[FreshMint]

So.... I thought up a very simple (likely conservative) method to estimate cutting forces so some of you can size your stepper motors accordingly.

This question has been posted time and time again, and when i was looking for the answer I found tons of conflicting results, equations and conclusions.

So here is a derived solution and equation to this problem that should be used only to ball park....

http://aimenshawki.com/?p=234

Let me know what you think!

[louieatienza]

So.... I thought up a very simple (likely conservative) method to estimate cutting forces so some of you can size your stepper motors accordingly.

This question has been posted time and time again, and when i was looking for the answer I found tons of conflicting results, equations and conclusions.

So here is a derived solution and equation to this problem that should be used only to ball park....

Aimen Shawki | An Engineer's Online Resume

Let me know what you think!

I saw your equations, and it reminded me of my days sitting in calc 1,2, 3 and differential equations (none of which I remember!) and the "drool" reflex and glazed eyes almost kicked in...

There's more to the forces involed besides tool deflection. The machine will flex as well and your actual "tool deflection" may be larger than the equations might show. You also have to take into account how much force is needed to move the axis at the speeds you need for the tasks at hand; you need to consider the efficiency of your drive system (ballscrew, anti-baclash nuts, plain nuts); you need to know how much force is needed to overcome any "stiction" in your linear bearings.

Another point to consider is acceleration. Since part accuracy and surfaace finish are somewhat linked to acceleration (and generally the higher the better) I suppose the "inertia" of the moving components need to considered as well if a more thorough analysis is to be done.

Though simplistic, I feel the equations Mariss Freimanis [of GeckoDrive] has on the GeckoDrive website give a pretty good start.

[FreshMint]

I saw your equations, and it reminded me of my days sitting in calc 1,2, 3 and differential equations (none of which I remember!) and the "drool" reflex and glazed eyes almost kicked in...

There's more to the forces involed besides tool deflection. The machine will flex as well and your actual "tool deflection" may be larger than the equations might show. You also have to take into account how much force is needed to move the axis at the speeds you need for the tasks at hand; you need to consider the efficiency of your drive system (ballscrew, anti-baclash nuts, plain nuts); you need to know how much force is needed to overcome any "stiction" in your linear bearings.

Another point to consider is acceleration. Since part accuracy and surfaace finish are somewhat linked to acceleration (and generally the higher the better) I suppose the "inertia" of the moving components need to considered as well if a more thorough analysis is to be done.

Though simplistic, I feel the equations Mariss Freimanis [of GeckoDrive] has on the GeckoDrive website give a pretty good start.

The equations and solution presented is by no means the entire equation or forces needed to size motors...but cutting forces are something that one has to consider in order to properly size the motors.

I thought it was pretty clear that the article is just about estimating cutting forces...nothing else?

[louieatienza]

The equations and solution presented is by no means the entire equation or forces needed to size motors...but cutting forces are something that one has to consider in order to properly size the motors.

I thought it was pretty clear that the article is just about estimating cutting forces...nothing else?

I believe there are charts for the cutting forces needed for different materials, though there is probably more discussion on this in the MetalWorking section. After re-reading the article i stand corrected.

I think however the more common parameter is how much horsepower is needed for a certain tool to produce the recommended chiploads, and there are charts for that; and if those guidelines are met then tool deflection is kept to a minimum and then the equation boils down to how much force is needed to push that spindle around. Onsrud's catalogs have some useful charts regarding chiploads, material removal rates, and horsepower ratios.

[wizard]

Is bending the tool really the objective here? I don't really know so that is why I ask. I was always under the impression the goal was to get the right chip load on the tool. For most of us that would happen well before we started to bend the tool significantly. In other words we are horsepower limited. The other thing is how does direction come into play here?

I suppose if one where building a large machine, overflowing with power, this approach might mean something. In a sense what you offer up is a maximum power number but I'm not convinced it is a number often seen in the real world. It would be more interesting to have a quick way to figure load based on the maximum spindle horse power available and for a given maximum tool.

As it is I'm tired so maybe this will be more interesting after some sleep.

So.... I thought up a very simple (likely conservative) method to estimate cutting forces so some of you can size your stepper motors accordingly.

This question has been posted time and time again, and when i was looking for the answer I found tons of conflicting results, equations and conclusions.

So here is a derived solution and equation to this problem that should be used only to ball park....

Aimen Shawki | An Engineer's Online Resume

Let me know what you think!

[louieatienza]

Is bending the tool really the objective here? I don't really know so that is why I ask. I was always under the impression the goal was to get the right chip load on the tool. For most of us that would happen well before we started to bend the tool significantly. In other words we are horsepower limited. The other thing is how does direction come into play here?

I suppose if one where building a large machine, overflowing with power, this approach might mean something. In a sense what you offer up is a maximum power number but I'm not convinced it is a number often seen in the real world. It would be more interesting to have a quick way to figure load based on the maximum spindle horse power available and for a given maximum tool.

As it is I'm tired so maybe this will be more interesting after some sleep.

This is a point I made as well. You'd have to build a beast of a machine, and use it to move an underpowered spindle? To note, commercial CNCs can have spindles from 10hp to over 140hp(!) and can still run a 1/16" ball endmill (or smaller) for finishing work!

I have to look for it, I have a PDF that has Onsrud's recommended horsepower factors for different materials which is pretty helpful. But agree here with wizard. Most of us here in the DIY router section are horsepower (or probably torque and speed) limited as far as metal milling is concerned. Most metal mills have higher power at lower RPMs and are ridgid enough to cut at bit diameter depth per pass or more (up to over 2XD with high speed toolpaths.) This would not be an easy task for a router, or even a Chinese spindle.

[Drassk]

I don't think the fact that most people are horsepower limited is relevant here. That's like saying engine calcs for sports cars are pointless because most people drive commuter cars.

In my opinion, any machine should have enough moving mass that it trumps the machining forces for stability (the load on the servo should be highly predictable). It's not very expensive to strap a 3 or 5HP spindle on a light router frame, and so it would be good to know if the tool forces are significant compared to the force of moving the spindle around. Me, I'm putting 400lbs on my table and at least 200lbs on my spindle for my new machine so that I'll know the cutting forces don't matter

[louieatienza]

I don't think the fact that most people are horsepower limited is relevant here. That's like saying engine calcs for sports cars are pointless because most people drive commuter cars.

In my opinion, any machine should have enough moving mass that it trumps the machining forces for stability (the load on the servo should be highly predictable). It's not very expensive to strap a 3 or 5HP spindle on a light router frame, and so it would be good to know if the tool forces are significant compared to the force of moving the spindle around. Me, I'm putting 400lbs on my table and at least 200lbs on my spindle for my new machine so that I'll know the cutting forces don't matter

It depends on what you're cutting I guess. Your machine might be heavy relative to most builds here, but probably would be lightweight by commercial standards. The spindle alone on a commercial machine can weigh over 200lbs.

But your example supports my point that if the spindle is sized properly for the material being cut (and the machine can support it) then the concern should be more about how much force is needed to push the weight at the speed needed rather than how much force is needed to "bend" the tool.

Since most machines here are limited by their spindle horsepower and high speed, they are thus limited by their depth of cut, and thus less power is needed from the drive motors, and there is less deflection since the machine cannot be pushed that hard anyway. Unless of course the tool is pushed to the point that the spindle stalls while it's still being pushed around.

The other point is, as in the article, the endmill is stuck out relatively far from the collet with the cut depth relatively small, smaller than the bit diameter. Usually, either a "stub" or "necked" endmill would be used if the cut depth was that small, and usually that cut depth is small because there is not enough spindle power. Then the tool "has" to be pushed through the material, which I don't feel is the best way to machine most materials either...