So You Wanna Be A Guerrilla RTO, II

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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.

Question-
> How many minutes before sunrise and after sunset is the D layer of
> the
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
the top.
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 degrees.

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.

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19 thoughts on “So You Wanna Be A Guerrilla RTO, II

  1. I never got to respond to the question due to work- and the fact that the combination of morphine,oxycodone, and flexaril is not condusive to solving math problems.
    ( slipped and fell on some ice and broke my wrist/forearm-broke both ulna and radius just above where wrist bends.
    Shouldn’t have put my arm out to break the fall- my hard head would’ve probably broke the asphalt.
    I was close on how to do the math-only because I use geometry and trig often. Need both for carpentry work-especially figuring out custom rafter cuts.
    Hopefully I’ll only need pain meds another couple days so I’ll be able to do more than stare at the tee-vee or computer monitor.

    Liked by 1 person

    1. Oh Brother…man, you have my prayers. That’s rough, especially in the winter.

      If it makes you feel any better (lifts spirits, at least) when I was a teen my dad slipped and fell on the concrete behind our back porch after an ice storm. It was funny, at first. We picked at him, telling him to ‘toughen up’ and ‘quit being a baby’ like he used to tell us as kids…he sneezed and his three broken ribs popped loose.

      Since then, everytime something serious has happened, we’ll joke with each other and say, ‘toughen up!’

      Get well, prayers to you, and if you need anything at all either you or somebody close to you email me.

      Like

      1. Thanks,I’ll be okay,the pain is starting to back off some now,and I’m going 6 hours between doses instead of every 4 hours
        . The day I fell,I finished the rest of our accounts,shoveled snow,used snowblower,and salted the sidewalks. Had to go and salt all of them the next night too. I refuse to use salt spreaders-they always clog and you spend more time getting the things to work right than you do pushing them to spread salt. We just use the same bulk road salt we load the dump trucks with,shovel it into 5 gallon buckets,then I carry the bucket under one arm,and throw salt out with my free hand. My wife says it looks like I’m feeding chickens,but I can salt the 4-6′ wide sidewalks at our commercial accounts in half the time it takes if I use a salt spreader.
        I had to have been sleeping really good-fell asleep on couch in a room where the only heat is woodburner-it was cold enough I could see my breath. Got a good fire going again now,got a nice big chunk of red oak that was a fork in the tree-it will burn all night.
        Merry Christmas to you and yours!

        Liked by 1 person

    2. Ouch. been there, done that. Ice is great for hockey and in summertime drinks, but that’s about it. Heal quickly, Gamegetter, there are more questions coming down the line! I hope you have a Merry Christmas, sir!

      Keypounder

      Liked by 2 people

  2. Chris S

    Nice job figuring the time window. If the bad guys are reading it too, they have a couple of 30 minute or so windows for the DF teams to go to work on. Note that it doesn’t matter how obscure the digital mode is provided it is narrow band. They don’t need to decode it to DF it. If they hear a suspicious signal they can take a cut on it regardless. Trained operators could hear a signal and start taking DF cuts in a couple of seconds. That was in the 90’s. The newer software driven versions may be faster. The key to LPI is low power and a very fast mode (short time on the air), or a very wide bandwidth mode (not much in use on the amateur bands today).

    And they don’t need an adcock. The PRD-13 antenna is about a foot in diameter and 9 inches high (with a little flexible screw-in whip on the top – extra points if you know what that is). Back in the day when it was developed, it was used with a 3 channel receiver – a right and left channel to scan for and ID signals and a “center” that would do the DF at the push of a button. Those SOC guys got amazingly good at using it. It would a good thing to know what “the other side” is capable of doing.

    Liked by 1 person

    1. the window for operation depends on where your receiving station is, and that is not given. Perhaps you don’t need to know. 😉 if one assumes that the receiving station is north or south of you, then a conservative person would want some margin to cut down on DF. I’d go for at least half an hour, which means that the D layer is fully energized at least 500 miles in every direction. At the equinox, you’d have a 12 hour window.

      Liked by 1 person

      1. Yes it is.

        A very well-versed engineer buddy of mine regularly uses a homebrew 80m version originally made to isolate a neighbor’s noise interfering with the CW portion of the band. And a reliable Adcock array can be man-portable in place of a loop.

        But all these answers of course leave out the underlying tradecraft issues. A good OPFOR has been able to DF signals reliably for a LONG time, and yet…signals still get out. 🙂

        Then again, that wasn’t part of the question (or maybe it was).

        Like

      2. Chris S

        Pretty close. The little whip is an electric field probe, which is used in conjunction with the crossed ferrite rods to resolve the directional ambiguity. It’s not metal. It’s resistive, like carbon fiber. Metal scatters too much RF and destroys the accuracy on other bands. The extra resistance doesn’t matter since the antenna impedance is very high anyway and it feeds a high input impedance amp.

        Liked by 1 person

      3. Chris S

        To get back to the point, an entire wideband DF station can be man portable, small, and located nearly anywhere. It can be very fast at finding and DFing new signals of interest. So to improve your odds, I would say open the transmission time and frequency window as wide as possible, use the lowest power possible, the fasted mode possible and hope for a very low probability of intercept. I am assuming both receiver and transmitter have the right technology, will know the intended time and frequency of transmission and an accurate source of time. And I would not worry about distant fixed DF stations unless the enemy plans to use MLRS and take out a grid square.

        Liked by 1 person

  3. Hoss

    Just as a reference to help visualize the grey line propagation times which mark the beginning and end of the D layer propagation window, (as well as MUF which is not an issue in the hypothetical scenario proposed using 1.8mhz) check out this site: http://www.spacew.com/www/realtime.html
    Also, from the Solar Ham website, one can click on the Gobal D-Layer Absorption graphic to see current D-layer attenuation conditions. Obviously these references won’t always be available, but familiarizing oneself with the what D-layer looks like now will help in the future. For what it’s worth-
    Hoss

    Liked by 1 person

  4. Pingback: Brushbeater: Guerrilla RTO – Solution | Western Rifle Shooters Association

  5. shocktroop0351

    All this math is hard on my head, can’t we just start doing some cross border raids on the DF sites? Seriously though, thanks again for trying to teach us squishy faced knuckle draggers.

    Liked by 1 person

      1. I’ve been reading about the ionosphere for decades, and it is hard to recommend one particular source, but I will see whether I can compile a bibliography and shoot that over to NC Scout.

        Keypounder

        Liked by 1 person

  6. OK, here are some starting places for the aspiring propagation student-

    the following article is a very good starting place for those who have little to no prior knowledge-
    http://www.arrl.org/files/file/Technology/tis/info/pdf/8312011.pdf

    A USN basic course on radio propagation and antennas-
    http://www.globalsecurity.org/military/library/policy/navy/nrtc/14092.pdf

    http://hfradio.org/muf_basics.html

    http://www.sws.bom.gov.au/Educational/5/2

    http://www.arrl.org/propagation-of-rf-signals

    NOAA’s space weather website-
    http://www.swpc.noaa.gov/products-and-data

    http://www.greg-hand.com/hfwin32.html

    Liked by 1 person

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