So now that the shakeup has settled from the first post, we can discuss the solution.
It is interesting to see that surprisingly few people are actually
READING the question, and fascinating to note that nobody, so far, has
attempted the mathematical solutions, which ought to be straightforward.
> How many minutes before sunrise and after sunset is the D layer of
ionosphere fully illuminated by the Sun? (Assume only direct light from
the Sun; neglect atmospheric scattering effects)
This is a geometry and trigonomic problem. What are we given?
The D layer elevation is given as 35 miles to the bottom and 50 miles to
Your base is given as being 300 miles away.
Assume Slobovia is on the Equator.
Assume that the D layer is instantly ionized by the Sun.
What you are NOT given is the radius of the Earth- you have to look this
up. The equatorial radius of the Earth is 3963.2 miles, give or take.
For purposes of this analysis, we’ll neglect the minor effects of
mountains or terrain irregularities, and assume that Slobovia is just
above sea level.
In order for the D layer to be ‘fully illuminated’ the bottom edge of
the D layer has to be exposed to sunlight. This must occur when the Sun
is just visible at sunrise at some unknown distance to the east of us.
If we solve for that unknown distance, and we know the equatorial
circumference, we can arrive at the time before sunrise, or after
sunset, that the D layer is fully illuminated, and fully ionized.
If we imagine a right triangle, with the hypotenuse being the radius of
the Earth plus the height to the bottom of the D layer, then the three
sides have the following lengths:
Hypotenuse is a distance of 3963.2 + 35 miles = 3998.2 miles.
The adjacent side is simply the radius of the Earth, 3963.2 miles.
Now, with this preliminary work done, there are two ways to get a useful
answer. We’ll deal with the exact approach, which uses trigonometry,
first, then compare it to the approximate solution.
The cosine of an angle is defined as the ratio of the length of the side
adjacent to that angle to the length of the hypotenuse. the arccosine
is the angle given by a cosine. In our case, the cosine is
3963.2/3998.2 = 0.991246, and the angle whose cosine is that value is
7.59/360 = 0.021, which is both the fraction of Earth’s circumference
from your location to the sun-rise line, and the fraction of the Earth’s
period of rotation, nominally 24 hours.
0.021 x 24 hours works out to about 30 minutes 20 seconds before
sunrise. 0.021 x 24 hours works out to about 30 minutes 20 seconds.
If you want to know how far away the sunrise line is (which comes in
handy in a little bit) then you calculate the fractional circumference.
C=2 * Pi * R;
0.021 x 2 x Pi x R(eq) =0.021 x 2 x 3.1415928 x 3963.2 =~ 525 miles.
So let’s imagine that you forgot to bring your solar-powered scientific
calculator with you, or the Slobovian customs officials stole it from
you when you entered the country. Or you don’t know trig. Can we use a
simpler method to get a close approximation of the answer? Yup.
Take the same triangle, with the same two sides. Pythagoras said that
the square of hypotenuse equals the sum of the squares of the two sides-
A*A + B*B = C*C We know A and C, so this equation looks like
B*B = C*C – A*A. If we solve for B, we get 527.8 miles or 30 minutes
32 seconds. This is a difference of 12 seconds from the exact answer or
6 parts in a thousand. Given the variability of the height of the D
layer, and the real world effects of mountains, solar movement north and
south, and other variables, this is plenty close enough.
the second question is solved the same way, but the height above ground
is 50 miles, not 35. I will leave this exercise for the students at
present, and also leave the thunderstorm question for now.
So that’s (part of) the answer- those that said 30min or so, great job.
On to the next part, questioning modes, Peak Envelope Power (PEP) and so forth-
There’s a few issues which need to be addressed. Most ostensible is the fact that the Ham crowd which was so quick to chime in are focused on making many contacts over HF. This is not what we are doing.
We are sending a one way signal, on a predetermined frequency at a predetermined time, to a predetermined distance, and NO MORE. The atmosphere is going to act as a retarding barrier for what little signal we put out, with the time determined by our math problem.
Are we seeing now why it’s important? Unlike Amateur Radio, who’s objective is to make contacts, clandestine transmissions are meant to be heard by as few as possible save for the one you want to hear it. That being said, understanding how to build our antennas for directional use now becomes extremely important, as is knowing the direction of transmission, as is the time, as is the proper band, etc, etc, etc. I keep harping on the same antennas because they’re important.
The second issue is with modes. The guy who kept justifying SSB…not even close, bud. Not simply for the fact that now there’s a good chance they have a recording of your voice, more importantly SSB is far too susceptible to noise, and you have no way of knowing if your transmission was readable. THEY ARE NOT RADIOING YOU BACK. Difficult concept, I know, but clandestine TX is almost always one-way. So this leaves CW and Digital. CW is great because all it needs is a keyer and you’re good to go. It can be heard far below the noise floor (a serious issue for Phone Ops). It’s downsides is that it’s still very recognizable when heard, but otherwise, the original is still in fairly sound, logically speaking. Next comes digital, which offers some advantages. First, there’s some digital modes that are so obscure they’re never on the air. As a general rule, you want the narrowest one possible but with the fastest speed (I’m not telling you what to use here…that’s up to you), or using a mode so off the beaten path that no one will recognize it. Now before any caterwauling, if you use one mode once, and have done your part, it’ll get where it needs to go safely and without interference. Older, more obscure modes have advantages too- but this comes entirely from working knowledge.
Power Output, or PEP once radiated from out antenna, is related to the issue above. We should be looking at least at 5w or less, most favorably to 1w or less. CW and Digital work just fine with tiny amounts of power…Voice, not so much. With a clandestine transmission, as short and low power as possible…so to the 50w guy commenting…nope.
The DF issue is another issue raised ancillary to the modes/PEP question. DFing 160 is not particularly hard at high power, given the space for an adcock array, but at lower power when directed from, say, a resonant dipole Vee terminated with resistors, it becomes far more difficult. The first assumption is that there just happens to be a team on the ground actively looking for the transmitter- which, well, unless they had prior knowledge of your commo window (soft or undetected compromise, AKA an informant in the ranks), this is unlikely. Since our window is predetermined based on our math, well in advance of our transmitting time (why the whole comment about being tired and in the mud and not having time to calculate this stuff was absurd) and we don’t make a habit of transmitting every day at the same time (once a week or bi-monthly, coupled with other means such as dead drops) we should be good. And while there is vertical radiation off the ends of our dipole and we’re definitely still making noise, DFing a half watt signal using an obscure mode is a far cry from DFing commercial AM radio stations.
And there you have it- Low power and out of the box thinking, coupled with working knowledge.