Vector projection to determine elbow in fitted curve

Question

I have a graph that has been fitted with nonlinear regression. I would like to calculate the elbow in the fitted curve. The majority of 2nd differential methods fail to accurately capture this point, and a visual inspection seems to be the only recourse (not useful for automation). The closest thing to an automated "visual" approach would be to use vector projection to calculate the farthest data point from a line that connects the first and last points of the data set (see question mark below). Can this line, which is normal to the line connecting the first and last point, be calculated using R?

My nonlinear function is: result <- lm(y ~ x + I(x^2) + I(x^3) + I(x^4) + I(x^5), data = myData)

Answer 1

Give this a try. See if it works on your real data.

library(MASS)

# fake data
x <- 5:300
y <- (x - 0.03*x^2 + 0.02*x^3 + rnorm(length(x), sd=5000))/1000
myData <- data.frame(x, y)

# fitted curve (I used a simpler example)
result <- lm(y ~ x + I(x^2) + I(x^3), data=myData)
p <- fitted(result)

# line connecting endpoints of fitted curve
i1 <- which.min(x)
i2 <- which.max(x)
slope <- (p[i2] - p[i1]) / (x[i2] - x[i1])
int <- p[i1] - slope*x[i1]

# for every point on the predicted curve (xi, pi), the perpendicular line that goes through that point has
perpslope <- -1/slope
perpint <- p - perpslope*x

# the intersection of the perp line(s) with the connecting line is
xcross <- (int - perpint) / (perpslope - slope)
ycross <- slope*xcross + int

# the distance between the intersection and the point(s) is
dists <- sqrt((x - xcross)^2 + (y - ycross)^2)

# the index of the farthest point
elbowi <- which.max(dists)

# plot the data
eqscplot(x, y)
lines(x[c(i1, i2)], p[c(i1, i2)])
points(x[elbowi], p[elbowi], pch=16, col="red")
lines(x[order(x)], p[order(x)], col="blue")
lines(c(x[elbowi], xcross[elbowi]), c(p[elbowi], ycross[elbowi]), col="red")