help related to frequency of schmitt oscillator
Short explanation:
The \$C_T\$ is charged more or less slowly via \$R_T\$ until it reaches the OFF level of the Schmitt-Trigger-Inverter. Then the output turns low and \$C_T\$ is discharged via \$R_T\$ until the lower ON-level is reached and the whole process start over again.

This kind of osciallator is called Relaxation Oscillator, see e.g. at Wikipedia (paragraph about comparator based relaxation oscillator).

The frequency may differ depending on the actual levels of the ON and OFF thresholds of the 40106 Schmitt-Trigger-Inverter (see datasheet. As much as I know those values are not very precicely specified and may vary quite a bit. Probably your simulation used different threshold levels as the experiment description you linked). If you build this circuit in reality there will be another deviation from theory/simulation if you use small \$C_T\$s (just a few pF): the wiring itself represents a capacitance (maybe 5..20pF) wich inrceases the total capacitance. 

EDIT:
Calculating the frequency:
Either see example from Wikipedia article and adapt or see e.g. here:

\$f = 1 / T\$ and 
\$T = t_1 + t_2\$ (where \$t_1\$ is time for charging up and \$t_2\$ is time for charging down)

\$t_1 = τ \ln [((V_u - V_1) / (V_u - V_2)] \$ (time for charging up) and
\$t_2 = τ \ln[(V_2 - V_d) / (V_1 - V_d)]\$ (time for charging down)

(where \$V_u\$ is upper supply voltage, \$V_d\$ is lower supply voltage (in your case = 0V) and \$V_1\$ and \$V_2\$ are lower and upper thresholds, \$τ\$ is the time constant \$R_T C_T\$) 

EDIT:
Determining frequency dependency by experiment:
From theoretical consideration above you know that frequency will be a function of capacitance of the following form:

\$f(C) = \frac{k}{C_{stray} + C}\$  

Where \$k\$ and \$C_{stray}\$ are constants (but yet unknown).
If you measure the frequency of the real circuit for two different known capacitances \$C\$ you'll get two equations you can solve for both constants simultaneously (or you can do more measurements and do a least square fit).