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-M1 M0 max M0 M1 min M1min > 3/4 M1 ⇔ M1min > 75% of M1 Let ’ s use 70% of M1 for a 30% tolerance M0max < 3/2 M0 ⇔ M0max < 150% of M0 Let ’ s use 130% of M0 for a 30% tolerance M1max and M0min are not constrained by other values, (as well as LMmax and RMmin) let ’ s set all of them with a tolerance of 30%. Bit Mark 0:M0max = 1.3 M0 M0min = 0.7 M0 (M0 = 1 125 µs) Repetition Mark:RMmin = 0.7 RM RMmax = RM + Rt (RM = 2812.5 µs) (Rt = 803.6 µs) Gap:GapMin = 0.7 Gap1 GapMax = 1.3 Gap2 (Gap1 = 48937.5 µs) (Gap2 = 105187.5 µs) Note:The OnRise mode was too ristrictive with the leading/repeating marks ’ tolerances. With Rt at 1 1% of RM, Lt could not be higher than 8.33% of LM. Now both Rt and L T are at 28.6% Leading Mark:LMmax = 1.3 LM LMmin = RMmax + 1 (LM = 5062.5 µs) If both marks have the same tolerance percentage,then: Lt / LM = Rt / RM ⇔ Lt × RM = Rt × LM and because RM+Rt+Lt=LM ⇔ Rt = LM-RM-Lt Substituting into the previous equation we get: Lt × RM = (LM-RM-Lt) × LM and after solving for Lt: Lt=(LM² - LM × RM) / (RM+LM) Knowing that: Rt+Lt=2.25 LM=5.0625 and RM=2.8125 we get: Lt = 1.4464 and Rt = 0.8036 (Lt is 28.6% of LM and Rt is 28.6% of RM) Bit Mark 1:M1min = 0.7 M1 M1max = 1.3 M1 (M1 = 2250 µs) 0 LSB MSB MSB MSB Address ~Address LSB Command ~Command LSB MSB LSB 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 9ms x16 x24 x8 562.5µs 4.5ms Leading Mark (13.5ms) x (8x2 + 8x4) Address frame (27ms) x (8x2 + 8x4) Command frame (27ms) x 120 Signal frame (67.5ms) End(0.56ms) 67.5msSignal frame 108msRepetition period 108msRepetition period 40.5msGap1 108 - 1 1.25 = 96.75ms Gap2 1 1.25ms Repetition mark Repetitionmark x2 1.125ms x4 2.250ms 9ms Bit Marks’ tolerances NEC Protocol T olerances summary Repetition Mark lenght Leading Mark & Repetition Mark’ s tolerances x16 x5 562.5µs Repe tition Mark( 2.8125 m s) Start Pulse before RM Start Pulse before LM LM RM max Rt Lt LM mim RM 9ms x16 x9 562.5µs Leading Mark (5.0625ms) x (8x2 + 8x4) Address frame (27ms) x (8x2 + 8x4) Command frame (27ms) x 105 Signal frame (59.0625ms) 59.0625msSignal frame 108msRepetition period 108msRepetition period 48.9375msGap1 108 - 2.8125 = 105.1875msGap2 Repetition mark 1.125ms 2.250ms OnFall Mode 2.25ms M1 M0 max M0t M1t M0 M1 min Bit Mark 0:M0max = 1.3 M0 M0min = 0.7 M0 (M0 = 1 125 µs) Repetition Mark:RMmin = 0.7 RM RMmax = RM + Rt (RM = 2812.5 µs) (Rt = 803.6 µs) Gap:GapMin = 0.7 Gap1 GapMax = 1.3 Gap2 (Gap1 = 48937.5 µs) (Gap2 = 105187.5 µs) Note:The OnRise mode was too ristrictive with the leading/repeating marks ’ tolerances. With Rt at 1 1% of RM, Lt could not be higher than 8.33% of LM. Now both Rt and Lt are at 28.6% Leading Mark:LMmax = 1.3 LM LMmin = RMmax + 1 (LM = 5062.5 µs) If both marks have the same tolerance percentage,then: Lt / LM = Rt / RM ⇔ Lt × RM = Rt × LM and because RM+Rt+Lt = LM ⇔ Rt = LM-RM-Lt By substituting into the previous equation we get: Lt × RM = (LM - RM - Lt) × LM and after solving for Lt: Lt = (LM² - LM × RM) / (RM+LM) And knowing that: Rt + Lt = 2.25 LM = 5.0625 and RM = 2.8125 we get: Lt = 1.4464 and Rt = 0.8036 (Lt is 28.6% of LM and Rt is 28.6% of RM) As in the leading marks, if both bit marks have the same tolerance percentage, then: M1t = (M1² - M1 × M0) / (M0 + M1) And knowing that: M0t + M1t = 1.125 M1 = 2.250 and M0 = 1.125 we get: M1t = 0.75 and M0t = 0.375 (M1t is 33.33% of M1 and M0t is 33.33% of M0) The maximum tolerance for the bit marks is then 33.33%, but let's only use 30%. M1max and M0min are not constrained by other values, (as well as LMmax and RMmin) so let ’ s set all of them with a tolerance of 30%. Bit Mark 1:M1min = 0.7 M1 M1max = 1.3 M1 (M1 = 2250 µs) 0 LSB MSB MSB MSB Address ~Address LSB Command ~Command LSB MSB LSB 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 9ms x16 x24 x8 562.5µs 4.5ms Leading Mark (13.5ms) x (8x2 + 8x4) Address frame (27ms) x (8x2 + 8x4) Command frame (27ms) x 120 Signal frame (67.5ms) End(0.56ms) 67.5msSignal frame 108msRepetition period 108msRepetition period 40.5msGap1 108 - 1 1.25 = 96.75ms Gap2 1 1.25ms Repetition mark Repetitionmark x2 1.125ms x4 2.250ms 9ms Bit Marks’ tolerances NEC Protocol T olerances summary Repetition Mark lenght Leading Mark & Repetition Mark’ s tolerances x16 x5 562.5µs Repe tition Mark( 2.8125 m s) Start Pulse before RM Start Pulse before LM LM RM max Rt Lt LM mim RM 9ms x16 x9 562.5µs Leading Mark (5.0625ms) x (8x2 + 8x4) Address frame (27ms) x (8x2 + 8x4) Command frame (27ms) x 105 Signal frame (59.0625ms) 59.0625msSignal frame 108msRepetition period 108msRepetition period 48.9375msGap1 108 - 2.8125 = 105.1875msGap2 Repetition mark 1.125ms 2.250ms OnFall Mode 2.25ms 1.125ms