bezier

Smooth Hilbert curves

时光毁灭记忆、已成空白 提交于 2019-12-02 05:48:11
I'm trying to smooth out the path taken by a Hilbert curve . I can define the points and connect them by straight lines, but I want a path that doesn't take the edges so sharply. I attempted to connect the curve using Bezier curves of higher and higher orders but this doesn't work, there are always 'kinks' in the path as I try to reconnect them: I feel like this a solved problem, but I'm not searching for the right terms. How about using piecewise cubics for this... Does not really matter if BEZIER SPLINE or whatever. You just need to connect the patches with proper point call sequence which

How to convert image pixel into s-curve shape in canvas

不打扰是莪最后的温柔 提交于 2019-12-02 05:45:05
问题 I have such type of image I want that what we are seeing straight lines in image (actually which are pixel), it should get converted into the S-curve. I have implemented C-type of curve using canvas and its properties, but unable to do S-curve. Please help me for this. 回答1: If I understand you correctly you want each vertical line to follow a "S"? If that's the case you can use f.ex. Math.sin() combined with drawImage() and its clipping parameters to slice the image per pixel column while

Split one quadratic bezier curve into two

筅森魡賤 提交于 2019-12-02 05:14:05
问题 So I have an imaginary circle divided into multiple parts (I use 8 for simplicity, but in the end, I would like to divide it to 16 or 32 parts). Then I have N number of quadratic bezier curves, that is between 2 nearest segments. It may rest upon the circle or further from the center, but not nearer than the circle. I know how to find, what in witch line I should look for intersection in, but I do not know how to split it into two parts... I know, that if I looked for intersection of the line

Drawing arc with bezier curves

我们两清 提交于 2019-12-02 03:27:51
I am trying to draw an arc using bezier curves. I have learned that you can't draw a perfect circle using bezier curves but you can come close. Unfortunately the math is too complicated and I can't personally figure it out. I can create the A1 slice below as a triangle, but I can't figure out how to determine the control points. Also if I try drawing a slice out of a circle in the opposite direction, notice how the control points seem to point in the negative direction. So if I want a slice of a circle with a radius R and I have already computed the anchor points, how can I calculate the

How to draw a pie chart using pdfbox?

◇◆丶佛笑我妖孽 提交于 2019-12-02 03:01:10
问题 I have to draw a pie-chart using pdfbox. Let the data be: Subject Mark in Percentage Mark in Degrees Cumulative Degrees Sub-1 80 80 80 Sub-2 70 70 150 Sub-3 65 65 215 Sub-4 90 90 305 Sub-5 55 55 360 Let the radius and centre be 100 pixels and ( 250, 400). Let us take initial line parallel to x axis. Drawing initial line statement will be: contentStream.drawLine(250, 400, 350, 400); I stuck up with: a) finding x, y co-ordinates of point on the circle that is some degrees away from the initial

Split one quadratic bezier curve into two

痞子三分冷 提交于 2019-12-02 01:39:20
So I have an imaginary circle divided into multiple parts (I use 8 for simplicity, but in the end, I would like to divide it to 16 or 32 parts). Then I have N number of quadratic bezier curves, that is between 2 nearest segments. It may rest upon the circle or further from the center, but not nearer than the circle. I know how to find, what in witch line I should look for intersection in, but I do not know how to split it into two parts... I know, that if I looked for intersection of the line and curve I should get the point that the previous curve should end and the next should start, and

Nearest point on cubic bezier curve to a given point

牧云@^-^@ 提交于 2019-12-02 01:11:42
问题 I have a cubic-Bezier curve defined as A, B, C, D. Where A is the start, B and C are control points, and D is the end. I understand how to find the position at any value t, where 0 <= t <= 1, and that concept in general since it just uses a handful of calls to a linear interpolation function that result in the curve. (Can be visualized easily here on wikipedia just below the heading Higher-order curves) I am now looking to find a point on the curve that is closest to some point in space, P.

How to convert image pixel into s-curve shape in canvas

最后都变了- 提交于 2019-12-02 00:50:02
I have such type of image I want that what we are seeing straight lines in image (actually which are pixel), it should get converted into the S-curve. I have implemented C-type of curve using canvas and its properties, but unable to do S-curve. Please help me for this. If I understand you correctly you want each vertical line to follow a "S"? If that's the case you can use f.ex. Math.sin() combined with drawImage() and its clipping parameters to slice the image per pixel column while displacing the slice based on the sin(). The key formulas are: var step = Math.PI * 2 / w; this maps a full

jQuery animating along a sine wave

别说谁变了你拦得住时间么 提交于 2019-12-01 18:33:57
I've spent a couple days at this and I give up. I'm trying to get an object to animate along a sine wave infinitely. It should not end after the first period. Main Problem: The cycle ends at approx 1 1/3 Pi rather than just Pi. This extra movement ruins the animation. I'm stuck here: http://jsfiddle.net/WPnQG/12/ . After each period, it skips about 40 pixels and then continues along its path. This is the problem I can't get past -- the value it ends at and proceeds to restart at are not equal, so the object appears to skip around. Could anyone help a man out? Thanks I am using the jQuery Path

Calculate the length of a segment of a quadratic bezier

主宰稳场 提交于 2019-12-01 18:17:13
问题 I use this algorithm to calculate the length of a quadratic bezier: http://www.malczak.linuxpl.com/blog/quadratic-bezier-curve-length/ However, what I wish to do is calculate the length of the bezier from 0 to t where 0 < t < 1 Is there any way to modify the formula used in the link above to get the length of the first segment of a bezier curve? Just to clarify, I'm not looking for the distance between q(0) and q(t) but the length of the arc that goes between these points. (I don't wish to