Typical cutting force

[Evilness]

Hi there
Since a couple of weeks, I have been thinking about building my own CNC mill (thinking seriously that is ), and it didn't take long before I reached this forum and spent hours reading about all the homemade CNC machines here .

Something I haven't been able to find are typical cutting forces. And with that I mean the force the router exites (sideways) on the piece of material beïng milled. Ofcourse this depends on the speed, the material and the depth of the milling (and other factors I forgot to mention ). I want my machine to be able to mill copper and aluminium, maybe even steel, depending on how much harder that is to mill. What will then be typical values for given (conservative) speeds and depths?

[arvidb]

Hi Evilness,

I really hope you get a good reply, because I have been trying to get an answer to this for about six months now, without much luck :-(

It apparently all depends on things that's very difficult to define, like cutter shapness, rigidity of machine, cutter material, what kind of alloy you're cutting, what kind of lubrication you use and so on.

After doing some manual milling of aluminum, my advice is to really get the machine as rigid as possible. This is if you want long tool life and a smooth surface. That is, zero backlash components, and make use of U channel and the like, no plate material without reinforcement! (I'm building out of MDF, and making tubular structures for rigidity... not sure if MDF is going to be good enough, but it's what I've got!)

I guessed and set my goal to 1000 N of axial force for each axis, but I'm beginning to think this is much more than needed. Well, I hope someone else knows better!

Perhaps we can ask balsaman or kong, who have both cut aluminum with their machines, to measure their machine's axial thrust for us?

Also, the optimal spindle speed is much lower for aluminum than for wood. But you need about the same spindle power (more torque).

// Arvid

[HomeCNC]

Are you wanting a milling machine, or a CNC router? It would be easy to cut the material you listed with a converted mill like a Mill/drill. You would need to build a very ridid router frame. I have both, a home made CNC router, and a converted CNC drill/mill. I can cut some aluminum with my router, but not very well. I use my mill for metal and my router for wood. See my website for the mill

[Evilness]

What is the difference between a mill and a router? I thought those 2 were the same :P.

Well I want to build something like the machine kong built. So computer controlled and able to cut trough al/cu.

I think 1000N would indeed be quite a lot, and it would require quite powerful stepper/servo motors, controllers and linear guiding rails, thus making the whole machine more expensive. So I want to build a good machine, but not much stronger than it needs to be.

Isn't it possible for someone here to test the required torque, by disconnecting the motor and connecting a rod with a fixed length. At the end of the rod, you then pull, with a Newton meter between your hand and the rod. This way you can measure the required force needed to cut something. That can then be converted to torque, and knowing the characteristics of the used screw, you can make a rough calculation of the exited force. Is there anybody here who wants to do such an experiment?

An other way to get an indication is to hear about experience with motors. Has anyone had small motors that didn't work good, which they then replaced with bigger motors? That should at least give an impression on what kind of torque isn't enough, and what is.

And concerning rigidity: I have looked at a datasheet of Thomson linear bearings and steel rods, and there was information on how to calculate the deflection of the rods. I've put the data into a spreadsheet, and I can now insert length, distance between the bearings and applied weight. I can then read the deflection out of the table for every available rod diameter they have. You might find this useful when choosing a diameter for you guidance bars.
See the attached file. If you have any questions, plz post them or PM me .

[bb99]

Thanks for the spreadsheet. I'm sure I'll find some good uses for it.

--bb99

[HuFlungDung]

Interpreting deflection is a guessing game, too. There is no such thing as absolutely rigid. The repeated deflection caused by the cutter engaging the work leads to machine vibration, or ringing. So one of the sought after characteristics to dampen ringing is a material that is non-homogenous, such as cast iron, and various other modern day cast materials.

I do not know where you would come up with the acceptable deflection range of the tool, there are so many factors coming into effect between the cutting edge all the way back to the base of the machine.

Suffice it to say, that machine tools are heavy relative to what they accomplish. So, the more mass you can add to the machine, the more likely it will work as anticipated. "Toothpick engineering" simply will not work.

[Evilness]

It looks like it's a lot better to use supported rods. That way you might be able to use a smaller size (like 1/2" instead of 1"), thus saving on the bearings. I have ordered a catalogue from my local bearing store, so I'll b able to c the prices and compare.

BTW what is the conversion factor between oz-in and Nm? Here in Europe we work with the SI system but those guys at the other side of the ocean don't seem to b able to keep up with that :P.

[arvidb]

Originally posted by Evilness It looks like it's a lot better to use supported rods. That way you might be able to use a smaller size (like 1/2" instead of 1"), thus saving on the bearings. *snip*

Yep. The thing is, twice the flex doesn't just make the cuts half as good, it might make it very difficult to cut at all! So supported rods are way better! (Unless, of course, you are just cutting thin balsa sheet, or the like.)

BTW what is the conversion factor between oz-in and Nm? Here in Europe we work with the SI system but those guys at the other side of the ocean don't seem to b able to keep up with that :P.

1 Nm = 141.612 oz-in. (Source: Baldor Motors and Drives: Motion control application notes.)

One of the things that really bugs me about the imperial system is that it uses the same unit for both mass and force, although these are two very different things! Mass is measured in kilograms (kg) in the SI system, and pounds in the imperial. Force is measured in Newtons (N) in the SI system, and pounds (ok, sometimes pound-force, lbf) in the imperial system. I have read lots of posts where this has made a mess of things...

It's probably just that I haven't learnt it well enough, but I hope there will soon be a time when the imperial system will R.I.P.

(And I *had* to write something where I could use that smilie. )

// Arvid

[Evilness]

Yep. The thing is, twice the flex doesn't just make the cuts half as good, it might make it very difficult to cut at all! So supported rods are way better! (Unless, of course, you are just cutting thin balsa sheet, or the like.)

What do you mean? That open bearings are less precise? Or that smaller bearings lack precision?

edit: w8 I thought you said unsupported rod is better but I c not . Are there any disadvantages in taking thin supported rod vs thick unsupported?

[arvidb]

Originally posted by Evilness Are there any disadvantages in taking thin supported rod vs thick unsupported?

I don't know, perhaps thin rod (& smaller bearings) wear faster? In a hobby machine I'd take the thin supported rod anytime.

// Arvid

[tsalaf]

Conversion calculator is here http://www.export911.com/convert/conFac-N.htm

Steve