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  #1 (permalink)  
Old 06-30-2004, 07:57 PM
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Default 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
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Old 06-30-2004, 08:34 PM
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neilfj
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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.
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Old 06-30-2004, 09:19 PM
M_D M_D is offline
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M_D
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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.
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Old 06-30-2004, 10:24 PM
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neilfj
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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!
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Old 07-01-2004, 12:17 AM
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NeoMoses
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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.
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Old 07-01-2004, 09:42 AM
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neilfj
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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:

Quote:
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.

Quote:
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
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