Honda-SOHC
Other Stuff => Misc / Open => Topic started by: Oddjob on December 13, 2021, 11:16:29 PM
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I'm going to start zinc/nickel plating shortly and one of the things I'd like some help with is working out the area of where I'll be plating. Ideally I'd like a sort of gun which lasers the bolt etc and just gives a figure so I could add up all the parts I'd be plating in one go and then I can work out the ampage. Anyone know of anything that might be suitable or am I asking for the moon here. Can't be too expensive of course.
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Hi Oddjob
Is it just bolts
Best wishes Al
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If it was you could creat a excel formula that works out the shank surface area and one for the head and the the two formulas then add together?
All you then need is a vernier abs a pc or caculatot
Best wishes Al
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No it's all sorts of part mate, some linkage parts off the carbs etc might do the main stand etc.
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Trying to think that one through.
If you put (these are just examples I can immediately think of in convenient containers) a washing up bowl onto a large tray, fill the bowl right to the brim level with water, then put the items into the water together. The water now in the tray is that displacement of cubic centimetres of your items, measuring the tray contents with depth, length, width should give you surface area. At least I think that's right.
For tray, I'm thinking more along the lines of some 800mm x 600mm commercial baking trays I bought to work on things while containing all the parts and leakage while dismantling etc, not just a household tray.
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I have some experience of using 3D scanners for measuring surface area of aggregate particles in a laboratory. I know that 5 years ago you would have to pay at least £10k for one good enough to do what you are asking. I’m sure prices will be cheaper now but still a fairly heavy investment.
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Isn't it Archimedes principle? ;D
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I think it will work, the displacement method.
Try it theoretically, put a 10cm cube into a full container of water completely submerged, it'll displace it's exact volume in water. That water should now have the same surface area (the water displaced) as the original cube as both are effectively incompressable. Just put the displaced water in a container you can measure and complete the maths on.
It'll even do a the little rimple dimple detail for you.
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I think it will work, the displacement method.
Try it theoretically, put a 10cm cube into a full container of water completely submerged, it'll displace it's exact volume in water. That water should now have the same surface area (the water displaced) as the original cube as both are effectively incompressable. Just put the displaced water in a container you can measure and complete the maths on.
It'll even do a the little rimple dimple detail for you.
Hi
Would that not would work out volume not surface area?
Especially if some tubes were sealed
Best wishes Al
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I’m not sure that would work, as displacement is related to volume and not surface area. You could work out cubic centimetres by this method but not square centimetres. Two objects can have the same volume but different surface areas.
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I’m not sure that would work, as displacement is related to volume and not surface area. You could work out cubic centimetres by this method but not square centimetres. Two objects can have the same volume but different surface areas.
I'm kind of thinking out loud which with our collective thoughts may shed some light on a concise direction.
"Two objects can have the same volume but different surface areas." Would that not demonstrate different densities? Of material.
Trying to think of an example that would have different volume (irrespective of material density) and still be the same surface area. Isn't volume the space included within the specified surface area ? And so would change in a fixed relationship.
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Or, "The density of steel is in the range of 7.75 and 8.05 g/cm3 (7750 and 8050 kg/m3 or 0.280 and 0.291 lb/in3). The theoretical density of mild steel (low-carbon steel) is about 7.87 g/cm3 (0.284 lb/in3).
Density of carbon steels, alloy steels, tool steels and stainless steels are shown below in g/cm3, kg/m3 and lb/in3."
From this site https://amesweb.info/Materials/Density_of_Steel.aspx which would mean, if you weigh the components, each 7.87 gram would occupy a cubic centimeter, which gives the maths route to calculate total surface area for plating.
Which looks to me that you can reliably give a surface area for the components as mass of the low carbon steel is a constant.
Is that true ?
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Or, "The density of steel is in the range of 7.75 and 8.05 g/cm3 (7750 and 8050 kg/m3 or 0.280 and 0.291 lb/in3). The theoretical density of mild steel (low-carbon steel) is about 7.87 g/cm3 (0.284 lb/in3).
Density of carbon steels, alloy steels, tool steels and stainless steels are shown below in g/cm3, kg/m3 and lb/in3."
From this site https://amesweb.info/Materials/Density_of_Steel.aspx which would mean, if you weigh the components, each 7.87 gram would occupy a cubic centimeter, which gives the maths route to calculate total surface area for plating.
Which looks to me that you can reliably give a surface area for the components as mass of the low carbon steel is a constant.
Is that true ?
Unfortunately it's not that simple. Volume = mass/density. So two objects with the same mass and density will have the same volume. However, two objects with the same mass and density can have two completely different surface areas. For example, a cube of mild steel measuring 1cm x 1cm x1cm would indeed have a mass of 7.87 gm/cm3. However, if you hammered the cube into a plate say 2mm thick, it would have the same mass, the same density and therefore the same volume but a much larger surface area.
Probably the simplest and cheapest means of determining the surface area of a complex shape is to cover the surface area with tin foil and then flatten out and measure the tinfoil.
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Trying to get my mind around it :)
"For example, a cube of mild steel measuring 1cm x 1cm x1cm would indeed have a mass of 7.87 gm/cm3. However, if you hammered the cube into a plate say 2cm thick, it would have the same mass, the same density and therefore the same volume but a much larger surface area."
I don't think that holds true. If you hammered a 1cm cube it could never be 2cm thick. As you hammered it the top and bottom surface area increase but with the height reduced with every blow. It's still the same maths although not an even 1cm uniformity.
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Modified the post to say 2mm to correct the 2cm typo. Unfortunately, not before you had read it. :)
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To give a simpler example and always remembering that Volume = Mass/Density
Take a ball of dough which will have a given mass and density and therefore volume. If you rolled it out as flat as you can it would still have the same mass, density and volume as the ball but a much larger surface area.
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Trying to visualise it. And so if I had that notional 1cm cube in steel and milled a V shape into one surface, then mass is reduced but surface area changes by the difference between that one face being removed, plus the two triangles missing from the other two faces affected and compared to the two newly made facet of that V ? Maths should prove or disprove that.
If that's true, wouldn't the displacement method be more accurate, and certainly close enough for the purpose calculations needed for the original question.
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Rather than trying to visualise it. Try this exercise.
Take a steel cube 100mm x 100mm x 100mm and work out the surface are and volume.
Now slice the cube in half to give two pieces 100mm x 100mm x 50mm and now work out the total volume and surface area of the two pieces.
You will find that whilst the volume remains the same, the surface area will have increased by over 30%. So the displacement method for determining surface area isn’t remotely accurate even for non complex shapes.
I’ll let you do the actual maths which will clearly demonstrate the point
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What have I started :D :D :D :D
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;D interesting though.
Thanks for input from other posters, certainly makes you use your thinking power to try and figure something out.
Apologies if it's diverting your thread Ken :)
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Well you did ask for cubic centimetres (volume) and then went on to talk about square centimetres (surface area).
You are clearly a man of many dimensions.
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:) It can get quite complex, even more so if you put rugosity (surface roughness) into the equation.
But here’s a link to a useful guide and calculator from Caswell Plating that may be of help.
https://caswellplating.com/surface.html
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Rather than trying to visualise it. Try this exercise.
Take a steel cube 100mm x 100mm x 100mm and work out the surface are and volume.
Now slice the cube in half to give two pieces 100mm x 100mm x 50mm and now work out the total volume and surface area of the two pieces.
You will find that whilst the volume remains the same, the surface area will have increased by over 30%. So the displacement method for determining surface area isn’t remotely accurate even for non complex shapes.
I’ll let you do the actual maths which will clearly demonstrate the point
An alternative approach to visualising this is to consider the 10 x 10 x 10 cube (in cuboid form), then visualise yourself smashing the living sh*t out of it with a 10lb club hammer (repeatedly) and imagine the surface area of the resultant, flat, but larger surface area object. Simples.
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Now why didn’t I think of that :D
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Ken, I have found that the plating area only has to be approx. My instructions from the kit state:- (for example)
'Current approx 0.5 Amp - Plating area 2.5 - 5.00 square inches (optimum 4 square inches) - Typical part: Bolt 19mm dia x 50mm long'
This gives you some wiggle room - its other factors like temperature and mixture condition that have made the difference in finish in my experience.
I have made rough calculation on the area by breaking the object visually into sections and applying an area calculation to each section and adding together.
Always had reasonable results over several projects
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The owner of Gateros told me in a conversation that a 6mm x 50mm bolt would be 1/3 of an amp for 20 mins, plating companies only plate for 15 min he said so the extra 5 mins gives you a better coating whilst only being a few more microns thick. If the part isn't a critical fit part you can leave it for a good while and really pile the plating on but not if you have threads etc.
Getting ready for my first attempt this weekend I reckon. Some of the parts I'm going to plate are unusual, we'll see how they come out.
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Hi, Ken.
Screws arrived thanks.
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I have always worked on 100mA /square inch of surface area for BZP. I got that figure from a 1943 Canning book.
One of these days I am going to make a DIY mini barrel plater as loads of tiny bits are a PITA using the wire 'hanging' method.
You basically have a little plastic barrel with lots of holes in it rotating slowly at an angle in the electrolyte and metal cathode 'danglers' strategically placed to contact the steel parts as they rotate.
I have discussed this with PM 's with Ken but I made a big mistake and tried out 'YouPlate's' ready mixed chemicals as opposed to the 'Gateros' stuff I have used for ten years. I got a really dull finish with the Youplate stuff and no amount of playing around with settings/brighteners will achieve a bright finish ... so voted with my feet and gone back to the Gateros stuff. Well we all make mistakes :-[ :-[
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Just to throw another spanner in the works with the water displacement method. When measuring the surface area of the water you will
get vastly different results just by adding a drop of soap to break the surface tension of the water yet still have the same volume of
water.