Official Q&A Thread: 7.5 Era

ForcestormX, on Sep 03 2012 - 17:10, said: I think, expanding on this, that the calculation for effective armor for situations where the opponent has angled his plate in front of you in addition to the natural angle of the plate, would be
T = X / (cos(Y-Z)*cos(B)); where B is the angle between the centerline along the tank facing you and the line extended from your gun. In other words, the amount the other tanker has angled his tank.
So a Type 59 who has angled his hull 10 degrees would have T = 100 / (cos(60-7)*cos(10)) = 100 / (0.6018 * 0.9848) = 100 / (0.59266) = 168.73 mm
This is assuming that normalization is in only one axis. If it is in fact in two axis, then the thickness would be T = X / (cos(Y-Z)*cos(B-Z)), making for that same 10 degrees
T = 100 / (cos(60-7) * cos (10-7)) = 100 / (0.6018 * 0.9986) = 100 / 0.6010 = 166.40 mm
Am I on the right track Vallter?