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We have an older Biesse Rover with a C-Axis, equipped with a fixed inclinable aggregate set to 45 degrees. In Inventor, we've drawn a 3/4" thick panel and put a 3/4" chamfer along the edge at X=0 using the Pocket2D strategy, and to achieve this, we've set the Tool Orientation. This works nicely, and in simulation the tool is at the correct angle. Everything looks great.
Where I am stuck in the post is that I am getting incorrect results for my X and Z in the onLinear() entry function. It appears that the Inventor Tool Orientation is implemented in the post by tilting the workPlane, which equals [[0.7071, 0, -0.7071], [0, 1, 0], [0.7071, 0, 0.7071]], and that makes sense, because I have tilted the tool in X and Z, but in onLinear(), for instance, Z goes to zero and stays there. But I need Z to increase as X increases to affect my 45 degree slope.
I feel like I need to transform the points in onLinear() in the inverse direction of the tilt of the workPlane, and so I have tried this (and many other things) ...
function onLinear(x, y, z, feed) {
// simplified
// get the work plane
var w = currentSection.workPlane;
// get workplane inverse and multiply by my point
var v = w.getTransposed().multiply(new Vector(x, y, z));
// resulting vector does not yield the points I need
writeBlock('testing', v);
}
... but I am well and truly stumped. If anyone can offer some insight into how to approach this problem, it would be greatly appreciated.
Thanks,
Mark
Solved! Go to Solution.
