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Anyone else confused by this? Can someone please explain?

You deploy in a triangle that is 3' on one side and 5' on the other.


Automatically Appended Next Post:
Note: While that is not the most mathematically correct way to do it, it is the easiest.

Got it. Thank you for clearing that up.

Math.
If you did 12" from each corner, you have a right triangle with two 12" sides off the 90 degree angle, making your long leg only 22", meaning your not 12" off the center line.
To get a 24" separation, ie 12" off center line on both sides, you have to measure in 17" from each corner for the start points of your triangles.

Numbers rounded for ease of posting.

Hrm.....I'm not a math kind of guy. =p

I drew a line from the bottom left of the table to the top left of the table. Then I drew a perpendicular line 12" away from it to both sides. Then I drew two more lines perpendicular to the two 12" marker lines.

The end result was two triangle deployment zones with a 24" open space between them. Someone draw a picture for us!

Not a graphics kinda guy but I can explain the math.........

Pythagorean theorem, Asquared + Bsquared = Csquared.
We know our C value is 24, 12" off the center line to each side. 24 squared = 576. So Asq+Bsq (equal sides as it's a right triangle)=576. So Asq =288, square root of 288 = 16.97. So each side on our short legs is just under 17".

Well don, I think we still aren't correct. What you are saying would work on a square board. This is like a SAT question here...

Basically, the SIMPLEST way is to do just as they instruct: Draw the line corner to corner and measure 12" off it (mark these with dice). Bad math will make some armies able to deploy closer than 24" away, but then... castling is really simple here as you can get your guys a significant distance away in the back corner.

However, the mission frowns on castling as you gain a significant amount of battle points for having scoring units in your opponents deployment zone :-)

Can someone draw a picture? Might be eaiser for when we get to the stores on Saturday.

Fearspect wrote:Well don, I think we still aren't correct. What you are saying would work on a square board. This is like a SAT question here...

.

Whoops, you're right, we're not working with right triangles off the baseline corner. OK, I'll be back.

If you ignore my paint-suck, this should do.

That's awesome!!! Thanks for that. A picture is truly worth a 1000 words.

: )

That is correct. Just start from the center, 12" both sides and work your way down.

Print this pic off to do you and your opponent a big favor and save the arguing.

I am thinking I am going to cut some string for the FLGS. that should avoid a whole bunch of measuring and arguments to boot. Kudos on the diagram though, it does make it abundantly clear when descriptions sometimes fail. And it saved me from trying to do the math myself to figure out how long to cut the string.

Not a fan of this set-up. but it could be worse. Getting a massacre could be pretty hard, unless you table someone.

Mondo was originally corect. The diagram is way off. It doesn't matter if you have a square table, 6x4 or 8x4, the triangle in the corner formed by the two deployment zone angles will form an isosceles triangle. The distance apart will be radical 2 times as big as the inches from the corner. 17radical2 is 24.04.

Ooop,nevermind. The diagram is correct.

They shoulda just done it cleanse 2 style, 18" in from the corners. Much easier to implement and understand.