Prob?

[Japperf]

Ok, so assuming I have no way to work out surface area on my items and a batch constists of many items of varing shapes/sizes etc, when does the run stop if using CC? Don't want to burn the items.

Cheers

[sswee]

If you have no clue as to the surface area of the parts you are doing, how are you planning to anodize in CC. You need to know the SA to know what current to use and knowing what current to use will determine how long to run for the desired thickness of coating. It is all integral. If you need help figuring SA, that can be done. It's a time consuming task and does not have to be exact but needs to be fairly close.
SS

[sage]

I have a question / comment on this multiple / variable sized part issue. I've always simply calculated the total surface area of all the parts, large and small. As mentioned there is no way around that with CC.
But, I was of the opinion that they will sort out their own current based on their individual surface areas. A part twice as large as another will have half the resistance and will therefore take twice the currrent per sqin and it will all sort itself out. At least this has always (seemed) to work for me but I always used to anodize to substantial mils - maybe 1-2mils.
I have recently found that if I'm doing the LCD process at the lower limits like 4.5 A/sqft that there becomes some variability in the dying results between parts I never used to have. I'm assuming becasue there are still minor differences in the balance between different sized parts due to maybe surface finish, distance from cathode - or something else and that tips the parameters to below satisfactory.
What is everyones take on this variable sized part issue?

[RedRiver]

OK...I may not follow you exactly but I'll share my comments all the same.

When you have several different shapes and sizes of parts in your batch, you are working with the total surface area of all the parts combined (batch surface area). From that number you can calculate your set current accordingly, based on your desired current density. When it comes time for the power supply to deliver the current, it doesn't decide which part requires 2 amps and which part requires 1/2 an amp to anodize. Infact, the resistance is just like the batch surface area in that it is uniform. The entire system is one cell or loop. It doesn't have branches or areas with more or less current. If for example, your set current is 5 amps for the batch, then every square inch of surface area is receiving the same amount of current (5 amps / batch surface area).

Its a difficult thing for me to explain...I hope this helps and doesn't make it more confusing. I know what I am trying to say inside of my own head, however, I realize it may not relay that way.

- Dan

[acidrain]

RedRiver, I agree with your explanation, except if it is a mixed alloy batch. Mixed batches will always have variations of color depth, and even hue to a degree. You would have the same variations if they were ano'd separately, so it's not worth worrying over. You'll simply have to adjust the color with dye times as discussed in previous threads. I digress.

To answer your question, there is a way around measuring surface area... I haven't measured surface area for quite some time using a modified CV ano method. Caswell has suggested I write a procedure for this, but I've been reluctant to go outside of the CC method as I suspect there will be many questions or problems associated with tank variables, but here it is in a nut shell:

First and foremost, you absolutely have to maintain repeatable ano conditions. This means exact tank temperature, known alloys, good agitation, and use a CC/CV power source.

With your tank temp exactly as you wish (I use 68F), ano a known alloy sample (same as the parts to be ano'd) of exact dimensions. I use a 6x6x1/4in piece of 6061, since I ano mostly 6061. Calculate the required amps at the CD you wish to ano at. Run the part in CC, and carefully graph the volts curve.
Study the volts curve, and note the beginning volts, and how long it takes to stabilize. Mine stabilize fully within 5min at 13.1 volts.
Now you can start your odd shaped parts run in CV (amps knob all the way up, Volts knob slowly adjusted to volts) at your "normal" volts (keeping the tank temp exact, and using good agitation). Continue the run in CV just for the beginning time period (so the current stabilizes), then note the amps at the end of the stabilization period.
Here is the modified part... start out in CV, but switch to CC for the remainder of the run after the current stabilization period:
With the power still on, slowly turn down the amps knob until voltage just begins to drop (or CC power light comes on).
Now turn the volts knob all the way up. Final adjustment of the amps knob to get your "normal" volts may be required after messing with it.
This adjustment method will seamlessly switch from CV to CV.
Finish the run in CC without further adjustment of the power for the time calculated to give you the millage at the CD you calculated in your beginning experiments.
The volts curves can be graphed for all of your alloys, then the appropriate beginning volts for each alloy can be used.
Periodic re-checking your numbers may be necessary, as tank condition changes.
Make sense?

[cameraman]

RedRiver, you're right, the power supply isn't going to sort out where the current goes - but the parts will. The supply voltage is going to be the same everywhere, but the resistance is different all over the place, so the current distribution is ultimately going to be determined by that.


I believe the attached diagram is an accurate model of the anodizing 'circuit'. For those who don't read schematics, the multiple barred object on the left represents the power supply. The squiggly lined things with arrows through them (labelled P1, P2, Pn) are variable resistors, and represent the parts. The squiggly lines below those are fixed value resistors, and represent the resistance of the solution. Technically the solution's resistance will vary during the course of the run due to the aluminum dissolving into it, but in relative terms we can ignore it - with agitation and the amount of the increase as compared to the change that we'll see from the part, it's negligible.

Each of the parts is a branch in the circuit. sage is right - as the part's size increases, its overall resistance decreases. The overall resistance of the solution from the cathode to each of the parts may be the same or it may be different - it depends on the distance and other factors.

Just to make it easier to wrap our brains around, let's assign values. Let's say that we've got two parts and they are hanging an equal distance to the cathode (side-by-side), so we can use the same value for each of the R's. Let's say that's 2 ohms. Now let's say that each of the parts is identical in size, and each has an overall resistance of .1 ohms.

We set the CC power supply to 6 amps.
E = IR
I = 6 amps
The resistance for the first part's path is P1 + R1. The resistance for the second part's path is P2 + R2. The overall resistance of the circuit is parallel between the two paths. A parallel resistance is (R1 * R2) / (R1 + R2):
R = (2.1 * 2.1) / (2.1 + 2.1) = 1.05 ohms
So E = 6 * 1.05 = 6.3 volts.

We apply the same formula to one of the branches to determine its current:
E = IR
E is 6.3 volts, R is 2.1 ohms, so I = 6.3 / 2.1 = 3 amps
So each part will have 3 amps flowing through it.

Now let's double the second part's resistance to .2 ohms.
E = IR
I = 6
R = (2.1 * 2.2) / (2.1 + 2.2) = 1.0744
So E = 6.45 volts
For the .1 ohm part, the current is:
6.45 / 2.1 = 3.07 amps
and for the .2 ohm part:
6.45 / 2.2 = 2.93 amps

So what we see is that the doubled part doesn't get double the current. That's because overall it contributes so little to the circuit. However, we do see that the current is always going to take the path of least resistance.

What's wrong with all this? Firstly, the resistance numbers are made up (the rest of the math is real). However, we know that the solution presents a much much higher resistance in the circuit than the metal does, at least before the anodic layer gets involved. The actual resistance of the bare aluminum is much lower than depicted, but if I'd used the real numbers I would have lost you in the zeros. Essentially the conclusion is the same, it's just even less of a difference than we see above - but there is a difference.

Here's where it gets trickier, though. The other thing wrong with the model is that it depicts the part as a single homogenous resistance. It needs to be broken down further - that model can also be applied to a single aluminum part. With a conventional electrical device such as a resistor, there's one way in and one way out. With anodizing, however, you've got electrons streaming from the cathode looking for a way in to the anode, and they're all looking for the easiest way in.

What that translates to is billions of branches all over the surface area of the part, and it's dynamic - as one point anodizes, the electrons find easier ways in. More current flows down the larger surface area path because there's more spots for the electrons - which is definitively lower overall resistance when we zoom back out to the part level.

On Monday I anodized 10 parts all at once. Total surface area was 101.3 in² and I anodized at 5 amps/ft², so I used 3.5 amps. The largest of the parts was 74 in² and the smallest was .86 in² (actually it was smaller than that because I masked part of it and ignored the difference). I dyed them three different colors so I can't say whether they would have come out the same shade, but I do know that if 3.5 amps were flowing through the .86 part it would have looked pretty nasty coming out of the ano tank.

sage, I would suspect other factors first - like you said, surface finish and distance to cathode - I'd think those would be most prevalent. Coming in close behind would be the particular amount of agitation each is seeing - e.g. a bigger part might be shadowing a smaller from your mixer.

[RedRiver]

Originally Posted by cameraman

RedRiver, you're right, the power supply isn't going to sort out where the current goes - but the parts will. The supply voltage is going to be the same everywhere, but the resistance is different all over the place, so the current distribution is ultimately going to be determined by that.

Thank you...Your words are much clearer and more detailed than mine and infact, this is what I would have liked to have said. Thank you.

Originally Posted by RedRiver

Infact, the resistance is just like the batch surface area in that it is uniform.

As I reflect back on my post, I'm not really sure even myself what I was trying to relay with the above comment. Reading it for a second time, it is obviously incorrect and miss-leading. I've had a lot on my mind as of late and I apologize for the error. Cheers!

[sage]

Boy, that was a really good exchange of information. Thanks to all.
I think we all might have been saying the same thing but envisioning the process differently in our minds eye. It's nice to see - for me, that I'm not thinking alone.
Cameraman - It's too bad you didn't dye the parts in your experiment all at the same time, and the same color. This would also have answered a few questions and added some more proof to the theory since the dye and the shade achieved would have been a pretty good gauge to the anodized layer of each part. I was planning to perform exactly this experiment and haven't gotten around to it yet. If you have occasion to repeat it as described I'd really like to know how it turns out.
i.e different sized parts, anodized together, and dyed together.

Good work none the less.
The reason I suspected that the surface finish and other parameters have a lot to do with the current is as follows:
I anodized two identical parts, polished exactly the same. For reasons I won't go into one of the parts surface got messed up but they were identical when they came out of the dye. I decided to strip the one part and try to re-do it with exactly the same A/sqftas the first and try to dye it the same too. I couldn't get them to match, becausue my process isn't really too well controlled. I decided the only way I was going to get them the same was to strip them both (one for the second time) and re-do them. Upon stripping the piece that was being stripped for a second time looked different (very dark) in the stripper. After stripping them I polished them both again. They looked identical after polishing so I lost track of which was which. During anodizing one of the parts created bubbles alarmingly more than the other but eventually they settled in to be the same. When I went to dye them I lost track of them again but one of them took the dye very quickly to the shade I required. The second took about 3x the time to get to the same shade. Luckily I got them to match (good enough). The moral of the story is maybe that stripping messed up the alloy or the surface on a microscopic level or something.
What a crazy process. So many variables. How do we ever manage all of them.

Sage

[cameraman]

RR, no problem, I thought that was where you were trying to head and it got kind of turned around somewhere in the middle.

I've had a lot on my mind as of late
Man, I hear you.

If you have occasion to repeat it as described
I'll keep it in mind. My next couple of runs are likely to be multiples of the same part but when I do the bigger one I'll stick in a small piece of scrap if nothing else. Same though, if you get to it first, let us know.

[RedRiver]

Originally Posted by RedRiver

I've had a lot on my mind as of late

Originally Posted by cameraman

Man, I hear you.

Well, I don't mean to highjack anyone's thread or share what I am about to share with you people in search of pitty...But here we go :

Most of you have probably seen my anodizing line...I have it posted here from start to finish. In any event, its a 35 gallon line with a dozen tanks and all the required hardware. It was a business that I had been working on for two years and researching for over three. My younger brother owns a custom machine shop and wanted me to start this business with him and he would employ me in his machine shop while it took off. Fifteen months after moving into his shop, one month after the completion of the line's construction and about one week before some big jobs were slated to roll through the door, my brother and I had a fight. He fired me from his machine shop and called back all of his loans within my company. I couldn't pay back the loans (just opened the doors for business) so the company went broke. How's that for brotherly love?

He asked me to help him while he was going through a divorce and some health problems and as an added bonus he would aid me in constructing my own anodizing shop by giving me a place to put it. All I had to do was quit my job of tens years and give up my security, retirement savings plan, dental, health and eye care package, my top dollar wage and the potential to move up the corporate ladder. No biggie, right?

The worst part is, in Canada (where I am from) because I worked for a family member and/or myself am I not entitled to employment insurance or government assistance. What does this mean? It means I need a job ASAP! The bills are coming in but the income has stopped.

Got a lot on my mind? Yeah...You could say that!

The good news...I personally owned enough of the equipment that I can re-open another business at home once I find full time employment. This time I am going to do what I initially wanted to do and build a "hobby business" in my garage. All I need now are tanks and heaters...

Sorry for highjacking!! Its hard not to vent sometimes!

- Dan