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I am following this design for a simple decametric radio telescope.

diagram of telescope

I understand the principle that a radio wave hits the metal inducing an electric current that can be measured, but I do not understand the meaning behind the structure.

  • Why does there need to be a copper circle and a wire mesh to detect the waves?
  • Why does the copper need to be a circle with a gap?
  • How would one determine the sizes of the loop and mesh required to measure radio waves with wavelengths that are tens of meters?
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    $\begingroup$ I wasn't familiar with the term "decametric", which means having a wavelength of a few tens of meters. A decameter is ten meters. $\endgroup$ – rclocher3 Dec 26 '17 at 17:46
  • $\begingroup$ That plywood should be 60-70 centimetres, no? 60-70mm is only 2.5-3 inches $\endgroup$ – Scott Earle Dec 27 '17 at 1:24
  • $\begingroup$ Also, the photo of the people standing in the freezing cold playing with their home-made antenna and all holding cups of tea, is about the most British photo I have ever seen! $\endgroup$ – Scott Earle Dec 27 '17 at 1:34
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Why does there need to be a copper circle and a wire mesh to detect the waves?

The electromagnetic radio field exerts a force on the charges in the antenna, creating a very small voltage and current measured by the radio. The parts must be good conductors for the charges to be easily moved without much loss.

The copper circle is the basic antenna, and the mesh is a reflector, making the antenna more directional. Were it not directional it would still work, but it would be even more difficult to separate astronomical sources from terrestrial ones.

Why does the copper need to be a circle with a gap?

The gap is the feedpoint, where the radio attaches to measure the charge movement in the antenna. Without a gap, how would the radio attach? There are other possible ways, but a gap works fine and is easy to build.

How would one determine the sizes of the loop and mesh required to measure radio waves with wavelengths that are tens of meters?

With antennas, all dimensions are relative to wavelength. So if you wanted to build a similar antenna for half the frequency or twice the wavelength, double the dimensions. The spacers would become 60cm long, and the loop diameter 106cm.

That said, 21 MHz is a wavelength of approximately 1.4 meters. All the dimensions in this antenna are significantly smaller than that, making it an electrically small antenna, and it may work just as well for a wide range of frequencies with no modification.

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    $\begingroup$ 21 MHz is 1.4 meters? ~300/21 MHz = ~15 meters. Also note that the design connects the center of the coax to one end of the loop and the braid to the mesh "reflector"... $\endgroup$ – Glenn W9IQ Dec 25 '17 at 16:01
  • $\begingroup$ This helped a lot thank you, how does the mesh act as a reflector and how does this distinguish astronomical from terrestrial signals? $\endgroup$ – cal Dec 25 '17 at 16:04
  • $\begingroup$ @cal well apparently I had my head on backwards when I wrote this, and really my answer is BS and should be deleted. But alas, you've already accepted it, so you're not likely to get better responses... $\endgroup$ – Phil Frost - W8II Dec 26 '17 at 13:44
  • $\begingroup$ @PhilFrost-W8II which part was off? $\endgroup$ – cal Dec 26 '17 at 19:55
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    $\begingroup$ This is a complicated way of making a short dipole antenna. There should be an isolator, an unun on the coax near the mesh because otherwise the coax would be the dominating place where currents go. There is also a great risk for conducted intreference on the cable. The antenna is simply two modest capacitors to infinity separated by 30 cm. $\endgroup$ – sm5bsz Dec 26 '17 at 23:46
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This is a complicated way of making a short dipole antenna - an E-field sensor. There should be an isolator, an unun on the coax near the mesh otherwise the coax would be the dominating place where currents go. There is also a risk for conducted interference on the cable. The antenna is simply two modest capacitors to infinity separated by 30 cm. It might resonate near 21 MHz The capacitance between the two parts plus the capacitance to infinity form C. The wire to the copper part forms L. For a better E-field sensor look here https://www.youtube.com/embed/ItLkn8r4s3E Change the toroid and remove the tuning cap to move the resonance from MW to 21 MHz.

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  • $\begingroup$ Thank you for the answer, Leif. Your comment ought to be deleted, since it's redundant. $\endgroup$ – Mike Waters Dec 27 '17 at 0:38

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