Paineel Faction, Evil Races, Paineel Key and keyz, Heretic Faction, Guard MR

[Vineyea]

Ethnically I'm a lizard. Sssipp sipp sippp.

Regarding evil races and the paineel faction grind: It is necessary to raise faction above doubious in the warrens or in the (under utilized) stronebrunt mountains.

When my level 20 [iksar] necromancer achieved apprehensive faction I regarded the faction grind dubiously. It was then, therefore, that I sought to charm the undead guards in the paineel newbie area such that I might get the -amiable- faction that is required to request a key. For 4 or 5 or as many mana bars, the dominate undead spell (effective to level 32) did not charm the green con Guard Heridion.

I employed the assistance of the Dawn Believers: Darkin, a 58 enchanter attempted to tash the undead, after learning that it truly is undead, since expulse undead works... We found that lowering the magic resitance of the guard does work, we had a successful charm ratio of about 1/9. The key, thus, was obtained.

When a magician was invited to malo (lvl 55), we found the ratio to be something like ½ to ¼ over three mana bar trials or something like that. The lesson here is that once sufficient faction is obtained, particularly powerful magic resist debuffs are required for charming.

[Swish2]

Newbie area guards are something else. Harder to do things to.

The alternative here is to get the enchanter to go down the lift, charm one of the guards at the base of it (the lowest of all inside Paineel) and pull it up to you to take the charm. You'll find they're much easier to work with.

You don't even need the enchanter, just someone who has a key already that can tag one of them.

[Vineyea]

Guards will not issue keys during combat. It is the case in paineel, however, that undead guards do not issue faction hits. Once the lower lift guard is charmed and brought to the surface it may be "charm traded" (?) before it paths back.

At level 23 dominate undead was sufficient to charm and obtain a new paineel key from the guard (in one try). The method is simpler and I can confirm it.