Conversely, the Iowa's 25 mm lower hull plate is overmatched by nearly every battleship in its matchmaking spread, so shells that penetrate that plate won't have its trajectory altered. What does this mean? The 38 mm lower hull plate (which is historically accurate) on the Missouri means that no gun in the game can overmatch that plate, so every shell that penetrates it will have its trajectory altered by the normalization angle. The following diagram visually illustrates what I mean. For instance, the 406 mm shells have a normalization angle of 6 degrees, meaning that if the shell strikes a plate with 20 degrees obliquity, in other words, 20 degrees from the normal (or perpendicular) vector, then ( if the shell doesn't overmatch the plate) upon impact the trajectory of the shell will be changed such that it exits with 14 degrees obliquity.

To my understanding, WOWS penetration mechanics is such that any shell that doesn't penetrate via overmatch has its exit obliquity reduced by a small angle I believe this is what the game refers to as 'normalization' and the specific angle is generally by shell caliber. I strongly suspect that the 38 mm hull plate on the Missouri (versus 25 mm on Iowa) is responsible for this, due to shell normalization. Let me try to illustrate this below. Namely, the Iowa feels moderately easier to citadel than the Missouri, enough to be noticeable. Over all the experience I've had shooting Iowas and Missouris, there feels like a difference in terms consistency and reliability in hitting their citadels, even though their citadel heights are identical, i.e.

If it's demonstrated that this isn't the case in WOWS, then disregard what I've written here. Foreword: I am operating under the assumption that shell normalization affects the exit trajectory after a non-overmatching penetration.

World of warships missouri citadel