Thats what i meant, in general. I remember doing a lot of experiments and always there was some smallish problem. And a lot of broken bits and melted aluminum :-). Then i read somewhere, /i don't know where exactly, but it was a person doing alu jobs everyday/ that speed 8000RPM is the key. And so it was, at least for me.
Again as you say- quite generally speaking.
13-02-2014 #12Spelling mistakes are not intentional, I only seem to see them some time after I've posted
Thanks Eddy. I misunderstood what that was for and this afternoon (before reading your reply) I experimented with boosting the voltage by 10% in the region of 4500rpm [default was 5% at 720rpm] and also had the DOC set to about 1.5mm. It happily cut a slot 50mm long (6mm 2F Carbide bit) without problems.
Now I'm confused. If the boost is momentary until the speed is reached it wouldn't have affected the torque delivered during the cut since I allowed the tool to reach the required speed before starting the cut. Perhaps everything was cold (tool and workpiece) and I would need to machine for longer to see if this was actually better.
Anyway before I could experiment further I had to take the machine out of action to send one of the ballscrews away for straightening so I'll have another play when it returns.
Thanks for the advice - I don't think things are so drastic as I first thought.
The low speeds you're suggesting imply you're trying to use HSS tools, but from latter posts maybe not? If so switch to carbide, then you can use around 12-13krpm with a 6mm cutter. A splash of something wet can make a huge difference...
A vector drive will get a bit higher torque at low speeds, but we want more than that. In general we're still after a high power at low speed, so even when you have rated torque available across the full speed range, power is still torque times angular velocity, so the power will be decrease proportionately with speed. e.g. If you have a motor rated for 3000rpm and 3kw, then with a vector drive the best you can do without exceeding the ratings is get the same torque at say 300rpm, so you'd only have 0.3kW available at this speed. With a V/f drive you'd get a bit less than 0.3kW, but in the whole scheme of things that's not really a significant difference compared to the difference in output power compared to rated. It's only at very low speeds (like a few Hz) that the difference in torque output is very large.
I expect the issue with getting the vector drives to work with these spindles is that the spindle motor's parameters (time constants etc) are going to be quite different from your average 'low speed' induction motor. To control an induction motor with vector control, a model of the motor is used within the drive to calculate the rotor flux orientation, which is necessary to ensure constant torque and stable operation. This model requires certain parameters, so the manufacturers tend to ship the VFDs with the PI constants for the current control set to values which work for typical motors within the power rating of the drive, as the parameters over a small range of motor power ratings usually don't vary by much and you could damage the drive if these were set incorrectly.
Having said that, there is one thing you can do to get more torque from the spindle at low speed, if you're feeling adventurous. I accept no responsibility if you try it...I posted it a long time ago and nobody noticed, maybe for the better:
Originally Posted by Jonathan
I've left my VFD set that way and it doesn't seem to have caused problems. You can get away with it to an extent since a lot of the extra heat is generated on the rotor and although that is poorly cooled, it's essentially just a few lumps of metal so it doesn't matter so much (compared to the stator) if it gets very hot. There rotor resistance is a function of temperature and the former is often inferred as part of the vector control algorithm anyway (to maintain accurate rotor flux orientation), so I intend to use the vector drive I've made to do some further testing in this area to more accurately determine the temperature (as I clearly have full control over my drive's control structure and parameters).
Multiply the torque-speed graph for the 1.5kW spindle by 2.2/1.5=1.47 and you'll be pretty close.
Last edited by Jonathan; 15-02-2014 at 01:58 AM. Reason: grammar
By drumsticksplinter in forum Gantry/Router Machines & BuildingReplies: 8Last Post: 14-03-2013, 11:19 PM
By mZeles in forum Projects, Jobs & RequestsReplies: 4Last Post: 27-12-2012, 07:17 PM
By Steff in forum Metalwork DiscussionReplies: 7Last Post: 14-11-2012, 11:48 AM
By ukpete in forum Tool & Tooling TechnologyReplies: 10Last Post: 16-08-2011, 06:23 AM
By Lee Roberts in forum Metalwork Project ShowcaseReplies: 12Last Post: 15-07-2009, 11:00 PM