As I am really interested who will win the battle for Taj, I tried to calculate when each team will finish it. For Spulla it seems to be turn T189 (T184 the begin to research Nat, finished EOT187, able to build on T188 and finished EOT188 ).
Nakor: They had on T183 according to Sulla 204 hammers in Taj. On T184 I am missing how many hammers they put in it, but I guess it is safe to assume they have put their base without chops into it, so 45 hammers. As they show 9 turns to go and have a production of 117 + 8x45 hammers that equals 477 hammers. 204 + 45 + 477 = 726. So the assumption of no chop on T184 seems correct (as then there would be one turn less needed to get it finished) and they will have 366/700 of Taj on T185 with two more chops to come. Well at least I guess that as Nakor wrote they will get another one in next turn and then another one some turns after. But I am unsure as he might have meant with the chop coming in next turn the one we can already see in the picture.
Assuming for now that there are two more to come: I have no clue how many hammers a chop will give but the furs probably 30 and the one third ring 16? Together with base production and multipliers this should give 135 and 93. So:
EOT always:
T184: 366/700
T185: 501/700
T186: 546/700
T187: 639/700
T188: whipped and finished
T189: coin flip with Spulla
So Nakor either has to get the second chop done 1 turn after the first (having by EOT186 594 hammers into the wonder which should be whippable by T187 with 8 (?) pop, landing the wonder T188 ) or will get into a coin flip with Spulla.
Only question now: Is my math right? Did I mess up with how many hammers chops give?
Nakor: They had on T183 according to Sulla 204 hammers in Taj. On T184 I am missing how many hammers they put in it, but I guess it is safe to assume they have put their base without chops into it, so 45 hammers. As they show 9 turns to go and have a production of 117 + 8x45 hammers that equals 477 hammers. 204 + 45 + 477 = 726. So the assumption of no chop on T184 seems correct (as then there would be one turn less needed to get it finished) and they will have 366/700 of Taj on T185 with two more chops to come. Well at least I guess that as Nakor wrote they will get another one in next turn and then another one some turns after. But I am unsure as he might have meant with the chop coming in next turn the one we can already see in the picture.
Assuming for now that there are two more to come: I have no clue how many hammers a chop will give but the furs probably 30 and the one third ring 16? Together with base production and multipliers this should give 135 and 93. So:
EOT always:
T184: 366/700
T185: 501/700
T186: 546/700
T187: 639/700
T188: whipped and finished
T189: coin flip with Spulla
So Nakor either has to get the second chop done 1 turn after the first (having by EOT186 594 hammers into the wonder which should be whippable by T187 with 8 (?) pop, landing the wonder T188 ) or will get into a coin flip with Spulla.
Only question now: Is my math right? Did I mess up with how many hammers chops give?