eldin raigmore wrote: ↑21 Jun 2021 19:02
Salmoneus wrote: ↑21 Jun 2021 13:28
…., as you've written it it's actually a quatrain with an AABB rhyme scheme. In addition, it has internal rhyme in an ABBA pattern. That seems to describe it, to me.
Thanks! I think I was struggling to say exactly that. I thought of “caesura” and of “internal rhyme” but I wasn’t sure I was using either term correctly.
In this case it's both.
Internal rhyme is rhyme that isn't end-rhyme. Of course, the definition of a line, and hence of end-rhyme, is subjective. One reason for the distinction is that internal rhyme is often less regular than end rhyme.
But if the internal rhyme is consistent, then I guess we just have to bow to authorial intent as suggested through visual line breaks. Although in your case, given that ABBA and ABAB quatrains are both extremely common, and an ABCBCDAD scheme is much less common, my instinct would be to automatically see your stanza as a quatrain with internal rhyme. In addition, trimeter is much more common in English verse than hexameter, which reinforce the quatrain interpretation. But of course, like two rival phonemic analyses, there's ultimately no matter of fact about it.
Meanwhile, a caesura is a rhythmic pause in the middle of a line, forced by syntax. You happen to have caesuras with internal rhymes attached.
Also relevant: if you systematically divide lines of a poem in half, each half can be called a 'hemistich'. Hemistiches can have their own rhymes, or often end with caesuras, or may just in some other way be poetically distinct - Anglo-Saxon poetry had alliteration in each hemistich that didn't have to match between hemistiches.
Salmoneus wrote: ↑21 Jun 2021 13:28
There are 4,140 possible rhyme schemes for an eight-line stanza, 715 of which involve every line rhyming with at least one other line. It should be no surprise that the vast, vast majority of them have not been given specific names.
How do you calculate those numbers?
I let wikipedia tell me. The mathematical term is 'Bell number' - the number of ways a set of a given size can be subjected to partition.
But you can work them out yourself - it's simple, albeit time-consuming.
Consider the quintain. Let's say: for each line, a 'set' is a combination of rhymes up to and including that line. Then divide sets by their highest letter: an A set has no letter higher than A in it, a B set has no letter higher than B, and so on. No new line can contain a letter more than 1 higher than the highest letter of the existing set (a set that only contains A and B rhymes cannot suddenly have a D rhyme in the next line, because that would be called C instead). We'll go through line by line. Let's say that "set(2)" means "set up to and including the second line) and so on.
- the first line must be A
there is 1 A-set(1)
- the second line must be A or B
for each A-set(1), there is one A-set(2) and one B-set(2)
- after a A-set(2), the third line must be A or B. But after a B-set(2), the third line may be A, B, or C.
for each A-set(2), there is one A-set(3) (it must end A), and one B-set(3).
for each B-set(2), there are two B-sets(3) (it can end A or B), and one C-set(3)
so in total there is 1 A-set(3), 3 B-sets(3), and 1 C-set(3)
- repeat for the fourth line
for each A-set(3), there is one A-set(4) , and one B-set(4) 
for each B-set(3), there are two B-sets(4) , and one C-set(4) 
for each C-set(3), there are three C-sets(4) , and one D-set(4) 
- that's 1 A-set(4), 7 B-sets(4), 6 C-sets(4), and 1 A-set(4). So for the fifth line:
for each A-set(4), there is one A-set(5) , and one B-set(5) 
for each B-set(4), there are two B-sets(5) , and one C-set(5) 
for each C-set(4), there are three C-sets(5) , and one D-set(5) 
for each D-set(4), there are four D-sets(5) , and one E-set 
- so there are 52 possible rhyme schemes in a quatrain.
I'm sure you can see the pattern. Alternatively you can write out the Bell number dumbly though an algorithmic calculation diagram called a Bell triangle.