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Mathias · April 10th, 2019, 07:35 AM

Also, double check your math

cost = (uval - 1) / 2

So, at a uval of 0, the cost = -1/2 (but then you apply a minimum)
At a uval of 2, the cost = 1/2 (and the minimum applies)
at 4, the cost = 3/2
At 6, the cost = 5/2

That doesn't match the costs you were describing.

Mathias Senior Member Lone Wolf Staff Join Date: May 2005 Posts: 13,213 April 10th, 2019, 07:35 AM	Also, double check your math cost = (uval - 1) / 2 So, at a uval of 0, the cost = -1/2 (but then you apply a minimum) At a uval of 2, the cost = 1/2 (and the minimum applies) at 4, the cost = 3/2 At 6, the cost = 5/2 That doesn't match the costs you were describing.
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