Plotting multiple pm3d surfaces each having its own palettes in gnuplot

The development branch of gnuplot supports multiple named palettes. The method shown here, however, works on earlier versions of gnuplot also. It uses the fill style to provide a color (rather than the pm3d palette), and shows how to define the fill colors so that they mimic set palette defined(). This demo constructs only one mapping, but you could define several mappings each with its own array of colors and mapping function to use them.

This demo is extracted from the full demo for named palettes in the development branch. If you are interested you can find the full demo here: Version 5.5 named palette demo

#
# Demonstrate construction and use of a separate palette
# Ethan A Merritt - May 2020
#
# Method 1:
#       The first method works also in 5.2 but requires "lc rgb variable"
#       rather than the more natural "fillcolor rgb variable".
#       "set pm3d interpolate" breaks the color mapping of this method

#
# This creates a palette equivalent to
#       set palette defined (0 "dark-blue", 1 "white")
#
array blues[256]
do for [i=1:256] {
    blues[i] = int( (0x7f + (i-1)/(255.) * 0xffff80) );
}

#
# This is the equivalent of
#       set cbrange [-1:1]
blues_min = -1
blues_max = 1

#
# This function maps z onto a palette color
#
blues(z) = (z <= blues_min) ? blues[1] \
         : (z >= blues_max) ? blues[256] \
         : blues[ floor(255. * (z-blues_min)/(blues_max-blues_min)) + 1]

foo(x,y) = sin(x*y)

set samples 41
set isosamples 41
unset colorbox
set cbrange [-1:1]
set xrange [0:5]; set urange [0:5]
set yrange [0:5]; set vrange [0:5]

set title "Use hand-constructed 'blues' palette via rgb variable"

splot '++' using 1:2:(foo($1,$2)):(blues(foo($1,$2))) with pm3d fillcolor rgb variable \
           title "pm3d using 1:2:3:4 with pm3d fillcolor rgb variable"

Maybe you could define the palette differently with set palette defined but then you probably would have to combine your 3 palettes into 1 palette and you would lose "color resolution", as far as I know a palette has 256 color steps. To be honest, I haven't thought about this in detail.

I checked again the code you referenced... apparently an additional line will do the "trick". Then you can plot with pm3d.

set pm3d depthorder 

Code: (slightly modified code from here: https://stackoverflow.com/a/57501649/7295599)

### multiple "palettes" within one splot command
reset session

set samples 101,101
set isosamples 101,101 

f(x,y) = sin(1.3*x)*cos(0.9*y)+cos(.8*x)*sin(1.9*y)+cos(y*.2*x)
set table $Data01
    splot f(x,y)
unset table

g(x,y) = y
set table $Data02
    splot g(x,y)
unset table

h(x,y) = 0.5*x
set table $Data03
    splot h(x,y)
unset table

Zmin = -3
Zmax= 3

set xrange[-5:5]
set yrange[-5:5]
set zrange[Zmin:Zmax]
set hidden3d
set angle degree

Frac(z) = (z-Zmin)/(Zmax-Zmin)
# MyPalette01
Red01(z) = 65536 * ( Frac(z) > 0.75 ? 255 : int(255*abs(2*Frac(z)-0.5)))
Green01(z) = int(255*sin(180*Frac(z)))*256
Blue01(z) = int(255*cos(90*Frac(z)))
MyPalette01(z) =  Red01(z) + Green01(z) + Blue01(z) 

# MyPalette02
Red02(z)   = 65536 * int(255*Frac(z))
Green02(z) = 256 * (Frac(z) > 0.333 ? 255 : int(255*Frac(z)*3))
Blue02(z)  = (Frac(z) > 0.5 ? 255 : int(255*Frac(z)*2))
MyPalette02(z) =  Red02(z) + Green02(z) + Blue02(z) 

# MyPalette03
Red03(z)   = 65536 * (Frac(z) > 0.5 ? 255 : int(255*Frac(z)*2))
Green03(z) = 256 * (Frac(z) > 0.333 ? 255 : int(255*Frac(z)*3))  
Blue03(z)  = int(255*Frac(z)) 
MyPalette03(z) =  Red03(z) + Green03(z) + Blue03(z) 

set pm3d depthorder
unset colorbox 

set view 44,316
splot $Data01 u 1:2:3:(MyPalette01($3)) w pm3d lc rgb var notitle, \
      $Data02 u 1:2:3:(MyPalette02($3)) w pm3d lc rgb var notitle, \
      $Data03 u 1:2:3:(MyPalette03($3)) w pm3d lc rgb var notitle
### end of code

Result: