Re: [CAD_CAM_EDM_DRO] Re: More on servo motor selection

ym_wong@... wrote:

> --- 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?

Given a specific tool at a specific speed, spindle horsepower sets some
upper limit on the force needed to feed the work into the spinning tool.
There are a million variables here, but it seems like with a sharp tool,
it would be unlikely that the linear force would be more than the torque
needed to turn the tool, at the radius of the tool. I could be all wet on
this supposition, but I don't have a much better idea of how to calculate
the feed forces.

Jon