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How fast can I Machine?

This is not an easy question to answer. The first step is to determine how much coolant pressure and flow are available to make the proposed cut. Enter that information in the calculator below and press the calculate button to see your available
horsepower.
kilowatts.

Once the horsepower has been determined we can apply the following rule of thumb to estimate a maximum metal removal rate.

Once the kilowatts have been determined we can apply the following rule of thumb to estimate a maximum metal removal rate.
In mild steel we can remove about 1 cubic inch of material per horsepower per minute.
In mild steel we can remove about 22000 cubic milimeters of material per minute per Kw.

About 3 times that much in aluminum and half that in heat treated 4140 or stainless steel. If your material is not included in our calculator or you wish to be more precise you can use the machinability rating from a source like the Machinery Handbook where B1112 steel has a machinability of 1.0  The calculator will allow you to type in a specific machinablility number.

Links to on-line sources with machinability charts.

Machinability considers a lot of factors besides the power required to remove material. Things like abrasiveness and surface finish. But the accepted method of calculating a material removal rate per unit of power would be to multiply the available power by the machinability factor of the material. To do otherwise can get extremely complicated. We could consider cutter rake angles and lead angles, coolant lubricity and a laundry list of other factors but as this is not intended to be a machining course we will keep things as simple as possible. This should provide us with an acceptable starting point.

Just because we have enough horsepower to run an 1/8" End mill 5000 inches per minute does not mean the cutter would not break.
Just because we have enough horsepower to run a 3 mm End Mill 100 meters per minute does not mean the cutter would not break.
Nor does it mean we can ignore all the machining rules we normally follow such as surface speed, chip load per tooth, part rigidity, surface finish etc. At the same time if we have limited power we may not be able to machine at the maximum capability of the cutter without stalling the head.

 Coolant Flow gpm Coolant Pressure psi Width of Cut in Depth of Cut in Tool Diameter in Select Material Mild Steel 1.0 Stainless Steel 0.5 Aluminum 3.0 Cast Iron 0.5 Other Machinability Factor Speed rpm Available Power hp Available Torque inch/lbs Metal Removal Rate inches³ Maximum Feedrate inches/min Surface Feet sfm
 Coolant Flow litres/min Coolant Pressure bar Width of Cut mm Depth of Cut mm Tool Diameter mm Select Material Mild Steel 1.0 Stainless Steel 0.5 Aluminum 3.0 Cast Iron 0.5 Other Machinability Factor Speed rpm Available Power kw Available Torque N/m Metal Removal Rate mm³ Maximum Feedrate mm/min Tool Velocity M/min

 Coolant Flow gpm Coolant Pressure psi Desired Feedrate inch/min Drill Diameter inches Select Material Mild Steel 1.0 Stainless Steel 0.5 Aluminum 3.0 Cast Iron 0.5 Other Machinability Factor Speed rpm Available Power hp Available Torque inch/lbs Required Power hp Required Torque inch/lbs Surface Feet sfm
 Coolant Flow l/min Coolant Pressure bar Desired Feedrate mm/min Drill Diameter mm Select Material Mild Steel 1.0 Stainless Steel 0.5 Aluminum 3.0 Cast Iron 0.5 Other Machinability Factor Speed rpm Available Power kw Available Torque N/m Required Power kw Required Torque N/m Tool Velocity M/min
Milling

To determine our allowable metal removal rate we need to multiply available Kilowatts by the machinability factor of the material we wish to cut and multiply that by 22000.

Examples:
1.5 Kw in Mild Steel would be 1.5 × 1 × 22000 = 33000 millimeter³ per minute metal removal rate.

1.5 KW in Aluminum would be 1.5 × 3 × 22000 = 99000 millimeter³ per minute metal removal rate.

2.0 Kw in Stainless steel would be 2.0 × .5 × 22000 = 22000 millimeter³ per minute removal rate.

(Note: normally we would also multipy the result by the efficiency of the machine being used but the specifications provided for a Titespot head and used to calculate power have already factored this in.)

With the allowable metal removal rate determined we can use the following formula to calculate a maximum feed rate for milling. Again we can not exceed this feed rate because we don't have enough power to do so. Other machining factors are not considered.

W × D × F = Mr

Where:
W=Width of cut
D=Depth of cut
F=Feed rate in mm per minute
Mr=Metal removal rate

Example: We want to run a 12mm end mill 6mm deep in in steel with 2.0 Kw.
W=12
D=6
Mr=44,000 ( 2Kw × 1.0 machinabiliy factor × 22,000 )

Remember all those algebra classes~!!!

W × D × F = 44000      (width × depth × feed can not exceed 2.0)
12 × 6 × F = 44000  (12mm diameter mill × 6mm deep × feedrate = 44000 or less)
12 × 6 × F ÷ 12 = 44000 ÷ 12   (divide both sides by 12 to get F alone)
6 × F = 3667 (result)
6 × F ÷ 6 = 3667 ÷ 6   (divide both sides by 6mm to get F alone)
F = 611 (result)
So we should have enough horsepower to feed a 12mm mill 6mm deep at 611 mmpm.

Milling

To determine our allowable metal removal rate we need to multiply available horsepower by the machinability factor of the material we wish to cut.

Examples:

1.5 HP in Mild Steel would be 1.5 × 1 = 1.5 inch³ per minute metal removal rate.

1.5 HP in Aluminum would be 1.5 × 3 = 4.5 inch³ per minute removal rate.

2.0 HP in Stainless steel would be 2.0 × .5 = 1.0 inch³ per minute removal rate.

(Note: normally we would also multipy the result by the efficiency of the machine being used but the specifications provided for a Titespot head and used to calculate power have already factored this in.)

With the allowable metal removal rate determined we can use the following formula to calculate a maximum feed rate for milling. Again we can not exceed this feed rate because we don't have enough power to do so. Other machining factors are not considered.

W × D × F = Mr

Where:
W=Width of cut
D=Depth of cut
F=Feed rate in inches per minute
Mr=Metal removal rate

Example: We want to run a 1/2" end mill 1/4" deep in in steel with 2.0 HP.
W=.50
D=.25
Mr=2.0

Remember all those algebra classes~!!!
W × D × F = 2.0      (width × depth × feed can not exceed 2.0)
.50 × .25 × F = 2.0  (.50 diameter mill × .250 deep × feedrate = 2.0 or less)
.50 × .25 × F ÷ .50 = 2.0 ÷ .50   (divide both sides by .50 to get F alone)
.25 × F = 4.0 (result)
.25 × F ÷ .25 = 4 ÷ .25   (divide both sides by .25 to get F alone)
F = 16    (result)

So we should have enough horsepower to feed a 1/2 mill 1/4 deep at 16 IPM.

Drilling

Drilling will be similar to milling. Our formula for metal removal rate of course will be different.

Metal removal rate = Drill dia² x π ÷ 4 × feed rate.

Example:

A 1/4" drill feeding at 5 IPM will remove .245 inches³ of material per minute.

.250² × π ÷ 4 × feed
(.250² × 3.1416) ÷ 4 × 5 = .245 inches³

Now that we know the metal removal rate we can use this formula to determine the horsepower required for drilling.

Mr ÷ Mf = Hp

Where:

Mr = Metal removal rate
Mf = Machinability factor
Hp = Horsepower required

Example: We want to drill a 1/4 dia. hole at 15 IPM in Aluminum which has a machinablility factor of 3.

Mr = .735 ( .250² × π ÷ 4 × 15 )
Mf = 3.0

.735 ÷ 3 = .306 Horsepower required.

Example 2:  We want to drill a 5/16 dia. hole at 2.5 IPM in Stainless which has a machinablility factor of .50

Mr = .191 ( .312² × π ÷ 4 × 2.5 )
Mf = .5

.191 ÷ .5  = .382 Horsepower required.
Drilling

Drilling will be similar to milling. Our formula for metal removal rate of course will be different.

Metal removal rate = Drill dia² x π ÷ 4 × feed rate.

Example:

A 6mm drill feeding at 125 mmpm will remove 3534 cubic millimeters of material per minute.

6² × π ÷ 4 × feed
(6² × 3.1416) ÷ 4 × 125 = 3534mm³

Now that we know the metal removal rate we can use this formula to determine the Kilowatts required for drilling.

Mr ÷ (Mf × 22000) = Kw

Where:

Mr = Metal removal rate
Mf = Machinability factor
Kw = Kilowatts required

Example: We want to drill a 6mm dia. hole at 400mmpm in Aluminum which has a machinablility factor of 3.

Mr = 11309 ( 6² × π ÷ 4 × 400 )
Mf = 3.0

11309 ÷ (3 × 22000) = .17 Kilowatts required.

Example 2: We want to drill a 6mm dia. hole at 65mmpm in Stainless which has a machinablility factor of .50

Mr = 1837 ( 6² × π ÷ 4 × 65 )
Mf = .5

1837 ÷ (.5  × 22000) = .167 Kw required.

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