...understanding the transfill math for dissimilar tanks...
Sounds mysterious. Or impossibly complicated. Really isn't. Suppose you will dive your old-school LP 67, but you have two HP 120's that you will bring along with you additionally. Then think this way:
Caveat: I know nothing about old-school LP 67's. For this example let's assume that a LP 67 will hold 67 cu ft of gas at a service pressure of 1,980 psig (= 1,800 psig + 10%).
1. Step #1. The internal volume of an HP 120 is calculated by solving 14.7 * 120 = 3,500 * x for x. (Boyles Law.) This yields x = 0.5040 cu ft.
2. Step #2. Similarly, the internal volume of this old-school LP 67 is calculated by solving 14.7 * 67 = 1,980 * x. (Boyles Law.) This yields x = 0.4974 cu ft.
3. Step #3. By "equalizing" an "empty" LP 67 and a full HP 120, you will be putting 120.4974 cu ft (= 0.4974 + 120) of gas into a 1.0014 (= 0.4974 +0.5040) cu ft fixed volume. To find the resulting pressure, solve
14.7 * 120.4974 = x * 1.0014 for x. (Boyles Law.) This yields x = 1,769 psig.
That is, after "equalizing", the pressure in both the old-school 67 and the
first HP 120 will be 1,769 psig.
4. Step 4. So, use the
second HP 120 to top up the old-school 67 to it's final 1,980 psig service pressure. Then go do your second dive!
When you finish your dive, can you again fill your old-school LP 67, again treating your two HP 120's as if they were mini-cascade tanks, for another--the third--dive? How many times can you do this? The answers to these questions are provided by computations as straightforward as those shown above.
(Never underestimate the power of simple middle school algebra for these types of scuba problems. You never have to resort to "tank factors" and other concepts of dubious "added value." Fundamentals.)
Safe Diving,
rx7diver