A simple question with a horrendously complicated answer.
Short of carrying out elliptical integrals, the best I can suggest is the Ramanujan approximation according to which:
Perimeter = pi*{3(a + b) - sqrt[10ab + 3(a2 + b2)]} where a and b are the semi-major and semi-minor axes.
Substituting a = 15 mm and b = 6 mm gives
Perimeter = 69.04 mm.
A quick and easy, but roughly approximate method is as follows:
The ellipse is between a circle of diameter 30 mm and one of 12 mm. Averaging these two gives a diameter of 21 mm. A circle with a 21 mm diameter has a circumference (perimeter) of 65.97 mm.
72mm
28mm
283mm
49mm
52mm
Since a square has four sides, just multiply the side length by four: 12mm x 4 = 48 mm
12*2+20*2 = 64 mm
-- Find the circumference of a full circle with a diameter of 12 mm.-- The perimeter of the semi-circle is(1/2 the circumference of the full circle) + (the diameter).
12mm + 12cm
1.2 cm in 12mm
12mm 12mm
A standard HSS drill bit in 12mm is all that's needed.
yes,12mm= approx 1/2inch.
If it is a square, divide 12 mm by 4. Obviously it is a 3 mm per side. You don't have enough info for a rectangle or other shape unless it is a triangle which would be 4 mm for an equilateral triangle.
12mm gives more spin on the cue ball
10mm or 12mm metric 10mm or 12mm metric