[H8crilA]

It can bounce off of meteor trails or the moon. It's been done many times in the ham community. Signals do bounce off of aircraft too, there are even passive military radars that track airborne objects using DVB-T or other strong signal reflections. It could be used if the conditions are just right.

All of this has really nothing to do with LoRa and everything to do with physical properties of the emissions (frequency, power, polarization) and physical properties of whatever happens to exist in the world at the time.

[wl]

The path loss for moonbounce is approximately 390 dB. You're not doing that by accident. In addition to using weak signal modes, you'll need directional antennas and lots of power to overcome that.

[drmpeg]

It's not that high. You have to use the radar equation with the moon being a very lossy but huge target cross section. The path loss at 432 MHz and average moon distance of 384,400 km is 261.6 dB.

[extraduder_ire]

Would it be closer to the other value if you factored in the atmosphere, and whatever else is in the way?

As an aside, would radio via the moon ever be practical? Without pumping out so much transmit power that you pick up the signal anyway, without pointing a receiver at the moon. I think it would be neat, if only for the sake of novelty.

[drmpeg]

No, there's very little loss from the atmosphere at 432 MHz. The 390 dB figure is a misunderstanding of how the path loss is calculated. The poster "wl" took the one way path loss (195 dB) and doubled it. But the moon is a huge reflector and provides "gain".

Here's a C program with the correct equation.

  #include <stdio.h>
  #include <stdlib.h>
  #include <math.h>

  int main(int argc, char **argv)
  {
    double  d, f, lambda, loss;

    if (argc != 3) {
      fprintf(stderr, "usage: moon <km> <frequency(MHz)>\n");
      exit(-1);
    }

    d = atof(argv[1]) * 1000.0;
    f = atof(argv[2]);

    lambda = 299792458.0 / (f * 1000000.0);
    loss = (0.065 * (1.738e6 * 1.738e6) * (lambda * lambda)) / (631.65468167 * (d * d * d * d));
    printf("EME path loss = %f dB\n", 10 * log10(loss));
    return 0;
  }

1.738e6 is the radius of the moon in meters and 0.065 is the reflection efficiency.

Satellites make moonbounce impracticable.

[BenjiWiebe]

It's still not going to happen accidentally.

[dbrgn]

Yes, of course this is related to the physical properties of the emissions, but LoRa uses specific frequencies of the ISM band (868 MHz in Europe), and this discussion was about the reasons why that specific LoRa transmission worked even though there wasn't a line of sight. Airplanes could be an explanation, although your sibling comment thinks there isn't sufficient transmission power for it to be feasible.

[H8crilA]

Ah. It's most likely tropospheric ducting, I thought the top comment explained it.