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ronbarnes
08-04-2007, 02:01 AM
I am finaly to the point of wiring up my spindle (HSD 4 hp) single phase on my new 4x8 standard and the vld has a big orange cable that runs to spindle and a black cable that (i think goes to 220 plug in) my question is how to wire fan? and isn't their a wire that goes from the vld to my black control box? It's actually a black cpu box. the fan is for 24 volt so do I just run into the vld with that?

thanks Ron

Brady Watson
08-04-2007, 03:01 AM
Ron,
There is a separate line for the cooling fan, which attaches to a dedicated 24v power supply in the control box. It should be off to the side away from the rest of the electronics, if I remember correctly.

-B

dvanr
08-04-2007, 03:13 AM
http://www.shopbottools.com/files/SBG00318070629HSDManual.pdf

The documentation covers most of it, except for the little grey 24VDc module. The Module needs to be mounted inside the box. I wired mine so that it always runs when the control box is on.
Check out Support , Documentation
There is an install doc for the Colombo as well

ronbarnes
08-05-2007, 12:31 AM
Thanks guys, I got the fan going and it runs and I also got the spindle running but I guess we can not control on/off or speeds on the keyboard? I say this only because I don't have anything left to wire up(from the vld to control box). I tried the dry run or test spin but nothing happened and that is after I installed the usb port disc and drivers disc. I did have a bit of a hard time trying to find a com port as com 4 in the instructions was not connecting and so it went to a com 3 but still had or am having issues finding it. Any thoughts ? I wanted to also add that after 2 days I got the x-car lined up but when I engaged the pinion on the x and y motors I get a very erratic response when I push the axis along by hand, It is as if it's skipping or hangs up as I push it?????Thanks Ron

myxpykalix
08-05-2007, 03:29 AM
Other have cautioned against pushing the carriage while gears engaged for possible damage to the electronics.
Having said that when you pushed it could you tell if it was a consistent noise that occurred with the equal revolution of the gear regardless of its position in the X or Y position? If this noise occurred when it was on opposite ends of the table then it sounds like it might be the gear on the motor as opposed to the track gear.
Slop that baby up with plenty of grease.(thats not the first time i've heard that!!)

Towersonline
08-05-2007, 10:20 AM
Ronald, I also am in the process of assembling my PRS. I'm at the point that the x-car is on the rails and the rails are squared up. I can push the car up and down the rails with one finger and wery little restance. The one problem I am having is that one of the wheels is high and its high down the length of the rail. This tells me that the gantry has a slight twist or the end plates are not square. You might want to drop your motors and be sure the x-car slides up and down the rails smoothly.

ronbarnes
08-05-2007, 11:41 AM
Bill and Jack,
Thanks for the reply. If I disengage the motors on all axis everything runs perfect, it is only when the motors are engaged that it does that. Now when I push the x or y car I don't need to push it more that a very few inches and it hangs up. I had forgotten about (the warning ) on pushing the car when the motors are hooked up. oops, Power is off of course. I did notice that when power is on the x and y axis is rock solid on the rails and offer zero play of any type. Bill-as far as the x car and wheels I had to work on that for 2 solid days easy and finally lossened up the rails and gantry gussets both then started at one end and clamped the car over every bolt then tighten down and I also shimmed as needed. After that and only after that did my car fit on those rails. I did have a few small high spots say about a 1/32" but with just a bit of hand preseeure downward it was fixed. And as I mentioned with the motors disengaged she runs smooth (not smooth like a Chinese built lathe but like a swiss time piece) ha. So now I will stew over the fact I may have ruined my motors dry running them??great

richards
08-05-2007, 11:45 AM
Ron,

I turn on my Colombo spindle by adding this line at the beginning of all of my sbp files:

SO, 1, 1
PAUSE 4

After that line is read by the control software, a prompt pops up that tells me to push the START button on the control box. Then, when I push the ENTER key, the program pauses for four seconds to let the spindle get to speed.

I turn the spindle off by adding this line to the end of each sbp file:

SO, 1, 0

I can also start the spindle by entering keyboard mode (K key) and toggling output # 1 ON. As long as output #1 is on, each sbp file will automatically prompt me to push the START button each time a file is run.

You can expect erratic/rough movement when pushing the gantry by hand. (As Jack pointed out, it's not a good idea to do that except for very short distances because the motors act as generators and some stepper drivers can be destroyed by the voltage that is generated.) A stepper motor, when off, aligns itself with magnetic poles inside the motor. At rest it tries to stay lined up with the magnetic pole. When you turn the motor by hand, you will feel a very definite 'cogging' as the motor goes from pole to pole. That roughness is normal and has nothing to do with a motor's ability to run smoothly when its under the control of a stepper driver (Gecko or Oriental Motor, depending on the model of your machine).

Without special add-on electronics, you cannot change the speed of the spindle from the Shopbot keyboard. You have to do that directly from the VFD front panel. On my Delta VFD for my Colombo, I can enter the desired RPM directly. On other VFD controllers you might have to enter a frequency instead of an RPM. If that is the case for your VFD, just make a simple chart that lists the frequency in one column and the resulting RPM in another column. Remember that line frequency (60hz or 50hz - depending on which part of the world you live in) multiplied by the frequency number that you enter into the VFD equals RPM. For example 60 X 250 = 15,000 RPM, 60 X 225 = 13,500 RPM.

I'm sitting at a Linux computer so I can't walk you through the steps of checking which port has been assigned to your USB to serial device. But if I remember correctly, for Windows XP, you click Start, click Control Panel, click System, click Hardware, click Device Manager, click ports (COM and LPT), and then right click on the USB serial port to see its COM assignment. My memory isn't very reliable, so someone else may need to post the correct sequence of key strokes.

Good luck.

Edited:
Ron, you posted while I was composing. Your motors can handle manual movement with the power off, but some stepper drivers cannot. The PRS has either the Gecko stepper drivers or the Oriental Stepper drivers. Both stepper drivers are robust and can probably handle an occasional manual 'opps'. If the stepper motors move properly with power applied, you've been given a 'get out of jail free card' - this time. ;>)

rhfurniture
09-05-2007, 07:02 AM
Mike,
As a just about to buy a spindle person myself (in a 50 Hz electric area), I have been trying to sort out a lot of confusion arising from your last post in this topic:

"Remember that line frequency (60hz or 50hz - depending on which part of the world you live in) multiplied by the frequency number that you enter into the VFD equals RPM. For example 60 X 250 = 15,000 RPM, 60 X 225 = 13,500 RPM."

Surely spindles are simple 2 pole motors, that turn once per electric cycle. So at 60 Hz or 3600 cycles/min it will turn at 3600rpm. So if the vfd say's 250 Hz, then the motor will turn at 250x60(secs)=15000 rpm whatever the input frequency. Please correct me if I am wrong.

R.

richards
09-05-2007, 09:13 AM
Ralph,

Here's a quote from www.patchn.com/vfd.html (http://www.patchn.com/vfd.html) that explains how it is supposed to work:

"Basic Theory

The basic equation for a 3 phase electric motor is: Speed = (120 * F) / number of poles where:
120 = electrical constant, F = frequency and # of poles is determined at motor construction ie: a 2 pole, 4 pole or 6 pole machine. If we look at a 2 pole machine and 60 HZ supply, the speed calculates out to 3600 RPM. The only way to vary the speed is to change the F in the equation. We can accomplish this with a Variable Frequency Drive (VFD)."

It's important to note that the 120 in the example above is the AC cycle X 2. Remember the sine wave passes through the zero point 2X every cycle, once when going from + to - and once when going from - to +, otherwise it wouldn't be alternating current.

There's some more information from en.wikipedia.org/wiki/Variable-frequency_drive (http://en.wikipedia.org/wiki/Variable-frequency_drive) that explains it even better.

EDITED: Because the spindle motors are 2-pole motors, it is possible to just multiply the AC line frequency by the frequency that you set into the VFD to compute the desired spindle speed.

rhfurniture
09-05-2007, 09:37 AM
I still do not understand. If I set my VFD to 300, then according to:
"Speed = (120 * F) / number of poles"
my 2 pole spindle should turn at 120*300/2=18000rpm
However, as I am on 50Hz electric, according to:
"multiply the AC line frequency by the frequency that you set into the VFD"
then it should turn at 50*300=15000rpm.
My logic indicates the former, and it is just an accident using the USA frequency that allows the latter to work in America. But I am probably mis-understanding something.
Are you saying that the 120 constant is the input frequency to the VFD X 2, rather than 60 seconds (Hz are per second, rpm per minute) x 2 (number of poles needed for a full motor turn on 1 full cycle)? If that is the case, then the no-load speed of my 2 pole planer motor should be 100*50/2=2500 which it isn't - it is 3000.

richards
09-05-2007, 09:57 AM
Ralph,

The 120 in the formula used above is the AC line frequency X 2.

Speed = (line frequecy * 2 * F) / number of poles

The line frequency is a constant supplied by your power company. In the USA, the line frequency is 60 hertz. Where you live, the line frequency is 50 hertz. So the equation that you would use is:

Speed = (50 * 2 * F) / number of poles

Since the number of poles equals 2 and the line frequency multiplier equals 2, those numbers in the formula cancel each other out so that the short method of writing the forula becomes:

Speed = line frequency * F

Here in the USA that means that 60 hertz * 300 = 18,000 RPM. Where you live that means that 50 hertz * 300 = 15,000 RPM.

rhfurniture
09-05-2007, 10:45 AM
Mike,
I still do not see it. I really would like to get to the bottom of this. The logic to me say's that the VFD puts out a frequency that determines the speed of the motor. Now VFD's are sold in the UK as accepting a 50Hz OR 60Hz input. Surely the output frequency displayed by the VFD is accurate? That the 2 pole motor will turn once for every cycle it receives seems logical to me, and ties in with the fact that a 2 pole induction motor driven at 50Hz will turn at near 50 times/second = 3000 times/minute. In that case a synchronous motor driven at 300Hz should turn at 300 times/second = 18000 times/minute.
I have looked carefully through the Wiki, and cannot find any suggestion that:
"The 120 in the formula used above is the AC line frequency X 2."
I do not understand why the supply frequency should affect the rpm when there is an inverter in between. Also, my supplier has told me that he supplies hundreds of small inverters to machine builders, so that their machines can be shipped to anywhere in the world and still turn at the same rate.

R.

richards
09-05-2007, 11:48 AM
Ralph,

Maybe my understanding of how a VFD works is faulty, but I think that the line frequency (either 50 or 60 hertz) is the time base (or clock frequency) upon which everything is based. So, for a standard induction motor, (2 X line frequency X seconds in a minute) / motor poles = motor speed. Carrying forward the assumption that the same formula is used by a VFD controller, except that the 'seconds in a minute' multiplier is replaced by a 'VFD frequency', the new formula becomes (2 X line frequency X VFD frequency) / motor poles = motor speed.

My understanding is that the clock pulse generated by the line frequency is multiplied by the PWM circuitry of the VFD to generate the spindle speed. Perhaps there is circuitry/programming within the VFD to handle the difference between a 50 hertz line frequency and a 60 hertz line frequency.

I'm afraid that I just don't have an absolute answer. The books and manuals and websites that I visited this morning don't seem to give any information other that that which I cited earlier. The cloudy area is whether a formula for a 'standard' induction motor can be used for a VFD driven spindle. I just don't know.

Here is one more link www.aosmithmotors.com (http://www.aosmithmotors.com/html/motorDoctor/Determiningspeed.htm) that defines speeds that can be obtained from an induction motor at different line frequencies.

"Let’s begin the discussion by describing what causes an AC induction motor to run at a particular speed. Every AC induction motor has poles, just like a magnet. Unlike a simple magnet, these poles are formed by bundles of magnet wire (called windings) wound together in slots of the stator core. In most cases, you can look inside the motor and count the number of poles in the winding: they are distinct bundles of wire, evenly spaced around the stator core.

The number of poles, combined with the alternating current line frequency (HZ), are all that determine the no-load revolutions per minute (RPM) of the motor. So all four-pole motors will run at the same speed under no-load conditions, all six-pole motors will run at the same speed, and so on.

The mathematical formula to remember in helping make these calculation is the number of cycles (HZ) times 60 (for seconds in a minute) times two (for the positive and negative pulses in the cycle) divided by the number of poles.

Therefore, for a 60 hertz (cycles) system, the formula would be:
60 x 60 x 2 = 7,200 no-load RPM divided by the number of poles.

For a 50 hertz system, the formula would be:
50 x 60 x 2 = 6000 no-load RPM divided by the number of poles."

It shows the 50 hertz, 60 hertz formula - but for standard induction motors. Sorry, but that's all that I've been able to find - and it doesn't really answer the question.

EDITED: I forgot to mention that the Delta VFD-B has a setup parameter where you can enter the line frequency of your power source. I assume that by entering the correct number into that parameter, the VFD will give the correct RPM reading regardless of whether the line frequency is 50 hertz or 60 hertz.

rhfurniture
09-05-2007, 12:37 PM
Ah, now that makes more sense. I have downloaded the manual to my intended inverter, and there is an input frequency setting. This is described as "base frequency", but it is limited to 50Hz or 60Hz. If this is used as the VFD system clock, then setting it to 50 in UK would give an accurate result (rpm = 6 X displayed frequency), both in output and displayed frequency. To quote from the manual:
"In the past, AC variable speed drives used an open loop (scalar) technique to control speed"
Reading on, I think that it is similar these days, but with more sophistication to develop different torque and power curves.
So for us 50Hz'ers we need to check that the VFD setting is right. I do notice from the literature that they have a different distribution for Europe and USA, so the presets for this setting may well differ. I am buying mine (Hitachi) from a UK dealer, so I will chat to him about it at some stage.
BTW, I think a synchronous motor (eg spindle) is just an induction motor that is not allowed to be lazy and drop a bit of speed.
R.

rhfurniture
09-05-2007, 01:20 PM
Sorry,
(rpm = 6 X displayed frequency)
should have been:
(rpm = 60 X displayed frequency)

rhfurniture
09-06-2007, 07:12 AM
Sorry again,
It turns out that a HSD/Fimec spindle is actually asynchronous, which I guess makes it like a regular induction motor.