Why does my math not jive with 720 Calculator?

[RedRiver]

I've always anodized using the 720 rule. Its pretty basic really but I have since decided that understanding the Peak Anodic Resistance method may be beneficial as well. To my understanding the formula to calculate PAR is :

PAR = 0.95 / Surface Area x Total Current

Having said that, I generally anodize with a current density of 12 ASF. I run a 9.4% electrolyte acid concentration which maintains a working temperature of 69-72F. In order to achieve an anodic thickness of 1 mil, I anodize for a duration of 60 minutes.

If for example I were to anodize a part with a total surface area of .383 sq.ft, I calculate PAR to occur at 11.4 volts. My math is as follows :

Total Current = ASF x Surface Area
Total Current = 12 x .383
Total Current = 4.596 amps

PAR = 0.95 / Surface Area x Total Current
PAR = 0.95 / .383 x 4.596
PAR = 11.4 volts

Anodizing Duration = 720 / ASF x mils
Anodizing Duration = 720 / 12 x 1
Anodizing Duration = 60 minutes

My questions :

1. How come when I plug my criteria into the 720 calculator located HERE, the calculator shows that my PAR occurs at 30 volts? Do I not have the correct formula for calculating PAR? There is quite a difference between 11.4 and 30 volts!

2. I've read and read that just because PAR is reached, the anodizing may not be complete. Several things such as electrolyte temperature and agitation may allow PAR to be reached prematurely. I assume that if the working parameters are held constant (electrolyte concentration, temperature, agitation, ect...) that in theory PAR should be achieved at the same time as the calculated Anodizing Duration (my last formula, taken from the 720 rule)?

3. Is it possible to calculate the anodizing duration required using PAR? If so, how?

5. I also understand that some of you wait for the voltage to drop by 10-12% from PAR thus indicating that the dissolution is beginning to overtake the creation of the anodic layer. Is this recommended? I realize that some dissolution is a good thing - is this why this method is recommended?

6. In the PAR formula (PAR = 0.95 / Surface Area x Total Current) I realize that 0.95 represents 0.95 Ohms per 1 square foot. I am also aware that temperature and agitation will alter the resistance, however, is this resistance of 0.95 Ohms proportional to the current density utilized or the alloy being anodized? If so, is it enough to be concerned about?

I appreciate all your help folks. Though I've been anodizing for quite some time now, I still have boat loads to learn!

- Dan

[RedRiver]

Looking back at it this morning (it was a long day yesterday) it appears that the 720 Calculator accessable via the link above uses the formula :

PAR = 0.95 / Surface Area x ASF

I was calculating PAR with the "set current" or total amps required rather than the current density. At this point, because the 720 calculator is so popular, I must assume that the 720 Calculator is correct and I am wrong...?

It just seems like an awefull lot of volts (30) to anodize at this current density. I suppose we all learn something everyday...

- Dan

For sale : 1 barely used Encyclopedia Britannica set. Like new condition but no longer needed. Just got married and apparently the wife knows everything.

No...I didn't just get married nor do I have any books for sale...I just thought it was funny. It came to mind so I shared it...Have a great weekend all!

[acidrain]

Your acid concentration sounds stronger than the recommended 1:3 battery acid/water formla for LCD anodizing. The PAR calculator assumes 1:3 ratio... an electrolyte that required higher voltages to achieve the same amps.
Maybe I'm off base on that... anybody else chime in?

[RedRiver]

Hmmmm....Well, I am using a more concentrated electrolyte. 9.4% acid by volume if I recall (been a really long day) but then again, I run with a current density of 12 ASF. It is my understanding that this electrolyte and this current density work well together (and I have had really good results time after time). I realize that this current density is not considered LCD, however, the formula that the 720 calculator uses to calculate PAR should still be the same even with high current densities. One question, when using an electrolyte with an acid concentration of 9.4% by volume, will the 0.95 Ohms that the 720 calculator uses as the constant value for resistance remain the same or will the resistance change with a different acid concentration? If it changes, does anyone know off hand what the resistance would be in a tank with a 9.4% acid concentration, operated at 70F with a current density of 12 ASF?

Thanks again all...Trying to calculating PAR is giving me a headache (and I only want to know how to calculate it out of pure intrest...Go figure!)

[destroyer125]

The link for the 720 calculator seems to be broken for me

[sswee]

After writing this, I went back over everything and found the problem and where the .95 ohms came from. It was in a calculation step I had been skipping because I already new the answer. I hope there is still something in here you can use. SS

1. How come when I plug my criteria into the 720 calculator located HERE, the calculator shows that my PAR occurs at 30 volts? Do I not have the correct formula for calculating PAR? There is quite a difference between 11.4 and 30 volts!

The 30V is arrived at from Fibergeeks tests which determined the average ohms per square foot to be 2.5 ohms. A CD of 12 ASF x 2.5 ohms per square foot = 30V peak voltage. The tests were run at a 3:1 acid ratio.

2. I've read and read that just because PAR is reached, the anodizing may not be complete. Several things such as electrolyte temperature and agitation may allow PAR to be reached prematurely. I assume that if the working parameters are held constant (electrolyte concentration, temperature, agitation, ect...) that in theory PAR should be achieved at the same time as the calculated Anodizing Duration (my last formula, taken from the 720 rule)?

PAR does not coincide with the anodizing duration calculated by the 720 rule. The 720 rule calculates a run duration based on a desired coating thickness. It is true that PAR is sometimes reached before a run duration is complete. This is due to the enlargement of the pore size even though the coating continues to grow in thickness. If all the parameters were held constant, then PAR would be achieved at approximately the same time on different runs of the same material type. Different materials have different voltage curves under the same parameters. All the runs I have made confirm this.

3. Is it possible to calculate the anodizing duration required using PAR? If so, how?

Not to my knowledge. You can anodize according to PAR but I don't see how it could be calculated with so many parameters to figure in. The results I achieved running by par were not consistent by any means. I was told it was due to the accuracy of the equipment I have.

5. I also understand that some of you wait for the voltage to drop by 10-12% from PAR thus indicating that the dissolution is beginning to overtake the creation of the anodic layer. Is this recommended? I realize that some dissolution is a good thing - is this why this method is recommended?

Yes because the enlargement of the pore size is causing the drop in resistance even though the thickness of the coating still grows.

6. In the PAR formula (PAR = 0.95 / Surface Area x Total Current) I realize that 0.95 represents 0.95 Ohms per 1 square foot. I am also aware that temperature and agitation will alter the resistance, however, is this resistance of 0.95 Ohms proportional to the current density utilized or the alloy being anodized? If so, is it enough to be concerned about?

Where did you get the figure of .95 ohms? It is not familiar to me at this moment. According to my understanding, the resistance per foot squared is proportional to all the parameters of the circuit but not enough to be concerned with. If I can remember how it was told to me, The value of PAR is inversely related to the size of the part. This is another reason calculating anything by PAR would be difficult.
The way I was taught to calculate everything before there was a calculator is the same as the calculator gives. Maybe something is off on the calculator you have. If anyone wants a copy of the calculator on MS Excel or the other one just email me and I'll send you a copy.
Sorry if some of this is not clear. It is in my head but hard for me to sometimes convey to another.
SS

[RedRiver]

I appreciate your time SS. As I am currently running out the door (to carry a washing machine into my sister-in-law's basement of all things), I don't have the time right now to look over your response. Before I go, if you haven't answered this already, I have a question for all :

Using an electrolyte with a 9.4% (by volume) sulfuric acid contend and a temperature of 70F, if I were to use a current density of 12 amps to anodizing a part, what it the calculated PAR?

Thanks for the help...

[RedRiver]

I suppose a surface area is required in my above question, so to add to my above post (since I can't seem to edit it) :

Using an electrolyte with a 9.4% (by volume) sulfuric acid content and a constant temperature of 70F, if I were to use a current density of 12 amps per square foot to anodizing a part with a surface area of 1.75 square feet, what would be the calculated PAR?

If you are able to, please show me in longhand your mathimatical answer.

Thanks again!!

[sswee]

According to my notes:

Plugging your numbers into an equation for the 9.4% electrolyte -
For a current density of 12 ASF, and a thickness of 0.0005" (0.5 mils); 720 / 12 x 0.5 = 30 minutes.
The actual current you would apply is 1.75 sq,ft.(252" sq.) x 12 ASF = 21 Amps.
or 12A / 144 = .0833 per in^2 * 252 = 21A setting
The required peak voltage is 0.95 Ohms per sq.ft / 1.75 sq.ft. = .54 Ohms. .54 Ohms x 21 Amps = 11.34 Volts. This assumes no voltage drops due to bad connections, and you haven't increased the actual surface area with racking.

Excerpt from other post:
For 20 ASF, 9.4% electrolyte concentration by volume, and 70 deg. F., this resistance is about 0.95 Ohms per square foot. Its value is inversely proportional to surface area (2 sq.ft. = 0.475 Ohms). It varies a little with alloy as well.

The calculations in the 720 calculator are based on the 5% concentration. The resistance in the circuit is an average or approximate but there are no places to change this in relation to CD. Therefore, I would think that if it has been determined that .95 OSF is the resistance of 20 ASF with a 9.4% acid concentration, it should be close enough for calculations at other CD's with the same acid concentration. With all the different parameters effecting the circuit resistance, it would take several pages to figure it all out.

I'll keep reading through my note and looking around for more. Maybe Neilfj will stop in and help clarify it all.
SS

[RedRiver]

Excellent post! I appreciate the help immensly! When I get a few more minutes I will look it over again and attempt to apply it to my work. I'll let you know the outcome. Cheers!

Beers on me!

- Dan