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Does Whirlwind Attack allow you to attack an invisible creature you didn't know was there?

Take a medium fighter (F) under the effects of enlarge person, so they are now large and have a natural reach of 10. He is fighting goblins (G) and somehow two invisible ones (I) have joined the fight behind him without him knowing about them.

G · G · G ·
· · · · · ·
· · F F · ·
· I F F · ·
· · · I · ·
· · · · · ·

So when the fighter goes to Whirlwind Attack, he first goes to attack the 3 goblins he knows about, but would he be allowed to also attack the two he didn't know about? Whirlwind Attack only seems to care about being within reach. So are the attacks granted and done with concealment or are they denied because you didn't know to attack those squares?

Whirlwind Attack

Benefit: When you use the full-attack action, you can give up your regular attacks and instead make one melee attack at your highest base attack bonus against each opponent within reach. You must make a separate attack roll against each opponent.

Also, would you now know about the invisible creatures?

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    \$\begingroup\$ Oh, much cleaner battle field than what I had, thanks. I'll keep the dots in mind instead of X's. \$\endgroup\$
    – Fering
    Dec 24, 2016 at 20:26
  • \$\begingroup\$ These particular dots are hard to type, but if you search for "unicode middot" in Google there are pages you can copy it from. They do make for a very nice text grid though, don't they? \$\endgroup\$ Dec 24, 2016 at 20:32
  • \$\begingroup\$ Oh much better, and all I have to do is copy your dots. I was using spaces but SE has some funny behaviors for spaces and stuff \$\endgroup\$
    – Fering
    Dec 24, 2016 at 20:35

1 Answer 1

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You can only attack invisible opponents that you know that exists

You will notice that Whirlwind Attack will only hit those who are your opponents, meaning that it doesn't work like a fireball and hit everybody inside your reach. Instead, you will hit those you want to hit, your companions are excluded from your Whirlwind Attacks (unless you want to hit them and declare so).

Likewise, if they know that there is an invisible opponent somewhere near him, he may attempt to hit him, but you roll the d% in secret and only tell him if he missed or hit something.

Invisible creatures have Total Concealment agaisnt your attacks.

Total Concealment: If you have line of effect to a target but not line of sight, he is considered to have total concealment from you. You can't attack an opponent that has total concealment, though you can attack into a square that you think he occupies.

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  • \$\begingroup\$ "An attack roll represents your attempt to strike your opponent on your turn in a round." (paizo.com/pathfinderRPG/prd/coreRulebook/combat.html) Citation from your answer: "you can attack into a square that you think he occupies" Does it mean that I can attack into each square within my reach with a single Wirlwind attack? I.e. without any evidence of anybody's presence in them. \$\endgroup\$
    – Ols
    Dec 25, 2016 at 0:31
  • \$\begingroup\$ You can only attack a square you believe there is something invisible. \$\endgroup\$
    – ShadowKras
    Dec 25, 2016 at 2:07
  • \$\begingroup\$ For that, the invisibility special ability gives us a bunch of ways to discover the location of an invisible enemy. d20pfsrd.com/gamemastering/special-abilities#TOC-Invisibility \$\endgroup\$
    – ShadowKras
    Dec 25, 2016 at 2:09
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    \$\begingroup\$ @ShadowKras Then shouldn't a creature with the feat Whirlwind Attack convince itself that there's an enemy in every square it threatens just in case? I mean, the creature uses the same amount of resources either way, so why not? \$\endgroup\$ Dec 25, 2016 at 5:22
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    \$\begingroup\$ @ShadowKras No. There are two quotes above. One says an atack is always made against an opponent (at least no other valid target is named). The second says an attack into a square is valid. Either we agree that a square can be an opponent, or the term "opponent" cann't be taken literally and the word "opponent" in Whirlwind description is not an argument. \$\endgroup\$
    – Ols
    Dec 25, 2016 at 14:53

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