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 change the amount of power available. You are
just trading off rotational speed with torque. Based on this
calculation, seems like you can only expect to be able to cut at 1/3
the rate.

However, this does raise a question.. how does the spindle horsepower
come into the equation? Can someone elaborate on that?

Thanks.
ym