David Higgon's solution and mine reduce to a good approximation to
Length ~ (w*A/(4*pi)*sqrt(A^2 + 1) = (R/2)*sqrt( (2*pi*R/w)^2 + 1)where w is the distance between successive turns, A is the total rotation angle of the spiral in radians and R is distance of the outer end from the centre. To make this on topic, can we please have a dial based on an Archimedian spiral! Les Cowley
