Math

A project in the shop just got complicated. The project is in a three axis Matsuura vertical mill with a manual a and b axis; tilt rotary table that is, calibrated in degrees min sec. The part has several facets that form compound angles. Two of the angles only require the fixture to tilt but the last facet requires movement in all five axis , even though it is purely positional 4th and 5th. My question is: which mathematical strategies best apply them selves to solving the system? Because of the rotatory table atop the tilting fourth the usual Signe plate calculations; obviously, do not apply. What do the NC control computers use to calculate the movement of the fourth and fifth axis?
Thank you for any suggestions.
Rex

Do your math for a single angle and then take that answer and use it in the next calculation. This is the method I use to do compound angle roll dimensions on a double sine plate set up. Not a compound sine plate but two sine plates one atop the other with the tilt axis perpendicular to one another.

Not sure if this will help or not but it is difficult with next to no information.

Joe

Its still all just trigonometry... With a little geometry thrown in for good measure...

You need to break it down into pieces and do it step by step, and it is a bit more complicated than a simple sine calculation, but
not by all that much.

You could make a CAD model of your setup, including the indexers/rotary tables. Then measure the change in coordinates from within the CAD program.

Best would be use a program like HSMWorks to generate the code off of a model of the entire setup, indexers included. Then set the work offset from a known datum on the indexer or table, and let the CAM program figure out the coordinate shifts for you based on the CAD model. This could be considered a poor man's 3+2 machining.

I'm not saying you shouldn't know how to do the shop math (trig and geometry) necessary to figure this out, but for a job like this there are more reliable, faster ways to get the same result.

Yes, it is simple geometry. This problem requires two rotations with respect two lines: the center of a and the center of rotation of b. However the goal is to get a plain that is "somewhere" in the work cube to be parallel to the table. I have a cad model drawn and am going through the laborious process of crunching the numbers but I can not help but think that there is a more elegant mathematical solution.
Thanks to all.

Rex Rat --

The first question you'll need to answer is the orientation of the cut-plane relative to the machine's (X,Y,Z) coordinate frame. Will the cut-plane be parallel to the table top, which is to say in the X,Y plane of a vertical-spindle machine, in the Y,Z plane, or in the Z,X plane?

The next question is how the rotational positioner needs to be set to orient the workpiece so that the nominal surface to be cut is parallel to the machine's cut-plane. You'll need to know a) how the workpiece coordinate frame is oriented relative to the coordinate frame of the rotational positioner's base, when both positioner axes are set to zero, b) how the position base coordinate frame -- which does not change position as the positioner angles are changed -- is oriented relative to the machine's linear coordinate frame, and c) how the surface to be cut is oriented with respect to the workpiece's coordinate frame.

Once you know those things, it's a matter of doing the trigonometry or linear algebra needed to determine the Tilt Position and Rotary Table Position need to be set to position the to-be-cut surface parallel to the machine cut-plane.

Three more hints: 1) Consider the rotational positioner's Tilt and Rotary Table angles as a Spherical Pair, with one axis fixed with respect to the machine, the other axis fixed with respect to the workpiece. 2) Your rotational positioner's rotation axes, if perpendicular to each other, can be oriented in one of two ways, either a) Rotary over Tilt, with the Tilt axis fixed with respect to the machine and the Rotary axis moving in a plane perpendicular to the Tilt axis, or b) Tilt over Rotary, in which case the Rotary axis is fixed with respect to the machine, and the Tilt axis rotates in a plane perpendicular to the Rotary axis. Either one will work, but the algorithms needed to derive the (Rotary Angle, Tilt Angle) pairs for the different configurations are different. 3) Be sure you understand the "sign convention" of the Rotary axis . . . does the indicated position increase or decrease as the table top rotates clockwise? . . . and do you understand how the workpiece orientation changes relative to the machine axes as the Rotary axis position value increases and decreases?

John

I need to maneuver the plane to the XY plane. The tilt axis is parallel with Y and is 4.0311" below the local work coordinate system, has 90deg of freedom from 0 being parallel with the mill table and tilting to the right from the operators prospective. 0deg is the initial condition. It is the base of the fixture.

The rotary table's center is X0 Y0 in the local system. It is calibrated to increase if rotated counterclockwise and has 360deg of freedom. 5deg is the initial condition for the part to be parallel with the X axis.

the plane in question may be defined by these coordinates in the local system:
1)X8.0Y0.84Z1.43
2)X-5.75Y0.62Z3.0
3)X-5.75Y-3.0Z2.0
4)X8.0Y-3.0Z0.370

Rex Rat --

I'm "up to my eyeballs" in a special project, and Excel and my computer had an argument yesterday and still haven't patched up their differences. So I'm not going to be able to explain, in detail and with real numbers, how to do this job. But, I can try to map out a path for you to take.

First, let's assume that when your Tilt Angle and Rotary Table Angle are both zero, the Tilt Axis is parallel to the Y(Machine), and the Rotable Table axis is parallel to the Z(Machine), and the workpiece is mounted on the Rotary Table so that the workpiece axis system is precisely parallel with the Machine axis system.

Now for some arithmetic. The four in-workpiece-plane points you provide Cartesian coordinates for are at the corners of an approximate rectangle. We need to determine how their plane is oriented, and that orientation needs to be described in terms of a Direction (in mathematical jargon, a Direction is commonly described as "A vector of undefined magnitude".), in the workpiece coordinate reference frame. Being neither a machine programmer nor a CAD jockey, I would first define Vector X', which begins at one of the four known points, and ends at the known point diagonally opposite the first point, in the rectangle formed by the four known points. Then I would define Vector Y', which begins at one of the two remaining known points, and ends at the forth known point.

The notation I'd use would start this way: Point 1 known location in workpiece coordinates is (x1, y1, z1), Point 2 known location in workpiece coordinates is (x2, y2, z2), and so on.

Vector X' would be (P3 to P1) = (x1-x3, y1-y3, z1-z3), and Vector Y' would be (P2 to P4) = (x4-x2, y4-y2, z4-z2).

Next, calculate the Cross Product of Vector X' x Vector Y'. The result will be a Vector that is perpendicular to both Vectors X' and Y', and is therefore perpendicular to the plane to which the workpiece is to be cut.

The Cross Product Vector, like the X' and Y' Vectors, will be a Cartesian (x,y,z) triple in the workpiece reference frame, and has both a Direction and Magnitude. It needs to be analytically adjusted to force its Magnitude to 1, making it a Unit Vector. You will need to calculate the Magnitude of the Cross Product Vector as the square root of the squares of the x, y, and z components . . . in other words, squareroot (x times x + y times y + z times z). The x of the Perpendicular Unit Vector will be the x of the Cross Product Vector divided by the Magnitude of the Cross Product Vector, the y of the Perpendicular Unit Vector will be the y of the Cross Product Vector divided by the Magnitude of the Cross Product Vector, and the z of the Perpendicular Unit Vector will be the z of the Cross Product Vector divided by the Magnitude of the Cross Product Vector.

There is one little problem . . . we don't know if the Perpendicular Unit Vector is the PUV going into the workpiece, or thePUV coming out of the workpiece. So we need to develop a second, polar opposite, Perpendicular Unit Vector, by simply inverting the sign of the original Perpendicular Unit Vector's x, y, and z components.

At this point, it's worth verifying that the two Perpendicular Unit Vectors have Magnitudes of 1.000000. If they don't, go back and check all your arithmetic to find your error or errors.

Now you should project the two Perpendicular Unit Vectors onto the workpiece X,Y plane. Take the ArcTangent of the PUV y component divided by the PUV X component -- if your Excel is working, the magic function and syntax is "=Atan2(value-of-PUV-x-component,value-of-PUV-y-component)*180/pi()" to provide the answer in decimally-divided degrees. If you're using a calculator, you'll have to keep track of the signs of the components to keep the quadrants of the plane straight.

Convert the in-X,Y-plane-projection value of the PUVs to Degree/Minute/Second format, and then turn your rotary table so that the OUTBOUND Perpendicular Unit Vector lies in the Z(Machine),X(Machine) plane AWAY from the hinge of the Tilt Plate.

Now calculate the Tilt Angle by taking the ArcCosine of the z component of the OUTBOUND PUV, and set the Tilt Plate. This rwill rotate the workpiece in the machine Z/X plane, and make the PUV square to the Machine X/Y plane.

I hope that my hurried description is clear enough for you to follow. As I said at the beginning, I'm swamped with a special project right now, and won't be able to get back online before NEXT weekend, which is going to be too late to answer questions. Having said that, about all I can add is Good Luck!

John

I use quaternions for all that stuff.

I'm sure a common core approach to the problem would be helpful. Let me know what century your solve the problem. Oh, and it doesn't matter if you get the right answer as long as you know the simplified procedure...

John; many thanks, I think the math will fall into place :-).
Rich, Alright! I have been looking for this term! I will have to cozy up to quaternions; no doubt ad nauseam, to develop this little piece of software I have started to write.
Stuart, I fail to see how this discussion of applied mathematics has any relevance to a political hot button issue. Thank you for your input.
This forum is a place rich in learning thank you for help :-)
Rex

Rex Rat --

Any news to share?

John