MeshFunctions::invmeshf that resists other solutions

Question

I'm trying to base mesh on the slope of the function. That's why there is a Cross of two partial derivatives.

With[{R = Function[If[Abs[#1] < \[Pi]/4, 1, Sec[Abs[#1] - \[Pi]/4]]]},
  With[{F = 
    Function @@ {{\[CurlyPhi], 
       h}, {Sin[\[CurlyPhi]] R[\[CurlyPhi], h], 
       Cos[\[CurlyPhi]] R[\[CurlyPhi], h], h}}}, 
  ParametricPlot3D[
     F[\[CurlyPhi], h], {\[CurlyPhi], -(\[Pi]/2), \[Pi]/2}, {h, -1, 
      1}, MeshFunctions -> #1, 
     Mesh -> {{0}}] &@{Function @@ {{\[CurlyPhi], h}, 
      Norm[{#1, #2, #3}] & @@ Cross[\!\(
\*SubscriptBox[\(\[PartialD]\), \(h\)]\(F[\[CurlyPhi], h]\)\), \!\(
\*SubscriptBox[\(\[PartialD]\), \(\[CurlyPhi]\)]\(F[\[CurlyPhi], 
          h]\)\)]}}]]

The graph is plotted, but the mesh is not, and there is an error message:

MeshFunctions::invmeshf: "MeshFunctions->Function[{φ,h},Sqrt[Abs[Cos[<<1>>] If[<<3>>]+If[<<3>>] Sin[<<1>>]]^2+Abs[-Cos[<<1>>] If[<<3>>]+If[<<3>>] Sin[<<1>>]]^2]] must be a pure function or a list of pure functions."

Quite suddenly, when R is replaced by Function[1], the error disappears. But I don't need 1.

I tried renaming the arguments to the mesh function (to no avail), I turned the option into an argument to ParametricPlot3D& (to no avail), I turned Function[] into Function@@{} (to no avail). Now I'm at a loss.

Answer 1

Sorry for my careless.

Maybe Abs can't be differential in Mathematica,so we have to use RealAbs instead.

And it seems that we need to set Mesh -> {{1}}.

With[{R = If[RealAbs[#1] < Pi/4, 1, Sec[RealAbs[#1] - Pi/4]] & }, 
 With[{F = 
    Function @@ {{φ, 
       h}, {Sin[φ]*R[φ, h], 
       Cos[φ]*R[φ, h], h}}}, 
     (ParametricPlot3D[
      F[φ, h], {φ, -(Pi/2), Pi/2}, {h, -1, 1}, 
      MeshFunctions -> #1, MeshStyle -> {Thick, Red}, 
      Mesh -> {{1}}] & )[
       {Function @@ {{φ, h}, (Norm[{#1, #2, #3}] & ) @@ 
       Cross[D[F[φ, h], h], 
        D[F[φ, h], φ]]}}]]]