Gallery, Projects and General > How do I??
Maths help please: Approximating a Flat Iteratively
efrench:
Your 'X' is my 'c'. A right triangle is always formed from a line perpendicular to the center of the flat, the end of the flat, and the axis of the lathe. As the cutter is moved tangent to the flat, the distance 'b' changes, but 'a' doesn't. 'b' varies by the stepover amount. 'c' squared is simply 'a' squared plus 'b' squared. Use sine/cos to determine the spindle angle.
RussellT:
My mathematician son says
"The last post in the thread (from early this morning) I don't fully follow, I think because it's assuming more understanding of the setup than I have, but the penultimate post seems fairly clear or at least the question in the diagram does and simple trig on the two right angled triangles with common edge L and angles A and B leads me to believe
X = R (cos A/cos B)
Does that help?"
I did discuss the setup briefly with him because I'm not convinced by your suggestion that allowance for cutter diameter is easy, it's complicated by the fact that the cutter doesn't just cut on the centre line and so you need to align a tangent of the cutter with the spanner flats.
Russell
awemawson:
Russell please thank your son, I think he's cracked it :thumbup:
I need to model the cutter / tangent issue and get my head round that bit, but to my simplistic way of thinking an infinitely small cutter just slides along the flat with no issue so a 10 mm cutter just slides along 5 mm further in so X needs 5 mm adding :scratch:
djc:
--- Quote from: awemawson on June 13, 2019, 04:50:00 PM ---I need to model the cutter /
tangent issue and get my head round that bit, but to my simplistic way of thinking an infinitely small cutter just slides along the flat with no issue so a 10 mm cutter just slides along 5 mm further in so X needs 5 mm adding
--- End quote ---
Your diagram, whether enhanced or no, does not capture what is going on because your polygon edge is always vertical. This will not be so. It means the terms you want us to use are not helpful. This is also why you are incorrect in your constant cutter offset.
Your part geometry is rotating around your coordinate system (which is fixed). Your cutter is not rotating relative to your coordinate system.
I am sure you have CAD. Draw a pentagon with its right side vertical inscribed in a circle. Draw the perpendicular bisector of this vertical line, starting at the line and going to the centre (so there are six lines so far on the drawing). Draw your cutter touching the edge of the pentagon. Draw a standard set of axes over this in a different colour. Lock the axis layer. Rotate the six lines on the pentagon layer some arbitrary angle (if your software will do rotate and copy at the same time, so much the better). Move the cutter circle horizontally only until it touches the edge of the pentagon again. The point of tangency will now be above the horizontal centreline of the cutter.
PDF attached. Diagram numbers relate to previous post.
awemawson:
I do take your point, but I suspect that at a practical level (and particularly where the cutter diameter is small in comparison the the stock diameter) the error between tangential contact and diametrical contact will be acceptable.
This is a work in progress and entirely experimental, and I suspect that I would do best to get the simple case working then apply any sophisticated corrections later as things evolve dependant on the results that I get. And those corrections will probably be based on your calculations for which I thank you :thumbup:
In built into the Siemens 820T controller (1990 remember) is an arcane mathematical series of '@' commands that theoretically should allow for all the trig required, but is a nightmare to get going and understand, and I suspect that they will take a considerable proportion of the controls CPU time so best to keep it minimalistic initially !
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