Place the balls in a triangular shape forming an equilateral triangle.
Labeling the balls at the "points" of the triangle a, b and c (going clockwise), draw a line from the midpoint of a to the midpoint of b. Do the same with b-c and c-a.
The result is an equilateral triangle with 8" sides.
Using the formula in post #12 the triangle would have an area of 27.71 sq"
Discounting balls a, b and c for the moment and only considering the remaining 9 balls that comprise the outside perimeter of the triangle. When the 8" triangle was drawn it also passed through the midpoint of these balls resulting in 50% of the ball to lie outside the triangle. 9 x 50% = 4.5 balls
Since we drew an equilateral triangle the corner angles would be 60 degrees. This would mean that 300 degrees of balls a, b, c are outside of the triangle. 3 x 300 = 900 degrees. A circle has 360 degrees so 900/360 = 2.5 circles or balls in this case.
Add the 4.5 that we calculated earlier to the 2.5 for the corner balls and we come up with the equivalent of 7 balls laying outside the triangle.
Area of 1 ball is 3.14 sq"
7 x 3.14 = 21.98 sq"
21.98 + 27.71 = 49.69 sq"
The shape would be triangular with rounded corners and corrugated sides.
i like it oldman,
only thing now is, is the shape he's looking for a triangle? must admit i haven't tried many other shapes but all have been bigger than the triangle
only reason i used straight edges for my sides is because tomk said to draw a box around the balls when i placed them in a straight line.
so what about mine then?
it's a frame shaped exactly ( well maybe not quite exactly.. ) like yours except it has rounded corners.
hey oldman work out what the area is for those triangle-ish corner bits and add them to my answer for me, my brains gone dead.
I probably should have called it a "simple" shape. When you round the corners… it becomes a "complex" shape. Yet circles and ovals are also simple shapes.
Terry,
The "extra bit" that needs added to each end of your 8" line can be found be multiplying the radius of the ball by the tangent of 60 degrees.
The "extra bit" that needs added to each end of your 8" line can be found be multiplying the radius of the ball by the tangent of 60 degrees.
thanks.
60 tan x 1 = 1.732
so i make that 5.196 for all 3 corners,
5.196 + 54.855 = 60.051 square inches
doesn't seem correct to me somehow because it's greater than the 60 square inches a straight line gives you, where i thought it would be slightly smaller.
That calculation gives you the amount you need to add to each end of your 8" line in order to calculate the area of the triangle… that is if that is what you are trying to do.
So… rather than a triangle with each side 8" long… you would have a triangle with each side 11.4641 inches long.
it depends oldman, most of the older racks i've seen were wooden triangle but the newer plastic ones do have the rounded corners, i'd no idea what the measurements were tho.
figure 8 shape or 2 hexagons joined by last ball in the middle.
2
3
2
1
2
3
2
haven't got the time to work out the area yet to check but counting the spaces between the balls looks like it should be slightly smaller than a triangle.
only thing now is, is the shape he's looking for a triangle? must admit i haven't tried many other shapes but all have been bigger than the triangle 
only reason i used straight edges for my sides is because tomk said to draw a box around the balls when i placed them in a straight line. 
it's a frame shaped exactly ( well maybe not quite exactly.. 
Terry,
The "extra bit" that needs added to each end of your 8" line can be found be multiplying the radius of the ball by the tangent of 60 degrees.