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  • Fibergeek
    replied
    Thanks, Acidrain.

    I should point out that Acidrain is anodizing for a layer thickness of 0.56 mils which results when anodizing for the CD and time he mentions. This is very appropriate for good dyeing characteristics without risking the evil effects of excessive dissolution. Thicker coatings require more advanced cooling and agitation requirements, and improve the dyeing characteristics very little if at all.

    Anodizing is science and math, its got nothing to do with luck. You do your homework like Acidrain did, and you won't need any luck.

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  • acidrain
    replied
    I think you have the concept wrong. You need to control the amps being delivered, not just let it fly and measure what is being drawn.
    This what I do, and it works every time:
    Calculate the surface area in inches.
    Divide by 144 to convert sq. inches to sq. feet.
    Multiply the sq. ft. of surface area by your target current density. Caswell suggests 4.5 amps per sq. ft. C.D.
    Anodize your parts at the calculated amps needed.
    I have had great luck anodizing for exactly 90 minutes at that C.D.
    If you are having difficulty anodizing at that C.D., and your electrolite is the correct ratio (1 part comercial battery acid, 3 parts distilled water), check your connections and read the stickie "the importance of a good connection".
    Hope that helps. Good luck.

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  • engle_andrew
    started a topic is this correct?please answer

    is this correct?please answer

    will this equation tell me the current density of my system? Amps being pulled or drawn/square foot of part = actual current density of set up.. So for examplepulling 2 amps on .294 square foot part = 6.8actual current density. And is the ideal current density for .294 square foot =1.76 amps for anodizing for like 105 minutes. Thank you for doing the math...
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