Machining 3d ball mold

[cbradshaw]

I am trying to machine a 2 inch diamater mold for a plastic ball. It machines fine except it has all of the steps from the cutting tool. Can someone tell me how to cut this so that it will require mimimal sanding? I am cutting this in red urethane plastic board

Thanks,

Carroll Bradshaw

[landdesign]

I'd use a smaller step over in the tool pathing and as small of ball end cutter as you can. 1/16th ball end with a 10-15% step over should give you a pretty sanding free surface it will take longer to machine, but thats what the make coffee for


Erik

[stevem]

Use the largest diameter ball end bit you can find. Use a small stepover space.
After cutting in the horizontal, try going over the same area with a radial machining strategy for further smoothing.

[Brady Watson]

If you have ArtCAM, do a spiral or offset machining strategy. This will eliminate most of the 'moire' lines created from a raster strategy toolpath.

-B

[jsfrost]

Erik's suggestion of a very small bit with modest stepover seems to be in contradiction with Steve's large ball end/smaller stepover. Which is right? After some thought, I think they both are, depending on specific details.

In the general case of a finely detailed 3D shape, the smallest bit possible gives the most faithful reproduction of the target shape. And the smallest stepover gives minimum rills (Paul Zank's term for the unwanted ripples between adjacent raster or spiral bit paths.) Both small bit size and small stepover add machine time, so one must sometimes trade quality for cost.

However, where the target shape is well behaved, use of a large diameter bit can give smaller rills and yet faithfully reproduce the target shape. For this to work, the target shape must have a slope that is, everywhere in the toolpath,less than the slope of the bit. If Carroll's "plastic ball" has a well behaved shape, Steve's answer possibly with Brady's may be the better choice. If not, go with Erik and Brady.

Correctly recognizing which case applies can dramaticly shorten cut time. Incorrectly picking a larger bit leads to missing essential detail.

[stevem]

The best bit to machine a 2" diameter ball mold is a 2" diameter ball mill. You just plunge it 1" deep into each half of the mold. This would give you a very smooth finish with the shortest machining time.

Of course, the above would be a little tricky in steel, but would work well in foam.

[btk]

Smallest bit with Samll Stepover will give highest resolution (short of using a 2" bit as Steve suggested).


Brian

[gerryv]

I'm just learning so find this an interesting debate. Isn't resolution machine dependent rather than bit dependent? (No, come to think of it, that's accuracy I guess.) However, if smoothness is the desired outcome I'll cast my vote with Steve here as too much resolution, like with printed photos can make the lines (rills in this case) more noticable I'm thinking because of the sharper radius? I'm still only at 0.5 cents rather than 2 cents worth though, so watching with interest to see how close I'm am... or not. :-)

[gpari]

You want to use the largest bit you can. Your limiting factor is the smallest inside radius of the part you are machining. Your bit must match or be less than this radius.

To see this concept, think of the extreme cases. Case 1 has already been given. A 2" diameter bit, one pass. Case 2 would be a REALLY small diameter bit, say .010" Imagine using a .010" stepover (pretty small, right?) You would end up with .005" ridges. Not bad, but we can do better


A .500" ball end with .010" stepover yields .00005" ridges!

Gabe "I think I did the math right, using a 99 cent Office Depot calculator" Pari

[landdesign]

ok... I love a good discussion...

I guess we first need to make sure we are all on the same page. when I program in EnroutPro, I give the step as a % of the bit with that I am using. ie.. small bit, smaller step. I very rarely use a set step distance of in Gabe's idea of .01. and I would never want to challenge Gabe's 99 cent Office Depo calculator but I did work it out in cad.

6835 as you can see in the photo the highth and distance of the surface ripple if you use a % of the bit width is much smaller with the smaller bit. I completly concead it might not be profitable because of the added machining time to use my method. thoughts?

Erik

[garyb]

Just to clarify; the lines left behind caused by the stepover of ball cutters being called ridges and rills in this thread are actually called "Cusp" which are singular points of a curve.
here's the wiki link or you can find it under basic machining by Onsrud.
http://en.wikipedia.org/wiki/Cusp_(singularity)

[jsfrost]

Eric has the math right, but each party in the discussion may be using different terminology.

A small bit with a given percentage of bit diameter as stepover will have less rill height than a larger bit with the same percentage stepover.

However a small bit with any stepover measured in absolute terms (inches, centimeters etc) will have taller rills than a larger bit with the same absolute stepover.

In the absence of software, bit quality or machine issues, a larger bit will always give a smoother surface at the same absolute steopover. In the case of an irregular 3D surface with significant height change over short distance in X or Y or both, the big bit will cut so smooth that little detail is retained. So, in general, the smallest bit and stepover gives the best results.

For cutting something like a concave surface to fit a large diameter ball, a very large bit, with radius less than or equal to the ball radius, with modest stepover will cut smoother than a very small bit with far less absolute stepover. And in this special case, there is no detail, so none is lost.

[btk]

The major issue depending on CAM software and strategy you are using would be on flat areas vs. sloped areas.
On a flat area, and a fixed step-over, you will always be removing the same amount of material on each pass and can achieve a smooth surface with either bit by making the stepover infinitely small.
However in a steep concaved/convexed sloped area, using a fixed stepover would actually tend to create steps/ridges when using a larger bit on a fixed stepover,
A true 3-D strategy would adjust for the slope and decrease stepover when the slope is greater and would compensate for wall collisions, etc by constantly finding tangent points between the two radiuses, etc.
Therefore, using a smaller bit and smaller stepover would tend to minimize such collisions as the smaller a bit gets, the closer it would get to having no wall collisions on a steep slope/fixed stepover scenario.