Perpendicular Lines at Regular Intervals along Lines with Multiple Nodes - r

Question

I am attempting to sample along multiple lines (roads) at regular intervals and am struggling to obtain exact perpendicular angles for each road segment. I have split each road into points giving the node at which each line changes orientation and what I have so far creates a point within a straight segment of each road and appears to be working fine.
This is the code I am using to produce perpendicular angles for each node segment.
# X and Y for 3 points along a line
road_node <- matrix(
c(
381103, 381112, 381117,
370373, 370301, 370290
),
ncol = 2,
)
road_node <- as.data.frame(road_node)
angle_inv <- c()
for (i in 2:nrow(road_node) - 1) {
n1 <- road_node[i, ]
n2 <- road_node[i + 1, ]
x <- as.numeric(n1[1] - n2[1])
y <- as.numeric(n1[2] - n2[2])
ang <- atan2(y, x) + 1 / 2 * pi
if (!is.na(ang) && ang < 0) {
ang <- 2 + ang
}
angle_inv <- rbind(angle_inv, ang)
}
Where road_node gives the coordinates of each node.
From this I take the mid points and the inverse angles to create two points either side of the mid points, to produce a line segment.
# X Y and Angles (angles for one segment are the same
mids <- matrix(
c(
381374.5, 381351.0, 381320.5,
371590.5,371560.0, 371533.590,
2.3, 2.3, 2.3
),
nrow = 3,
)
mids <- as.data.frame(mids)
pts <- c()
for (i in 1:nrow(mids)) {
x1 <- mids[i, 1] + 10 * cos(mids[i, 3])
y1 <- mids[i, 2] + 10 * sin(mids[i, 3])
x2 <- mids[i, 1] - 10 * cos(mids[i, 3])
y2 <- mids[i, 2] - 10 * sin(mids[i, 3])
p1 <- cbind(x1, y1)
p2 <- cbind(x2, y2)
pair <- rbind(p1, p2)
pts <- rbind(pts, pair)
}
Some line segments appear to be correctly perpendicular to the node they are associate with, however some are not. Each appear to correctly share the same length.
I believe the problem lies with either how I am selecting my angles using atan2, or with how I am selecting my points either side of the node segment.

Answer 1

Firstly, there's no need to use trigonometry to solve this. Instead you can use the inverse reciprocal of the slope intercept form of the line segment equation, then calculate points on a perpendicular line passing through a give point.
See Equation from 2 points using Slope Intercept Form
Also your mid points appear incorrect and there are only 2 mid points as 3 points = 2 line segments.
This code appears to work fine
# Function to calculate mid points
mid_point <- function(p1,p2) {
return(c(p1[1] + (p2[1] - p1[1]) / 2,p1[2] + (p2[2] - p1[2]) / 2))
}
# Function to calculate slope of line between 2 points
slope <- function(p1,p2) {
return((p2[2] - p1[2]) / (p2[1] - p1[1]))
}
# Function to calculate intercept of line passing through given point wiht slope m
calc_intercept <- function(p,m) {
return(p[2] - m * p[1])
}
# Function to calculate y for a given x, slope m and intercept b
calc_y <- function(x,m,b) {
return(c(x, m * x + b))
}
# X and Y for 3 points along a line
road_node <- matrix(
c(
381103, 381112, 381117,
370373, 370301, 370290
),
ncol = 2,
)
road_node <- as.data.frame(road_node)
perp_segments <- c()
for (i in 2:nrow(road_node) - 1) {
n1 <- road_node[i, ]
n2 <- road_node[i + 1, ]
# Calculate mid point
mp <- mid_point(n1,n2)
# Calculate slope
m <- slope(n1,n2)
# Calculate intercept subsituting n1
b <- calc_intercept(n1,m)
# Calculate inverse reciprocal of slope
new_m <- -1.0 / m
# Calculate intercept of perpendicular line through mid point
new_b <- calc_intercept(mp,new_m)
# Calculate points 10 units away in x direction at mid_point
p1 <- rbind(calc_y(as.numeric(mp[1])-10,new_m,new_b))
p2 <- rbind(calc_y(as.numeric(mp[1])+10,new_m,new_b))
# Add point pair to output vector
pair <- rbind(p1,p2)
perp_segments <- rbind(perp_segments,pair)
}
This is how it looks geometrically (image)
I hope this helps.
Edit 1:
I thought about this more and came up with this simplified function. If you tink of the problem as a right isosceles triangle (45,45,90), then all you need to do is find the point which is the required distance from the reference point interpolated along the line segment, then invert its x and y distances from the reference points, then add and subtract these from the reference point.
Function calc_perp
Arguments:
p1, p2 - two point vectors defining the end points of the line segment
n - the distance from the line segment
interval - the interval along the line segment of the reference point from the start (default 0.5)
proportion - Boolean defining whether the interval is a proportion of the length or a constant (default TRUE)
# Function to calculate Euclidean distance between 2 points
euclidean_distance <-function(p1,p2) {
return(sqrt((p2[1] - p1[1])**2 + (p2[2] - p1[2])**2))
}
# Function to calculate 2 points on a line perpendicular to another defined by 2 points p,p2
# For point at interval, which can be a proportion of the segment length, or a constant
# At distance n from the source line
calc_perp <-function(p1,p2,n,interval=0.5,proportion=TRUE) {
# Calculate x and y distances
x_len <- p2[1] - p1[1]
y_len <- p2[2] - p1[2]
# If proportion calculate reference point from tot_length
if (proportion) {
point <- c(p1[1]+x_len*interval,p1[2]+y_len*interval)
}
# Else use the constant value
else {
tot_len <- euclidean_distance(p1,p2)
point <- c(p1[1]+x_len/tot_len*interval,p1[2]+y_len/tot_len*interval)
}
# Calculate the x and y distances from reference point to point on line n distance away
ref_len <- euclidean_distance(point,p2)
xn_len <- (n / ref_len) * (p2[1] - point[1])
yn_len <- (n / ref_len) * (p2[2] - point[2])
# Invert the x and y lengths and add/subtract from the refrence point
ref_points <- rbind(point,c(point[1] + yn_len,point[2] - xn_len),c(point[1] - yn_len,point[2] + xn_len))
# Return the reference points
return(ref_points)
}
Examples
> calc_perp(c(0,0),c(1,1),1)
[,1] [,2]
point 0.5000000 0.5000000
1.2071068 -0.2071068
-0.2071068 1.2071068
> calc_perp(c(0,0),c(1,1),sqrt(2)/2,0,proportion=FALSE)
[,1] [,2]
point 0.0 0.0
0.5 -0.5
-0.5 0.5
This is how the revised function looks geometrically with your example and n = 10 for distance from line: