# Optimization with Matrix value constraint

I am trying to find the orientation and position of a polygon resting on a curved surface under gravity.The polygon MUST not be penetrated by any part of the surface below. Here is an example for a square shape.

The curved surface is defined by a 2D Matrix: Tdata. The problem can be formulated to a constrained Optimization problem. The initial square can be approximated by five points(four vertices and one center point):

ConVer = {{l/2, w/2, 0}, {l/2, -w/2,0}, {-l/2, -w/2, 0}, {-l/2, w/2,0}, {0, 0, 0}};


Where l is the length and w is the width of the Square.

The square's orientation is parameterized by three angles:

Rθ = RotationMatrix[θ, {0, 0, 1}];
Rψ = RotationMatrix[ψ, {0, 1, 0}];
Rϕ = RotationMatrix[ϕ, {1, 0, 0}];


Then the parameterized square is

For[i = 1, i < 6, i++, ConVer[[i]] =Rθ.Rψ.Rϕ.ConVer[[i]] + {x, y,z}];


Where {x ,y ,z} is the coordinates of the center point of the square.

The optimization uses NMinimize function:

x = 1; y = 1; θ = 0;
NMinimize[{z,
Tdata[(ConVer[[1]])[[1 ;; 2]]] <= (ConVer[[1]])[[3]] &&
Tdata[(ConVer[[2]])[[1 ;; 2]]] <= (ConVer[[2]])[[3]] &&
Tdata[(ConVer[[3]])[[1 ;; 2]]] <= (ConVer[[3]])[[3]] &&
Tdata[(ConVer[[4]])[[1 ;; 2]]] <= (ConVer[[4]])[[3]] &&
Tdata[(ConVer[[5]])[[1 ;; 2]]] <= (ConVer[[5]])[[3]]},
{{ψ, -Pi/4, Pi/4}, {ϕ, -Pi/4, Pi/4}, {z, 0, 1}}];


When Tdata is a 2d function this works well, but as to 2d Matrix error occurs! Why is that? I also tried to solve this problem in Matlab using 'fmincon', and Matlab can return desired result.

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You need to leave a blank line before code snippets, otherwise they are treated as part of the text of the previous paragraph. – Rahul Nov 11 '13 at 8:26
What type of error? – cormullion Nov 11 '13 at 8:33
NMinimize::bcons, which means my constraints are invalid – novice Nov 11 '13 at 8:36
Please give an example on how do you want to define Tdata[] as a "2D matrix" – Dr. belisarius Nov 11 '13 at 13:40
The curved surface is defined by Tdata as follows:f[{x_, y_}] := ( Cos[x] + Sin[y])/3; Tdata = Flatten[ Table[Flatten[{i, j, N[f /@ {{i, j}}]}], {i, 0, 6.5, 0.5}, {j, 0, 6.5, 0.5}], 1] – novice Nov 12 '13 at 0:41