Help Calculating Cutting Forces for Chamfering OP

# Thread: Help Calculating Cutting Forces for Chamfering OP

1. Hot Rolled
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## Help Calculating Cutting Forces for Chamfering OP

Hey Gang - I know it's kind of a weird question, but if we could keep this thread from turning into a "well why would you need to know that?" dogpile, it would be appreciated. And yes, I understand by saying that, at least one person will feel the need to comment exactly that, so 'thank you' for that ahead of time.

I'm looking for help figuring out the cutting forces (the resolved forces in X, Y, and Z) for a chamfering OP.

Let's say, for the sake of this discussion, I want to put a .010" chamfer on a square edge. The effective area of the cut (area of a triangle) is 0.00005 in^2, and let's say I want to do it quickly, at 800 ipm. That gives me a MRR of 0.04 in^3/min, in tool steel with a K factor of 1.56.

So my HP Consumption is 0.025 (or effectively nothing.)

And this is where my understanding falls flat. Assuming this is an infinitely long straight line, how do I figure out the resolved forces in X, Y, and Z directions? Cutter geometry should impact this significantly, correct? But if I were to assume some sort of mythical cutter that put all of the cutting force along just one axis, how much force would it be? If I'm using a 4 flute tool versus a 1 flute tool (like a PCD ball endmill), the cutting force isn't impacted at the same feeds and speeds, correct? If I'm using a carbide burr tool that doesn't have a specified flute count, is the cutting force the same/similar?

Thanks for the help!

2. Stainless
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Work out torque from power and rpm. and from cutter radius and torque you can work out "mythical" minimum force that you would experience.

3. Hot Rolled
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Originally Posted by MattiJ
Work out torque from power and rpm. and from cutter radius and torque you can work out "mythical" minimum force that you would experience.
Ah! There we go - that makes perfect sense.

Thanks MattiJ!

4. Diamond
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Originally Posted by Johnny SolidWorks
Hey Gang - I know it's kind of a weird question, but if we could keep this thread from turning into a "well why would you need to know that?" dogpile, it would be appreciated. And yes, I understand by saying that, at least one person will feel the need to comment exactly that, so 'thank you' for that ahead of time.

I'm looking for help figuring out the cutting forces (the resolved forces in X, Y, and Z) for a chamfering OP.

Let's say, for the sake of this discussion, I want to put a .010" chamfer on a square edge. The effective area of the cut (area of a triangle) is 0.00005 in^2, and let's say I want to do it quickly, at 800 ipm. That gives me a MRR of 0.04 in^3/min, in tool steel with a K factor of 1.56.

So my HP Consumption is 0.025 (or effectively nothing.)

And this is where my understanding falls flat. Assuming this is an infinitely long straight line, how do I figure out the resolved forces in X, Y, and Z directions? Cutter geometry should impact this significantly, correct? But if I were to assume some sort of mythical cutter that put all of the cutting force along just one axis, how much force would it be? If I'm using a 4 flute tool versus a 1 flute tool (like a PCD ball endmill), the cutting force isn't impacted at the same feeds and speeds, correct? If I'm using a carbide burr tool that doesn't have a specified flute count, is the cutting force the same/similar?

Thanks for the help!
if i understand you, cutting force is not constant and if you have a 1 tooth cutter its more a whack whack whack. many a flycutter job on a bridgeport mill you can just stall the motor as when it comes around you got inertia and motor hp. you can exceed motor hp and eventually spindle stops
.
basically older ironworkers used a flywheel so a lower hp motor was used. the chopping of steel is a big sudden force and the flywheel supplies it without totally slowing to a stop
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with milling cutters 2 flute cutting force still varies per rev. a gradual build up then force stops then next flute starts force build up. with 3 flute or higher it tends to be a smoother action
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as for which direction cutting forces are thats harder to predict. with big octagon insert facemill the cutting forces push down and to the side. some say it helps push part down or hold it down when roughing.