Anodizing 720 Rule Thickness Calculator
Metal Finishing Engineering: Calculate process time, anodic oxide thickness, dimensional penetration/build-up, and chiller refrigeration load via the 720 Rule.
Anodizing Process Parameters
Cycle Duration & Dimensional Impacts
The Empirical Rule of 720 in Aluminum Anodizing
Anodic oxidation is an electrochemical conversion treatment that transforms the surface of aluminum alloy components into a hard, corrosion-resistant, porous aluminum oxide ($\text{Al}_2\text{O}_3$) ceramic matrix.
1. Mathematical Form of the 720 Rule
To produce a specified coating thickness in mils ($0.001\text{ in}$), the required anodizing process duration $t$ in minutes is:
Time (min) = (720 · Thickness [mils]) / Current Density [ASF]
where $\text{ASF} = \text{Amperes per Square Foot} = I / A_{\text{ft}^2}$.
2. Dimensional Growth & Bore Shrinkage
Because aluminum metal is oxidized in situ, the original component dimension is displaced:
- External Dimensions (Shafts, Plate Thickness): Increase by $2 \times T \times (\% \text{buildup})$. For a 50/50 ratio and 2.0 mil coating, a shaft diameter grows by $+2.0\text{ mil}$ ($+0.0020\text{ in}$).
- Internal Dimensions (Bores, Reamed Holes, Threaded Pockets): Decrease by $2 \times T \times (\% \text{buildup})$. A precision $0.5000\text{ in}$ bore will shrink to $0.4980\text{ in}$.
Frequently Asked Questions
What is the "Rule of 720" in aluminum anodizing?
The Rule of 720 is an empirical anodizing guideline stating that 720 Ampere-minutes of electrical charge per square foot of anode surface area are required to produce 1.0 mil ($25.4\ \mu\text{m}$) of porous anodic aluminum oxide coating in a sulfuric acid electrolyte bath.
How does anodizing affect machined dimensions and bore tolerances?
Unlike electroplating which only adds material outward, anodizing converts parent aluminum into aluminum oxide ($\text{Al}_2\text{O}_3$). As the oxide is less dense than raw aluminum, roughly 50% of the total coating thickness penetrates inward below the original surface, while 50% builds outward. An outside diameter increases by $2 \times (\text{buildup})$, whereas internal hole diameters shrink by $2 \times (\text{buildup})$.
Why do Type III Hardcoat baths require high chiller capacity?
Type III hard anodizing runs at high voltages ($30\text{--}75\text{ V}$) and high current densities ($24\text{--}36\text{ ASF}$) at near-freezing temperatures ($32\text{--}40^\circ\text{F} / 0\text{--}4^\circ\text{C}$). The electrical energy ($P = V \cdot I$) dissipated directly into the bath generates tremendous heat that must be continuously removed by refrigeration chillers to prevent oxide dissolution and burning.