HSM Support Forums>HSMAdvisor Support Forum>Workpiece and Spindle Cutting Forces

Hello, It is not a stupid question but one that was asked and answered before: https://zero-divide.net/?page=forums&shell_id=170&article_id=4513 So it is not just tangential force. Obviously whatever force is excerted is transferred to the spindle and the equal force is applied to the work piece. I do not really have anything to add over that. Feel free to ask clarifying questions. Regards!

Is the force applied to the spindle a lateral (linaer) force or a rotational force (torque)? I.E. does it try to displace the spindle or rotate it?

Quote:&QUOT;GMACK&QUOT;

Is the force applied to the spindle a lateral (linaer) force or a rotational force (torque)? I.E. does it try to displace the spindle or rotate it?

It is a combination if all forces acting upon the cutter. Some or it goes to trying to pull cutter out, some goes to try to twist it out and some goes to try to snap it.

Quote:&QUOT;ELDAR GERFANOV&QUOT;

Is the force applied to the spindle a lateral (linaer) force or a rotational force (torque)? I.E. does it try to displace the spindle or rotate it?

It is a combination if all forces acting upon the cutter. Some or it goes to trying to pull cutter out, some goes to try to twist it out and some goes to try to snap it.

I'm trying to understand what effect the cutting forces have on spindles, not cutters or workpieces. Intuitively it seems that, since its spindle torque that produces virtually all the cutting forces, there really shouldn't be any significant lateral (radial X or Y axis) forces on the spindle. I realize that there could be significant axial (Z axis) forces though, especially with spiral cutters. Does this seem correct to you?

Quote:&QUOT;GMACK&QUOT;

Intuitively it seems that, since its spindle torque that produces virtually all the cutting forces, there really shouldn't be any significant lateral (radial X or Y axis) forces on the spindle. I realize that there could be significant axial (Z axis) forces though, especially with spiral cutters. Does this seem correct to you?

Nope. As spindle is tryin to rotate the cutter, the lateral force (x and y) is about the same as the torque, applied to the radius of the tool. If it were not, deflection due to lateral force would not have been the issue.

Quote:&QUOT;ELDAR GERFANOV&QUOT;

Intuitively it seems that, since its spindle torque that produces virtually all the cutting forces, there really shouldn't be any significant lateral (radial X or Y axis) forces on the spindle. I realize that there could be significant axial (Z axis) forces though, especially with spiral cutters. Does this seem correct to you?

Nope. As spindle is tryin to rotate the cutter, the lateral force (x and y) is about the same as the torque, applied to the radius of the tool. If it were not, deflection due to lateral force would not have been the issue.

But, isn't the lateral force on the cutter and workpiece caused by the rotating cutter's edge(s) hitting the workpiece? Doesn't that force come from torque provided by the spindle? Wouldn't increases in cutting force just increase spindle torque, rather than try to move it laterally? How does that differ from drilling holes with a hand held power drill - where only torque and axial forces are felt on the drill motor?

I lack the proper scientific background to properly explain how the cutting forces combine. All I am saying is F = Fz+Fxy and Fxy = Q×r ×cos(a/4) Where F is the total force, Fz and Fxy are Axial and lateral force. Q is the torque and r is the radius and a is the engagement angle. So yeah in case if a drill tour engagement angle is 360 that will give the tangential force = 0 In normal milling your tangential force will from 1 to 0.70 of the torque times radius.

Quote:&QUOT;ELDAR GERFANOV&QUOT;

I lack the proper scientific background to properly explain how the cutting forces combine. All I am saying is F = Fz+Fxy and Fxy = Q×r ×cos(a/4) Where F is the total force, Fz and Fxy are Axial and lateral force. Q is the torque and r is the radius and a is the engagement angle. So yeah in case if a drill tour engagement angle is 360 that will give the tangential force = 0 In normal milling your tangential force will from 1 to 0.70 of the torque times radius.

Thanks! I guess that milling is more like a grinder than a drill because of the limited engagement angle - but, isn't Fxy = Q/r (xcos(a/4?) rather than "Q×r ×cos(a/4)"? My original thinking was clearly wrong because it would mean that there would be no lateral force on vehicles from the torque applied to the road by its tires.

Hi, Good catch it should have been Q/r ×cos(a/4) But I just made it up literally on the go. Are you sure you are not my competition's spy?

Quote:&QUOT;ELDAR GERFANOV&QUOT;

Hi, Are you sure you are not my competition's spy?

I'm sure - Unlike you, he clearly doesn't welcome feedback. He told me this last June "Gerald, here's the thing--I'm not really here to teach everyone the detailed cutting physics of how G-Wizard works. Sorry, but there just aren't enough hours in the day. You can take my word that this approach works or not, but I'm just not interested in diving down into futher details."

Your "competition" is quite the salesman (http://thinkitmakeit.us/podcast-episodes/) - buyer beware!

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