plunge feed rates

[dougm]

I am looking for a way to figure chip load

in this sample program I am using a 1/2- 2 flute endmill driven tool

in the turret of a lathe to plunge 2" deep

.220 off centerline

main spindle (chuck) is turning 200 rpm

driven tool is turning 3500 rpm

at 30IPM

then the same thing at

at .650 off centerline

 

how do I adjust either the main spindle speed or feed

to get the same plunge angle and chip load

at different diameters ?

and formulas or thoughts

Thanks

 

 

%

O0000

G20

(PROGRAM NAME - CTX TEST )

 

( 1/2 endmill right angle driven tool .5 DIA. parallel to main spindle )

G28 U0. V0. W0.

N100

G18

G97 S200 M03 T0101

S3500 M130

G0 X.440 Z.25 M108

Z.1

G94 G1 Z-2. F30.

G0 Z.25

G0

X1.3

Z.1

G1 Z-2.

G0 Z.25

M109

M30

%

[JDowe]

Interesting question

 

OOPS - The solution I was going to propose didnt

look right

 

Will look again

[JDowe]

Consider it like a thread path

 

(lathe rpm) X (Inches per rev)

------------------------------

(Mill rpm) X (Number mill teeth) = Chip load

 

Top divided by bottom = chip load

 

The part diameters do not appear to matter

[dougm]

JDOWE

Thanks for the reply

it seems to me the part diameter

should matter because the more you

move away from centerline the longer the cut distance gets per lathe revloution ?

[YoDoug®]

First calculate time in cut.

 

2" plunge at 30IPM. = 1/15 minute in cut.

 

Second calculate the volume of material being removed.

 

.650 off center line = 1.8 OD/1.3 ID groove.

Volume of groove = 4.084ci

 

That means in 1/15 of a minute you are removing 4.084 ci of material or approx 61.26cim

[dougm]

I never thought of using ci of material

thanks guys

 

I use this to punch holes alot lately

we machine only plastic here and this

process works great on gummy materials

no long stringy shavings

Doug

[JDowe]

If I understand correctly chip load you need is on the end of the milling cutter

If the the tool moves .1 in the Z direction and

there is one cutting edge chip load is .1

if there are two cutting edges and the part makes one rev while the cutter makes one rev

the chip load is .05 because the tool will move

.05 for each half spindle rev so for each spindle rev there will be two cuts of .05 or

.1 per spindle rev if there are 4 cutting edges

each cut will be .025 because the spindle will

make 1/4 rev for each cutter edge - the diameters

do not affect the chip load on end of the cutter in the z direction - the faster the part rpm

the higher the chip load (if you are programming in ipr)- the faster the the

cutter rpm - the more cutting edges - the lighter the chip load on the cutter face

[JDowe]

Correction to previous - post if you are programming in ipr the part rpm (spindle speed)is irrelevant only the number of teeth on the

cutter and its rpm affect chip load on the

cutter front

 

IPR

-------------------------- = Chip Load

Cutter rpm * number of teeth

[DavidSV]

Chip load is IPM/(teeth x RPM).

 

To get your IPM you look at the speed the EM is traveling.

 

200(rpm)x1.8(bigger diameter being cut)/3.82=94.24 IPM

 

94.24/7000(teeth per minute) is a chip load of .0134 per tooth

 

200x.94/3.82 = 49.21 IPM for the small pass for a .007 chip load.

 

So, to answer the question, to keep the chip load the same, you should slow the chuck down so the EM is going about 50 IPM, or 106 RPM. Then you should slow the plunge rate down by the same rate to 16 IPM