By Kevin L'Heureux
Polygon has released news that Zen studio's will be releasing a new stand alone version of the Star Wars pinball table.
Zen Pinball developer Zen Studios will launch Star Wars Pinball as a standalone pinball title May 14 on PlayStation 3 and PlayStation Vita via PlayStation Network, the studio announced today.
The game will launch with the three tables that were released earlier this year in the Star Wars Pinball pack for Zen Pinball 2 and the other Zen Pinball titles — Star Wars Episode V: The Empire Strikes Back, Boba Fett and Star Wars: The Clone Wars. PSN users who already bought that $9.99 table pack will be able to download Star Wars Pinball at no extra cost.
Zen Studios is planning to release seven additional Star Wars tables in the future, said community manager Bobby Loertscher in a comment on the PlayStation Blog. All Star Wars tables can be imported from Zen Pinball 2 into Star Wars Pinball and vice versa, and because both titles support Sony's PS3/PS Vita Cross-Buy system, the tables can be played within both games on both platforms.
Read here via Polygon
Zen Pinball developer Zen Studios will launch Star Wars Pinball as a standalone pinball title May 14 on PlayStation 3 and PlayStation Vita via PlayStation Network, the studio announced today.
The game will launch with the three tables that were released earlier this year in the Star Wars Pinball pack for Zen Pinball 2 and the other Zen Pinball titles — Star Wars Episode V: The Empire Strikes Back, Boba Fett and Star Wars: The Clone Wars. PSN users who already bought that $9.99 table pack will be able to download Star Wars Pinball at no extra cost.
Zen Studios is planning to release seven additional Star Wars tables in the future, said community manager Bobby Loertscher in a comment on the PlayStation Blog. All Star Wars tables can be imported from Zen Pinball 2 into Star Wars Pinball and vice versa, and because both titles support Sony's PS3/PS Vita Cross-Buy system, the tables can be played within both games on both platforms.
Read here via Polygon
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