Another newbie - surface area to anodizing time?

[kumaichi]

Hi,

Another newbie here probably asking the same questions (I searched the forum but couldn't find the answer.) I apologize ahead of time for any redundancy.

Anywho, I purchased the big anodizing kit from Caswell and got everything setup this past weekend per the instructions. Now, I'm ready to anodize something . My question is, the time to anodize the part is based off of 1 square foot. Well, my initial pieces are very small, 5 square inches. Is there some kind of formula I can apply so I know how long to anodize a 5 sq" piece? Also, as I add more parts, the same part many times, how would I recalculate the anodizing time?

Thanks in advance for any help you can provide. I can't wait to try this out .

Have a great day,

Craig

[NeoMoses]

144 sq. inches per square foot. Therefore, 5 sq. inches = 5/144 sq. ft.

that is, 5 sq. inches = 0.0347 sq. ft.

So, if anodizing at 4.5 amps/sq. ft., you will need 0.156 Amps.

When anodizing at constant current density, you do not need to vary the time. 90-120 minutes at 4.5 ASF should work well.

[kumaichi]

NeoMoses,

Thanks for the reply. So it looks like the determining factor is the amps being applied to the part. Thanks for that information.

I've been doing a lot of reading and it looks like I have to figure out how many amps my system runs at, is that correct? Bascially, take a 144 sq" piece of aluminum and anodize it, looking to see how many amps are drawn while anodizing the part, is that correct? If I'm not supposed to set the amps on the rectifier, what would the settings on the rectifier be? Also, I have the 5 gallon buckets, would it be ok to cut the piece in half and anodize both halves to get my 1 sq'?

Thanks again for all your help with this.

Craig

[Fibergeek]

The determining factors are the surface area of the work, and the current density you wish to anodize at (3 to 6 amps per square foot for LCD).

From these you calculate the applied current as NeoMoses explained, and you calculate the anodization time. We recommend that you use the 720 Rule to calculate anodization time.

In a thread started by GrandSlam (further down this page) M_D explains the 720 Rule and how to apply it.

In the example here (.0347 sq.ft. at 4.5 Amps per sq.ft.) the anodization time is 69 minutes for a layer 1/2 mil (0.0005") thick, which is a good place to start.

[kumaichi]

Fibergeek,

Thanks for the info, I'll certainly read the post on the 720 rule to try and understand better. If I read your post correctly, all I really need to do is set my rectifier amps to 4.5 and then use the 720 rule to actually determine the duration of anodizing. This sounds much simpler than what I was thinking I had to do.

Thanks again for clearing this up. I'm waiting on my replacement rectifier from Caswell, supposed to have it on Friday, can't wait to try it out.

Craig

[Fibergeek]

all I really need to do is set my rectifier amps to 4.5

No. If you did that with your 5 sq.in. example, the current density would be 103 Amps per sq.ft.; way too high, and all you will get is a mess.

You calculate the current to apply like NeoMoses said.

[kumaichi]

Bing!!! A light bulb just went off . I went back and read about the 720 rule again and now it makes sense. I couldn't understand how NeoMoses got the equations.

Correct me if I'm wrong, please:
5 sq. inches / 144 sq. inches = 0.0347 sq. ft.
4.5a x 0.0347 = 0.156Amps

Applying 720 rule
720 x 1 mils = 720
720 / 4.5a = 160 minutes

So, I would set my rectifier to 0.156 amps and I would anodize for 160 minutes to get a 1 mil anodizing layer.

How do you know if your setup is anodizing at 4.5a per square foot? Is it just a given or is there some kind of test I need to do?

Sorry for all the confusion, still trying to grasp the concepts of the amps and anodizing time.

Thanks for your patience,

Craig

[Fibergeek]

Like you surmise, 160 minutes at 4.5 ASF will produce a layer 1 mil thick. Since dye (any dye) won't penetrate any deeper than 0.7 mils, there is no point in a layer 1 mil thick for most purposes. Notice that the basic LCD instructions call for 90 minutes at 4.5 ASF, which by the 720 Rule produces a layer 0.56 mils thick.

You should not attempt a thick layer until you have the whole process figured out and working. If you do, you may find out the hard way about the evil effects of excessive dissolution.

[M_D]

......How do you know if your setup is anodizing at 4.5a per square foot? Is it just a given or is there some kind of test I need to do?.....

First you need to figure the part surface area, and then derive the actual amp needed for the parts, like you figured in the 720 rule. If you figure it all out correctly, and have a way to accurately set the amperage output on the power supply, then you can assume you are anodizing somewhere close to the amps per square foot figure used in the 720 rule equations. There are two main areas that could skew the actual amps per square foot from the calculated, which would would be figuring the surface area incorrectly, and inaccurate current metering equipment.

If you don't have a power supply that is capable of constant current (constant amps, rather than constant volts) with a readout (or some reasonably accurate way to meter the output), then you are more or less guessing. Like everything else, some equipment is better than others, so the actually accuracy and ability to measure and meter current output will vary. If your part size isn't too small, the error will probably be acceptable. If you were trying to do very small parts in small or singular batches, you may need some high resolution equipment to be consistent.

[GramSlam]

I posted a new thread with regards to this topic. In the thread there is a link to a calculator I built in excel. Perhaps that can help out.

[DarkDesigns]

Re: Another newbie - surface area to anodizing time?

Just double checking my calculations ....

144 sq. inches per square foot. Therefore, 20 sq. inches = 20/144 sq. ft.

20 sq. inches = 0.138888 sq. ft.

4.5a x 0.138888 = 0.62499 Amps (625 miliamps) for a duration of 80 minutes.

720 = 1 Mil
360 = .5 mil

360 / 4.5a = 80 minutes in the bath

By upping the amperage to say 1.2 amps, will I (in theory) reduce the amount of time in the bath?? Say by approx. 1/2?? (80 / 2 = 40)

[acidrain]

Re: Another newbie - surface area to anodizing time?

Yes, that changes your current density. Try calcing it at 6 amps/sq. ft.