First, you should eliminate interbar gaps in order to see the cases correctly.Please see the image attached with the examples.
Note: i don`t mark translations - only the internals on my chart.View attachment 172910
Looking at the middle circle in your example, the Internal starts at 10:10 (Bar 1) and lasts for 7 inside bars, first testing the short lateral boundary before BO of the long lateral boundary on the 7th bar. The lateral ends with two closes outside the long boundary.
Bar1 - with a translation short (XR), There is one leg to the 10:10 bar.
Bar2 - SYM 2 FBP, 2 leg
Bar3 - XR, 2 leg
Bar4 - OB, STD (outside bar -stitch down), 3 leg
Bar5 - XB 2 FTP, 1 leg
Bar6 - OB, XR (outside bar - translation short, stitch down), 1 leg
Bar7 - OB, XB (outside bar - translation long, stitch up), 1 leg
Bar8 - XB, BO of lateral, 3 leg
From what I gather from your post, All the bars within a circle would add up to 1 bar. I don't understand how that connects with Jack describing legs of a bar, the dominant sentiment always being the second leg and how synthesizing all the bars of an internal can come up with one bar. Would you clarify?