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Is the water in Spyro the Dragon (Reignited Trilogy) meant to be this deadly, and how do I just take damage?

Arqade Asked by doppelgreener on December 30, 2020

I’m playing Spyro the Dragon, the very first game in the Reignited Trilogy. I haven’t played the original and I’m running into something that’s a bit confusing.

When Spyro touches deep water in this game (or ice water, or poison water), that’s an instant kill. He sinks and drowns, loses a life, and respawns. A couple of times while playing I’ve seen him bounce off the top and just take damage, but this doesn’t happen often and I’m confused as to how/when it will happen. (And I don’t really want to spend a ton of lives on experimentation.)

Part of me is wondering if this is a bug in the remaster, like when Crash Bandicoot 1 had his hitbox changed in the N. Sane Trilogy remaster making the platforming an awful lot harder. Then again, it’s a remaster of a game from 1998, and maybe that’s how the original actually worked.

So, is the water meant to be this deadly? What’s the pattern to actually surviving the water and just bouncing off the surface?

One Answer

Deep water deals damage on contact, and will instantly kill Spyro if he doesn't jump out in time or if he had no Sparx when he touched the water. Jumping out of the water is as simple as pressing the jump button.

I believe some of the water hazards where it's impossible to make your way back to land disable your ability to jump out. In those cases the water is an instant kill, same as a bottomless pit.

Correct answer by Wrigglenite on December 30, 2020

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