13

IMO, your question is too general to give a good, specific answer. Also, it would help if you elucidated the motivation for the question. I am a semi-retired electrical engineer, not a security or RF expert but I'll give it a go anyway. The first assumption is that the transmitting equipment is hidden from plain sight and other non radio-related detection. ...


11

Another possibility to look into is meteor burst communication. This is a well-established technology. You beam a VHF signal up into the sky, it bounces off a meteor trail, and your recipient picks up the reflected signal. Since transmission is upwards, picking up the transmitter's location would require being above the transmitter at the time (making ...


8

The story is about Eric LeMarque who got lost after snowboarding out of bounds in the Sierra Nevada mountains. Radio receivers usually contain some kind of oscillator. Detecting the unintentional radiation from the oscillator or the mixing products in the receiver is technically possible, but in modern receivers the power is so weak this power is so weak it ...


7

My first post in SO. Here's a C version... made this for an Arduino project. void calcLocator(char *dst, double lat, double lon) { int o1, o2, o3; int a1, a2, a3; double remainder; // longitude remainder = lon + 180.0; o1 = (int)(remainder / 20.0); remainder = remainder - (double)o1 * 20.0; o2 = (int)(remainder / 2.0); remainder = ...


5

at least in the world where I come from Assuming this is the world of arrogant computer security experts (sorry I'm a bit allergic to this): Well, unlike computers, physics wasn't designed by humans. So, any signal that contains significant information needs to contain significant power (that's a direct result of Shannon capacity), and if it has significant ...


5

This helped me. For anyone that needs it, here is a port to Java: public class Location { String latlon; String maidenhead; public Location(String p1, String p2) { float lat = -100.0f; float lon = 0.0f; try { lat = Float.parseFloat(p1); lon = Float.parseFloat(p2); maidenhead = latLonToGridSquare(lat, lon); } catch ...


4

Is this [detection of a receiver through its amplification of received signals] possible in practice? Aaaah, no. Not exactly like described in that thread. In more general terms, yes, it is. So, first of all: the point of radio is that the transmitted energy radiates away from the transmitter. That means that you can't know whether something absorbed that ...


3

The Maidenhead locator system (which you correctly tagged) is described in detail in this Wikipedia page. A short summary from that page is: To summarise: Character pairs encode longitude first, and then latitude. The first pair (a field) encodes with base 18 and the letters "A" to "R". The second pair (square) encodes with base 10 and the ...


3

if by "Grid" you mean "Maidenhead Locator" then here is a similar question, with some sample Python code as answer. The result from the code will be Lon = (start_lon) to (end_lon) Lat = (start_lat) to (end_lat) From that you can easily calculate the "middle" of the square. or you can take the code and adapt, here is the code from the above mentioned ...


3

When you fill out the paperwork at the time of taking the test you specify your permanent home address for your license. Thus, if you take the test in California but use a NY address then you will be assigned a call sign appropriate for your license level for New York which uses the 2 zone. Licenses like for a Technician license are issued usually as a 2x3 ...


3

I've included Locator based calculations in PyHamTools - an open source python Library, which is easy to use: It's as easy as this: from pyhamtools.locator import calculate_distance, latlong_to_locator locator1 = latlong_to_locator(48.52, 9.375) locator2 = latlong_to_locator(-32.77, 152.125) distance = calculate_heading(locator1, locator2) print("%.1fkm" %...


2

Yes, there is now. This is probably what you want. https://g7vjr.org/2019/08/google-earth-kmz-files-for-cq-zones-and-itu-zones/


2

There is not any such listing, as the definitions for CQ zones are a bit ambiguous. The official definitions can be found at the CQ Magazine WAZ award definition. Areas like the ocean aren't covered, otherwise it's a fairly simple geography lookup to get the areas defined.


2

Here. This is, I expect, what you want. https://g7vjr.org/2019/08/google-earth-kmz-files-for-cq-zones-and-itu-zones/


2

The Maidenhead system breaks down a grid into increasingly smaller chunks. The fourth pair is defined as an evenly spaced 100 square grid, or 10x10. This site has an excellent image that breaks that down for you: In this picture, the black box outline is MK80ht. You can see this box was further subdivided, and the user's location of Kalpathy, India is ...


2

I didn't see any answers that had the extended square as referenced in the Wikipedia article. My Clojure implementation (GitHub Gist Here) encodes the fourth and fifth pairs. (defn to-maidenhead [lat long] (let [long (-> long (+ 180) (/ 2)) lat (-> lat (+ 90)) funs [#(* 10 (mod % 1)) #(* 24 (mod % 1)) #(* 10 (mod % 1)) #(* 24 (mod % 1))] ...


1

I remember an episode of the live action TMNT series in which Mike had a pirate radio station. To make detection harder, the transmitter was in his truck. I believe everyone here has been thinking about stationary radio sources so far. By keeping your transmitter moving, and going dark as needed, you may avoid detection for a while longer. They will only be ...


1

There are several techniques. first physically hide the transmitter and antenna. While this is hard for large antennas, it becomes more feasible at higher frequencies. I read a story about a HAM radio fox hunt where the fox (transmitter / antenna) was hidden in a coed's swim suit as she was sunbathing in a park. The hunters knew the transmitter was in the ...


1

A recommendation for a formal definition of the 10 character system is available in the IARU-R1 VHF Handbook. Here's the relevant quote: 3.7 Proposals to Clarify and Standardise the IARU Locator, including higher Accuracy positioning The definition of the existing 8-character scheme should be extended by adding a further division into 24 lettered ...


1

I do a lot of off-network travel and one thing I found out is that you can cache Google Maps as described on this link. Then your Google Maps app will keep running as long as you have GPS (and if you don't you likely have bigger problems).


1

There is a Python package converting to/from Maidenhead/WGS84 Lat/Lon


1

C# version public static String LatLonToGridSquare(double lat, double lon) { double adjLat, adjLon; char GLat, GLon; String nLat, nLon; char gLat, gLon; double rLat, rLon; String U = "ABCDEFGHIJKLMNOPQRSTUVWX"; String L = U.ToLower(); if (double.IsNaN(lat)) throw new Exception("lat is NaN"); if (double.IsNaN(lon)) throw ...


1

Here's the docs on it. http://www.qrz.ru/vhf/qth_h.pdf And of course you were given this one. https://en.wikipedia.org/wiki/Maidenhead_Locator_System Take a look at http://aprs.fi/ http://aprs.fi/page/api


1

The solutions I've seen all seem to be rehashing the same code with little thought to what is actually happening. I decided to write a solution that solves the general case, making it a simple matter to extend the precision of the result to any length. Since I am an iOS developer, the solution is in Swift: private let upper = "ABCDEFGHIJKLMNOPQRSTUVWX" ...


1

Have a Look at numpy.meshgrid http://docs.scipy.org/doc/numpy-1.10.0/reference/generated/numpy.meshgrid.html


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