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Old 04-23-2017, 08:42 PM
Riemann Riemann is offline
Join Date: Nov 2015
Location: Santa Fe, NM, USA
Posts: 3,463
Remember that even when we're talking about confidence x coverage, we're still talking about the probability of a binary outcome for each "patch" within the area - i.e. over what proportion of the area is it expected to rain at all in the given time period (see below).

Given that, how can the numbers above from possibly make sense? How can ther be three individual hours when confidence x coverage is greater than the figure for the entire day?

Mathematically, PoP is defined as follows:

PoP = C x A where "C" = the confidence that precipitation will occur somewhere in the forecast area, and where "A" = the percent of the area that will receive measureable precipitation, if it occurs at all.
So... in the case of the forecast above, if the forecaster knows precipitation is sure to occur ( confidence is 100% ), he/she is expressing how much of the area will receive measurable rain. ( PoP = "C" x "A" or "1" times ".4" which equals .4 or 40%.)

But, most of the time, the forecaster is expressing a combination of degree of confidence and areal coverage. If the forecaster is only 50% sure that precipitation will occur, and expects that, if it does occur, it will produce measurable rain over about 80 percent of the area, the PoP (chance of rain) is 40%. ( PoP = .5 x .8 which equals .4 or 40%. )

Last edited by Riemann; 04-23-2017 at 08:44 PM.