If what you suggest is true, the added cell area would cause a hit of about 17% using the same footprint. The added height would give a credit of about 9%. So if the 50% physically larger 2170 cell yielded only 30% more energy, you would lose 8% of the 30% increased energy, and still get a net credit of 22% more power per same area footprint.
But that's only if you accept what you say, that '30% more energy per cell.' refers to any size of cell. It makes much more sense to credit that 30% increase only to cells of the same size as the 18650, and correspondingly more energy to bigger ones. Don't you agree? That is what energy density means.
The 2170 is a bigger cell, so logically, increased energy density of 30% implies that much greater energy per volume. This means that if the cell is 50% larger, then the greater energy per (2170) cell is not 30%, but 45%. Additionally, considering the original pack footprint, this would be compounded by the added height of 9%, bringing the total added energy per pack volume to well over 50%.
If I'm missing something, please give a reference to the 2170 cell as actually losing energy density per volume - i.e. only 30% more energy for 50% greater volume. Straubel talks greater density, so less density is doubtful.
