7
\$\begingroup\$

Let's say a very old (true) dragon gets turned into a vampiric dragon. Would they naturally advance to ancient after the requisite number of years, or would they remain very old indefinitely?

If they cannot progress naturally, is there a way that exists for them to intentionally bring themselves to the next age category?

\$\endgroup\$
3
\$\begingroup\$

Bad news Jim, it's not a dragon anymore...

Since to create a vampire dragon we need to apply the acquired template Vampiric Dragon from Draconomicon, here is what happens:

Once you take a Vampiric Dragon Template, your monster loses the type Dragon and gains the type Undead, it will also lose all subtypes as per Vampiric Dragon's description.

Therefore, after applying the template, the monster is not a Dragon anymore, so it can't gain abilities through aging because it now progresses as an Undead following the descriptions of Vampiric Dragon Template (Pg 195-197 of Draconomicon).

\$\endgroup\$
2
\$\begingroup\$

The main problem I can see here is that according to the rules you can't create vampiric dragon because you can create vampires only from humanoid and monstrous humanoid.
But if you really want to do that. I don't think that they should get next age level because they are undead so what is age for them?

Draconomicon update:
FWIK dragon gains its age levels because its type is Dragon. In the Draconomicon there is this note:

Size and Type: The creature’s type changes to undead, and it loses any subtypes it had in life.

So I am still convinced that the dragon is not able to age anymore.

\$\endgroup\$
  • 1
    \$\begingroup\$ There is a specific template in the Draconomicon for Vampiric Dragons. \$\endgroup\$ – StephenTG Jun 26 '17 at 13:55
  • \$\begingroup\$ @StephenTG I made an edit to my answer. \$\endgroup\$ – Artholl Jun 26 '17 at 15:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.