"Proctologically Violated©®" wrote:
Awl--
Volume-wise, of course.
Opinions?
I read some time ago somewhere that it was drilling.
This surprised me, as I thought the drill point, being essentially zero sfm,
would act as a limiting factor.
PV:
This subject came up years ago. Here are the pertinent parts of a post
I made in Jan 2003.
================================================== ==========================
I also was taught that drilling removes more material, when an
apprentice decades ago. Is it still true today? I don't know, but
let's play with some theoretical figures.
pi * D^2 * f * N
MMR When Drilling = ------------------
4
Whe MMR is the Material Removal Rate in cubic inches per minute
f is the feed rate in IPR (inches per revolution) IPR X RPM =
IPM
N is the spindle speed in RPM
SFPM * 12
RPM = -----------
pi * D
IPM = RPM * Chip Load * Number of Flutes
2" Dia. Drill
525 SFPM in Aluminum
.0065 Chip load per flute
I'm trying to keep everything as close to 10 HP as I can for
comparison
purposes. So plugging those numbers in, I get:
1000 RPM at 13 IPM
pi * 4 * 13 (IPM)
----------------- = 40.8 MMR (cubic inches per minute removed)
4
HP = MMR * UHP (unit horsepower - number taken off chart - .25 for
aluminum)
So 40.8 * .25 = 10.2 HP
2" Dia. Inserted End Mill
7200 RPM (3769 SFPM) in Aluminum
.007 Chip load per flute
IPM = RPM * Chip Load * Flutes
7200 * .007 * 2 = 100.8 IPM
Now the MMR formula for milling is:
MMR = W * H * F
Whe W = Width of cut
H = Depth of cut
F = Feed Rate in IPM
So if we take a 2" inserted end mill at 2" Width of Cut and .200
Depth
of cut at 100 IPM we get:
2 * .200 * 100 (IPM) = 40 MMR (cubic inches per minute removed)
And HP = MMR * UHP so 40 MMR * .25 = 10 HP
So it appears that "theoretically" in Aluminum the Drilling and
Milling
MMR rates remove similar amounts of material at the same 10 HP
In 1020 steel the rates would be (I'll just give the results)
remember
I'm trying to keep the horsepower consistent at around 10 HP for
comparison purposes).
2" Drill, 76 SFPM, 145 RPM, .026 per flute, 7.55 IPM
= an MMR of 10.26 cu.in./min and 9.9 HP (UHP taken off chart is
1.6
for 1020 steel)
A 2" inserted end mill, 4 flutes, 365 SFPM, .0055 per flute
= 697 RPM and 15.3 IPM
A 2" Width of cut * .2 Depth of cut * 15.3 IPM gives:
An MMR of 6.12 cu.in./min. at 9.8 HP
So it appears that "theoretically" in 1020 steel more material
could be
removed by drilling (10.26 cu.in./min.) than by milling (6.12
cu.in./min.) at the same 10 HP. (Unless I've slipped a digit somewhere)
The difference between cutting steel and aluminum is that there
doesn't
really seem to be an upper SFPM limit to cutting aluminum. Whereas in
steel the inserts degrade quite rapidly if you exceed recommended SFPM
limits to any significant degree.
================================================== ==========================
--
BottleBob
http://home.earthlink.net/~bottlbob