Page 1 of 3 123 LastLast
Results 1 to 10 of 22

Thread: One for the mathematically minded...

  1. #1
    Join Date
    Jun 2015
    Posts
    440

    Default One for the mathematically minded...

    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

  2. #2
    Join Date
    Dec 2009
    Location
    Central London
    Posts
    8,845

    Default

    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.

  3. #3
    Join Date
    Jun 2013
    Posts
    668

    Default

    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.

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

  4. #4
    Join Date
    Apr 2010
    Location
    Exeter, Devon
    Posts
    1,803

    Default

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

  5. #5
    Join Date
    Jun 2015
    Posts
    440

    Default

    Quote Originally Posted by Patstone View Post
    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

  6. #6
    Join Date
    Aug 2010
    Location
    England
    Posts
    1,902

    Default

    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

  7. #7
    Join Date
    Jul 2015
    Posts
    83

    Default

    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.

  8. #8
    Join Date
    Jul 2014
    Location
    Rural Somerset, between Yeovil and Shepton Mallet
    Posts
    201

    Default One for the mathematically minded...

    Quote Originally Posted by Goldsmith View Post
    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
    Last edited by BarryM; 30-11-2015 at 01:26 PM.
    Barry the Flying Silversmith👍

  9. #9
    Join Date
    Jun 2015
    Posts
    440

    Default

    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

  10. #10
    Join Date
    Jun 2013
    Posts
    668

    Default

    Quote Originally Posted by metalsmith View Post
    The surface area of a sphere is ...
    blah blah blah ... you lost me

    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.

Bookmarks

Posting Permissions

  • You may not post new threads
  • You may not post replies
  • You may not post attachments
  • You may not edit your posts
  •