Zinc Plating & Amps Query

[XK120DHC]

In the subject of this post I wrote "query" as opposed to "question".. simply, because I know the amp requirement to zinc plate steel..
My "query" deals with "batch plating" a "pile" of nuts and washers..
Anyone involved in restoring a car or motorcycle has probably run into the "minor" problem that not all the nuts and washers, etc. are exactly the same size.. and especially those of us involved in electroplating parts where amps per square inch can be somewhat "critical" at least, and "picky" at most..
I'm dealing with a 50+ year old British car (that right there says I'm nuts!) and being "typically anal-retentive" about this vehicle, am in the process of calculating the surface area of every individual item I plan on Zinc Plating (I have the "old kit") and Black Chromating. My problem is, I have a small number of nuts and washers whose ASI differs by five to ten milliamps per piece and I just don't know if this minor, almost negligible amount will have an effect on the quality of the plate.
I might add that I have built my own small Barrel Plater (5"dia) just to handle these hundreds of washers, etc. (some washers only have an outside diameter of 1/8th inch).
Additionally, as an example, I have a small bunch of "2BA" nuts that calculate out to 30ma, 32ma and 36ma per piece.. there are 1, 4 and 2 pieces of each respectively.. (I also have some 3BA and 4BA washers that require 14ma and 18ma I need to "combine" with other items, as well) I would hate to have to plate each of these items individually (not to mention trying to regulate such a small amount of amps) if I could combine these pieces (along with others that are "close" in size), this would save a great deal of time..
So, the "query" is: What, if any, is the "allowable" milliamp difference (percentage?) that is acceptable and still achieve a quality plate?
Any and all help appreciated.
Thanks

[cprocess]

If you have calculated your milliamps per part, I assume you figured it by milliamps per square inch. Calculate your total square inches of all parts combined, then multiply by your milliamps per square inch. This will be the total miliamps you will need to process.
Depending on the type of current supply you are using, you may be able to set it at a certain voltage and allow the amps to adjust themselves. If you have calculted the milliamps for a certain part, see how many volts are required to get this milliamp reading. Then, when plating the batch of parts, adjust the current supply to that same voltage and the milliamps should adjust themselves.

Hope this helps.

[XK120DHC]

Originally Posted by FinishingTalkdotcom

If you have calculated your milliamps per part, I assume you figured it by milliamps per square inch. Calculate your total square inches of all parts combined, then multiply by your milliamps per square inch. This will be the total miliamps you will need to process.
Depending on the type of current supply you are using, you may be able to set it at a certain voltage and allow the amps to adjust themselves. If you have calculted the milliamps for a certain part, see how many volts are required to get this milliamp reading. Then, when plating the batch of parts, adjust the current supply to that same voltage and the milliamps should adjust themselves. Hope this helps.

I appreciate the reply..
Yes, I have calculated the surface area of each piece and the milliamps required for that one piece and, of those I have multiple pieces, the milliamps required for the entire lot..
My Power Supply is an older analog 12v25a unit (like the units Caswell is selling--in fact, this unit might be a Caswell offering from years ago).. the "Volts" are only "adjustable" by moving the positive lead to a different "lug" on the unit (the lugs are 5v-7v, 8v-10v and 10v-12v). I have fairly infinitely adjustable amps (and milliamps, I've fitted an in-line milliammeter).. In short, I can "Set the Amps and the Volts will follow".
I'm clear on plating a "batch" of twenty or so of the same size pieces, my query is more about a "mixed lot" of pieces where each would require a different amount of amps/milliamps.. If plated together, what is the "allowable" difference, between the pieces, either in surface area or amps required, to be plated together?? And, we are not talking about pieces that are "close in size", but those that are noticeably different.. i.e. a 3/16" OD washer and a 1/2" OD washer or a 3/8" tool-size nut and a 9/16" tool-size nut.
I would hate to mess-up the plating and have to repeat.. these fifty year old washers, etc. cannot stand much abrasive or acidic cleaning and the bolts with the proper Headstamps cannot be replaced..

[seanc]

Cahrles:

Size disparity between parts is irrelevant. You simply need the total surface area of all parts combined.

eg, take a 9" part & a 1" part together. The 9" part will attract 90% of the current, the 1" part 10%. Each will get the same ASI.

You'll likely never need more than the 5-7 volt setting, as long as it allows you to set the appropriate current.

Sean

[XK120DHC]

Originally Posted by seanc

Cahrles:

Size disparity between parts is irrelevant. You simply need the total surface area of all parts combined.

eg, take a 9" part & a 1" part together. The 9" part will attract 90% of the current, the 1" part 10%. Each will get the same ASI.

You'll likely never need more than the 5-7 volt setting, as long as it allows you to set the appropriate current.

Sean

Sean;
Thanks for the reply.. And the Plating Information.. In reading about other folks problems with "Low Current Density Areas", I wasn't certain if the zinc plating on a larger part would be "thin" while the plating on a smaller part would burn..
I have my PS set on the 5-7 volt range.. In the little plating I've done so far (not Zinc), I "set" the amps to whatever is requred and the Volts read whatever the volts read ;-}.. generally about Two Volts (Five Amps for a 16.5 sq in Copper Tube in Nickel Plating Solution).
Thanks.. again.. You are a helluva contribution to the Caswell Forum!! (I've probably read all your posts.. and have learned a lot.)