06 November, 2011

Multiple 28 MHz propagation paths and excellent conditions

Final 'F' in UA3MIF. Notice echoes of final dot
marked by arrows. Press image to enlarge.
There were excellent conditions during this weekend's Ukrainian contest. I was active on 10 m and noticed several reverberant signals coming from Russia and Kazakhstan around 11-12 UTC on 6. Nov.

Here is an example from UA3MIF (28 MHz, 11.30 UTC 6. Nov 2011, press link to hear audio) as received on my vertical end-fed dipole (omnidirectional) for 10 m and the K3.  See the picture for the final 'F' in his call sign and what seems to be multiple echoes. The final dot starts at 3.247 sec and multiples seem to occur at 3.322 and 3.373 seconds, i.e. after 75 and 126 ms.

Although 126 ms is almost 138 ms, the round-the-world travel time, I could need some help to interpret these delays based on the locations of the transmitter relative to me near Oslo, Norway. It was easier for a similar reverberant signal from JA3YBK some years ago.


  1. Hello Sverre,

    I saw your interesting entry on the received signal of UA3MIF.

    According Wiki, the meter was originally defined as 1/10,000,000 of the distance from the equator to the North Pole through Paris, France. So the distance around the world is 4 * 10000 = 40000 km.
    The speed of light c= 300000 km/s So the time for a signal to travel around the world will be t = s/c
    t = 40,000,000 / 300,000,000 = 133 ms (all in meters)

    The distance between you and Vladimir is about 1600 km. JO59 --> KO98 To find the distance, I use: http://f6fvy.free.fr/qthLocator/fullScreen.php

    The signal needs 1,600,000 / 300,000,000 = 5.3 ms to travel this distance. 1,600,000 meters divide by 300,000,000 meters per second (the speed of light)

    The signal travels to you in 5.3 ms, but also travels the other way around the world to reach your antenne from the other side.

    The first signal will reach you in 5.3 ms and the other signal will reach you after 133 ms. The time difference between the signals is 133 - 5 = 126 ms This is what you have seen. hi

    Thank you for the entry. 73, Bert

  2. Hi Bert,

    Thanks for your comment. Yes you are right, this is the propagation the other way around. It really tells us that 10 m was booming. But what is the second signal, the one coming after 75 ms? In the case of the JA3YBK signal in 2003 (link at the end of the main article) I was able to understand both echoes, but that is not so straightforward here it seems.