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