Out of curiosity we did a couple types of analysis of the Peltz oscillator.
Here's a simple AC linear analysis and the net result shown below using a frequency independent transconductance model for the transistors.
Vout/Ve = -gm1*Re*Z*(gm1 + gm2)/(1-gm1*gm2*Z*Re), where Z is the parallel impedance of the LC network {Z = jwL/(1-w^2*LC)} as seen from the collector of Q1, where Q1 is common base and Q2 is common collector (emitter follower), and Vout is from the collector of Q1, and Ve is the emitter junction. This shows the gain goes to infinity (necessary for oscillator and remember linear AC analysis) when 1-gm1*gm2*Z*Re goes to zero, or gm1*gm2 = 1/(Z*Re) and at a frequency of where 1-w^2*LC goes to zero, or w = Sqrt(1/(LC)), or f = 1/(2*pi*Sqrt(LC)).
Since gm is Ic/Vt, where Vt is kT/q, then oscillations occurs when 1-(Ic1/Vt)(Ic2/Vt)*Z*Re goes to zero, or Ic1*Ic2/Vt^2 = 1/(Re*Z). Since Ic1 ~ Ic2 because Q1 and Q2 both conduct same current thru Re and since they have similar Vbe and Vce assuming high beta.
Ic = Vt(Sqrt(1/(Re*Z))
So the minimum Vee to sustain oscillations can be found by including the drop across Re and Vbe. This makes some sense because the transistor voltage to current conversion, or gm, must overcome the emitter resistor Re and LC tank losses.
For a 2N3904 transistor with Is of 1fa, Re of 510 ohms, Z at resonance of ~900 ohms (L = 449.6uH, C = 214.7nF), we get a collector current of Ic = Vt(Sqrt(1/(Re*Z)) ~ 39.2ua for a Vbe of Vt*Ln(Ic1/Is) or ~ 648mv, assuming junction temperature is ~35C.
So Vee is 2*Ic*Re + Vbe or ~ 688mv which is minimum Vee to sustain low frequency oscillations with Re, L and C. This agrees with a measurement of 692mv at a frequency of ~15.94KHz (calculates as 16.2KHz).
Best,