Quote - Why is it that values of 2(for Value_1,) and 1(for Value_2) cause more of a dissplacement than values of 1, and .5

That's pretty simple. Let:

M mean the value of the displacement map image  - we know that 0 <= M <= 1.
A mean a Math:Subtract with Value_1=1 and Value_2 = .5. So A(M) = M - .5
B mean a Math:Subtract with Value_1=2 and Value_2 = 1. So B(M) = 2M - 1

Now you asked why does B cause more displacement than A?

So - this is quite simple.

B(M) = 2M - 1 # by definition

divide both sides by 2

B(M)/2 = M - .5

since we defined A(M) = M - .5, we can substitute

B(M)/2 = A(M)

now multiply by 2

B(M) = 2 A(M)

So - we conclude that B(M) is always twice A(M).

As a little check, we examine the boundary cases where M = 0 and M = 1.

B(0) = -1, B(1) = 1
A(0) = -.5, A(1) = .5

Looks right.

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