Depth of cut vs stepover

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I’m trying to think of ways to speed up my cutting time for a part. I’m using a 1/4" EM. Currently have the DOC set at .1875", feed rate 125 IPM, 40% stepover. My question is - if I change the stepover to 20% or 15%, I should be able to go to .375" DOC and bump my IPM to like 175? I never really see people talking about adjusting stepover, just DOC and feed rates so I wanted to ask.

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Depending on material and bit… A 1/4 hogging 1/2" at 100 is feasible with full width cuts… But am extra bit and push the machine till the bit breaks then back off a bit :slight_smile:

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I have been roughing a lot of pine and walnut lately with a 2 flute 1/4"EM. Here is what I found:
Walnut - speed=23000 RPM, feed=350 IPM, ramp feed=100 IPM, stepover=.240", DOC=.100", climb milling. This gives a feed per tooth of .0076" and uses .84 HP.
The cutter will only get up to 350 IPM if there is a straight cut over 1". If there are shorter lines or a lot of curves the actual fee rate will be 50 - 200 depending.

To answer your suggestion of smaller stepover and deeper DOC:
I have also tried in walnut: speed=23000 RPM, feed=470 IPM, ramp feed=100 IPM, stepover=.100", DOC=.233", climb milling. This gives a feed per tooth of .0087" and uses .84 HP. This i felt was too aggressive when the cutter was starting a pass and was cutting full width - otherwise it worked fine.

Pine: speed=23000 RPM, feed=250 IPM, ramp feed=100 IPM, stepover=.200", DOC=.200", climb milling. This gives a feed per tooth of .0054" and uses .33 HP. Pine does not cut as clean and you have to be more careful of breaking chunks off - thus the slower feed rate.

In general the deeper DOC then the shallower stepover you will need. I like to keep the feed per tooth up close to .010" for long tool life so with smaller step overs that will mean higher feed rates. But you have to be careful of the areas where the cutter will burry itself to full width then it may be too much feed. That is why I decided that .240" wide stepover and .100" DOC was best. The limiting factor with feed then was the HP which I like to keep around .8

This all probably seems aggressive - and it is because I was getting tired of roughing taking so long. Cutters seem to be lasting longer since they are actually cutting for less time and taking more in each bite. Router is working hard but never gets hot - or even warm at the bearing. I like to keep the RPM high to give lots of cooling. I never leave it alone though with all the horror stories.

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One of the articles I found interesting…

https://www.cnccookbook.com/2-tools-calculating-cut-depth-cut-widthstepover-milling/

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Usual preface, I’m with PreciseBits so while I try to only post general information take everything I say with the understanding that I have a bias.

Just to get it out of the way, all of the below will first be effected by the tool geometry. Things like rake, helix, edge radius, and carbide grade change all these factors. e.g. like for like a higher rake tool will take less force for the same cut, or a higher helix tool will put more forces into the Z and create more sheer.

I’ll also get the simple ways of getting this done faster. Get a tool with more flutes. Like for like switching from a 2 flute to a 3 flute will let you go from 125IPM to 187.5IPM with the same forces. You can also increase the RPM and feed proportionally if the tool is aggressive enough. These change other things too but you’ll see some of that in the below (if anyone can get through it).

In general you can increase any of those parameters until you reach the limit of the machine, material, or tool. However, they come at the cost of increased cutting forces and therefore increased tool/machine deflection (bending) even before you reach the limit.

A simple way to think of it is to treat them in a like for like as proportional. e.g. the forces will be the same for cutting twice as deep at half the feed. This is not technically correct as the helix of the tool (flute twist) changes the cutting forces, force direction, and sheer. So if you have a tool with a helix you will take slightly less force cutting twice as deep at half the feed. With stepover you also have the issue of chip thinning. Chip thinning happens when you are taking less than half the bit diameter as a stepover. When you do that you are not cutting the chipload (feed) you think you are. It gets worse the lower you go.

Let’s use that with some of the examples you have. You are currently cutting with 0.1875" pass, 40% step over, and 125IPM. I need a RPM and flute count for this so I’m going to make them up and say we are at 12KRPM (~800SFM) with a 2 flute cutter.

We also need the chipload for this so feed / RPM / flutes which gives us 0.0052". If you don’t know, chipload is the thickest part of the chip and without chip thinning, how far forward the tool is moving for each rotation of a flute. This is reduced by chip thinning to 0.0051".

So now let’s look at your proposed cut of changing to 15-20% stepover, 0.375" pass and 175IPM.

    The stepover is 50%-37.5% of the proposed original.

    Now we need to know what that stepover does to the chipload. So changing nothing else a 20% stepover changes our true chipload to 0.0042" (82% of the original) , and 15% changes it to 0.0037" (72%). The engagement time is different too. I’ll leave that for now though.

    The pass increase is simple as you want to double it. So you are functionally doubling the force depending on helix angle. If we say that it’s decently aggressive maybe only a 90-95% increase. This can also change your engagement time with the flutes. By how much and if it’s an advantage or disadvantage depends on flute count, helix, material, machine, etc.

    The feed increase is a 40% jump so that’s simple too.

So let’s use all this now and call the original a cutting force of “1” and take the reductions and increases.

    So for the 20% stepover we have 0.42 (1 * 0.5 * 0.82 or base * stepover reduction * chipload reduction). For the 15% we have 0.27 (1 * 0.375 * 0.72).

    Now let’s take those and double them for the for the pass depth. So now we have 0.84 for the 20% and 0.54 for the 15%.

    Now the 40% increase for the feed takes us to 1.176 for the 20% and 0.76 for the 15%.

    So the 20% is more aggressive and the 15% is less aggressive.

Again this is a simplified way of looking at it. There are other things that change that can effect the cut. As an example the peak forces will go higher or lower depending on the axis in the original vs the 20%. Additionally, the peak forces are held for much longer in the 20% and 15% than the original (dependent on helix). Another issue with this is that with the chip thinning you are almost certainly under the chipload where you will get the best cut (tool geometry and material dependent).

Hope that’s useful and is not just a bunch of technical babble. Let me know if there’s something I can expand on or help with.

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