Most of the current RT workloads are developed with a focus on what Nvidia hardware can and cannot do. So I think it's not (yet) the right time for such an absolute statement as your "uring is definitely more capable RT wise than RDNA2. Period."
For example Doom Eternal Benchmarks say otherwise. Don't be blinded by the better Ampere, RDNA2 can be quite competitive with Turing, even on a 550-vs-1100-comparision of RX 6800 vs. 2080 Ti.
(anecdotal) proof:
View attachment 5643
source:
https://www.pcgameshardware.de/Doom...ng-RTX-Update-DLSS-Benchmarks-Review-1374898/
Since RX 6800 is quite faster than 2080 Ti in non-RT workloads, it's probably not showing a complete picture of their relative RT performance by simply comparing total rendering performance.
A better comparison (RT-wise) would be "how much more time each card spent on RT each frame." Unfortunately, there's no pure "RT off" benchmark in the review above, so it's difficult to make such comparison.
However, we know from other reviews that RX 6800 is slightly faster (< 10%) than 2080 TI with RT off at 4k in Doom Eternal, and quite a bit faster (~20%) at 1440p. So if RX 6800 has a similar performance to 2080 TI with RT on, it's probably due to its RT units not as fast in this game.
Or, using another example, it's quite likely that 3070's RT units are better than 2080 TI's RT units. However, in the benchmark above, 3070 (with a lower texture settings, likely due to memory size constraint) actually performs similar to a 2080 Ti. If we follow the same logic, we'd conclude that 3070 has a similarly performing RT units as 2080 TI, but that's likely to be wrong. However, if we consider that 3070 is actually slower (~ 6% at 1440p and ~10% at 4K, with comparable texture settings) than 2080 Ti with RT off in Doom Eternal, it makes more sense now.
[EDIT] There are also the "overlapping" factor to be considered, as some of the RT computations can be done concurrently with traditional rendering works, but we have even less information on this and it probably reasonable to assume the overlapping portion is more or less proportional to the non-overlapping portion.