Posted by: davidhayden | August 13, 2011

The Importance of Feedrate calculations and how to make them


One of the most important calculations you need to understand as a machinist or CNC programmer is the feedrate calculation.

Why are feedrate calculations so important?

Feedrate, the speed at which the cutting tool engages the part, is directly related to safety, productivity, tool life, surface finish, part quality and so on.  Let’s look at these individually.

  • Safety – Feedrates determine cutting forces.  The faster you attempt to remove the material the greater the cutting forces.   So, the faster you cut the part, the more rigid you set up needs to be.  Anyone who has ever been around a machine when a cutting tool crashed into a part at rapid travel knows the damage that can be cause by excessive feedrates.
     
    Also, combined with tool geometry, the feedrate you use affects the chips that are made during the cutting process.   If the feedrate is too slow the cut material will not break into small chips and will create long stringers that whip around the spindle.   Stringers are very dangerous and can severely cut people, remove limbs, and so on.
      
  • Productivity- obviously for production machining you want to cut the part as fast as possible without endangering people or machines or diminishing the quality of the part.  Some may disagree, but the fact is few things cost more spindle time.  
       
    For round numbers imagine you borrowed$100,000 for a machine at 10% interest. Just the interest on the machine is costing you $1.60 per hour.   If you only run the machine 1 shift per day, the cost is really $4.80 per hour. . . just to pay the interest.   Now if you want a 5 year pay back on the machine, that adds another $9.60 to the cost.
      
    To break even on the $100,000 machine loan at 10% interest, that machine needs to earn $14.42 per hour or about a quarter per minute.   And that does not even include wages, benefits, profit, G&A, marketing etc.Are you starting to get the picture.
     
  • Tool Life – But what about tool life you ask?  Don’t aggressive feedrates reduce tool life? They can and the tooling cost is a consideration, but compared to inefficient spindle time, the cost is minor.
      
    The faster you cut the more pressure you put on the tool cutting edges and those forces tend to wear the tool more.   By the same token, cutting some materials too slow will work harden the material in front of the tool causing very rapid tool wear because of having to cut hard material.
      
  • Surface Finish – Surface finish is directly related to feedrate and tool geometry.  Cut too fast with a tool nose radius that is too small and you will get a course finish.  A perfect example is threading on a lathe.   The tool is moving very fast in relationship to the spindle speed.   A face mill moving too fast in relation to spindle speed tends to leave a grooved pattern.
        
    Surface finish is also directly related to cost.   If the customer requirement allows for a 125 finish and your cutting process is giving them a 32 or 60 finish, you may be cutting too slow, wasting machine time, and increasing your cost of production.
      
  • Part Quality – Customers don’t usually want low quality parts.  But what does that mean?  If a customer requirement is for a 125 surface finish, is a part with a 32 finish higher quality?  Print tolerances and surface finishes, when stated correctly, give the customer exactly what they need at the best possible price.   If the requirement is for a 125 finish, the customer is saying they don’t want to pay for the additional cost to achieve a 32 finish.  Now they may prefer a 32 finish, they just don’t really need it or want to pay extra to get it.
      
    So cut as aggressively as you can to meet the requirement and charge accordingly.   That means, if you are giving the customer a 125 finish, they should be paying less money they would to a vendor giving them a 32 finish.   If the 32 finish is that important to the customer, they need to update their requirements.
      
    The cutting forces associated with machining can also distort parts or introduce residual stresses.  So cutting too aggressively may distort the part in such a way that it diminishes overall quality.  It may take some trial and error finding the optimum feedrate.

Whoa you say, there is a lot more to it that feedrates!

You are correct.  Feedrates are only one variable in a fairly complex process.  All of the following come into play when machining a part.

  • Cutting speed
  • Feedrate
  • Depth of Cut
  • Machinability of the material (i.e. less machinable = more difficult to cut)
  • Rigidity of the set up
  • Rigidity of the machine
  • Horsepower of the spindle
  • Geometry of the cutting tool

All of these variables affect the efficiency of the machining process and quality of the part.  But for this article the focus is limited to feedrates.  But there is a way to think about it to help keep it straight in your mind.  I promise after this section, I will show how to do the feedrate calculations but you need to have the right mindset so keep you safe and productive.

When you think of productivity, think of Cubic Inches of Material Removal Per Minute  Let’s call this the Material Removal Rate or MRR

  • MRR is a function of Depth of Cut, Feedrate and Cutting (spindle) speed.
  • The greater the MRR, the greater the cutting forces.
  • The less machinable the material the greater the cutting forces
  • The greater the cutting forces, the greater the requirement for horsepower and rigidity.
  • Tool geometry affects cutting forces. 
    • Positive cutting geometry tends to reduce cutting forces but the tool is weaker and often chip control is sacrificed. 
    • Negative cutting geometry typically increases cutting forces but has the benefit of stronger tools and better chip control.
    • Material properties and finish requirements often define the optimum tool geometry for the process.

How feedrates are calculated.

G95 Cutting in units per revolution

The simplest feedrate calculation is the G95 or feed per revolution (IPR) used mostly for lathe machining. 

FORMULA: Desired chip load * number of cutting edges

There really is not much calculation to do because you typically have one cutting tip cutting so if you want .030 chip load that equates to .030 feed per revolution of the spindle your program would include a G95 to set the cutting mode to IPR.  Then on the cutting line, you would have an F.03

EXAMPLE:  G00 G80 G95       . . . start-up block to set modal conditions
                         X### Z###              . . . positioning move to X, Z start of cut on a lathe
                        G01 Z-1.250 F.03   . . . cut Z-1.25 inches at .03 inches per revolution

NOTE:  if you are programming in Metric Mode, you would use mm/minute.  For example, your feedrate would be F1.5 to feed 1.5 milimeters per minute.

EXAMPLE:  G00 G80 G95     . . . start-up block to set modal conditions
                         X### Z###            . . . positioning move to X, Z start of cut on a lathe
                        G01 Z-125.0 F1.5  . . . cut Z-125.0 mm at 1.5 mm per revolution

Suppose you are drilling though.  In this case you have 2 cutting tips so to maintain a .030 chip load, you would program G95 with a feedrate of F.060

EXAMPLE:  G00 G80 G95     . . . start-up block to set modal conditions
                         X### Z###            . . . positioning move to X, Z start of cut on a lathe
                        G01 Z-1.250 F.06 . . . drill Z-1.25 inches deep at .06 IPR  (.03 per cutting edge)

G94 – Cutting using units per minute instead of units per revolution

For mills, routers, drilling machines and some older machines, the typical feed rate is expressed in units per minute.  In other words, a feedrate might be 10 inches per minute or 250 mm per minute.  This can be a little confusing because, what we are interested in is our actual chip load.

FORMULA: Desired chip load * number of cutting edges * RPM

Lathe Example:  .015 chip load * 500 RPM * 1 cutting edge = 7.5 IPM
                        G00 G80 G94         . . . start-up block to set modal conditions
                         G97 S500               . . . set spindle in direct spindle speed mode and RPM at 500
                         X### Z###            . . . positioning move to X, Z start of cut on a lathe
                        G01 Z-1.250 F7.5 . . . cut Z-1.25 inches at 7.5 IPM to achieve .015 chip load

Lathe Drilling Example:  .015 chip load * 500 RPM * 2 cutting edges = 15.0 IPM
                        G00 G80 G94            . . . start-up block to set modal conditions
                         G97 S500                  . . . Set Spindle in direct spindle speed mode and RPM at 500
                         X### Z###               . . . Positioning move to X, Z start of cut on a lathe
                        G01 Z-1.250 F15.0  . . . Cut Z-1.25 inches at 15.0 IPM to achieve .015 chip load/edge

Milling Example:  .005 chip load * 2000 RPM * 4 cutting edges = 40.0 IPM
                         G00 G80 G94            . . . start-up block to set modal conditions
                         G97 S2000                . . . Set Spindle in direct spindle speed mode and RPM at 500
                         X### Y###               . . . Positioning move to X, Y start of cut on a lathe
                         Z-###                          . . .  Positions end mill to proper depth to begin cut

                        G01 X2.5 F40.0        . . . Cut along X axis for 2.5 inches at 40 IPM to achieve .002 chip
                                                                         load per edge

Again, if programming in metric mode, units would be changed to millimeters.

G93 is the code for Inverse Time feed rates and is beyond the scope of this article but for a good discussion on the subject check out this article on Eng-Tips forum  http://www.eng-tips.com/viewthread.cfm?qid=200454&page=7

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