Electrical – LC tank Q factor

First all below concerns only CMOS gates with switching point at 1/2 of Vcc. So, not HCT series. HC series are OK and 4xxx too.

At first, R1 does not affect the frequency at all. It is placed there in order to make the input current of the inverter (through the protection diodes) to not affect the work of the schematic. That is why it should be much bigger than R2.

The frequency of the schematic is \$F=\frac{1}{2.2.R_2.C_1}\$

How it is derived?

^{simulate this circuit – Schematic created using CircuitLab}

At first notice that the voltage in points 2 and 3 can be only 0 or Vcc.

The schematic turns in the other stage when V1 is equal to the half of the power voltage.

When the second gate output flips from 1 to 0, the capacitor is charged to -0.5Vcc (the left plate is negative), so V1 becomes -0.5Vcc and starts to increase because R2 is connected to Vcc: $$ \tau = R_2.C_1 $$ $$ V_1 = V_{cc}.(1-e^{-\frac{t}{\tau}})-\frac{V_{cc}}{2}.e^{-\frac{t}{\tau}} = V_{cc} - \frac{3.V_{cc}}{2}.e^{-\frac{t}{\tau}} $$ The switching of the schematic will happen when V1 becomes equal to Vcc/2, so: $$ \frac{V_{cc}}{2} = V_{cc} - \frac{3.V_{cc}}{2}.e^{-\frac{t}{\tau}} $$ Or: $$ \frac{V_{cc}}{2} = \frac{3.V_{cc}}{2}.e^{-\frac{t}{\tau}} $$ $$ \frac{1}{3} = e^{-\frac{t}{\tau}} => 3 = e^{\frac{t}{\tau}} $$ $$ ln 3 = \frac{t}{\tau} $$ $$ t = \tau.ln 3 = R_2.C_1.ln 3 = 1.098612289.R_2.C_1 $$

This is the half of the period (because the schematic switches exactly on the half of the Vcc), so the period: $$ T = 2.t = 2.197224577.R_2.C_1 $$

BTW: This oscillator has very high frequency stability, both, by the temperature and by the Vcc. This way its use have to be encouraged for all schematics where quartz stability is not needed.

What you have is a series RLC circuit. One of the most common introductory circuit in EE. For example, look under Series RLC Circuit in this wikipedia page:
en.wikipedia.org/wiki/RLC_circuit

These are equations copied from the same page for the under-damped case. Given R, C along with measured \$\omega_d\$, L can be solved from these equations. (You don't even need to measure the attenuation factor \$\alpha\$.)

If the RLC do want to oscillate, the node where D1 cathode is connected will go negative. Therefore, depending on how D1 is driven, it may conduct during the negative half cycle and kill the oscillation.