I gotta disagree Marc.. In rock climbing one end is tied to you, one end to your belayer... un hook your belayer, attach the rope to you and take the same fall, its still the same factor yes?
Not in my view. First, you have a fall factor 1 (max), second, you get a fall factor 2.
In this example, it's like you pass from my case B (srt with ground anchor) to my case C (ddrt). In case B, the high tie-in point acts as a redirect, all the rope work in the same way, move around the tie-in point and stretches with one full load on its full length. It's the principle in its book.
It's different in ddrt, the high tie-in point acts as a fixed anchor, both half rope pulls on each side in an opposite way (rope's axis wise) with no move in the middle, like two ropes share the same load. So half a load for each half rope. But because there is no move in the middle, it's the same as if you have half the length of an other rope with twice the strength.
The rule to define the fall factor is too simple to be always right (kindly said for "incorrectly stated") and doesn't apply
as is for the ddrt.
For the math with ddrt, you have to take a virtual rope, which has half length of the real one ( this gives the right fall factor) and twice the strength (this gives the stretch capability).
In Scotts example, falling from 3' above the TIP you take a 6' fall on 6' of rope. Factor 1.
Factor 1 only in srt with ground anchor if you are 3 ' above ground.
If you are 30' above the ground, srt with ground anchor , the rope length is 33', for a 6' fall, so ff=0.18, the kindest
If you are srt with anchor at the tie-in point, the rope length is 3', fall is 6 ', so ff = 2
If you are ddrt 3' above the TIP, the virtual (stronger) rope is 3' too, fall is 6 ', so ff = 2 (and a stronger chock loading).
A drawing should be easier to follow than my text.