This approach won’t work. SVG doesn’t have conical gradients. To simulate the effect, you would have to fake it with a large number of small line segments. Or some similar technique. Update: Here is an example. I approximate the 360deg of hue with six paths. Each path contains an arc which covers 60deg of the … Read more
Use the quadratic Bézier formula, found, for instance, on the Wikipedia page for Bézier Curves: In pseudo-code, that’s t = 0.5; // given example value x = (1 – t) * (1 – t) * p[0].x + 2 * (1 – t) * t * p[1].x + t * t * p[2].x; y = (1 … Read more
I’ve written some quick-and-dirty code that estimates this for Bézier curves of any degree. (Note: this is pseudo-brute force, not a closed-form solution.) Demo: http://phrogz.net/svg/closest-point-on-bezier.html /** Find the ~closest point on a Bézier curve to a point you supply. * out : A vector to modify to be the point on the curve * curve … Read more
I needed to implement this for work as well. The fundamental concept you need to start with is that the main difference between the regular Catmull-Rom implementation and the modified versions is how they treat time. In the unparameterized version from your original Catmull-Rom implementation, t starts at 0 and ends with 1 and calculates … Read more
OK, I think I found the answer. First the solution: cov_x*s_sq is simply the covariance of the parameters which is what you want. Taking sqrt of the diagonal elements will give you standard deviation (but be careful about covariances!). Residual variance = reduced chi square = s_sq = sum[(f(x)-y)^2]/(N-n), where N is number of data … Read more
Think I got it: set.seed(1) df <- data.frame(PF = 10*rnorm(1000)) ggplot(df, aes(x = PF)) + geom_histogram(aes(y =..density..), breaks = seq(-50, 50, by = 10), colour = “black”, fill = “white”) + stat_function(fun = dnorm, args = list(mean = mean(df$PF), sd = sd(df$PF)))