## What is Metal Removal Rate ?

Material removal rate (MRR) is the volume of material removed per time unit during machining operations such as milling, turning, drilling, and grooving. It is designated by the letter Q and measured in Cubic Inches per minute or cubic centimeters per minute.

## how to calculate MRR ?

The Metal Removal Rate is calculated by multiplying the area of the chip’s cross-section by the linear velocity in the direction perpendicular to it.

Let’s look at a simple milling application as an example:

• The chip Area is Ap x Ae
• The Perpendicular Speed is the table feed (Vf).

(In a full plunging operation, the chip area would be π x D2 / 4, and the speed would be feed in the spindle’s direction.)

The result is multiplied by a constant depending on the units used (Metric/Inch) to get the final result in either cubic inches or cubic centimeters.

$$\large MRR = \text {Chip Area} \times \text { Perpendicular Speed} \times \text { Unit Constant}$$
$$\begin{matrix} \large MRR = \\ \large \text {Chip Area} \times \\ \large \text { Perpendicular Speed} \times \\ \large \text { Unit Constant} \end{matrix}$$

## Metal Removal Rate Formulas

( For detailed explanations check the sections below on each application)

Application Metric [Cubic Cm] Inch [Cubic Inch]
Milling $$\LARGE \frac {A_p \times\, A_e \times\, V_f }{1,000}$$
$$\large A_p \times A_e \times V_f$$
Turning $$\large A_p \times F_n \times V_c \$$
$$\large A_p \times F_n \times V_c\ \times 12$$
Drilling $$\LARGE \frac {D \times\, F_n \times\, V_c }{4}$$
$$\large D \times F_n \times V_c\ \times 3$$
Grooving $$\large W \times F_n \times V_c \$$
$$\large W \times F_n \times V_c\ \times 12$$

Units used in the above table:

• Ap, Ae, D, W – mm or Inch
• Vf – mm/min or inch/min
• Vc – m/min or feet/min (SFM)
• Fn – mm/rec or Inch/rev
• MRR – Metal Removal Rate CM3/min or Inch3/min

## MRR formulas explained

As explained in the introduction, Metal Removal Rate is defined as:

$$\large MRR = \text {Chip Area} \times \text { Perpendicular Speed} \times \text { Unit Constant}$$
$$\begin{matrix} \large MRR = \\ \large \text {Chip Area} \times \\ \large \text { Perpendicular Speed} \times \\ \large \text { Unit Constant} \end{matrix}$$

We will break down this basic formula for the main machining applications

### Metal Removal Rate in Milling

$$\begin{matrix} &\text{Chip Area}& & \text {perpendicular Speed} & & \text {Unit Constant}\\ \large MRR = &\overbrace{A_p\,\times\,A_e} &\times&\overbrace{V_f} &\times&\overbrace{K} \end{matrix}$$

• In imperial units, all the parameters are in Inches, therefore K=1 to get the final result in Inch3.
• In metric units, Ap and Ae are in mm, while Vf is in meters. therefore, K=0.001 get the result in Cm3.

$$\large MRR\,[\frac {Cm^{3}}{min}] = \LARGE \frac{A_p\,\times\,A_e\,\times\,V_f}{1,000} \\$$

$$\large MRR\,[\frac {Inch^{3}}{min}] = A_p\,\times\,A_e\,\times\,V_f$$

### Metal Removal Rate in Turning

• Ap – Depth of cut in mm or inches.
• FnFeedrate n in mm or inches.
• VcCutting Speed in m/min or feet/min (SFM).
• MRR – Metal Removal Rate in CM3/min or Inch3/min

$$\begin{matrix} &\text{Chip Area}&&\text {perpendicular Speed}&&\text {Unit Constant}\\ \large MRR = &\overbrace{A_p\,\times\,F_n}&\times&\overbrace{V_c} &\times&\overbrace{K} \end{matrix}$$

• In imperial units, the speed is given in SFM and K equals12 to convert the speed into Inches/min and get the final result in Inch3.
• In metric units, K=1 to get the result in Cm3

$$\large MRR\,[\frac {Cm^{3}}{min}] = A_p\,\times\,F_n\,\times\,V_c$$

$$\large MRR\,[\frac {Inch^{3}}{min}] = A_p\,\times\,F_n\,\times\,V_c\,\times\,12$$

### Metal Removal Rate in Drilling

• D – Drill diameter in mm or inches.
• FnFeed per Revolution in mm or inches.
• Vc – Max Cutting Speed in m/min or feet/min (SFM).
• MRR – Metal Removal Rate CM3/min or Inch3/min

$$\begin{matrix} &\text{Chip Area}&&\text {perpendicular Speed}&&\text {Unit Constant}\\ \large MRR = &\overbrace{D\,\times\,F_n\,\times\,0.5}&\times&\overbrace{V_c\,\times\,0.5} &\times&\overbrace{K} \end{matrix}$$

• The chip area is the radius (D/2) of the drill times the feed per revolution.
• The speed starts with zero at the center of the drill and reaches its maximum at the OD. Therefore, we use the average speed, which is Vcmax/2.
• In imperial units, the speed is given in SFM, and  K equals12 to convert the speed into Inches/min and get the final result in Inch3.
• In metric units, K=1 to get the result in Cm3

$$\large MRR\,[\frac {Cm^{3}}{min}] = \LARGE \frac{D\,\times\,F_n\,\times\,V_c}{4} \\$$

$$\large MRR\,[\frac {Inch^{3}}{min}] = D\,\times\,F_n\,\times\,V_c\,\times\,3$$

### Metal Removal Rate in Parting and grooving

• W – Width of Groove in mm or inches.
• Fn -Feedrate in mm or inches.
• VcCutting Speed in m/min or feet/min (SFM).
• MRR – Metal Removal Rate in CM3/min or Inch3/min

$$\begin{matrix} &\text{Chip Area}&&\text {perpendicular Speed}&&\text {Unit Constant}\\ \large MRR = &\overbrace{W\,\times\,F_n}&\times&\overbrace{V_c} &\times&\overbrace{K} \end{matrix}$$

• In imperial units, the speed is given in SFM and K equals12 to convert the speed into Inches/min and get the final result in Inch3.
• In metric units, K=1 to get the result in Cm3

$$\large MRR\,[\frac {Cm^{3}}{min}] = W\,\times\,F_n\,\times\,V_c$$

$$\large MRR\,[\frac {Inch^{3}}{min}] = W\,\times\,F_n\,\times\,V_c\,\times\,12$$

## What is MRR used for ?

The Metal Removal Rate is used for two main purposes:

#### 1) Estimating the machine power consumption for a given set of machining conditions

Each raw material has a Specific Cutting Force property, designated by Kc. The constant is in pressure units (Force per Area) and is usually listed in Mpa (N/mm2) or N/Icnh2. The specific cutting force indicates how much force is needed to shear a chip from the raw material, and multiplying it with the Metal Removal Rate yields the required Machining Power. This method of machining power calculation is an indirect estimation; however, due to its simplicity and decent accuracy, it is the most widely used way to compute machining power.

#### 2)Comparing the productivity of two machining processes

Suppose we need to mill a cube in the dimensions of 1″ X 1″ X 1″ with a 0.5″ diameter endmill. Two workers suggest different approaches to perform the task.

• Worker #1 r suggests to use a 1/2″ 4 fulte endmill with cutting conditions: Ap=0.5″, Ae=0.25″, fz=0.004 Inch/rev and VC=300 SFM.
• Worker #2 suggest to use a 1/2″ 6 fulte endmill with cutting conditions: Ap=0.5″, Ae=0.1″, fz=0.005 Inch/rev and Vc=350 SFM .

To evaluate which options yield the best productivity we can compare the MRR of both options:

Worker #1:

$$\large n = \, \LARGE \frac {V_c \times\,12}{\pi\times\,D} =\,\frac{300\times12 }{3.1415\times0.5} = \large\,2,293 \,\,RPM$$ $$\large V_f = F_z\times n \times Z = 0.004 \times 2,293 \times 4 = 37\, Inch/min$$ $$\large MRR = A_p \times A_e \times V_f = 0.5 \times 0.25 \times 37 = 4.6\,\, Inch^{3}/min$$

Worker #2:

$$\large n = \, \LARGE \frac {V_c \times\,12}{\pi\times\,D} =\,\frac{350\times12 }{3.1415\times0.5} = \large\,2,675 \,\,RPM$$ $$\large V_f = F_z\times n \times Z = 0.005 \times 2,675 \times 6 = 80\, Inch/min$$ $$\large MRR = A_p \times A_e \times V_f = 0.5 \times 0.1 \times 80 = 4.0\,\, Inch^{3}/min$$

By comparing the MRR value of the options, we can see that the approach suggested by worker #1 provides higher productivity.

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