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View Full Version : Swiftstride - Failed Charges - Use 2D6 or 3D6?

Spinocus
19-10-2010, 01:10
I fear this is a silly question with a painfully obvious answer, so much so that I am almost too embarrassed to ask. But ask I must...

I have played a few 8th ed games but none of them featured cavalry until this one. Buddy of mine played HE, I played Skaven and another friend watched/'moderated'. Dragon Princes charged my Clanrats 18" away but rolled low and failed the charge. The rules for failed charges are explicit, take the highest value of the 2D6 rolled and move that distance. However Swiftstride says relevant units must roll 3D6 and drop the lowest value for charges, fleeing & pursuing. My friends said since the Swiftstride rule did not mention failed charges the value should the highest of the three dice. I was of the mindset that a failed charge logically falls under the category of charges and the two highest values should be used. Also speaking simply in terms of 'reality' it makes sense that mounted or flying units with high movement values should take longer to stop than lower ones.

Am I missing something or is my logic sound and theirs flawed? Which is it, take highest of 2D6 or two highest of 3D6?

Ramius4
19-10-2010, 01:13
You move the single highest die of the 3D6. Not 2D6.

Synnister
19-10-2010, 01:14
Agreed. highest die of the 3d6

Lord_Elric
19-10-2010, 01:21
_+1.................(to the above answer not the dice roll lol_)

TheDarkDaff
19-10-2010, 01:28
Agree with the others. Swiftstride let's you roll 3d6 and drop the lowest result (which leaves you with a 2d6 charge). You simply then take the highest remaining d6 as the move distance for the failed charge.

Spinocus
19-10-2010, 17:58
Great, thank you all for the speedy responses!

stripsteak
19-10-2010, 22:47
I'm confused how this would matter. it should be the same value if you take the highest of the 3d6, or the highest of 2d6 (after 3d6 drop lowest)...you drop the lowest so you should be leaving the highest one behind anyway.

Scriboergosum
19-10-2010, 22:52
I think the OP wanted to take the 2 highest scoring dice, add them together, and then moved the failed chargers that far, so if the 3 dice came up 2, 3 and 5, the chargers would move 3'' + 5'' = 8'' in total as a result of their failed charge.

stripsteak
19-10-2010, 22:58
I think the OP wanted to take the 2 highest scoring dice, add them together, and then moved the failed chargers that far, so if the 3 dice came up 2, 3 and 5, the chargers would move 3'' + 5'' = 8'' in total as a result of their failed charge.