[Amps] Calculating ripple % ?

[Jim Thomson]

There appears to be 2 ways to calculate ripple  %.   There  also appears to be no consensus as to which method is correct. 

One method uses the  peak to peak value of  AC ripple, as seen on a scope.....then  divide by  loaded B+  (then  multiply that result  by  100).   This  is with a full bore load  on the B+ supply, like key down. 

The other method uses the RMS value of the AC  ripple, then   divide by  loaded  B+  value.    (than again,  multiply that result by 100).  Again, with a full bore load. 

Problem is,  most simple  AC meters..and also cheaper  DMMs  will not measure  RMS  AC voltage.   These lower cost type meters assume the AC  is a pure sine wave.   On a  simple  B+  supply, with 
full wave rectification and a C  filter, the   AC  component  looks more like a saw tooth  wave form. 

Then I see there is various formulae  used to find the RMS value of the typ saw tooth wave form...with  varying results. 

These days, with something like  my Fluke  87, it will read true  rms, so the saw tooth wave form is a moot point.   A true  RMS  DMM, like a fluke 87  etc,  will easily read the rms  component. 

But regardless of which method is used, the calculated  ripple %    difference between  each method is huge. 
EG:     8.75%  using the  P-P method  equates to just  3.1% if using the RMS  method.

So which method is... the  real deal ??   Which one should we be using ?    I suspect the P-P method is in vogue by some folks..since back in the day, it was..and still is, easy to measure.   The  formulae used to find the rms value of a saw tooth 
wave form are  plenty, varied....and all over the map..hence the use of a simple scope to measure  P-P.   Im inclined to believe the  RMS method is probably the real  deal.... but I may well be  wrong. 

Jim   VE7RF