Hi Lynette, the surface area for a sphere is

surface area = 4(pi)r2

for 3" diameter hemisphere = a little over 14" squared(14.13" sqr)

area of a circle = (pi)*(radius squared)

===> 14.13" sqr= (3.14)*(radius squared)

this would be the equivalent of a 4-1/4" diameter disc

I'd probably make it a bit larger just in case.

Hope this helps

Jake Balcom
Mettle Design
Lincoln, NE

Lynette,

When raising a bowl. The dimension used is usually figured by adding the hight to the width of the finishe bowl. In your case 4.5" would probably work just fine.

Best,
Fred

Fred Zweig
Metalsmith

Mr. Balcom's answer is way cooler. But then I love the math thing :-)

Gerald Boggs

Lynette, actually, all the previous answers are incorrect if you are, in fact, going to make this hemishpere by sinking.   Since sinking is a stretching process, you would start with a disc approximately the diameter you want your hemisphere to be, plus a little bit for later trimming.  The stock should be considerably thicker than the finished dimension - since you'll be making that 7 square inches of disc stretch to cover twice that area, you should start with metal twice as thick as you want the finished piece to be, plus about one gauge for finishing.  So, if you want to end up with a 18 gauge (.040") thick hemisphere that is 3" diameter, you should probably start with a disc that is 3-1/16" diameter by about 11 gauge (.090") thick.  Then start hammering.

If you're actually planning to raise the hemisphere, use the diameter plus height rule and use the gauge you want to end up with, i.e. 4-1/2" diameter 18 guage sheet.

Ric

Rich is right about the size. He noticed that you said "sink' not raise. Big difference.

marilyn

Thanks for all the help! I'll give it a go and see where I end up! :o) :o) :o)