Do I need a counterweight?
Still waiting to be able to get a Super X2 Plus mill for conversion so I've been doing a bit more reading and planning.
I've read that a number of people have reported problems with the Z axis on both the X1 and X2 mills. The general opinion seems to be that you need to rig up some sort of counterbalance system to offset the weight of the head. I am thinking that the problem isn't so much the weight of the head as the tilting force created by the leadscrew not being vertically in line with the centre of gravity of the head.
What torque is needed to raise the weight of the head? Torque is given by:
Torque = (W x g x P)/2 x π x η)
where W = weight of head
g = acceleration due to gravity
P = leadscrew pitch
η = leadscrew efficiency
For the X2 mill with an 18 kg head and a 5 mm pitch ballscrew this works out at 0.15 N.m. With a 2 mm pitch acme screw (50% efficieny) we get 0.11 N.m. This should be well within the capability of an NMEA 23 motor, so driving the weight of the head is not a problem High acceleration could cause lost steps as a result of the inertia of the head but a counterweight can only increase the inertia and people report a definite improvement with the counterweight.
I think the problem lies with the design of these small mills. If you look at big mills with a rising table the lifting screw is always centred under the table for good reason. In these small mills the screw or rack is well behind the centre of gravity of the head. With the X1 it is even situated behind the column! The lifting force from the leadscrew combined with the weight of the head will create a considerable tilting force on the head. This must be countered by forces on the dovetails which will tend to lock things up.
So I am considering using a Z leadscrew fixed as nearly as possible above the C of G of the head. This will probably require modification of the motor and belt drive to accommodate it.
What does anyone think? Is this a silly idea or might it work?
Russell.
Re: Do I need a counterweight?
Quote:
Originally Posted by
russell
considerable tilting force on the head. This must be countered by forces on the dovetails which will tend to lock things up.
That conclusion sounds plausible to me. A real life example is if I put my very heavy milling vice on the end of the table, so it's outside the area of contact of the dovetails then I can't get as high a feedrate as when it's in the middle.
However you've got the formula slightly wrong...
Quote:
Originally Posted by
russell
What torque is needed to raise the weight of the head? Torque is given by:
Torque = (W x g x P)/2 x π x η)
where W = weight of head
g = acceleration due to gravity
P = leadscrew pitch
η = leadscrew efficiency
Weight is measured in Newtons (W=mg if you like), so that should be mass in the formula, and you missed a bracket, i.e.:
Torque = (m x g x P)/(2 x π x η)
Where m=mass
So that implies, using your numbers T=(18*9.81*0.005)/(2*pi*0.5)=0.28Nm
(You seem to have used mass anyway in the formula but dropped one of the 2's -somewhere)
0.28Nm sounds good, and if you're using a ballscrew the efficiency is more like 90% ... but of course that completely ignores friction of the slides and acceleration of the head and screw. Acceleration of the head is easy, instead of lifting just m*g it's now F=ma+mg, so substitute that in and you get:
Torque = (m*P)(a+g)/(2 x π x η)
Not sure what's reasonable to expect for acelleration here, say 0.5ms^-2? That makes the torque 0.30Nm ... i.e. negligible difference. Can include the acceleration of the screw by knowing it's length and diameter, and thus it's moment of inertia ... but again that wont make much difference here.
To work out the friction on the dovetail slide we need the coefficient of friction of the slide ... about 0.2, and the contact force upon it which is not straightforward as we don't know where the centre of mass of the head is among other things. It will be proportional to that distance, so yes minimise it if you can. I'll think about it.
Ideally we'd work that out, then look up the speed/torque curve of a selection of motors and get one that has the calculated torque (plus a bit to make sure) at the required feedrate. If that doesn't get what you want then use pulleys to change the ratio.
I'd probably just try the motor you're going to use for X/Y (probably a 3Nm Nema 24?) and see what happens ... then you know which to by and only waste a small amount on postage.
Re: Do I need a counterweight?
Thanks for the comments Jonathan,
Yes, I did incorrectly refer to mass as weight which, while not mathematically the correct term is the common usage (e.g. My weight is 60 kg).
I used 95% as the efficiency for ballscrews and that gives the 0.15 N.m I quoted. The 0.5 (50%) figure was for an acme screw which gives the 0.11 N.m figure.
I agree that the force due to acceleration is not significant in such a small machine.
Yes, it's a bit difficult to work out the friction effect. I don't yet have a machine to measure but I would guess that the length of the dovetail on the head is about the same as the distance from the column to the C of G of the head. The head will tend to rotate about the lower contact region on the head. We can simplify the calculation by assuming a small contact area at the base and a small contact area at the top of the dovetail slide (I don't think this is an unreasonable simplification). So, at the top of the slide there will be a force of about 18 kg (176 N) pulling the head from the column. With the 60 degree dovetail the force normal to the surface will be 18/sin(30) = 36 kg. With your figure of 0.2 for the coefficient of friction this will increase the torque required to lift the head by 40% - hardly a show stopper. I wonder if some flexing of the components gives rise to a stick slip type of friction effect which makes things worse?
Russell.
