I know the 'nominal' voltage of 110/115/120 is both based on Vrms and that stability @ any of those levels is pretty much a fairy tail.
I also know that most rectifiers will output ~130-150V after re-assembling the waveform & filling a comditioning cap.
What I'm not 100% sure on, and don't really feel like digging out my soldering iron & finding a few diodes to test it today, is the peak-to-peak voltage of the 110VAC 'hot' line.
Place a high resistance (perhaps 1M ohm) from your + to - output. Now there is a closed circuit which includes 2 diodes and a resistor. Where is the source in that loop ? (You didn't say you wanted voltage to GND)
Then you edited in a capacitor, which is ok.
LT Spice simulator might or might not be trusted.
Along with seeing no source in your loop, LT Spice agrees that voltage is zero.
The voltage at the cathode of D1 (my circuit), with respect to ground is 340 volts peak to peak. ( I used 120 VRMS ).
The OP did state to use 110 VRMS. The voltage at the cathode of D1 (my circuit), with respect to ground is then 310 volts peak to peak.
I know the 'nominal' voltage of 110/115/120 is both based on Vrms and that stability @ any of those levels is pretty much a fairy tail.
Stability is not based on Vrms. Stability is determined by the ability of generators to maintain voltage despite fluctuating demand. Voltage is measured in Vrms. Utility companies usually comply with their specified voltage with ±10%. It's not a fairy tail. It's a requirement.
I also know that most rectifiers will output ~130-150V after re-assembling the waveform & filling a comditioning cap.
A full-wave bridge rectifier will give out a peak DC voltage of \$ \sqrt 2 \cdot V_{RMS} - 2 ~diode~drops \$ (typically 0.7 V each).
I never heard of a 'conditioning' cap. The smoothing cap will maintain the DC output at peak DC voltage on no load but voltage will droop between peaks depending on load.
What I'm not 100% sure on, and don't really feel like digging out my soldering iron & finding a few diodes to test it today, is the peak-to-peak voltage of just the 'hot' 110VAC line. I.E. Will the circuit below have ~130-150V, or ~260-300V across the + and - output pins?
simulate this circuit – Schematic created using CircuitLab
Figures 1 and 2 based on original question, "Why is there no voltage reading between Probe+ and Probe-?".
Your circuit will have 0 V between Probe+ and Probe- as there is nothing to induce a potential difference between the probes. Even if you connect NODE2 to mains it just bounces the whole circuit up and down with respect to ground or neutral but still does not induce any potential difference between the two probes.
Edit after OP edit including deletion of the original wonky schematic:
What I'm not 100% sure on, and don't really feel like digging out my soldering iron & finding a few diodes to test it today, is the peak-to-peak voltage of the 110VAC 'hot' line.
\$ V_{P-P} = 2 \sqrt 2 \cdot V_{RMS} = 2 \sqrt 2 \cdot 110 = 311~V\$
Your diodes and capacitor are now irrelevant to this question unless you intended this:
Figure 2. Positive and negative rails generated by half-wave rectifiers.
This circuit will give you the peak to peak voltage of the mains between the two probes.
You don't have to test anything, all you need to know is that the peak-to-peak value of the nominal (RMS) mains voltage is:
$$ V_{PP} = V_{RMS}\times 2\times \sqrt{2}$$
whether the output is taken across the entire secondary of the distribution transformer or from the center tap (Neutral) to either end, and what's a fairy \$ \style{color:red;font-size:100%}{tale} \$ is that, in the long term, the mains voltage varies more than about +/- 10% of nominal.
