4 ms·
The entire HF band, including the parts already used for something, is only 27 MHz of bandwidth, it's full of noise, and at any given time only a fraction of it
by kragen 1y ago
The entire HF band, including the parts already used for something, is only 27 MHz of bandwidth, it's full of noise, and at any given time only a fraction of it can propagate to the other side of the world, dependent on time of day and, literally, sunspots. This antenna has 1100 MHz of bandwidth, the analog front end has 40 MHz for any given conversation, and noise levels are much lower. It could conceivably deliver Shannon bit rates one or more orders of magnitude higher. But it only works when the moon is visible to both sides of the connection.
I discussed these possibilities and some more challenging ones in 02013 in https://dercuano.github.io/notes/ultraslow-radio.html https://dercuano.github.io/notes/ultraslow-radio.html, although I was considering laser moonbounce rather than phased-array microwave moonbounce because of the higher antenna gain available.
- dreamcompiler 1y agoI didn't see any discussion on your page of the retroreflectors the Apollo astronauts left on the moon. These were put there for distance measuring but they might be useful for laser-based communication too. Caveat: Retroreflectors only reflect in the same direction as the incoming beam. But I'd guess that imperfections in their construction together with the roughly 1 degree of arc spanned by 2 stations on opposite sides of the earth might make this idea practical with a better S/N than using only the lunar surface as a reflector. But I don't know. They might be a lot more precise than 1 degree.
- kragen 1y agoHey, thanks!
- dreamcompiler 1y agohttps://en.wikipedia.org/wiki/List_of_retroreflectors_on_the_Moon https://en.wikipedia.org/wiki/List_of_retroreflectors_on_the...
- kragen 1y ago1 degree and 54 minutes, it turns out. Amazing when you think about that, isn't it?
- dreamcompiler 1y agoI was thinking earth was about twice the diameter of the moon, hence 1 degree. Turns out it's closer to a factor of 4. Should have looked it up.
- kragen 1y agoI used units(1): : yeso; units Currency exchange rates from FloatRates (USD base) on 2025-10-06 3749 units, 113 prefixes, 120 nonlinear units You have: 2 arcsin(earthradius/moondist) Unknown unit 'arcsin' You have: 2 asin(earthradius/moondist) You want: dms 1 deg + 53 arcmin + 57.540656 arcsec
- atomicthumbs 1y agoThe APOLLO lunar laser ranging experiment uses a 3.5 meter telescope as a laser turret and manages to get about 2,400 photons back from those retroreflectors every half an hour, and it's a challenge just to find the things as the spot's a few km wide by the time it gets to the moon. Good luck.
- kragen 1y agoYeah, I was just doing some calculations on this. You'd think that with 3.5 meters you could do better than a few kilometers, wouldn't you? Is something wrong with their telescope? I don't know what wavelength they're using, but at 555nm, 1.22λ/d would be 0.193 microradians, which, unless I'm doing the math wrong, works out to a 74-meter Airy-spot radius at the distance to the moon. At that sort of size, you'd think the majority of the photons in the desired wavelength band would be from their laser rather than stray Earthshine. I was doing calculations based on λ = 350nm and a 500-mm reflector, and no retroreflector, and getting rather sad estimates of 3 joules of light transmitted per returned photon (per receiver). While that's clearly a feasible commnications system, it's going to be pretty limited in bandwidth. I'm not sure if C-band radio is better?