How to apply perspective transform to Bezier curve? - graphics

Should I apply transform to Bezier control points only? Will this be correct?
Or should I add some corrections?
Transform is 2D->2D with matrix
a b c
d e f
g h 1

No, it is not correct. You can apply affine transform to Bezier control points and obtain new Bezier curve.
But perspective transformation transforms polynomial curve (traditional Bezier) into rational curve. It may be described by rational Bezier curve or NURBS (short reference p.111)
So it is possible to represent usual Bezier curve as rational curve (it is always possible), apply persp. transform to control points of rational curve (using weights in homogeneous form), and draw new rational curve.

Related

Closest point on a B-Spline Curve?

This thread asks how to get the closest point on a Bezier curve given an arbitrary point on the same plane: Closest point on a cubic Bezier curve?
How can I accomplish the same thing but for a B-SPline curve?

How to decompose affine matrix?

I have a series of points in two 3D systems. With them, I use np.linalg.lstsq to calculate the affine transformation matrix (4x4) between both. However, due to my project, I have to "disable" the shear in the transform. Is there a way to decompose the matrix into the base transformations? I have found out how to do so for Translation and Scaling but I don't know how to separate Rotation and Shear.
If not, is there a way to calculate a transformation matrix from the points that doesn't include shear?
I can only use numpy or tensorflow to solve this problem btw.
I'm not sure I understand what you're asking.
Anyway If you have two sets of 3D points P and Q, you can use Kabsch algorithm to find out a rotation matrix R and a translation vector T such that the sum of square distances between (RP+T) and Q is minimized.
You can of course combine R and T into a 4x4 matrix (of rotation and translation only. without shear or scale).

Finding tangent point coordinates on a circle?

Given
x1,y1,
d,
r,
and x2,y2
How would I find x3,y3 and x4,y4?
Construct two equations.
At first, tangent vector P31=(x3-x1, y3-y1) should be perpendicular to radius-vector P32=(x3-x2, y3-y2), so dot product of these vectors is zero.
At second, squared length of P31 is expressed through d and r using Pythagorean theorem.
Then solve system of two equations and get two solutions for both tangent points.
Note that this task has little relation to programming - it is just math.

Convert quadratic curve to cubic curve

Looking at Convert a quadratic bezier to a cubic?, I can finally understand why programming teachers always told me that math was so important. Sadly, I didn't listen.
Can anyone provide a more concrete - e.g., computer-language-y - formula for converting a quadratic curve to a cubic? Understanding that there's some rounding errors possible, which is fine.
Given a quad curve represented by variables:
StartX, StartY
ControlX, ControlY
EndX, EndY
And desiring StartX, StartY and EndX, EndY to remain the same, but to now have Control1X, Control1Y and Control2X, Control2Y of a cubic curve.
Is it...
Control1X = StartX + (.66 * (ControlX - StartX))
Control2X = EndX + (.66 * (ControlX - EndX))
With the same essential functions used to calculate Control1Y and Control2Y?
Your code is right except that you should use 2.0/3.0 instead of 0.66.
You avoid most rounding errors by using
Control1 = (Start + 2 * Control) / 3
Control2 = (End + 2 * Control) / 3
Note that line segments are also convertible to cubic Bezier curves using:
Control1 = Start
Control2 = End
This can be handy when converting a complex path mixing various types of curves (linear, quadratic, cubic).
There's also a basic transform for converting elliptic arcs to cubic (with some minor unnoticeable errors): you just have to split at least the arc on elliptic quadrans (cutting the ellipse first on the two orthogonal axis of symetries, or on arbitrary orthogonal axis passing through the center if the ellipse is a circle, then representing each arc; when the ellipse is a circle, the two focal points are confused on the same point, the center of the circle, so you can use any direction for one of the orthogonal axis).
Many SVG renderers do that by adding an additional split on octants (so that you get also precise position not only for points where the two main axis are passing through, but also for two diagonal axis which are bissecting (when the ellipse is a circle) each quadrant (when the ellipse is not a circle, assimilate it as a circle flattened with a linear transform along the small axis only, you do the same computation), because octants are also quite precisely positioned:
cos(pi/4) = sin(pi/4) = sqrt(2)/2 ≈ 0.71, and because this additional splitting will allow precise rendering of tangents on points crossing the diagonals at 45 degrees of the circle.
A full ellipse is then converted to 8 cubic arcs (i.e. 8 points on ellipse and 16 control points): you'll almost not notice the difference between elliptical arcs and these generated cubic arcs
You can create an algorithm that uses the same "flattening error" computed when splitting a Bezier to a list of linear segments, which are then drawn using the classic fast Bresenham algo for line segments; a "flattenning" algorithm just has to measure the relative deviation of the sum of lengths of the two straight segments joining the two focal points of the ellipse to any point of the generated cubic arcs, as this sum is constant on any true ellipse: if you make this measurement on the generated control points for the cubic arcs, the difference should be below a given percentage of the expected sum, or within an absolute distance precision, and can be used to create better approximation of control points with a simple linear formula so that these added points will be on the real ellipse.
Such transform of arbitrary paths is useful when you want to derive other curves from the path, notably the curves of "buffers" at a given distance, notably when these paths must be converted to "strokes" with a defined "stroke width": you need to compute two "inner" and "outer" curves and then concentrate on how to converting the miters/buts/squares/rounded corners, and then to cut long miters at a convenient distance (matching the "miter limit" factor times the "stroke width").
More advanced renderers will also use miters represented by tangent circles when there's a corner between two arcs instead of two segments (this is useful for drawing cute geographic maps)...
Converting an arbitrary path mixing segments, elliptic and bezier arcs to only cubic arcs is a necessary step to compute precise images without excessive defects visible when zooming in. This is then necessary when your "stroke" buffers have to take some effects (such as computing dashes), and then enhancing the result with semi-transparent pixels or subpixels to smooth the rendered strokes (smoothing is easy to computez only when everything has been flattened to line segments, and alsos may be simpler to develop if it only has to manage paths containing only cubic beziers: it can easily be parallelized if needed and accelerated by hardware). Bezier arcs are always interesting because drawing them is fast and requires only basic arithmetics, and the time needed to draw them is proportional to the length of the curve with every point drawn with the same accuracy level.
In summary, all curves are representable by cubic Bezier arcs with a maximum measurable deviation allowed (you can set this maximum deviation to one half pixel, or one subpixel if you first scale up the measurement grid for half-toning or subpixel shading, and then represent accurately every curve with a reasonnaly fast rendering, and get accurate rendering at any zoom level with curves smoothed everywhere, including with half-toning or transparency levels when finally drawing the linear strokes with the classic Bresenham algorithm using fast integer-only arithmetics). These rendered curve will all have the correct tangeants everywhere, without any unexpected angles visible on approximation points, and the remaining control points in the approximation will make also a good smooth rendering of the curvature everywhere (i.e. radius of the tangeant circle), so you can use this approximation as well to derive other measurements such as acceleration, inertial forces, or magnetic effects of paths of charged particles).
If you ever need higher precision, use Bezier arcs with degree 4 (i.e. with 3 control points between points on curve) to get smoothed derivation at a supplementary degree (e.g. gradients of forces), or just split the cubic arcs with additional steps further, until the derivation is smooth enough (but using degree-4 Bezier arcs requires much less points curves and less control points for the same accuracy tolerances, than when using cubic Bezier only).

How to calculate control points for cubic curve, which approximates an elliptic arc?

Almost all vector graphics applications (like Corel) approximate elliptic arcs with several cubic Bezier curves. I need to add similar functionality to my application. So my question is: how to calculate control points of that Bezier curve?
There are lots of pages explaining how to do this. This paper by Don Lancaster, for example, gives control parameters for divisions of an ellipse into between 2 and 8 cubic splines, with a detailed analysis of the 4-spline case.

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