Choosing a Band for a 2000 km Midday Circuit
An operator wants a dependable daily circuit to a station 2000 km away, operating around midday in summer. Ionosonde data for the midpoint give foF2 = 5.5 MHz, with the reflecting layer at about 300 km height, and the circuit is expected to work as a single hop at a low take-off angle. The station has a tri-band antenna covering 40, 20 and 15 m, and 100 W. On the first attempt on 15 m at midday nothing at all is heard; on 40 m at midday signals are weak and fluttery. The operator has been told to "just use the highest band that works".
1. Estimate the maximum usable frequency for the path and choose a band, with reasons.
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Geometry
A single hop of 2000 km has legs of 1000 km. With the layer at 300 km:
tan(theta) = 300/1000 -> elevation angle theta = 16.7 degrees.
The angle of incidence measured from vertical at the layer is phi = 90 - 16.7 = 73.3 degrees.
sec(phi) = 1/cos(73.3) = 3.46
MUF
MUF = foF2 x sec(phi) = 5.5 x 3.46 = about 19 MHz
Band choice
- 20 m (14 MHz) is below the MUF and is the highest band that will be returned - the best choice, and the lowest loss of the three below the MUF.
- 15 m (21 MHz) is above the MUF: the wave penetrates the layer and escapes, which is exactly why nothing was heard.
- 40 m (7 MHz) is below the MUF and will propagate, but at midday the D layer absorbs heavily on a 2000 km path, so weak, fluttery signals are the expected result. It is the night band for this circuit.
So the "highest band that works" rule is fine - the error was picking a band above the MUF rather than the highest one below it.
2. Later in the afternoon foF2 falls to 3.0 MHz. What changes?
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Recompute
The geometry is fixed by distance and layer height, so sec(phi) stays at 3.46.
MUF = 3.0 x 3.46 = about 10.4 MHz
Consequences
- 20 m (14 MHz) is now above the MUF and the path closes - the band appears dead, not noisy.
- 40 m (7 MHz) becomes the highest usable band, and D-layer absorption is falling as the sun drops, so 40 m improves at the same time.
- Watch the lower limit too: it is absorption that sets how low you can go (the LUF), so at midday 40 m is marginal while at dusk it is the workhorse.
Practical point
A band that "worked yesterday at noon" can be dead today at the same time. The MUF follows the ionisation, and the operating response is to move to a lower band, not to keep calling on a closed one.
3. The operator proposes to raise power to 400 W to make 15 m work at midday. Comment.
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Why it will not work
15 m at 21 MHz is above a MUF of about 19 MHz. Above the MUF the ray is not returned at all - the ionosphere is effectively transparent at that frequency for that geometry. Extra power is not reflected; it simply goes out into space. Increased power can make a marginal path just under the working MUF more readable, but it cannot move the MUF from 19 MHz to 21 MHz.
What it does buy, at a cost
- Much greater field strength near the station, with the EMR and interference consequences that follow.
- More stress on the amplifier and the antenna system, and a higher electricity bill for nothing on this path.
Better responses
- Operate on 20 m while the MUF supports it, and move to 40 m when foF2 falls.
- Improve the antenna (gain, take-off angle, height) so that a band below the MUF is comfortably workable - the MUF limits which band you may use, but antenna and power determine how well you hear and are heard on it.
- Schedule around the ionosphere: the same 2000 km path on 15 m may be easy later in the day when the MUF rises.