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Consolidated circular arc construction methods into Arc3d (#7239)
Co-authored-by: dassaf4 <68340676+dassaf4@users.noreply.github.com>
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common/changes/@itwin/core-geometry/saeed-torabi-circularArcMethods_2024-10-04-20-42.json
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| { | ||
| "changes": [ | ||
| { | ||
| "packageName": "@itwin/core-geometry", | ||
| "comment": "", | ||
| "type": "none" | ||
| } | ||
| ], | ||
| "packageName": "@itwin/core-geometry" | ||
| } |
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| # Create Circular Arc using Start, TangentAtStart, and End | ||
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| Below we explain the algorithm used in `Arc3d.createCircularStartTangentEnd` to create a circular arc using start point, tangent at start, and end point. | ||
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| We first set up a frame of three perpendicular unit vectors using `Matrix3d.createRigidFromColumns`: | ||
| * `frameColX` lies along `tangentAtStart`, | ||
| * the circle normal lies along the cross product of `tangentAtStart` and the vector `startToEnd`, and | ||
| * `frameColY` lies along the cross product of the normal and `tangentAtStart`. | ||
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| We seek a formula for the radius of the circle. From the radius, we can find the circle center by moving a distance of radius from `start` along `frameColY`. From the center and the two input points, we can compute the arc sweep. Then we are done. | ||
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| The key to finding the radius is the inscribed right triangle pictured below, with one vertex at `start`, with hypotenuse along a diameter, and with leg along `startToEnd`. We know this is a right triangle from the inscribed angle theorem of classical geometry. | ||
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| Suppose `v = startToEnd`, `w = frameColY`, `r` is the radius we seek, and $\theta$ is the right triangle's angle at `start`. Then with | ||
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| $$v\cdot w = ||v|| ||w|| \cos\theta = ||v|| \frac{||v||}{2r} = \frac{||v||^2}{2r},$$ | ||
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| we have: | ||
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| $$\frac{v\cdot v}{2v\cdot w} = \frac{||v||^2}{2\frac{||v||^2}{2r}} = r.$$ | ||
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core/geometry/internaldocs/figs/Arc3d/createCircularStartTangentEnd.png
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| # Finding Roots of a Quartic Polynomial | ||
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| In `AnalyticRoots.appendQuarticRoots` we compute the solutions of the general quartic equation: | ||
| $$a_4x^4 + a_3x^3 + a_2x^2 + a_1x + a_0 = 0.\tag{1}$$ | ||
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| Without loss of generality, assume $a_4 = 1$. We further simplify by performing the substitution $x = y - a_3/4$ to obtain the depressed quartic: | ||
| $$P(x) = x^4 + px^2 + qx + r.$$ | ||
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| Note that the solutions of $(1)$ are obtained by subtracting $a_3/4$ from each root of $P$. | ||
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| Most classical solutions to $(1)$ derive a so-called _resolvent_ cubic polynomial $R$ from the depressed quartic, and use one of the roots of $R$ to construct a pair of quadratic equations whose roots are the roots of $P$. We will show this, using a resolvent cubic of this form: | ||
| $$R(y) = y^3 - \frac{1}{2}py^2 - ry + \frac{1}{2}rp - \frac{1}{8}q^2.$$ | ||
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| Let $x$ be a root of $P$. Then, introducing a quantity $y$ to complete the square, we have the following equivalences: | ||
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| $$\begin{align} | ||
| \notag{}P(x) = 0 &\Leftrightarrow x^4 = -px^2 - qx - r \\ | ||
| \notag{}&\Leftrightarrow (x^2 + y)^2 = -px^2 - qx - r + 2yx^2 + y^2 \\ | ||
| &\Leftrightarrow (x^2 + y)^2 = (2y - p)x^2 - qx + (y^2 - r)\tag{2} \\ | ||
| &\Leftrightarrow (x^2 + y)^2 = \left(\sqrt{2y - p}x - \frac{q}{2\sqrt{2y - p}}\right)^2\tag{3} \\ | ||
| &\Leftrightarrow \frac{q^2}{4(2y - p)} = y^2 - r\tag{4} \\ | ||
| \notag{}&\Leftrightarrow q^2 = 4(y^2 - r)(2y - p) \\ | ||
| \notag{}&\Leftrightarrow 0 = 8 R(y), | ||
| \end{align}$$ | ||
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| where in $(3)$ we have rewritten the right hand side of $(2)$ as a square (since the left hand side of $(2)$ is a square), and in $(4)$ we have equated the constant terms on the right hand sides of $(2)$ and $(3)$. Thus it is seen that the quantity $y$ is actually a root of the resolvent. We can compute the roots of $R$ with `AnalyticRoots.appendCubicRoots`. | ||
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| Lastly, choose a root $z$ of $R$. We need only one, so `AnalyticRoots.mostDistantFromMean` picks one employing a numerical stability criterion. Then from the above equivalences, and taking the square root of both sides of $(3)$, we have two quadratics in $x$: | ||
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| $$\begin{align} | ||
| \notag{}R(z) = 0 &\Leftrightarrow x^2 + z = \pm\left(\sqrt{2z - p}x - \frac{q}{2\sqrt{2z - p}}\right) \\ | ||
| &\Leftrightarrow x^2 \pm\sqrt{2z - p}x + z \mp\sqrt{z^2 - r} = 0,\tag{5} | ||
| \end{align}$$ | ||
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| where we have used the equality $(4)$ to simplify the constant term. We solve these two quadratics with `AnalyticRoots.appendQuadraticRoots`, yielding 0-4 values of $x$. By the above equivalences, each of these is a root of $P$. | ||
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