Re: [CAD_CAM_EDM_DRO] Wood working Question !! forces, steppers and speeds calculated

Thank you Bob for your advice!!! I will have it under consideration to check my numbers.
Regards
Maxi

"gcode fi (hanermo)" <yahoog@...> escribió:
Good advice from Bob.

Maxi -
Please check your numbers !
Could you do this by hand pushing - is this realistic ?

You state 12 metres / minute on a 5 mm depth of cut in mdf, with 10 mm
cutter.
Thats 200 mm or 20 cm per second.
This is approximately what wood machining centers produce, using 5hp and
10 hp spindles (colombo or perske) and bigger feed (servo) drives in the
kW range.

I believe 1 HP spindle may not be able cut it and you need a much bigger
spindle motor.
This comes from my general knowledge of the forces, and what other,
commercial cnc machines use, this was not calculated.

On the other hand, lets see...

Speed:
With your lead of 5 mm per turn, at 1 rpm you move 5 mm per minute.
To move 12 m / minute, you need to go 12000 / 5 = 2400 rpm
This is (barely) possible with geckos and a high-end drive when
freewheeling.
Using say 68 V and Geko 203V drivers on Nema 34 motors you can get this
high rpm with good motors.

However, note that stepper torque falls off at high revs
At best, you will have about 20% of torque available at 2400 rpm.

It may work, with a 1:1 gearing, for woodcutting at these speeds.
Where a wood router has linear power requirements the stepper has
inverse power output.

You need more power the faster you cut.
A stepper produces less power the faster you go.

Thrust:
A 5/8" or 15 mm ballscrew with 400 oz-in stepper, no gearing (1:1),
produces thrust of 326 kG at 0 rpm.
Therefore, at 2400 rpm or 12 m/min, you get about 65 kG.
Should work, if 65 kG of force is enough to cut 20 cm in 1 sec.

Here its written out.

Torque (lb/ft or foot pounds) = L (lead in inches) x P (load in pounds)
/ 5.65
Therefore P = T / L x 5.65

T: 400 oz-in = 25.2 lb-in of torque.
L: 5 mm / turn = 5 mm / 25.4 inches = 0.196 in
The 5.65 number is 2 x pi x 90% efficiency.
A ballscrew is 90% efficient.

So force or thrust in lb
P = 25.2 lb-in / 0.196 in x 5.65 = 726 lbs = 329 kG

This is your force at 0 rpm.
At 2400 rpm you may get 20% of thrust, or 0.2 x 726 = 145 lbs = 65 kG

Spindle rpm and power design:
looking at wood router chip loads in mdf from onsrud,
https://www.onsrud.com/xdoc/ChipMDF
You wood look for (7/16 in or about 10 mm) 0.02 - 0.04 chip load per tooth.

At 12 m / min thats 472 ipm or inches/minute (7.8 ips or inches/second).
Feedrate = RPM x # of cutting edges x chipload
472 = RPM x 2 x 0.02 - 0.04
So rpm = 11800 which sounds reasonable.

(I used 30-00 as a sample, from onsrud, for roughing /finishing).

If you wish to look up on ballscrews design info/specs see
http://www.rockfordballscrew.com/ball-screws.htm

So, in conlusion:
About 400 oz/in steppers as a minimum, they must be modern
square-bodied, and must have as flat a torque curve as possible.
Note the torque at 2400 rpm !
You will need to have 50-70 V as your voltage, and gecko 203 drivers. It
will not work with smaller drivers with no mid-band resonance
compensation, or with lower voltages.

Since larken and others who make commercial routers, with similar
cutting speeds and capacities start at 2-3 HP, and have options in the
7.5 HP range, I believe you cannot cut at 12 m/min at 1 HP. By cutting
at lower diameter, say 2.5 mm, it should work fine. You will then need
to do 2 passes.

Robert Colin Campbell wrote:

>
> Maxi,
>
> The smallest stepper motor that I would use on a CNC router would be
> about
> 350 oz-in.
>
>
>
> ----- Original Message -----
> From: "Maxi Bertotto" <tiwanacote@...
> <mailto:tiwanacote%40yahoo.com.ar>>
>
> Subject: [CAD_CAM_EDM_DRO] Wood working Question !!
>
> Hi!!!
> Somebody know how could I calculate the cut forces for wood working
> (MDF)??.
> I will use a 1HP spindle, feed of 12000mm/Min, 10mm diameter tool and 5mm
> depth with ball screw of 5mm/turn. May be somebody can tell me how many
> aprox. torque I need for the steppers...
> Thankyou!!!
> Maxi
>
> .
>
>

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