One for the mathematically minded...

[ajda]

Anyone know a clever formula for calculating the diameter of the rim of a dome/bowl from the diameter of a flat disc - or vice versa? Specifically, if I want to make a small sphere, say 15mm across, from two hemispheres formed by doming two flat discs cut from 0.4 or 0.5mm sheet, how wide should I cut those two discs to start with? I'm sure I could do it by trial and error, but just wondered if anyone knew a way of working it out accurately to begin with.
Alan

[Dennis]

As the disk is stretched during doming, it would be difficult to make a universal table Alan

A quick experiment with copper of the required thickness will give you the answer in no time, and allow you to keep a record for the future. For instance:

Disk 0.7mm thick and 15.0 mm in diameter becomes 12.0 mm in diameter when made into a half sphere. (guestimate only)

I do know that it is tricky to obtain a true half sphere, where the hight is equal to the radius, unless you take your punch around the sides of the dome and not just the base. Dennis.

[metalsmith]

The surface area of a sphere is 4 x pi r^2 hence that of a hemisphere is 2 x pi r^2.
Surface area is fine for the conceptually infinitesimally thin bowl.
Keeping the units constant for r, above, and thickness, multiply the SA by the desired finished thickness.
This gives you a volume.

For your starting circle you should decide on either the starting diameter or the initial thickness. Maybe you are limited by the thickness of the sheet, or the design instructions you have.
Divide your vol by your constraint and you have the other parameter. All that remains then is to raise it into shape. :-O

A rule of thumb I have is the starting diameter is twice that of the desired finished article.

[Patstone]

I'll bet you are glad you asked that question.

[ajda]

I'll bet you are glad you asked that question.
I think I'll put my head down now and look again in the morning...! My brain gets tied in knots at this time of night, but those are the the kind of answers I was hoping for - thanks Dennis and metalsmith.
Alan

[Goldsmith]

To make things simpler. The formula for working out the circumference of a sphere is Pi (3.142) x diameter, so each half sphere is half this total.
For a 15mm. diameter sphere. 15mm x 3.142 = 47.13mm (the whole sphere), divide by 2 = 23.56mm (half a sphere). So as a rough guide 2 discs 24mm. diameter will shape to a 15mm diameter sphere.

I hope this makes sense.

James

[more534]

It's 4:25 am.

I've just begun my first cup of coffee.

I might have to skip work and go back to bed after poking my nose into this topic. :D

[BarryM]

To make things simpler. The formula for working out the circumference of a sphere is Pi (3.142) x diameter, so each half sphere is half this total.
For a 15mm. diameter sphere. 15mm x 3.142 = 47.13mm (the whole sphere), divide by 2 = 23.56mm (half a sphere). So as a rough guide 2 discs 24mm. diameter will shape to a 15mm diameter sphere.

I hope this makes sense.

James

I understand what you are doing here but it does not take into account any stretching or thinning of the disc when using a doming punch, nor does it take into account the thickness of the metal.
Your formula works for a zero thickness material - but it is a good place to start to get an approximate size of the disk.
Stretching will require a smaller disk, whereas thickness will require a bigger one - how these two interact in real life depends on the type of metal, its thickness etc.

Sent from my iPad using Tapatalk

[ajda]

Thanks for all the help - I think I have the info I need now. I'll be enlisting the help of Mr Pi and will post photos... eventually... I hope.... It's a project involving three colours of 18ct gold, but following the wisdom of Dennis I will work first in copper and/or silver.
Alan

[metalsmith]

The surface area of a sphere is ... blah blah blah ... you lost me:-O

It's not all that hard if taken one step at a time. Hope this helps those who find the maths challenging.

The surface area of a 1) a sphere and 2) a hemisphere. First find the multipliers (using pi = 3.1416)

1) 4 x pi = 12.566
2) 2 x pi = 6.283
These numbers are constant - they don't change for any other examples.


We'll assume that the intention is to produce a hemi-sphere. For a sphere just follow twice over!
Keep the 3 decimals for now. Since we're multiplying, any error will also be multiplied so keeping precision will keep the error small for the moment.

Decide on a desired diameter for the bowl - say 100 (mm!) therefore radius is 50.
From the surface area of a sphere, we need to square the radius - multiply it by itself
3) 50 squared = 50 x 50 = 2500

4) Multiply the results of 2) with the results of 3) to get a surface area -15,708 square millimetres

Decide on a desired finished thickness for the bowl. I'll use 3. We keep units the same - mm here...
From 4) x desired thickness

5) Thickness x surface area = 3 x 15708 = 47,124 cubic millimetres

Suggest that I make the bowl from 5mm sheet so starting with 5mm and hammering out to 3mm.
To get a surface area of our starting circle, divide the volume by our sheet thickness

6) 47,124 / 5 = 9,424.8 square mm

Then to get the dimensions of the circle, use the equation for the area of a circle = pi x r^2 . To start, divide by pi:

7) 9424.8/3.1416 = 3000

The answer to 7) is the radius squared, so to go the other way and find the radius ...
find the square root of 3000

sqrt(3000) = 54.77

Since we've been working all along in mm then we can dispense with the decimals - but round up since you can't hammer what isn't there.

Diameter of 5mm thick circular metal disc = 55mm

This will produce a hemisphere 'about' the desired diameter. By this, the finished bowl (if ideally hemispherical) will be half of the finished thickness (3mm), 1.5mm in - and 1.5mm out from the ideal 100mm design line.

[Goldsmith]

Just for interest, I was taught by goldsmiths and silversmiths who made a lot of beautiful pieces as they were all ex Garrard the Crown Jewellers craftsmen. None of them used mathematical formulas when working out metal sizes. When making a bowl from a design, the disc size was calculated by just adding the diameter of the bowl's top to the depth of the bowl, so for a 6 inch bowl that was 3 inches deep we would cut a 9 inch disc. The shaped bowl would be then filed to it's required size when tidying up it's top edge.

James
James Miller FIPG.

[Gemsetterchris]

Probably since most apprentices were not very good at maths.
Those who were went on to better paid jobs :)
I like the sound of that simpler method.

[metalsmith]

Just for interest...

James
James Miller FIPG.

Hi James,

I am very aware of your background. It's ok, I recognise that working out the metal is the easy bit, whether you do maths or not, it is raising it into a bowl that is the skill that takes time to master.

In my last post, my final point that the result of a 3mm thick bowl would be about the centre - i.e. 1.5 mm each way was intended a pun. If one was to raise a bowl to that level of precision, i.e. via the artisanal method, without engineering, that would be a mighty achievement indeed! It seems to have been missed as an attempt at humour. My last attempt at bowl raising only went so far before it resumed the shape of a lump of metal, to be reborn at a future attempt, but I had fun trying.

[more534]

blah blah blah ... you lost me:-O

It's not all that hard if taken one step at a time. Hope this helps those who find the maths challenging.

The surface area of a 1) a sphere and 2) a hemisphere. First find the multipliers (using pi = 3.1416)

1) 4 x pi = 12.566
2) 2 x pi = 6.283
These numbers are constant - they don't change for any other examples.


We'll assume that the intention is to produce a hemi-sphere. For a sphere just follow twice over!
Keep the 3 decimals for now. Since we're multiplying, any error will also be multiplied so keeping precision will keep the error small for the moment.

Decide on a desired diameter for the bowl - say 100 (mm!) therefore radius is 50.
From the surface area of a sphere, we need to square the radius - multiply it by itself
3) 50 squared = 50 x 50 = 2500

4) Multiply the results of 2) with the results of 3) to get a surface area -15,708 square millimetres

Decide on a desired finished thickness for the bowl. I'll use 3. We keep units the same - mm here...
From 4) x desired thickness

5) Thickness x surface area = 3 x 15708 = 47,124 cubic millimetres

Suggest that I make the bowl from 5mm sheet so starting with 5mm and hammering out to 3mm.
To get a surface area of our starting circle, divide the volume by our sheet thickness

6) 47,124 / 5 = 9,424.8 square mm

Then to get the dimensions of the circle, use the equation for the area of a circle = pi x r^2 . To start, divide by pi:

7) 9424.8/3.1416 = 3000

The answer to 7) is the radius squared, so to go the other way and find the radius ...
find the square root of 3000

sqrt(3000) = 54.77

Since we've been working all along in mm then we can dispense with the decimals - but round up since you can't hammer what isn't there.

Diameter of 5mm thick circular metal disc = 55mm

This will produce a hemisphere 'about' the desired diameter. By this, the finished bowl (if ideally hemispherical) will be half of the finished thickness (3mm), 1.5mm in - and 1.5mm out from the ideal 100mm design line.

dear god, my head just exploded.

I think I like the "take the depth, take the width, add together, and voila" school of measuring. :)

[ajda]

OK, so now I've got my Christmas list for the old man in red and white:

1. Bethlehem Champion Torch and Sequal Quad oxygen system for glassworking;
2. a blank cheque made out to Durston;
3. Alan Turing's brain (not in a jar, but live and kicking, to plug into my own like an external hard drive);
4. the intuition, instinct and experienced hands of a seasoned Garrard craftsman.

I don't think that's too much to ask, is it? Just have to hope now that Father C agrees I've been a good boy...

Thanks again for all the input.

Alan

[metalsmith]

OK, so now I've got my Christmas list for the old man in red and white...
Alan

like

oh my

:sleigh:

:xmaswave:

[RayGunGothic]

Hi folks, (new member to forum and coming in late to these discussions) - I recently had the same question and was looking for an online calculator... couldn’t find anything useful apart from this discussion- so as both a silversmith and a programmer I wrote my own. Workbook calculates the circular disk required to raise a spherical bowl of given dimensions.

Forum won’t let me post urls yet... so please search for “disk to raised bowl material calculator” on observablehq dot com (or google for that matter)

Happy for feedback and suggestions for improvements or other calculators.

Cheers, Tony

[Caro]

Hi folks, (new member to forum and coming in late to these discussions) - I recently had the same question and was looking for an online calculator... couldn’t find anything useful apart from this discussion- so as both a silversmith and a programmer I wrote my own. Workbook calculates the circular disk required to raise a spherical bowl of given dimensions.

Forum won’t let me post urls yet... so please search for “disk to raised bowl material calculator” on observablehq dot com (or google for that matter)

Happy for feedback and suggestions for improvements or other calculators.

Cheers, Tony

Having just tied my own brain in knots trying to work out the amount of metal I need to make a complex shaped bracelet, I appreciate any mathematical calculations I can find. I was hoping my geometry days were long behind me. Seems I need it at last!

I found your link but if it's this one it doesn't work - page not found error

https://observablehq.com/@neenahlight/disk-to-raised-bowl-material-calculator

[Dennis]

No offence intended, but people just starting out seek guidance from formulae, tables and recipes. For the most part this is quite unnecessary, when a quick experiment with a paper cut out, some card, or copper sheet will provide the answer.
Dennis.

[alastairduncan]

guidance from formulae, tables and recipes. For the most part this is quite unnecessary

I agree. I've raised a number of vessels with both hammer and spinning. I've never raised one with a spherical bottom. The methods I've used for estimating blank size has been stated by James. I've also used a piece of binding wire bent to the shape, trimmed and measured. The technique of raising can be applied in a number of different ways, it can stretch, compress or have no effect on the metal. This also needs to be taken into consideration when estimating.

[RayGunGothic]

[QUOTE=Caro;111535]... I found your link but if it's this one it doesn't work - page not found error

Thanks for looking Caro and for posting the link... try “@raygungothic” in place of “neenahlight”. Not sure why the old page is still returned by search.

I really appreciate the heuristics listed (e.g diameter + depth) - I was also looking for a upfront way to calculate the bowl volume (e.g. for a 25ml coffee measure spoon) and to take into account the change in material thickness due to hammering (or not, via a dapping block or spinning).

[Dennis]

Dear Ray,
I stick with what I wrote above.There is no need for formulae.
In this case get a cheap plastic measuring scoop https://www.ampulla.co.uk/shop/plastic/plastic-measuring-jugs-scoops-syringes/25ml-blue-pp-plastic-measuring-scoop/?gclid=CjwKCAjwj6SEBhAOEiwAvFRuKEHyix3UJuv4ysI3mj-_aKCx6yc4rV9wFYIF-UnSCqajpi6puwjUIRoCpk0QAvD_BwE
and copy it in your metal of choice. Save the maths. Dennis.