Determine Point Coordinates In 3D - math

I have a line that exists in 3D that is between two known points: {X1, Y1, Z1} and {X2, Y2, Z2}.
(X1,Y1,X1)----------(X2,Y2,Z2)
There is a point (Xd,Yd,Zd) on the line between those points at distance D from (X1,Y1,Z1).
(X1,Y1,X1)---D---(Xd,Yd,Zd)-----(X2,Y2,Z2)
How can I determine the coordinates of point (Xd,Yd,Zd)?


Assuming you want to move the distance D from point 1 to point 2 :
P1 = [ X1, Y1, Z1 ]
P2 = [ X2, Y2, Z2 ]
The line vector can be described as :
V = P2 - P1 = [ Xv = X2 - X1, Yv = Y2 - Y1, Zv = Z2 - Z1 ]
The line's length can be determined as :
VL = SQRT(Xv^2 + Yv^2 + Zv^2) // ^2 = squared
The line's versor aka the unit vector can be determined as :
v = V / VL = [Xv / VL, Yv / VL, Zv / VL]
The target point PD can be determined as :
Pd = P1 + D * v // Starting from P1 advance D times v
Please note that P1 and v are vectors and D is a scalar


First, determine the length of the line segment:
d=sqrt((X1-X2)^2+(Y1-Y2)^2+(Z1-Z2)^2))
You are moving D from P1=(X1,Y1,Z1) toward P2=(X2,Y2,Z2). This puts you at the point (X3,Y3,Z3):
{XYZ}3={XYZ}1+(D/d)*({XYZ}2-{XYZ}1})
Where you expand that into 3 equations, one for each of X, Y, and Z.
This works because you are D/d of the way between P1 and P2. Check: Say D=d. Then you should be at exactly P2.


Take the vector between the two points
<X2-X1, Y2-Y1, Z2-Z1>
Turn that into a unit vector pointing in the same direction but with length 1. You do that by dividing by the distance between the two points:
<X2-X1, Y2-Y1, Z2-Z1>
---------------------------------------
sqrt((X2-X1)^2 + (Y2-Y1)^2 + (Z2-Z1)^2)
Then multiply that by D and add to your original point to get the new point.
<X2-X1, Y2-Y1, Z2-Z1>
(X1, Y1, Z1) + D * ---------------------------------------
sqrt((X2-X1)^2 + (Y2-Y1)^2 + (Z2-Z1)^2)


This is a linear combination problem:
dist = distance(p1, p2)
distance D is given
f = D / dist (fractional coordinate of point D within LineSeg (p1, p2)
pD = LinearCombo (1-f, p1, f, p2) (coordinates of point distance D from p1)

Related

Julia BoundsError with less information - don't know why

I have the following code:
# package for ploting functions
using Plots
# use GR
gr()
# nb points to plot
nbPts = 22
# define polar coordinates of a 30 degree (pi/6) rotation
sine = sin(pi/6)
cosine = cos(pi/6)
# scale factor
scale_factor = 0.9
#---------------------------------------
# 1. PLOT POINTS USING ROTATION MATRIX
#---------------------------------------
# define Rotation matrix ( angle = pi/6, center = (0, 0) )
R = zeros(Float64, 2, 2)
R[1,1] = R[2,2]= cosine
R[1,2] = -sine
R[2,1] = sine
# Scale matrix
### ... <-- EXERCISE 4(c): define a uniform scaling matrix (use scale_factor)
# arrays of points coords
X_mat = zeros(nbPts)
Y_mat= zeros(nbPts)
# first Point (1,0)
X_mat[1] = 1.0
Y_mat[1] = 0.0
for i in 2:nbPts
prevPoint = [X_mat[i-1], Y_mat[i-1]]
#apply rotation to previous point to obtain new point
newPoint = R * prevPoint
### ... <-- EXERCISE 4(c): apply scaling matrix
X_mat[i] = newPoint[1]
Y_mat[i] = newPoint[2]
end
# plot points in blue
plt1 = scatter(X_mat, Y_mat, color=:blue, xlim = (-1.1, 1.1), ylim = (-1.1, 1.1), label=false, title="Rotation using matrices" );
#---------------------------------------
# 2. PLOT POINTS USING COMPLEX NUMBERS
#---------------------------------------
function ComplexProduct(z, w)
(((z[1]*w[1])+(z[2]*w[2])),((z[1]*w[2])+(z[2]*w[1])))
### ... <-- EXERCISE 4(b): implement complex product z * w
end
# first point: z = 1 + 0 * i
Z = ( 1.0, 0.0 )
# second point: w = cosine( pi/6) + sine( pi/6) * i
W = ( cosine, sine )
### ... <-- EXERCISE 4(c): apply scale_factor to W
# arrays of points coords
X_comp = zeros(nbPts)
Y_comp = zeros(nbPts)
# first Point (1,0)
X_comp[1] = Z[1]
Y_comp[1] = Z[2]
for i in 2:nbPts
prevPoint = (X_comp[i-1], Y_comp[i-1])
newPoint = ComplexProduct(prevPoint[1], prevPoint[2]) ### <-- EXERCISE 4(b): compute newPoint by applying rotation to prevPoint (use complex product)
X_comp[i] = newPoint[1]
Y_comp[i] = newPoint[2]
end
# plot points in red
plt2 = scatter(X_comp, Y_comp, color=:red, xlim = (-1.1, 1.1), ylim = (-1.1, 1.1), label=false, title="Rotation using complex numbers" );
# arrange and display
display( plot( plt1, plt2, layout = (1, 2), size=(600*2, 600) ))
The Error:
The Thing I want:
I have to implement a product of complex numbers and this should be used to calculate the rotation with complex numbers.
Should look like that:
What do I have to change so that the BoundsError is fixed?
Don't know what exactly i do wrong because of the poorly information i get from this error log.
Greetings and thanks for the help.

prevPoint[1] is a scalar while your function ComplexProduct expects something that has 2 elements. Perhaps you wanted to pass prevPoint instead of prevPoint[1]?
BTW you use incorrect naming pattern. CamelNaming is discouraged for Julia.
Your variable should be named prev_point and your function should be named complex_product.

Fixed the bug by changing the following code:
newPoint = ComplexProduct(prevPoint, W)
in line 92

3D surface plot in R, given x,y,z coordinates

I have the following data-set, and need to plot a surface based on this set of data (of 60 3D points). Here, X, Y is the horizontal plane coordinates, and Z is the vertical / height coordinate.
p = read.csv("points.csv")
PTS X Y Z
1 101 481897.9 5456408 94.18695
2 102 481888.8 5456417 94.30702
3 103 481877.0 5456410 94.29034
4 104 481879.9 5456425 94.25546
5 105 481872.7 5456424 94.09370
After looking through several posts and trying to use functions in several libraries, I still cannot figure out a way to properly plot the surface. I've tried the following:
library(plotly)
plot_ly( y= Y, x = X, z = Z, data=p, type = "surface") #returns empty graphic frame
PX = data.matrix(p$X)
PY = data.matrix(p$Y)
PZ = data.matrix(p$Z)
library(plot3D)
surf3D(PX, PY, PZ)
#returns: Error in if (is.na(var)) ispresent <- FALSE else if (length(var) == 1) if (is.logical(var)) if (!var) ispresent <- FALSE :
argument is of length zero
library(lattice)
wireframe(p$Z ~ p$X*p$Y, data = p) #returns just a cube
library(rgl)
surface3d(p$X,p$Y,p$Z)
#returns: Error in rgl.surface(x = c(481897.916, 481888.8482, 481876.9524, 481879.9393, : y' length != 'x' rows * 'z' cols;
#although there are 60 data points in the form (X,Y,Z) in the data set, with no points missing any coordinate
I must have been doing something horribly wrong here. Would anyone mind to point out what the mistake is?

You cannot make a 3D surface plot with this data because to do it you have to have Z value for each (X,Y) couple, like this :
X1 X2 X3 ... Xn
Y1 Z11 Z12 Z13 ... Z1n
Y2 Z21 Z22 Z23 ... Z2n
Y3 Z31 Z32 Z33 ... Z3n
. .
. .
. .
Ym Zm1 Zm2 Zm3 ... Zmn
For example you don't have Z value for (481897.9,5456417) couple.
So, all you can do is a scatter3d plot :
plot_ly(data = p,x = X,y = Y, z = Z,type = "scatter3d",showlegend = FALSE)

3D with value interpolation in R (X, Y, Z, V)

Is there an R package that does X, Y, Z, V interpolation? I see that Akima does X, Y, V but I need one more dimension.
Basically I have X,Y,Z coordinates plus the value (V) that I want to interpolate. This is all GIS data but my GIS does not do voxel interpolation
So if I have a point cloud of XYZ coordinates with a value of V, how can I interpolate what V would be at XYZ coordinate (15,15,-12) ? Some test data would look like this:
X <-rbind(10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,20,20,20,20,20,20,20,20,20,20,20,20,20,20,20,20,20,20,20,20,20,20,20,20,20,30,30,30,30,30,30,30,30,30,30,30,30,30,30,30,30,30,30,30,30,30,30,30,30,30,40,40,40,40,40,40,40,40,40,40,40,40,40,40,40,40,40,40,40,40,40,40,40,40,40,50,50,50,50,50,50,50,50,50,50,50,50,50,50,50,50,50,50,50,50,50,50,50,50,50,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,20,20,20,20,20,20,20,20,20,20,20,20,20,20,20,20,20,20,20,20,20,20,20,20,20,30,30,30,30,30,30,30,30,30,30,30,30,30,30,30,30,30,30,30,30,30,30,30,30,30,40,40,40,40,40,40,40,40,40,40,40,40,40,40,40,40,40,40,40,40,40,40,40,40,40,50,50,50,50,50,50,50,50,50,50,50,50,50,50,50,50,50,50,50,50,50,50,50,50,50,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,20,20,20,20,20,20,20,20,20,20,20,20,20,20,20,20,20,20,20,20,20,20,20,20,20,30,30,30,30,30,30,30,30,30,30,30,30,30,30,30,30,30,30,30,30,30,30,30,30,30,40,40,40,40,40,40,40,40,40,40,40,40,40,40,40,40,40,40,40,40,40,40,40,40,40,50,50,50,50,50,50,50,50,50,50,50,50,50,50,50,50,50,50,50,50,50,50,50,50,50)
Y <- rbind(10,10,10,10,10,20,20,20,20,20,30,30,30,30,30,40,40,40,40,40,50,50,50,50,50,10,10,10,10,10,20,20,20,20,20,30,30,30,30,30,40,40,40,40,40,50,50,50,50,50,10,10,10,10,10,20,20,20,20,20,30,30,30,30,30,40,40,40,40,40,50,50,50,50,50,10,10,10,10,10,20,20,20,20,20,30,30,30,30,30,40,40,40,40,40,50,50,50,50,50,10,10,10,10,10,20,20,20,20,20,30,30,30,30,30,40,40,40,40,40,50,50,50,50,50,10,10,10,10,10,20,20,20,20,20,30,30,30,30,30,40,40,40,40,40,50,50,50,50,50,10,10,10,10,10,20,20,20,20,20,30,30,30,30,30,40,40,40,40,40,50,50,50,50,50,10,10,10,10,10,20,20,20,20,20,30,30,30,30,30,40,40,40,40,40,50,50,50,50,50,10,10,10,10,10,20,20,20,20,20,30,30,30,30,30,40,40,40,40,40,50,50,50,50,50,10,10,10,10,10,20,20,20,20,20,30,30,30,30,30,40,40,40,40,40,50,50,50,50,50,10,10,10,10,10,20,20,20,20,20,30,30,30,30,30,40,40,40,40,40,50,50,50,50,50,10,10,10,10,10,20,20,20,20,20,30,30,30,30,30,40,40,40,40,40,50,50,50,50,50,10,10,10,10,10,20,20,20,20,20,30,30,30,30,30,40,40,40,40,40,50,50,50,50,50,10,10,10,10,10,20,20,20,20,20,30,30,30,30,30,40,40,40,40,40,50,50,50,50,50,10,10,10,10,10,20,20,20,20,20,30,30,30,30,30,40,40,40,40,40,50,50,50,50,50)
Z <- rbind(-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-5,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-17,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29,-29)
V <- rbind(0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,5,25,35,75,25,50,0,0,0,0,0,10,12,17,22,27,32,37,25,13,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,50,125,130,105,110,115,165,180,120,100,80,60,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0)

I had the same question and was hoping for an answer in R.
My question was: How do I perform 3D (trilinear) interpolation using regular gridded coordinate/value data (x,y,z,v)? For example, CT images, where each image has pixel centers (x, y) and greyscale value (v) and there are multiple image "slices" (z) along the thing being imaged (e.g., head, torso, leg, ...).
There is a slight problem with the given example data.
# original example data (reformatted)
X <- rep( rep( seq(10, 50, by=10), each=25), 3)
Y <- rep( rep( seq(10, 50, by=10), each=5), 15)
Z <- rep(c(-5, -17, -29), each=125)
V <- rbind(0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,5,25,35,75,25,50,0,0,0,0,0,10,12,17,22,27,32,37,25,13,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,50,125,130,105,110,115,165,180,120,100,80,60,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0)
# the dimensions of the 3D grid described do not match the number of values
(length(unique(X))*length(unique(Y))*length(unique(Z))) == length(V)
## [1] FALSE
## which makes sense since 75 != 375
# visualize this:
library(rgl)
plot3d(x=X, y=Y, z=Z, col=terrain.colors(181)[V])
# examine the example data real quick...
df <- data.frame(x=X,y=Y,z=Z,v=V);
head(df);
table(df$x, df$y, df$z);
# there are 5 V values at each X,Y,Z coordinate... duplicates!
# redefine Z so there are 15 unique values
# making 375 unique coordinate points
# and matching the length of the given value vector, V
df$z <- seq(-5, -29, length.out=15)
head(df)
table(df$x, df$y, df$z);
# there is now 1 V value at each X,Y,Z coordinate
# that was for testing, now actually redefine the Z vector.
Z <- rep(seq(-5,-29, length.out = 15), 25)
# plot it.
library(rgl)
plot3d(x=X, y=Y, z=Z, col=terrain.colors(181)[V])
I couldn't find any 4D interpolation functions in the usual R packages, so I wrote a quick and dirty one. The following implements (without ANY error checking... caveat emptor!) the technique described at: https://en.wikipedia.org/wiki/Trilinear_interpolation
# convenience function #1:
# define a function that takes a vector of lookup values and a value to lookup
# and returns the two lookup values that the value falls between
between = function(vec, value) {
# extract list of unique lookup values
u = unique(vec)
# difference vector
dvec = u - value
vals = c(u[dvec==max(dvec[dvec<0])], u[dvec==min(dvec[dvec>0])])
return(vals)
}
# convenience function #2:
# return the value (v) from a grid data.frame for given point (x, y, z)
get_value = function(df, xi, yi, zi) {
# assumes df is data.frame with column names: x, y, z, v
subset(df, x==xi & y==yi & z==zi)$v
}
# inputs df (x,y,z,v), points to look up (x, y, z)
interp3 = function(dfin, xin, yin, zin) {
# TODO: check if all(xin, yin, zin) equals a grid point, if so just return the point value
# TODO: check if any(xin, yin, zin) equals a grid point, if so then do bilinear or linear interp
cube_x <- between(dfin$x, xin)
cube_y <- between(dfin$y, yin)
cube_z <- between(dfin$z, zin)
# find the two values in each dimension that the lookup value falls within
# and extract the cube of 8 points
tmp <- subset(dfin, x %in% cube_x &
y %in% cube_y &
z %in% cube_z)
stopifnot(nrow(tmp)==8)
# define points in a periodic and cubic lattice
x0 = min(cube_x); x1 = max(cube_x);
y0 = min(cube_y); y1 = max(cube_y);
z0 = min(cube_z); z1 = max(cube_z);
# define differences in each dimension
xd = (xin-x0)/(x1-x0); # 0.5
yd = (yin-y0)/(y1-y0); # 0.5
zd = (zin-z0)/(z1-z0); # 0.9166666
# interpolate along x:
v00 = get_value(tmp, x0, y0, z0)*(1-xd) + get_value(tmp,x1,y0,z0)*xd # 2.5
v01 = get_value(tmp, x0, y0, z1)*(1-xd) + get_value(tmp,x1,y0,z1)*xd # 0
v10 = get_value(tmp, x0, y1, z0)*(1-xd) + get_value(tmp,x1,y1,z0)*xd # 0
v11 = get_value(tmp, x0, y1, z1)*(1-xd) + get_value(tmp,x1,y1,z1)*xd # 65
# interpolate along y:
v0 = v00*(1-yd) + v10*yd # 1.25
v1 = v01*(1-yd) + v11*yd # 32.5
# interpolate along z:
return(v0*(1-zd) + v1*zd) # 29.89583 (~91.7% between v0 and v1)
}
> interp3(df, 15, 15, -12)
[1] 29.89583
Testing that same source's assertion that trilinear is simply linear(bilinear(), bilinear()), we can use the base R linear interpolation function, approx(), and the akima package's bilinear interpolation function, interp(), as follows:
library(akima)
approx(x=c(-11.857143,-13.571429),
y=c(interp(x=df[round(df$z,1)==-11.9,"x"], y=df[round(df$z,1)==-11.9,"y"], z=df[round(df$z,1)==-11.9,"v"], xo=15, yo=15)$z,
interp(x=df[round(df$z,1)==-13.6,"x"], y=df[round(df$z,1)==-13.6,"y"], z=df[round(df$z,1)==-13.6,"v"], xo=15, yo=15)$z),
xout=-12)$y
# [1] 0.2083331
Checked another package to triangulate:
library(oce)
Vmat <- array(data = V, dim = c(length(unique(X)), length(unique(Y)), length(unique(Z))))
approx3d(x=unique(X), y=unique(Y), z=unique(Z), f=Vmat, xout=15, yout=15, zout=-12)
[1] 1.666667
So 'oce', 'akima' and my function all give pretty different answers. This is either a mistake in my code somewhere, or due to differences in the underlying Fortran code in the akima interp(), and whatever is in the oce 'approx3d' function that we'll leave for another day.
Not sure what the correct answer is because the MWE is not exactly "minimum" or simple. But I tested the functions with some really simple grids and it seems to give 'correct' answers. Here's one simple 2x2x2 example:
# really, really simple example:
# answer is always the z-coordinate value
sdf <- expand.grid(x=seq(0,1),y=seq(0,1),z=seq(0,1))
sdf$v <- rep(seq(0,1), each=4)
> interp3(sdf,0.25,0.25,.99)
[1] 0.99
> interp3(sdf,0.25,0.25,.4)
[1] 0.4
Trying akima on the simple example, we get the same answer (phew!):
library(akima)
approx(x=unique(sdf$z),
y=c(interp(x=sdf[sdf$z==0,"x"], y=sdf[sdf$z==0,"y"], z=sdf[sdf$z==0,"v"], xo=.25, yo=.25)$z,
interp(x=sdf[sdf$z==1,"x"], y=sdf[sdf$z==1,"y"], z=sdf[sdf$z==1,"v"], xo=.25, yo=.25)$z),
xout=.4)$y
# [1] 0.4
The new example data in the OP's own, accepted answer was not possible to interpolate with my simple interp3() function above because:
(a) the grid coordinates are not regularly spaced, and
(b) the coordinates to lookup (x1, y1, z1) lie outside of the grid.
# for completeness, here's the attempt:
options(scipen = 999)
XCoor=c(78121.6235,78121.6235,78121.6235,78121.6235,78136.723,78136.723,78136.723,78136.8969,78136.8969,78136.8969,78137.4595,78137.4595,78137.4595,78125.061,78125.061,78125.061,78092.4696,78092.4696,78092.4696,78092.7683,78092.7683,78092.7683,78092.7683,78075.1171,78075.1171,78064.7462,78064.7462,78064.7462,78052.771,78052.771,78052.771,78032.1179,78032.1179,78032.1179)
YCoor=c(5213642.173,523642.173,523642.173,523642.173,523594.495,523594.495,523594.495,523547.475,523547.475,523547.475,523503.462,523503.462,523503.462,523426.33,523426.33,523426.33,523656.953,523656.953,523656.953,523607.157,523607.157,523607.157,523607.157,523514.671,523514.671,523656.81,523656.81,523656.81,523585.232,523585.232,523585.232,523657.091,523657.091,523657.091)
ZCoor=c(-3.0,-5.0,-10.0,-13.0,-3.5,-6.5,-10.5,-3.5,-6.5,-9.5,-3.5,-5.5,-10.5,-3.5,-5.5,-7.5,-3.5,-6.5,-11.5,-3.0,-5.0,-9.0,-12.0,-6.5,-10.5,-2.5,-3.5,-8.0,-3.5,-6.5,-9.5,-2.5,-6.5,-8.5)
V=c(2.4000,30.0,620.0,590.0,61.0,480.0,0.3700,0.0,0.3800,0.1600,0.1600,0.9000,0.4100,0.0,0.0,0.0061,6.0,52.0,0.3400,33.0,235.0,350.0,9300.0,31.0,2100.0,0.0,0.0,10.5000,3.8000,0.9000,310.0,0.2800,8.3000,18.0)
adf = data.frame(x=XCoor, y=YCoor, z=ZCoor, v=V)
# the first y value looks like a typo?
> head(adf)
x y z v
1 78121.62 5213642.2 -3.0 2.4
2 78121.62 523642.2 -5.0 30.0
3 78121.62 523642.2 -10.0 620.0
4 78121.62 523642.2 -13.0 590.0
5 78136.72 523594.5 -3.5 61.0
6 78136.72 523594.5 -6.5 480.0
x1=198130.000
y1=1913590.000
z1=-8
> interp3(adf, x1,y1,z1)
numeric(0)
Warning message:
In min(dvec[dvec > 0]) : no non-missing arguments to min; returning Inf

Whether the test data did or not make sense, I still needed an algorithm. Test data is just that, something to fiddle with and as a test data it was fine.
I wound up programming it in python and the following code takes XYZ V and does a 3D Inverse Distance Weighted (IDW) interpolation where you can set the number of points used in the interpolation. This python recipe only interpolates to one point (x1, y1, z1) but it is easy enough to extend.
import numpy as np
import math
#34 points
XCoor=np.array([78121.6235,78121.6235,78121.6235,78121.6235,78136.723,78136.723,78136.723,78136.8969,78136.8969,78136.8969,78137.4595,78137.4595,78137.4595,78125.061,78125.061,78125.061,78092.4696,78092.4696,78092.4696,78092.7683,78092.7683,78092.7683,78092.7683,78075.1171,78075.1171,78064.7462,78064.7462,78064.7462,78052.771,78052.771,78052.771,78032.1179,78032.1179,78032.1179])
YCoor=np.array([5213642.173,523642.173,523642.173,523642.173,523594.495,523594.495,523594.495,523547.475,523547.475,523547.475,523503.462,523503.462,523503.462,523426.33,523426.33,523426.33,523656.953,523656.953,523656.953,523607.157,523607.157,523607.157,523607.157,523514.671,523514.671,523656.81,523656.81,523656.81,523585.232,523585.232,523585.232,523657.091,523657.091,523657.091])
ZCoor=np.array([-3.0,-5.0,-10.0,-13.0,-3.5,-6.5,-10.5,-3.5,-6.5,-9.5,-3.5,-5.5,-10.5,-3.5,-5.5,-7.5,-3.5,-6.5,-11.5,-3.0,-5.0,-9.0,-12.0,-6.5,-10.5,-2.5,-3.5,-8.0,-3.5,-6.5,-9.5,-2.5,-6.5,-8.5])
V=np.array([2.4000,30.0,620.0,590.0,61.0,480.0,0.3700,0.0,0.3800,0.1600,0.1600,0.9000,0.4100,0.0,0.0,0.0061,6.0,52.0,0.3400,33.0,235.0,350.0,9300.0,31.0,2100.0,0.0,0.0,10.5000,3.8000,0.9000,310.0,0.2800,8.3000,18.0])
def Distance(x1,y1,z1, Npoints):
i=0
d=[]
while i < 33:
d.append(math.sqrt((x1-XCoor[i])*(x1-XCoor[i]) + (y1-YCoor[i])*(y1-YCoor[i]) + (z1-ZCoor[i])*(z1-ZCoor[i]) ))
i = i + 1
distance=np.array(d)
myIndex=distance.argsort()[:Npoints]
weightedNum=0
weightedDen=0
for i in myIndex:
weightedNum=weightedNum + (V[i]/(distance[i]*distance[i]))
weightedDen=weightedDen + (1/(distance[i]*distance[i]))
InterpValue=weightedNum/weightedDen
return InterpValue
x1=198130.000
y1=1913590.000
z1=-8
print(Distance(x1,y1,z1, 12))

How can i get x or y between two points?

My question is about points is javafx:
I have two points for example:
Point p1 = new Point(2,2);
Point p2 = new Point(10,8);
And i draw a imaginary line between them, i wanna know if i say x = 8 what y is on the imaginary line or otherwise is i know y, what x is on the imaginary line.
I found a formula
x= x1 + blend * (x2 - x1),
and blend is the percentage on the line.
But i wanna have this kind of formula where i can give a Y.
It there some kind of formula to solve this problem?
Thanks in advance!

Since your imaginary line is an equation in the format y = ax+b, you can solve it like this http://www.wikihow.com/Calculate-Slope-and-Intercepts-of-a-Line
given
10,8
and
2,2
y = ax + b
so
8 = 10a + b
-
2 = 2a + b
------------
6 = 8a
a = 6/8 = 0.75
and
8 = 10a + b
8 = 10*(0.75) + b
8 = 7.5 + b
8 - 7.5 = b
0.5 = b
so
y = 0.75*x + 0.5

Scalar field visualisation in R

I have a table with 3 columns
x y f
-101.0 -101.0 0.0172654144157
...
x and y are coordinates. f is value.
I want to make a 2d picture, where x and y are coordinates and f is a colour. But I need this picture to be not a number of coloured points, but a continuous schedule.
Help me someone please

There are a couple of simple ways to do this if you have a regular grid with your data. Try:
require(ggplot2)
require(lattice)
# make some data
s = 100
i = 0.5
x0 <- 27
y0 <- 34
df <- expand.grid(x=seq(0,s,i), y=seq(0,s,i))
df <- transform(df, f = cos( 10*pi * sqrt((x - x0)^2 + (y-y0)^2)))
# try as points
ggplot(df,aes(x,y,color=f)) + geom_point()
# or as tile
ggplot(df,aes(x,y,fill=f)) + geom_tile()
# or even easier with lattice
levelplot(f ~ x * y, df)
Output examples:

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