Re: [CAD_CAM_EDM_DRO] More on servo motor selection

Drew Rogge wrote:

> I did some more thinking about what I expect from the little CNC mill
> I'd like to play with and I think that a .100" deep full width cut with
> a .500" dia. endmill at 30 ipm seems to be a resonable max. That cut
> removes .1 x .5 x 30 = 1.5 cu. in. of material per minute. At Jon's
> suggested rate of 3 cu. in. per hp for aluminum I need .5 hp to sustain
> the above rate. 1 hp = 746 watts so I need 373 watts. So given Mariss'
> calculation that, what seems to be the better motor I mentioned earlier
> (see message #30095), can generate 60 oz. in @ 2600 rpm which equals
> about 115 watts, I need at least a 3 to 1 reduction between the motor
> and the table. The table I'm using has a 10 pitch leadscrew so is this
> more than enough to cover the required reduction. In other words, does
> the leadscrew turn the 115 watts of the motor into 1150 watts?

Of COURSE not! How could a belt or screw increase power? You can increase
torque, or increase speed, but not both at the same time. So, a 10:1
belt reduction would provide 600 Oz. In., but at 260 RPM! The power
would still be the same, 115 W.

A 10 pitch leadscrew doesn't multiply ANYTHING by 10. What it does,
in fact, is have the effect of a drum wound with string, with a circumference
of .1 inch. how big would such a drum be? Circumference = Pi * D so,
it works out to Circ. / Pi = D, or .1 / 3.14 = .032" now, that's a pretty
small drum, not practical as a reality, but it has the same mechanical
effect as a .032" drum wound with string.

But, we're really going around in circles, here. Let's start from the
beginning. You want to use 1/2 Hp with up to 1/2" end mills, at what
RPM? In aluminum, 600 SFPM is recommended, so that comes out to about
4300 RPM. A dirty worst case cutting force calculation would show that
the torque to turn the tool was .5 Hp x 550 Ft-Lb/Sec = 275 ft-Lb/sec
A 1/2" end mill has a circumference of .5 X Pi = 1.571" or .131 Ft, and
is turning 4300 times a minute, or 71.7 times a second. So, the peripheral
velocity is 71.7 x .131 = 9.388 Ft/Sec. To get equivalent torque under these
circumstances, (275 Ft-Lb/Sec) / (9.388 Ft/Sec) = 29.3 Lbs. This is a totally
gross calculation, but I suspect the linear feed force should be less than
this for a sharp end mill fed at a reasonable chip load. It might be higher
for a dull mill fed at a high rate. I was using 1000 Lbs of linear force as
the goal for my system, a small Bridgeport mill.

OK, so now we have a number, but I'd recommend that you raise it to at
LEAST 100 Lbs linear force, to account for other tasks that need lower
speeds and more torque (boring, flycutting, etc.)

Now, how do we figure the torque needed from a motor to drive the table
against our made-up feed force?

Well, we know the leadscrew is 10 TPI. So, what torque does it take
for a .032" diameter drum (.016" radius) to apply a force of 100 Lbs to
the string? By similar triangles, 100 Lbs on a .016" arm = 1.6 In-Lbs.
That will be 25.6 In-Oz, which is your motor torque for direct drive.

> If I
> do add say a 2 to 1 reduction between the motor and the leadscrew do
> I now have over 3000 watts? With a 2 to 1 reduction @ 2600 rpm I can
> still run the table at 130 ipm which seems plenty to me right now.

No, you have 115 W. but, a motor will only need 12.8 In-Oz to deliver
25.6 In-Oz to the leadscrew. The motor will have to turn twice as fast
to get the same speed on the leadscrew.

Ah, now we have a speed requirement, too. Let's see how it works out.
130 IPM with a 10 TPI leadscrew - 1300 RPM on the leadscrew. That
sounds reasonable. Either way (direct or reduction) that would be
10.8 feet/minute or .181 Feet/second, and the force is 100 Lbs, so
that is 18.1 Ft-Lb/Second. Since a HP is 550 Ft-Lb/Sec, you need
.032 Hp. (As I noted earlier, I designed for 10 x greater force, so it
would be something like .32 Hp in my system. As I used 1/8 Hp continuous
rating motors, these numbers actually look reasonable!)

> So how does all this sound? Am I way off in left field confusing electrical
> hp with mechanical hp?

Electrical HP is identical to mechanical HP, it is power, or another unit
for Watts. 745.7 Watts = 1 (US) HP.

> Jon E., could you explain the process of computing the forces at the
> table for computing the

missing last words?
I hope I have done it, here. Someone should check my math, as there are a HELL
of a lot of unit conversions all over this, and a mistake could be very easy.

Jon