Re: [CAD_CAM_EDM_DRO] More on servo motor selection

Jon Elson wrote:

>
> Drew Rogge wrote:
>

[snip]

> > 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.
>

Ok bad choice of words but no need to shout Jon. I realize that
changing the the gear ratio between the motor and the leadscrew
doesn't increase the power of the motor but it does increase the
effective torque available at the leadscrew. I think one thing to
keep in mind here is that there are a whole bunch of people on this
list which don't have your level of experience. I'm just trying to
learn as I go along and end up with something to play with.

> 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.
>

Ok so then a 5 pitch leadscrew has the same mechanical effect as a .064"
drum. Assuming the string is strong enough for the load can't a .032"
drum lift a larger load than a .064" drum when both are being driven by
the same motor? Sure it will take longer to lift it but something has
to be given up somewhere.

> 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
>

Many thanks for all the math Jon. I'm going to have to go away and
digest it for awhile now. Hopefully I'll have a bit more knowledge
when I come back.

Drew

--
Drew Rogge
drew@...