0.5 gradient is an arbitrary aim that I just thought would be easy for people to wrap their head around.
Let me see if I follow the math:
Let EV = number of stops of overexposure.
2^EV = how much more light the film receives.
γ = target slope in the log-log plot; should be 0.65 but you're simplifying it to 0.5 to make it easier to test with an ND2 filter.
(2^EV)^(-γ) = change in transmitted light intensity through the negative.
Your test sets EV = +2 and aims to check γ = 0.5:
(2^2)^(-0.5) = 1/sqrt(2^2) = 1/2
So the overexposed film should allow 1/2 as much light through, which we can verify with an ND2 filter.
So far so good?
Assuming that I got this right, we could tweak EV to get closer to the ideal γ = 0.65.
(2^(1 + 1/3))^(-0.65) = 0.548412489847313
(2^(1 + 2/3))^(-0.65) = 0.47193715634084676
Therefore, the idea overexposure is somewhere between +1 1/3 and +1 2/3. Assuming that the camera does not permit you to adjust the exposure in increments smaller than 1/3 stops, we could consider a more complex test:
1) Take one exposure at EV = +1 1/3
2) Take one exposure at EV = 0
3) Take one exposure at EV = +1 2/3
Develop the film, then put an ND2 filter on top of the middle negative. Its density should be half-way between the density of the two neighboring negatives.
Did I get this right? It's a bit more complex than your version, but if I got it right, it would be quite precise at getting γ close to 0.65.
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EDIT: Reading
@Bill Burk 's last comment, it looks like the target is γ = 0.62, not 0.65.
(2^(1 + 2/3))^(-0.62) = 0.488579984217123
So... just use the original method but with EV = +1 2/3 instead of EV = +2 ?