I have a curve, derived from empirical data, and I can obtain a reasonable model of it. I need to identify a point (x, y) where the curve intersects a circle of known center and radius. The following code illustrates the question.
x <- c(0.05, 0.20, 0.35, 0.50, 0.65, 0.80, 0.95,
1.10, 1.25, 1.40, 1.55, 1.70, 1.85, 2.00,
2.15, 2.30, 2.45, 2.60, 2.75, 2.90, 3.05)
y <- c(1.52, 1.44, 1.38, 1.31, 1.23, 1.15, 1.06,
0.96, 0.86, 0.76, 0.68, 0.61, 0.54, 0.47,
0.41, 0.36, 0.32, 0.29, 0.27, 0.26, 0.26)
fit <- loess(y ~ x, control = loess.control(surface = "direct"))
newx <- data.frame(x = seq(0, 3, 0.01))
fitline <- predict(fit, newdata = newx)
est <- data.frame(newx, fitline)
plot(x, y, type = "o",lwd = 2)
lines(est, col = "blue", lwd = 2)
library(plotrix)
draw.circle(x = 3, y = 0, radius = 2, nv = 1000, lty = 1, lwd = 1)
To find the point of intersection algebraically, solve each equation for y, set the two expressions for y equal to each other, solve for x, and plug the value of x into either of the original equations to find the corresponding y-value. The values of x and y are the x- and y-values of the point of intersection.
The method for finding the intersection points of a line 𝑦 = 𝑥 𝑚 + 𝑏 and a circle given in general form is as follows: Substitute 𝑚 𝑥 + 𝑏 for 𝑦 in the equation of the circle, here 𝑥 + 𝑦 + 2 𝑥 − 2 𝑦 − 3 = 0 .
To obtain the point of intersection we can use the optim function in r to do so:
circle=function(x){
if(4<(x-3)^2) return(NA)# Ensure it is limited within the radius
sqrt(4-(x-3)^2)
}
fun=function(x)predict(fit,data.frame(x=x))
g=function(x)(circle(x)-fun(x))# We need to set this to zero. Ie solve this
sol1=optimise(function(x)abs(g(x)),c(1,5))$min
[1] 1.208466
Thus the two functions should evaluate to the same value at x=1.208466
..
To make it even more precise, you can use the optim
function:
sol2= optim(1,function(x)abs(g(x)),g,method="Brent",upper=5,lower=1)$par
[1] 1.208473
Now you can evaluate:
circle(sol1)
[1] 0.889047
fun(sol1)
1
0.8890654
circle(sol2)
[1] 0.889061
fun(sol2)
1
0.889061
From the above, you can tell that solution 2 is very close..
Plotting this point on the graph will be challenging since the draw.circle
function draws circles in proportionality with the zxes.. Thus changing everytime depending on how big the plot region is.
If you were to write your own circle function:
circleplot=function(x,y,r){
theta=seq(0,2*pi,length.out = 150)
cbind(x+r*cos(theta),y+r*sin(theta))
}
Then you can do:
plot(x, y, type = "o",lwd = 2)
lines(est, col = "blue", lwd = 2)
lines(circleplot(3,0,2))
abline(v=sol2,col=2)
points(sol2,fun(sol2),col=2,pch=16)
It's straightforward to find the intersection using functions from the sf
package.
Calculate the circle values (inspired by this answer and as done by @Onyambu)
circ <- function(xc = 0, yc = 0, r = 1, n = 100){
v <- seq(0, 2 * pi, len = n)
cbind(x = xc + r * cos(v),
y = yc + r * sin(v))
}
m <- circ(xc = 3, yc = 0, r = 2)
Convert the predicted values and the circle values to "simple features" (LINESTRING
), and find their intersection (a POINT
):
library(sf)
int <- st_intersection(st_linestring(as.matrix(est)),
st_linestring(m))
int
# POINT (1.2091 0.8886608)
Add the intersection to your plot:
plot(x, y, type = "o", lwd = 2)
lines(est, col = "blue", lwd = 2)
lines(m)
points(int[1], int[2], col = "red", pch = 19)
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