why does the break frequency equation involve pi?

so the equation for the break frequency is 1 / (2 * pi * R * C)

but my question is why would pi come into this?, i can see why R and C would come into this, but i cant immediately see why pi should

• So π is bothering you but 2 doesn't? – Dmitry Grigoryev Jan 13 '17 at 9:43
• Instead of "frequency" you must say "angular frequency" - however, in most cases one knows that the symbol "w (omoga)" means 2*Pif . Therefore: w=1/RC=2*Pif and f=1/(2*Pi*f). – LvW Jan 13 '17 at 9:44
• – Bimpelrekkie Jan 13 '17 at 9:45
• Same reason as the circumference of a circle of radius 1 = $2\pi$ – Andy aka Jan 13 '17 at 9:48
• Because you're using the wrong units for frequency. Using radians/second, you'd use 1 instead of 2*pi or 1/(2*pi) and the equations would be simpler. The reason for radians is simple : the length of 1 radian round the circle is the same as the length from its centre (radius) so you're comparing like with like e.g. real and imaginary components of impedance. Using the wrong units (complete cycles) you need a conversion factor equal to the number of radians in a full cycle ... 2*Pi. – Brian Drummond Jan 13 '17 at 11:25