View attachment 200426
Bar 1 : sent is long, as I put the carry over aside for now, I ignore what is on the left. P1 is assigned at bottom so long BM because of the sent.
Bar 2 : SYM, DEC volume, a first tape is built and it's a long one as that's what makes most sense considering both bar 1 and 2.
Bar 3 : LAT 3 and XR, close breaks out prior rtl on DEC volume so BO,T1 = FS EE = OOE resets on current bar = P1 assigned and BM short is placed at top. A new trend is beginning.
Bar 4 : XR, degap required, BO of LAT, rtl accelerates, volume is DEC so OOE advances : T1
Bar 5 : XB, neither close violates prior rtl nor prior BM so OOE advances, volume is INC : P2 : we have A-Band Pass.
Bar 6 : XR, rtl accelerates, OOE advances to repeat of P2 BUT it's a "just-after-P2 P2" being above P1 so EE Ag VEBO is located : P1 is assigned and new opposite BM is placed : a long one at bottom. Prior short trl is no longer considered cause it's part of the past that ends at the new long BM just placed.
Bar 7 : XR, close does not violate prior BM, just penetrates. A repeat long BM is placed at bottom as virtual pt1 of new long trend segment beginning underway. OOE advances , vol DEC : T1.
Bar 8 : XB, long rtl can now be drawn, OOE advances, volume is DEC : repeat of T1.
Bar 9 : XB, rtl accelerates, OOE advances : repeat of T1 AND there is acceleration so "NOT PP6".
Bar 10 : XB, rtl accelerates, degap required, OOE advances, volume is DEC so repeat of T1 with acceleration so again "not PP6".
Bar 11 : XB, rtl fans, volume is INC, OOE advances : P2 so A-band pass AND it is both after two consecutives T1's AND between them in value : PP2 EE -> P1 is assigned on next bar with opposite short BM placed at top.
Bar 12 : SYM, degap required, P1 is assigned due to prior EE, sent is congruent (short).
Bar 13 : XR, tape is built and a first short rtl appears, volume is INC, OOE advances : repeat of P1.
Bar 14 : XR, degap is required, volume is INC, OOE advances : repat of P1 AND that's the third P1 in a row AND there is acceleration -> PP1 EE so P1 is assigned on next bar with opposite long BM at bottom placed. OOE resets.
Bar 15 : XR, P1 is assigned due to prior EE, BM long.
Bar 16 : XR, close violates prior established BM -> BMrev = FS EE = P1 assigned on current bar with associated short BM at top. OOE resets. It has been impossible to draw a long rtl.
Bar 17 : XR, degap required, short rtl appears, volume is INC, OOE advances : repeat P1.
Bar 18 : SYM on DEC volume BUT the close breaks out prior rtl so if this takes precedence then -> BO,T1 = FS EE = P1 assigned on current bar with long BM. OOE resets
Bar 19 : XB, long rtl is created with the tape, volume is DEC, OOE advances = T1.
Bar 20 : XR, degap required, neither BO,T1 nor BM, AND this reminds me bar 6 to bar 7 move, so I'd say repeat BM long at bottom AND volume is DEC so OOE advances : repeat T1.
Bar 21 : XR, close is both breaking out prior rtl AND violating prior BM -> BO,T1 and BMrev EEs. Two reasons to assign on current bar P1 and short BM at top. OOE resets.
Bar 22 : XR, degap required, volume is INC, OOE advances : repeat P1.
Bar 23 : XR, degap required, close is out of prior rtl on DEC volume : BO,T1 and BM long is placed at bottom. OO resets.
Bar 24 : OB, a long tape appears so we have a long rtl congruent with the assigned long BM, VTP would say P1/T1 so PP4 EE so P1 assigned on next bar woth short BM at top BUT the close both crosses rtl and prior BM so BO,T1 and BMrev so P1 is assigned on current bar.
What I wonder here is : what tales precedence ? I'd say BO,T1 and BMrev do. So I'd say P1 is assigned on curreent bar only AND P1 is not assigned on next bar.
SO -> due to the nature of the OB, two opposite BM are placed at top and bottom and we have a short rtl built.
OOE resets.
Bar 25 : XR, rtl accelerates, prior long BM is violated so BMrev = FS EE = P1 assigned on current bar, BM short is placed, OOE resets.
Bar 26 : XR, rtl fans, volume is DEC, OOE advances : T1.
Bar 27 : XR, rtl fans, volume is DEC, OOE advances : repeat T1.