tiger Posted July 31, 2003 Report Share Posted July 31, 2003 Ok, after reading the FAQ now I'm wondering if I'm doing it right so.. If hero a receives 10 inches of knockback. He goes 2 inches into the a wall that has 4 Def 3 Body then travels 1 more inch. Is the damage Method A Inches traveled and then the wall = 1Dd(3"/2) + 7D6(Def+Bod)=8D6 or Method B Inches of KB + damge of wall 5d6(10"/2) + 7D6 (Def+Bod) = 12D6 I've been doing B but now I'm thinking it's A Link to comment Share on other sites More sharing options...
Steve Long Posted July 31, 2003 Report Share Posted July 31, 2003 The damage from KB depends on two factors: distance traveled and objects impacted. 1. If the inches of KB traveled are greater than the DEF + BODY of the object impacted, the KB damage equals the DEF + BODY. 2. If the inches of KB traveled are equal to or less than the DEF + BODY of the object impacted, the KB damage equals the inches traveled. 3. If the character doesn't hit an object (i.e., he hits the ground when he's KBed horizontally), he takes half the inches traveled in damage. The example on 5E 281, for instance, involves 11" KB and a DEF 6, BODY 2 wall. Thus, KB damage is 8d6 -- 6 + 2 -- even though KB is 11". To continue that example, if the wall was DEF 6, BODY 6, Mechanon would take 11d6 -- equal to the inches traveled, since they're less than 12 (6 + 6). Link to comment Share on other sites More sharing options...
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tiger
Ok, after reading the FAQ now I'm wondering if I'm doing it right so..
If hero a receives 10 inches of knockback. He goes 2 inches into the a wall that has 4 Def 3 Body then travels 1 more inch.
Is the damage
Method A
Inches traveled and then the wall = 1Dd(3"/2) + 7D6(Def+Bod)=8D6
or
Method B
Inches of KB + damge of wall
5d6(10"/2) + 7D6 (Def+Bod) = 12D6
I've been doing B but now I'm thinking it's A
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