How to calculate voltage after rectifier?
The sites formula is correct, but only under ideal conditions. The AC input must be a clean sine-wave. Some UPS's and DC-AC inverters put out a choppy sine wave that would make the 1.414 ratio of RMS value to peak value not true. 
This peak voltage assumes no load, whether a single diode is used or a bridge rectifier, plus capacitor of sufficient value to remove any AC ripple. Even a tiny load of .1% of capacity will drop the voltage by the amount the diodes dropped. So subtract .7 volts or 1.4 volts from the expected peak, and the numbers should match better.
With heavier loads a bridge or full-wave rectifier will provide the most current. At high current levels >10 amps the Vdrop across each diode can be 1 volt. The output voltage will drop as the load increases until a full safe load is reached.
By now the peaking effect is gone and the DC voltage is more like the AC-RMS value. The RMS-to-Peak formula is correct but only under ideal conditions, and as you might have guessed by now, real-world conditions dominate once you apply a load or use a inverter/UPS for AC power, making the RMS to Peak formula useless, especially under heavy loads.
@ElliotAlderson made an important observation;



  You are missing the point of the formula if you think it is wrong. The
  formula tells you the maximum dc voltage you can achieve from a given
  rms ac voltage. That value is very important when it is time to
  specify the voltage ratings of your components, such as the filter
  capacitor. The formula is very useful if you understand what it means.