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720 rule and time ????

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  • 720 rule and time ????

    ok guys i have a Q in this 720 rule and it's time
    I have an area of .13 sq/ft and I'm ussing LCD at 4.5 A/sq.ft.
    I have no problem anodizing for 60 mins
    but now I have to shorten that time
    I've been reading on this 720 rule
    correct me if I'm wrong
    .138 sq/ft x 4.5a = .625 AMPS on my rectifier
    720/4.5x.13 = 20.8 minutes on the tank
    correct me if I'm wrong but does this mean that I can use 20 mins and get the same result as I had before at 60 minutes

  • #2
    My Response is WRONG. I double counted the square footage. See response further down in thread.

    Not exactly.

    The entire formula is:

    720 min / 4.5 Current Density x .13 sq/ft x Desired thickness in mils = Anodizing time.

    To get a surface 1 mil thick (25.4 microns) you would anodize for 20.8 minutes. Typically 1 mil thickness is a VERY GOOD thickness as in anodizing, thicker doesn't equate to better, except in specific applications.

    If you anodized for 60 minutes, your thickness would theoretically be 3 mils (or 76.2 microns). You probably didn't get anywhere near this thickness as a number of other factors, such as dissolution rates, enter the equation.

    Generally speaking, if all conditions remain the same, you should get similar results at 20.8 minutes. There may be a slight variation in the deepness of the color as the layer would be thinner and absorb less dye but it is difficult to speculate how much of a difference (if any) there would be. The only way to be certain is to try it.

    Comment


    • #3
      I don’t figure surface into the 720 rule. I only use surface area to find the actual amps needed to get a certain current density per square foot.

      If I am shooting for .5mil and anodizing at 10 amp current density, I would figure it like this: 720 x .5 mils = 360 divided by 4.5 (current density per square foot) = 80 minutes. If I were trying for 1 mil I would double it to 160 minutes. It wouldn't matter if I were doing a little part or a large one, the size of the part is accounted for in the actual amperage setting.

      Edit- I started to give an example of 10 amps, and then actually used 4.5 for the example. If I were to do a load a 10 amps (which is what I use most often) then I would go 72 minutes if 1 mil were the target, and 36 is .5 mil were.

      Comment


      • #4
        I Screwed up...BIG TIME!

        I screwed up and double counted the surface area. M_D is correct. The surface area is already accounted for when the power supply amperage is set.

        To anyone following this thread, ignore this response. It is wrong!

        (PS - M_D - Sorry for doubting you and thanks for correcting my error.)




        M_D;

        I understand what you are saying and at first look it appears to make sense, but mathematically it doesn't work that way.

        Since it takes 720 minutes at 1 amp current density to grow 1 mil of surface per sq/ft of surface area.

        If you anodize at 4.5 amps, it would take 160 minutes to grow 1 mil of coating per sq/ft of surface area. (720 minutes divided by 4.5 amps).

        If you only want a .5 mil thickness, it would take 80 minutes per sq/ft of surface area. (160 minutes times .5 mils)

        If you only have a surface area of 0.10 sq/ft, it would take 16 minutes (160 minutes per sq/ft times 0.10 sq/ft).

        You have to cancel out the terms in order to get the correct answer. Under your formula, everything cancels out except the sq/ft, so your answer is 80 mins per sq/ft. It goes back to fractions and cancellation of terms.
        To express it in this light, it would be (just the final part of the equation for simplicity. Excuse the periods, it was the only to get proper placement):


        160 minutes per sq/ft.........................1
        ------------------------.....times.....---------------- [sq/ft cancel out] =
        ...........1...................................... ....10 sq/ft


        160 minutes..............................1
        ----------------.....times.....---------------- [1's cancel out] =
        ........1.....................................10


        160 minutes
        --------------- = 16 min
        ......10

        I hope this makes sense!

        Comment


        • #5
          The 720 rule: "It takes 720 Amp minutes per square foot to create 1 mil of anodized layer"

          You can see a bit more on it here: http://www.caswellplating.com/bbs/vi...;highlight=720

          As neilfj said, units are important. If you are CC anodizing at 4.5 amps/ft^2, the 720 rule says it wil ALWAYS take 160 minutes to achieve 0.001" of anodized layer. Your surface area is calculated in the current density; it's always accounted for. If you are going for a 0.0005" thick layer, 80 minutes is what you are aiming for.

          jtagger19, you can use a much higher current density than 4.5 A/ft^2 and anodize for a shorter time period to achieve the same anodized layer thickness, but you will not achieve the same results. Pore size is very important if you care about dyeing the part, and there is only a narrow range of current densities that will give you suitable dyeing characteristics. With the LCD kit, stick with 4.5-6 ASF if you want good results with the dyes.

          Comment


          • #6
            I was thinking about this after I posted. I've made an error in my formula, someplace. Don't go by postings until I've had a little time to verify a few things.

            Ok..checked things out and it turns out that I double counted the sq/ft in calculation. The sq/ft is already accounted for when the power supply amperage set as M_D correctly points out.

            I edited my previous post so that no one mistakenly follows the bad advice.

            The REAL formula is:

            720 minutes divided by Current Density times Desired Thickness in Mils.

            So, to achieve a coating thickness of 1 mil (25.4 microns) at 4.5 amp current density:

            720/4.5 x 1 = 160 min

            for 1/2 mil thickness (which is perfectly acceptable):

            720/4.5 x 0.5 = 80 min

            Now to jtagger19's specific questions:

            I have an area of .13 sq/ft and I'm ussing LCD at 4.5 A/sq.ft.
            I have no problem anodizing for 60 mins
            but now I have to shorten that time
            I've been reading on this 720 rule
            correct me if I'm wrong
            .138 sq/ft x 4.5a = .625 AMPS on my rectifier
            This is correct.

            720/4.5x.13 = 20.8 minutes on the tank
            correct me if I'm wrong but does this mean that I can use 20 mins and get the same result as I had before at 60 minutes
            You made a similar mistake to me by inserting the sq/ft of the workpiece into the equation. The size of the work piece is already taken into account when you set your rectifier to .625 amps.

            The actual formula for you would be:

            720/4.5 x 0.5 = 80 minutes (The 0.5 is the desired thickness of the coating in mils)

            You can reduce the time required by raising the current density, or by reducing the thickness of the coating you desire. Good dyeing can be achieved at about 1/3 mils or above, so I wouldn't go below 0.33.
            0.4 mils is a good number if time is a concern. It will give you a 10 micron thickness with good dyeing qualities if anodizing time is a major consideration.

            You could even bump your current density up to 6 amps and go with a thickness of 0.4 mils. That would drop your time to 48 minutes.

            720/6.0 x 0.4 = 48 min

            Comment


            • #7
              Re: 720 rule and time ?

              "You made a similar mistake to me by inserting the sq/ft of the workpiece into the equation. The size of the work piece is already taken into account when you set your rectifier to .625 amps.

              The actual formula for you would be:

              720/4.5 x 0.5 = 80 minutes (The 0.5 is the desired thickness of the coating in mils)

              You can reduce the time required by raising the current density, or by reducing the thickness of the coating you desire. Good dyeing can be achieved at about 1/3 mils or above, so I wouldn't go below 0.33.
              0.4 mils is a good number if time is a concern. It will give you a 10 micron thickness with good dyeing qualities if anodizing time is a major consideration.

              You could even bump your current density up to 6 amps and go with a thickness of 0.4 mils. That would drop your time to 48 minutes.

              720/6.0 x 0.4 = 48 min
              "



              but isnt that 80 min for 1 sq/ft?

              his area is only.13 sq/ft so wouldnt the time be alot less than 80 minutes?



              and above you said this for .10sf/ft
              If you anodize at 4.5 amps, it would take 160 minutes to grow 1 mil of coating per sq/ft of surface area. (720 minutes divided by 4.5 amps).

              If you only want a .5 mil thickness, it would take 80 minutes per sq/ft of surface area. (160 minutes times .5 mils)

              If you only have a surface area of 0.10 sq/ft, it would take 16 minutes (160 minutes per sq/ft times 0.10 sq/ft).


              so wouldn't his time be for .5mil, (80minutes * .13sq/ft ) = 10.4 minutes?

              or am i confused?
              Last edited by donafams; 10-26-2008, 10:18 AM.

              Comment


              • #8
                Re: 720 rule and time ?

                The 720 rule is primarily for time :

                Just think of it like growing a lawn ...the longer you leave it, the taller the sward of grass. If you are shining sunshine on it for 4.5 hours/day it will grow at one rate. If you shine sunshine on for 6 hours/day it will grow proportionately faster, and will take less days to get to a specific height/thickness. BUT how big the patch of lawn is, doesn't affect how fast it grows.

                Now, in terms of setting your current density ..... if you just want to illuminate one square foot you would need a bulb of say 30w; if you wanted to get the same illumination level for a 2 x 2 patch you would need a 120w bulb.

                ( and in fact the analogy works further - if you force the growth by giving it more and more light and elevated temperatures, it will grow faster but just have poor quality stems ... )

                So, the order of things is
                1) Choose current density in Amps per Square Foot (LCD is range 4.5 to 6.0)
                2) Based only on surface area, work out the constant current you need to use
                CC (amps) = area (sq ft) x current density (ASF)
                3) Decide on thickness of anodising layer you need
                4) Based only on thickness required and the CC you calculated in (2) above, use the 720 rule to calculate the time
                Time (mins) = ( 720 / CC ) x thickness required ( in thou. )

                hope that helps & doesn't muddy the waters !

                Dave

                Comment


                • #9
                  Re: 720 rule and time ?

                  VERY well put Dmiom!

                  I use 6 amps per sq. ft. (normally) because it grows the layer fastest, and is easy to calculate the time... for 1mil, its ALWAYS 120 minutes. For .5mil, it ALWAYS 60 minutes.
                  The only thing that changes is the amount of amps... it's calculated at: surface area (in sq. ft.) X 6 = amps needed.
                  I do things.

                  Comment


                  • #10
                    Re: 720 rule and time ?

                    so not matter what size the area is it will tkae 60 minutes for .5mil ?

                    Comment


                    • #11
                      Re: 720 rule and time ?

                      If you are anodizing at 6ASF then yes, it will always be 60 minutes for .5mil.

                      Comment


                      • #12
                        Re: 720 rule and time ?

                        ok im getting it.... so the only adjustment for size would be amps which would be

                        ex. a piece that is 3x2 would be 6x2, to cover both sides so 12sq".
                        12/144 x 6asf
                        .083 x 6asf
                        .5amps

                        that right?

                        Comment


                        • #13
                          Re: 720 rule and time ?

                          Absolutely - that's spot on

                          Comment


                          • #14
                            Re: 720 rule and time ?

                            so whats the largest area you can do with a 25amp cc power source?

                            also, which would be better for color, 4.5 or 6, or are they about the same?

                            Comment


                            • #15
                              Re: 720 rule and time ?

                              Originally posted by donafams View Post
                              so whats the largest area you can do with a 25amp cc power source?

                              also, which would be better for color, 4.5 or 6, or are they about the same?

                              To answer the questions in the reverse order

                              4.5 ASF vs 6 ASF

                              I'm sure others will contribute here; but just for colour (i.e. non-black) if everything else is controlled the same, I don't think we see much difference in the result. What you may need to consider is (a) the higher ASF gets the job done a bit quicker vs. the higher ASF introducing more heat into your ano tank, which you may then need to chill; and (b) the lower current density means that for a given size of rectifier, you can ano bigger area items. For black, our expeience is that we suspect the smaller pore size with 4.5 ASF can possibly contribute to a reluctance to take black, so any black we do is always done at 6 ASF

                              So, if you are using 6 ASF, a constant current source with max 25A would allow 25 / 6 = 4.16 sq ft
                              and if you are using 4.5 ASF, a constant current source with max 25A would allow 25 / 4.5 = 5.55 sq ft

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