HMR said:

It has nothing to do with TAS. The turn needs to be predicated off GROUNDSPEED.
No, it's TAS.
The airplane drifts with the wind during the turn. The groundspeed is not constant because the heading is constantly changing. Say for example that an airplane flying 120 KTAS has a direct tailwind of ten knots. As the standard rate turn begins to intercept the arc, the airplane drifts by
windspeed/120, or about 0.083nm per minute. Round that up to 0.1nm and that's about how far you'll drift during the intercept - tack it on if you're so inclined (the rule of thumb doesn't include this because it's not really worth the trouble.)
In reality, this doesn't matter much. Use groundspeed or TAS and you'll be "about" right. But the correct formula does call for TAS, not GS.
A proper formula would be GS/100 (not 200).
I disagree. The formula I've quoted works in all aircraft I've ever flown, and I've seen it in print (Mental Math for Pilots, IFR Magazine) more than once. Additionally, the concept appears in other forms. Transport Canada's Flight Training Manual advocates a DME arc intercept technique using 0.5% of GS (or TAS) for a standard rate (they call it 'Rate One') turn and 1.0% for a 1/2 standard rate turn. Since the Chickenhawk isn't likely to be going fast enough to ever require a half-standard rate turn for anything, 0.5% of 120 knots = 0.6nm (exactly the same result as the formula I published above.) An airplane with a groundspeed of 200 knots would require 1nm (0.5% of 200 = 1nm) for the lead-in, etc.
Another common rule of thumb is to use a 0.5nm lead-in for groundspeeds below 150 knots. It's not quite as precise - you'll overshoot a bit at the higher speeds - but it essentially works, and is pretty much in line with the formula I provided. As indicated above, add .1 or .2 to account for the roll-in. Add more if you're above 200 knots, but most of us aren't intercepting DME arcs much above that speed, whether we're flying jets or pistons.
In summary, your suggestion of TAS (or GS) / 100 would be accurate for half-rate turns, not standard rate. The piston driver would grossly undershoot, every time.
Don't believe me? Do the math: How long does it take to do a standard rate, 90 degree turn (including lag time to initiate the turn)? How far does an aircraft with a GS of 120kts travel in that time?
The math isn't as straightforward as you seem to think, and doesn't support your inferred conclusion. Take two airplanes flying 120 knots straight ahead; one begins a standard rate, 90 degree turn while the other continues. Start a stopwatch. The speed of the turning airplane relative to the level-wing airplane drops from identical (120 knots) at the beginning of the turn to zero at the thirty second mark. Had he continued straight ahead for one minute, he'd travel about a mile. But since he made the standard rate turn, he only covers about half that relative distance - 0.5nm. I.e., median average of 120 knots at the 0 second mark and 0 knots at the 30 second mark = 60 knots, which would cover 0.5nm.
FWIW- A 90 degree intercept seems to be the most common entry for DME arcs.
Right. But I'd suggest 80 degrees when intercepting from the outside, which is how most of us use DME arcs, most of the time.
Best regards,
-Ryan