Re: More on servo motor selection

--- In CAD_CAM_EDM_DRO@y..., Drew Rogge <drew@p...> wrote:

> Hi all,
>
> 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? 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.
>
> So how does all this sound? Am I way off in left field confusing

electrical

> hp with mechanical hp?
>
> Jon E., could you explain the process of computing the forces at the
> table for computing the
> Drew Rogge
> drew@p...

The leadscrew doesn't "create" more power. It just allows you to
trade off rpm with torque. Based on the above, it looks like you can
cut at 1/3 of the speed that you want, i.e. about 10ipm for a .1"
deep cut with a 0.5" diameter mill.

However, this does raise a question... how does the power of the
spindle motor come into play here? Can some guru please explain?

Thanks.
ym