Gallery, Projects and General > How do I??
Maths help please: Approximating a Flat Iteratively
RussellT:
Hi Andrew
I will pass on your thanks.
djc's diagram shows exactly what I was referring to with cutters, where the cutter is shown in his diagram it isn't cutting on the centre line. I understand your point about tool diameter, as you mention earlier an infinitely small cutter would solve the problem. The trouble with infinitely small cutters is that they're difficult to find and quite fragile. :lol: Of course an infinitely large cutter would also solve the problem but you may struggle to fit it in the lathe. :D
As a more serious question from my position of knowing very little about the capabilities of CNC lathes could this be done with a conventional lathe tool? I have in mind moving the tool in and out 6 times for each rotation of the work.
Russell
awemawson:
I've tried graphically to estimate the real life error using a piece of 25 mm stock and a 10 mm end mill. Cutting a hexagon, so six sided, the worst case occurs when the wanted flat is inclined at 75 degrees.
The discrepancy between tangential cutting and diametrical cutting works out at 0.1711 mm and obviously will reduce with larger stock and or smaller end mills.
So in a real life situation where I am putting two opposite spanner flats for an 18 mm spanner on a CAT40 pull stud for my vertical mill that represents a 0.95% error - so I don't think that the spanner will mind much!
That said, once I get his working I will try and perfect it as per DJC's musings, as it would be nice to get it spot on despite the fact that using finite angular movements rather than continuous cutting will always result is scalloping errors
Archie Opteryx:
Here's my attempt at the problem. This uses the tangent relationship between cutter and machined face and should be exact.
In the attached Figure 1, S is the centre of the spindle and C is the centre of the cutter. The blue line shows the face being machined, with the centre of the face rotated anticlockwise by an angle alpha from the line between S and C.
The distance from the centre of the workpiece to the face (half the "across flats" distance) is d. The radius of the cutter is r.
The distance between S and C, x, is given by the equation:
x = (d + r) / cos(alpha)
Note that alpha is zero when cutting the centre of the face. To machine a face, step the spindle angle (alpha) from -30 degrees to +30 degrees or so and bring the cutter in to a distance x calculated from the above equation at each step.
If you're making fewer than six faces, you'll need to vary alpha over a slightly larger range, since the face will be a bit longer.
awemawson:
--- Quote from: RussellT on June 14, 2019, 07:04:31 AM ---
As a more serious question from my position of knowing very little about the capabilities of CNC lathes could this be done with a conventional lathe tool? I have in mind moving the tool in and out 6 times for each rotation of the work.
Russell
--- End quote ---
Russell, the Traub WAS capable of that and they had a special software package available (at enormous cost!) for the Mitsubishi control to do exactly that. I suppose it depends how quickly the X axis slide can reciprocate and maintain an accurate position, or alternatively how slowly you can rotate the work and get a decent cut. More room for future experimentation, so no chance to get bored with the lathe just yet :lol:
Archie Opteryx:
Now to calculate the positions of the scallops and how deep they will be.
Fig 2 shows the same geometry as in the earlier post, but this time we are interested in the distance u between the centre of the machined face and the tangential point of contact with the cutter. This is the point where the cut is the deepest and corresponds to the bottom of the scallop.
A bit more simple trigonometry gives:
u = (d + r) tan(alpha)
It's a bit tricky to visualise the effect of the cuts in Fig 2, since the workpiece face is rotating. Fig 3 shows the situation from the point of view of the workpiece. Here a number of cuts have been made at different spindle angles alpha. I've used a small cutter and a large step in angle to show the cusps (pointy bits) between the cuts more clearly.
Due to the tan(alpha) term, the cutter positions become further apart as alpha increases, and the cusp height increases accordingly. However, as you can see from Fig 3 (which is roughly to scale), this effect is small for values of alpha likely to be used in practice.
Finally, how high will the cusps be?
Fig 4 is an enlarged version of part of Fig 3, showing two cutter passes spaced by w.
A bit more geometry gives the cusp height h above the bottom of the scallop as:
h = r - sqrt(r2 - (w/2)2)
It would be straightforward to put these equations into a program or spreadsheet to calculate values for conditions of interest.
Navigation
[0] Message Index
[#] Next page
[*] Previous page
Go to full version