Calculating Surface Area, the Easy Way.

[PrecisionPB]

Using the 720 calculator provided by the friendly folks here in the forums, we have stumbled across an easy way to get surface area. I dont know if its been discussed here, but its been working great for us.

First, connect your parts to the rectifier (we use a 20amp Hyui Yi, thanks Caswell). Turn your voltage all the way up, and your amperage all the way down. This sets the rectifier to constant current mode, in its lowest setting. Wait for your voltage/amperage to level out with the parts in the bath... when they steady (seems to take about 10 - 15 seconds with my rectifier) sprint into the other room where your handy desktop PC has the 720 calculator open and ready for bid'ness.

In the bottom portion of the calculator, enter the voltage and amperage of the rectifier at its "idle" setting into the according position. Take the surface area given, enter it into the top portion. Then its back to the rectifier with your new information, set amperage accordingly, and watch the clock.

These calculations have been as, or more accurate than my sketches in solid works. I hope this helps people out, if not, and your torch your first project....

[Fibergeek]

Sorry, PrecisionPB.
That's not going to work in the general case. It apparently is providing you with decent anodizing in your particular case, but watch out if you change something, like the SA of the part you want to anodize.

In a nutshell, you aren't anodizing at any particular predetermined current density, you are adjusting the time to accommodate the current density you think you have. Monitoring the voltage will not give an accurate measure of current density. The voltage can and will vary with almost everything in the tank. Monitoring the voltage is useful in repetitive anodizing, it will tell you something is wrong if an unexpected change in it occurs.

Mathematically, there isn't a way to calculate the applied current for a given current density without knowing the surface area. Short of a 3D scan of the part, or better yet, extracting it from a 3D rendering of the part if you have the 3D design software. There is no method I know of to "measure" it without resorting to geometry (and MATH, eeek!) and I have certainly looked.

[PrecisionPB]

Im not disputing your conclusion, since you certainly speak from experience... but something that made me scratch my head follows:

Example part A (including wires and plates), measured using with a CAD design program (lets say it sounds alot like solidworks), measures 50 sq/in.


Example part B (including wires and plates), "measured" using the process I described above also measures 50 sq/in.

Which brings me to two "why" questions.

Why are the measurements the same?

Why am I getting more consistent results?

[Fibergeek]

I also speak from the vantage point of mathematics and physics.

You are now getting consistant results because you fixed your dye contamination problem.

The voltage figure you want to use were based on measured values I derived last year; taken for various current densities and various alloys, they are only valid in my tank, with my cathodes, electrolyte mixture, anode connection method, spacing, temperature, etc. These voltages were expressed as a resistance per square unit area (Ohms per square foot). If by shear chance your tank's characteristics are close enough to mine, you will see agreement, but only then.

What you are proposing is just one step away from "voltage anodizing" if you fall into this you will find yourself thrashing around with inconsistant results just like a "professional anodizer", if they understood how electricity works they would realize the trap in this flawed reasoning.
The fatal flaw is that these "professionals", by casual observation seem to see that the bulk resistance (they don't even know enough to recognize what it is) looks pretty much the same regardless of tank characteristics, in fact this cannot be, they aren't looking hard enough. This can be proven mathematically but few have the math backround to understand the proof, so I won't bother.

[jtsuttle]

I have also done experimenting with determining surface area this way. There are a couple of problems that make it just too risky. One is the alloy. Different alloys have different current densities. Some alloys like 2024 may take 20 volts to produce a 12 amp current density wile others like 6061 may need only 18 to produce the same 12 amp current density (I'm just pulling numbers out of the air because I don't have my notes with me). Another problem is temperature. Temperature affects current density DRASTICALLY. In my experimenting, I concluded that if you are going to do it this way then it is better to use constant voltage (no more that 8 volts for a stable current reading) and use the current reading for calculations. After that you can switch back over to CC for the anodizing. You would have to take a current reading at the same time every time to be accurate because the current will be rising constantly as the anodize process carries on. Temperature needs to be held within a few degrees (+/- 2 F). If temp can't be held that close then you would also have to know current densities at different temps. You would also have to know the current densities for the particular alloy that you are doing, as they are all different. And finally my numbers for current density won't work for any one else because of differences in ano bath chemistry. You ano bath chemistry is changing slightly all the time from evaporation, rinse water drag in, dissolved alum, etc. so you would have to perform current density test on a regular basis to stay accurate.
It is not impossible to determine surface area this way; it just seems to me that it’s just not practical.
In my opinion there are just too many variables for this to have great success 100% of the time. But, I would love to be informed of any discoveries you may have while pursuing this.
I'm not trying to discourage you; I just thought you should be aware of the obstacles ahead.

[jcmmachining]

calculting

Hello can anyone help me here on how to calculate the the surface area on a round part that has a large radius on it. example 2" dia pc of 6061 3" long but has a .750 radius on the end of 1 side.

[sswee]

Calculating a part 2"dia x 3" OAL I get 25.1328" SA.
Calculating the part with one end radiused .75", I get 24.8773".
The way I was taught, if you are within 10 to 20% of SA you are ok. I would go with the first figure. You'll be only minutely high but not enough to bother anything.