Re: [CAD_CAM_EDM_DRO] More on servo motor selection

I have managed to find a book that we used to use for sizing drives "
Electrical Feed Drives for Machine tools" by Hans Gross (Siemens) ISBN
0-471-26273-0 and published by John Wiley, UK. If you are into the statics
and dynamics of sizing drives then this must be on your bookshelf. A quote
which indicates the problem of arriving at the cutting forces "The cutting
force depends on the variables of the cut. It is affected by the cutting
depth, surface speed, feed rate, tool geometry, the part-cutter combination
of materials, pretreatment of the part, coolant-lubrication material used
during machining and by the wear state of the tool. For the standard cutting
case many of these variables cannot be determined and can have a wide
variation. Therefore it is recommended the corresponding literature is
consulted."

Regards Peter

----- Original Message -----
From: "Peter Seddon" <peter@...>
To: <CAD_CAM_EDM_DRO@yahoogroups.com>
Sent: Tuesday, September 11, 2001 5:58 PM
Subject: Re: [CAD_CAM_EDM_DRO] More on servo motor selection


> Many years ago I used to be involved in dc servo drives for axis
> applications for machine tools - sadly I do not have my design data for
the
> selection of drives. However I do seem to remember that the main issue for
> servo drives is not the thrust to produce a desired depth of cut (that
tends
> to affect the spindle drive power). What affected the sizing was the
dynamic
> performance desired from the machine. The drive was part of a closed loop
> system with an encoder either on the leadscrew or a linear encoder on the
> machine slide. Depending upon the stiffness of the mechanics you could
> achieve a given position loop gain within the position loop. This position
> loop gain is effectively a measure of the stiffness of the positioning
> system and required that the servo drive could provide the necessary
> acceleration to maintain the following error. The acceleration of the
drive
> depended upon the the peak torque it can deliver, the rate at which the
> torque can be developed (this starts to bring into the equation the time
> constant of the motor and the peak current that the electronic drive can
> generate) and the inertia due to the weight of the mechanics to which the
> motor is connected. Finally we have to overcome the friction in the slides
> as well. Finally (did I ever say this was easy?) we have to consider the
> duty cycle under which the machine will operate - that is the type of
> machining it will undertake. A lot of rapid movements will be much more
> stressful than gentle steady machining.
>
> So what would I do for a typical drive on a hobby machine ?
>
> I would consider mainly the friction and acceleration loads which the
drive
> is going to see and how often they will see those loads (ie the duty
cycle).
>
> I wish I could remember more about the formula we used to use - they must
be
> on the net somewhere.
>
> Regards Peter
>
>
>
> So what does all this add up to.
> ----- Original Message -----
> From: "Jon Elson" <elson@...>
> To: <CAD_CAM_EDM_DRO@yahoogroups.com>
> Sent: Tuesday, September 11, 2001 6:44 AM
> Subject: 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
> >
> >
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