How does a radar signal bounce off of a distant object back to the exact spot that it was sent from?
In: Physics
Oh boy, literally my job in the Navy
So what people don’t realize is that a radar is constantly transmitting and receiving, just with transitions between the two being so tiny that we can’t envision it. It’s not like it’s “shouting” in a circle then being quiet the next 360 degrees “listening”. Light (or more accurately, RF energy) is fast, really fast, taking 12.26 microseconds (that’s 1/100,000th of a second) to travel one mile and back.
The classic “sweep” you see is a real thing (though modern radars don’t usually have a visible line because they’re not using CRTs, some repeaters still do though) but it isn’t like every other sweep is a “listen”, it is updating in real time (or within a few microseconds’ time) along the length of that sweep. If you took an ultra-slow-mo video of one of these older style displays, you would see the beam travel along the length of the sweep from center of the display (the ship or otherwise radar source) out to the edge of the display, leaving behind an illuminated dot or phosphor wherever a return is detected. before starting again from the center. This is why the sweep always looks kinda like it’s “bending” a bit, because the center gets “redrawn” before the edge is finished lighting up.
So if we wanna pretend the radar is a shouting guy, pretend he’s yelling at one precise fraction of a degree for a split-split-second, then listening for an echo. If it gets any noise back, the electronic wizardry in the console causes it to light up a dot on the display with an intensity proportional to how “loud” the echo was. And it’ll do that every fraction of a second through the rest of the rotation, dutifully drawing a dot or blob wherever it hears an echo.
Now, depending on how large the object is, it might have so much surface area that it bounces back a lot more RF over the length of it (considered from the point of radiation). This will make the dot become a blob as it is “hearing” a longer echo. Translate that through the length of the object (from front to back as the radar illuminates it) and, depending on the object’s radar reflectivity and size, you’ll see it appear to be a thicker blob in the center where there’s more surface area bouncing back more RF.
And depending on how big or tall the object is, you could have a radar shadow where you get absolutely *nothing* from “beyond” (further away, closer to the edge of the display) the object, say with a building. It’s interesting to see on a ship’s radars, you’ll have these *dummy thicc* solid returns (thick blobs) that are roughly square or rectangular and nothing, not even clutter or noise, past them, and you can tell you’re looking at a city downtown or something.
On the flip side, when we’re talking about a normal 2D radar, RF can sort of bounce or bend around a smaller object. This is why/how you can see multiple things from different distances at the same bearing, while some of the RF bounces back off the nearer contact some just goes right past to hit a further away object. Then, as I said, depending on how quickly RF gets back and how much gets back, you will see a return at a specific distance and… Solidness? Respectively.
Edit: also, most (all?) modern radars radiate a specific frequency of RF. The antenna will ONLY “listen” for that particular frequency (and no, this isn’t some secret navy tech, it’s how crowded fishing ports don’t have collisions every five seconds at night, or how you can race your friend’s RC car). So if you know the precise frequency of a radar, you can basically shoot noise at it from an emitter and that’ll just make a massive wall of noise in the direction the jamming is coming from. With that said, this also means that you can’t just shoot random RF at a radar antenna and expect to jam it, any more than taking a portable radio up to a radio tower for a particular station will prevent you or interfere with selection of another station/frequency — it’s simply not “listening” to that frequency, so regardless of signal strength it does nothing.
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