How this calculator works
An axial cut through the centre makes a rectangle of width 2r and height h. Its diagonal gives a radius through Pythagoras. Wall area, base area and volume give simpler inverse relations. Total closed area requires solving a quadratic; the implementation uses an equivalent expression that reduces subtraction error. Dividing closed area by volume gives 2/r + 2/h, so at a fixed height the ratio must exceed 2/h.
Formula
r = √[V/(πh)] = √(A_base/π) = A_wall/(2πh) = √(d² - h²)/2. For total area S: r = [-h + √(h² + 2S/π)]/2. For q = S/V: r = 2/(q - 2/h).
Worked example
With h = 4 m and axial diagonal √52 m, r = √(52 - 16)/2 = 3 m. One base is 9π m², the wall is 24π m², the total is 42π m² and the volume is 36π m³. Their ratio is 7/6 per metre.
Advertisement
Frequently asked questions
Which diagonal does this use?
The diagonal of the rectangular section passing through the axis. It spans the full diameter and the height, not a radius and the height.
Does total area include both ends?
Yes. Enter two circular ends plus the curved wall. For an open container, use its wall area or another mode instead.
Why does a small surface-to-volume ratio fail?
The two ends alone contribute 2/h to the ratio. The wall adds a positive 2/r, so a ratio at or below 2/h has no finite positive radius.