Is there a scanner that can work out cubic centimetres?

[Oddjob]

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.

[allankelly1]

Hi Oddjob

Is it just bolts

Best wishes Al


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[allankelly1]

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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[Oddjob]

No it's all sorts of part mate, some linkage parts off the carbs etc might do the main stand etc.

[K2-K6]

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.

[Charlie J]

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.

[K2-K6]

Isn't it Archimedes principle?   

[K2-K6]

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.

[allankelly1]

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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[Charlie J]

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.

[K2-K6]

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.

[K2-K6]

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 ?

[Charlie J]

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.

[K2-K6]

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.

[Charlie J]

Modified the post to say 2mm to correct the 2cm typo. Unfortunately, not before you had read it.