On 2008-04-22, JJ wrote:

I'm not sure what the units are (distance^4 ?)

Yes, the units are distance^4. The relevant formula for the moment of
inertia is actually I = 1/12 * b * d^3, but for comparison the factor
of 1/12 is unimportant.

or how the length of the span figures in

For a point load of fixed weight, deflection at midspan will vary as
L^3. For a distributed load, where the total load increases as the
length increases, deflection will vary as L^4.

but from the equation, it looks like using 4x6s should
provide almost four times the stiffness of 4x4s (3.5 * 3.5^2 = 150).

Yes, the exact ratio is (5.5/3.5)^3 = 3.88.

Anyone know how to calculate the weight that can be supported if the
weight is centered over a 14' span?

I'm afraid answering that question would take a bit longer than space
permits. If you know what you are doing, the span calculators at
www.awc.org are very useful, although they are for distributed loads,
not point loads. You should understand that allowable load may be
limited by deflection (where the deflection criteria is specified as
e.g. L/360, that midspan deflection should be less that 1/360 times
the span) or may be limited by strength.

Cheers, Wayne