three.js webgl custom shader sharing texture with new offset

Question

I am splitting a texture 1024 x 1024 over 32x32 tiles * 32, Im not sure if its possible to share the texture with an offset or would i need to create a new texture for each tile with the offset..

to create the offset i am using a uniform value = 32 * i and updating the uniform through each loop instance of creating tile, all the tiles seem to be the same offset? as basically i wanting an image to appear like its one image not broken up into little tiles.But the current out-put is the same x,y-offset on all 32 tiles..Im using the vertex-shader with three.js r71...

Would i need to create a new texture for each tile with the offset?

      for ( j = 0; j < row; j ++ ) {

           for ( t = 0; t < col; t ++ ) {

                customUniforms.tX.value = tX; 
                customUniforms.tY.value = tY;
                console.log(customUniforms.tX.value);
                customUniforms.tX.needsUpdate = true;
                customUniforms.tY.needsUpdate = true;
                mesh = new THREE.Mesh( geometry,mMaterial);// or new material
           }
       }

       //vertex shader :
       vec2 uvOffset = vUV + vec2( tX, tY) ;    

Image example:

Each image should have an offset of 10 0r 20 px but they are all the same.... this is from using one texture..

As suggested i have tried to manipulate the uv on each object with out luck, it seems to make all the same vertexes have the same position for example 10x10 segmant plane all faces will be the same

    var geometry = [
    [ new THREE.PlaneGeometry( w, w ,64,64),50 ],
    [ new THREE.PlaneGeometry( w, w ,40,40), 500 ],
    [ new THREE.PlaneGeometry( w, w ,30,30), 850 ],
    [ new THREE.PlaneGeometry( w, w,16,16 ), 1200 ]
];

    geometry[0][0].faceVertexUvs[0] = [];

    for(var p = 0; p < geometry[0][0].faces.length; p++){ 
        geometry[0][0].faceVertexUvs[0].push([
        new THREE.Vector2(0.0, 0.0),
        new THREE.Vector2(0.0, 1),
        new THREE.Vector2( 1, 1 ), 
        new THREE.Vector2(1.0, 0.0)]);  
    }

image of this result, you will notice all vertices are the same when they shouldn't be

Update again: I have to go through each vertices of faces as two triangles make a quad to avoid the above issue, I think i may have this solved... will update

Last Update Hopfully: Below is the source code but i am lost making the algorithm display the texture as expected.

            /*
            j and t are rows & columns looping by 4x4 grid
            row = 4 col = 4;
            */

            for( i = 0; i < geometry.length; i ++ ) {
            var mesh = new THREE.Mesh( geometry[ i ][ 0 ], customMaterial);
            mesh.geometry.computeBoundingBox();
            var max     =  mesh.geometry.boundingBox.max;
            var min     =  mesh.geometry.boundingBox.min;
            var offset  = new THREE.Vector2(0 - min.x*t*j+w, 0- min.y*j+w);//here is my issue
            var range   = new THREE.Vector2(max.x - min.x*row*2, max.y - min.y*col*2);
            mesh.geometry.faceVertexUvs[0] = [];
            var faces =  mesh.geometry.faces;

            for (p = 0; p <  mesh.geometry.faces.length ; p++) {
                var v1 =  mesh.geometry.vertices[faces[p].a];
                var v2 =  mesh.geometry.vertices[faces[p].b];
                var v3 =  mesh.geometry.vertices[faces[p].c];
                mesh.geometry.faceVertexUvs[0].push([
                new THREE.Vector2( ( v1.x + offset.x ) / range.x , ( v1.y + offset.y ) / range.y ),
                new THREE.Vector2( ( v2.x + offset.x ) / range.x , ( v2.y + offset.y ) / range.y ),
                new THREE.Vector2( ( v3.x + offset.x ) / range.x , ( v3.y + offset.y ) / range.y )
                ]);

            }

You will notice the below image in the red is seamless as the other tiles are not aligned with the texture.

Answer 1

Here is the answer:

      var offset  = new THREE.Vector2(w - min.x-w+(w*t), w- min.y+w+(w*-j+w));
      var range   = new THREE.Vector2(max.x - min.x*7, max.y - min.y*7);

if you could simplify answer will award bounty too: