Formula for voltage drop vs PCB trace width, temperature, current, and trace length
I'm going to have a stab at some maths :)

The DC resistance of a conductor - any conductor - is calculated as:

\$R_{DC} = \frac{{\rho}l}{A}\$

Where \$\rho\$ is the resistivity of the conductor in \$\Omega/m\$, \$l\$ is the length in meters, and \$A\$ is the cross-sectional area in m².

The thickness of 1oz copper is \$0.000034798m\$. Say you have a 3mm (or 0.003m) wide trace.  The cross-sectional area is (approximately, assuming a perfectly rectilinear cross-section) \$0.000034798 × 0.003 = 0.000000104m^2\$.  Resistivity of copper is \$1.68×10^{−8}\$ at 20C, and your trace is 100mm long (0.1m).

\$R_{DC} = \frac{1.68×10^{−8} × 0.1}{0.000000104} = 0.016153846\Omega\$ at 20C.

Ok, now for the tricky bit.  The temperature co-efficient (\$\alpha\$) for copper is 0.003862. 

\$R(T) = R(T_o)(1+\alpha{\Delta}T)\$

So for a temperature of 30C we have a \${\Delta}T\$ of 10C, or 10K (30 - 20 = 10, K = C + 272.15).

So \$R(30) = R(20)(1+0.003862×10) = 0.016153846×1.03862 = 0.016777708\Omega\$

So now solve Ohm's Law for voltage.  Say you have 100mA flowing through the trace.  That's \$V=RI\$, so \$0.016777708×0.1 = 0.001677771\$ or \$1.678mV\$ dropped across the trace at 30C.

Who says you need online calculators?

(Now, it's been about 20 years since I did this kind of thing at college, so I may be completely wrong ;) )