[R] Fitting a half-ellipse curve

[Michael Bedward]

Hello Niklaus,

I'm not sure if the following is the sort of thing you are looking for (?)

You can fit an ellipse to your data using a deterministic least
squares method. The following is a function that I use to do this...

fit.ellipse <- function (x, y = NULL)
{
# Least squares fitting of an ellipse to point data
# using the algorithm described in:
#   Radim Halir & Jan Flusser. 1998.
#   Numerically stable direct least squares fitting of ellipses.
#   Proceedings of the 6th International Conference in Central Europe
#   on Computer Graphics and Visualization. WSCG '98, p. 125-132
#
# Adapted from the original Matlab code by Michael Bedward
# <michael.bedward at gmail.com>
#
# Arguments:
# x, y - the x and y coordinates of the data points
#
# Returns a list with the following elements:
#
# coef - coefficients of the ellipse as described by the general
#        quadratic:  ax^2 + bxy + cy^2 + dx + ey + f = 0
#
# center - center x and y
#
# major - major semi-axis length
#
# minor - minor semi-axis length
#

  EPS <- 1.0e-8
  dat <- xy.coords(x, y)

  D1 <- cbind(dat$x * dat$x, dat$x * dat$y, dat$y * dat$y)
  D2 <- cbind(dat$x, dat$y, 1)
  S1 <- t(D1) %*% D1
  S2 <- t(D1) %*% D2
  S3 <- t(D2) %*% D2
  T <- -solve(S3) %*% t(S2)
  M <- S1 + S2 %*% T
  M <- rbind(M[3,] / 2, -M[2,], M[1,] / 2)
  evec <- eigen(M)$vec
  cond <- 4 * evec[1,] * evec[3,] - evec[2,]^2
  a1 <- evec[, which(cond > 0)]
  f <- c(a1, T %*% a1)
  names(f) <- letters[1:6]

  # calculate the center and lengths of the semi-axes
  b2 <- f[2]^2 / 4
  center <- c((f[3] * f[4] / 2 - b2 * f[5]), (f[1] * f[5] / 2 - f[2] *
f[4] / 4)) / (b2 - f[1] * f[3])
  names(center) <- c("x", "y")

  num <- 2 * (f[1] * f[5]^2 / 4 + f[3] * f[4]^2 / 4 + f[6] * b2 -
f[2]*f[4]*f[5]/4 - f[1]*f[3]*f[6])
  den1 <- (b2 - f[1]*f[3])
  den2 <- sqrt((f[1] - f[3])^2 + 4*b2)
  den3 <- f[1] + f[3]

  semi.axes <- sqrt(c( num / (den1 * (den2 - den3)),  num / (den1 *
(-den2 - den3)) ))

  # calculate the angle of rotation
  term <- (f[1] - f[3]) / f[2]
  angle <- atan(1 / term) / 2

  list(coef=f, center = center, major = max(semi.axes), minor =
min(semi.axes), angle = unname(angle))
}


There are quite a few functions / packages to draw ellipses in R, but
the following is will work directly with the output of the above
function...

get.ellipse <- function ( fit, n=360 )
{
  # Calculate points on an ellipse described by
  # the fit argument as returned by fit.ellipse
  #
  # n is the number of points to render

  tt <- seq(0, 2*pi, length=n)
  sa <- sin(fit$angle)
  ca <- cos(fit$angle)
  ct <- cos(tt)
  st <- sin(tt)

  x <- fit$center[1] + fit$maj * ct * ca - fit$min * st * sa
  y <- fit$center[2] + fit$maj * ct * sa + fit$min * st * ca
  cbind(x=x, y=y)
}

So if your data were in a matrix or data.frame X...

efit <- fit.ellipse( X )
e <- get.ellipse( efit )
plot(X)
lines(e, col="red")

Hope this helps,
Michael


On 29 September 2010 23:45, Niklaus Hurlimann <niklaus.hurlimann at unil.ch> wrote:
> Dear mailing list,
>
> I have following array:
>
>       X2                 Y2
> [1,] 422.7900      6.0
> [2,] 469.8007      10.5
> [3,] 483.9428      11.0
> [4,] 532.4917      25.5
> [5,] 596.1942      33.5
> [6,] 630.8496      40.5
> [7,] 733.2996      45.0
> [8,] 946.4779      32.0
> [9,] 996.8068      35.5
> [10,] 1074.3310  23.0
>
> I do afterwards the following:
>
> plot.new()
>
> plot.window(xlim=c(min(X1)-50,max(X1)+50),
> ylim=c(min(Y1)-25,max(Y1)+25))
>
> axis(2, cex.axis=1.2)
> axis(1, cex.axis=1.2)
>
> points(X2, Y2, type="p", pch=0, cex=1.2, col="black")
>
> title(main="Dyke Geometry Along Strike", cex.main=1.2, font.main=4)
> title(xlab="distance [m]", cex.lab=1.2)
> title(ylab="half-thickness [m]", cex.lab=1.2)
>
> box()
>
>
> I would like curve fitting where I proceeded with a non
> linear-regression with the according function below:
>
> nls(formula = Y2 ~ abs(b*abs(1-X2^2/a^2)^(1/2)), start = list( a=282,
> b=40))
>
> The formula should give the y-positive part of an ellipse in my opinion
> or I might be completely wrong.
>
> In the end I would like to fit a curve of half an ellipse through the
> data.  I might could do this as well with a 2nd order polynomial fit. I
> am grateful for any suggestions and comments to my problem.
>
>
> Cheers
>
>
>
> --
> Niklaus Hürlimann
> Doctorant-PhD
>
> Université de Lausanne
> Institut de Minéralogie et Géochimie
> L'Anthropole
> CH-1015 Lausanne
> Suisse
>
> E-mail: Niklaus.Hurlimann at unil.ch
> Tel:+41(0)21 692 4452
>
>        [[alternative HTML version deleted]]
>
>
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