this is clearer. I think in my mind I was thinking about the line from the point to the curve, but thinking about it the other way is simpler because it's more apparent that you're getting both the tangent and the incidence.

///
/// Result is array of t values such that the line from the given point
/// to the curve evaluated at that value is tangent to the curve.
pub fn point_tangents(&self, p: Point) -> ArrayVec<f64, 4> {
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Would tangents_though_point be a clearer name?

let (a, b, c, d_orig) = self.parameters();
let d = d_orig - p.to_vec2();
// coefficients of x(t) \cross x'(t)
let c4 = b.cross(a);
let c3 = 2.0 * c.cross(a);
let c2 = c.cross(b) + 3.0 * d.cross(a);
let c1 = 2.0 * d.cross(b);
let c0 = d.cross(c);
solve_quartic(c0, c1, c2, c3, c4)
.iter()
.copied()
.filter(|t| *t >= 0.0 && *t <= 1.0)
.collect()
}
}

/// An iterator for cubic beziers.
Expand Down