3 ms·

> With radar, it's actually worse! Received signal strength is proportional to the fourth power of distance (1/r^2 on the way to the target, then 1/r^2 on the w

> With radar, it's actually worse! Received signal strength is proportional to the fourth power of distance (1/r^2 on the way to the target, then 1/r^2 on the way back to the radar). I think that's specifically true when you're pointing a radar at an aircraft, but is not true for radar altimeters. When you're pointing a radar at an aircraft, the angular size of the aircraft is 1/r^2, so 1/r^2 of the emitted radiation hits the aircraft. Once emitted from the aircraft, the angular size of radar receiving dish is 1/r^2, so 1/r^2 of that radiation is detected. With a radar altimeter, you're pointing the radar at the ground. The angular size of the ground is 50% of the sphere, whether you're one foot above the ground or 100,000 feet. (...ok, maybe it's like 49.5% of the sphere...) As long as you're pointing it down-ish, 100% of the emitted radiation hits the ground. There is only one 1/r^2 relationship, on the return leg.

lutorm 5y ago
Well, 100% of the emitted radiation hits the ground in terms of total power, but the radar is looking for the _closest_ reflection, which is some relatively small solid angle straight down. It doesn't change your conclusion though, that solid angle is independent of the distance to the ground.
- nwallin 5y ago
  > Well, 100% of the emitted radiation hits the ground in terms of total power, but the radar is looking for the _closest_ reflection, which is some relatively small solid angle straight down. This fixed angle paints a larger and larger circle as altitude increases. If you double your altitude, you quadruple the size of the surface that's reflecting radar back at you. A bunch of stuff cancels out and you end up with 1/r^2 at the receiver, which is what we started with.
  - steerablesafe 5y ago
    An alternative explanation is that you treat the ground as a flat mirror, and the radar detects the distance of your mirror image. The distance between you and your mirror image is (2r), so the signal strength you receive from your mirror image is ~1/(2r)^2, still follows 1/r^2 .
fennecfoxen 5y ago
I’m not arguing with the idea most of the signal gets to the ground, but, I’m worried that you get specifics quite wrong. Planes don’t want a reflection from 50% of the sphere. They want it from directly below. Other land is further from the plane and obscures the measurement. But they’re not broadcasting to the whole sphere either, because the radar instrument is not an isotropic radiator. That antenna has gain. It’s louder / more sensitive in a specific direction. Probably by a lot. You can’t make perfect antennas, but you don’t need any signal at all from most of the sphere, so you can gain many decibels. The angle of the beam you do have is probably fixed, and being fixed would imply that the area swept by the core of the beam is indeed proportional to the square of the altitude. But I don’t think two trips through an n^2 distance is n^4 unless I’m missing some of the physics. Shouldn’t it be (2n)^2 and this still proportional to the square of the altitude not to the hypercube? (And affected by scattering and bad reflectivity from the terrain, which seems likely to be independent of altitude?) This is different for detecting random entities in free space where most of the first n^2 trip really is lost. So it’s like you say but for different reasons??!?!
mcguire 5y ago
The altimeter does not want to know the average distance to the ground over a several square mile region. It wants to know the distance to the ground directly under the aircraft.
deleted 5y ago
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lutorm 5y ago

Well, 100% of the emitted radiation hits the ground in terms of total power, but the radar is looking for the _closest_ reflection, which is some relatively small solid angle straight down. It doesn't change your conclusion though, that solid angle is independent of the distance to the ground.

nwallin 5y ago
> Well, 100% of the emitted radiation hits the ground in terms of total power, but the radar is looking for the _closest_ reflection, which is some relatively small solid angle straight down. This fixed angle paints a larger and larger circle as altitude increases. If you double your altitude, you quadruple the size of the surface that's reflecting radar back at you. A bunch of stuff cancels out and you end up with 1/r^2 at the receiver, which is what we started with.
- steerablesafe 5y ago
  An alternative explanation is that you treat the ground as a flat mirror, and the radar detects the distance of your mirror image. The distance between you and your mirror image is (2r), so the signal strength you receive from your mirror image is ~1/(2r)^2, still follows 1/r^2 .

fennecfoxen 5y ago

I’m not arguing with the idea most of the signal gets to the ground, but, I’m worried that you get specifics quite wrong. Planes don’t want a reflection from 50% of the sphere. They want it from directly below. Other land is further from the plane and obscures the measurement. But they’re not broadcasting to the whole sphere either, because the radar instrument is not an isotropic radiator. That antenna has gain. It’s louder / more sensitive in a specific direction. Probably by a lot. You can’t make perfect antennas, but you don’t need any signal at all from most of the sphere, so you can gain many decibels. The angle of the beam you do have is probably fixed, and being fixed would imply that the area swept by the core of the beam is indeed proportional to the square of the altitude. But I don’t think two trips through an n^2 distance is n^4 unless I’m missing some of the physics. Shouldn’t it be (2n)^2 and this still proportional to the square of the altitude not to the hypercube? (And affected by scattering and bad reflectivity from the terrain, which seems likely to be independent of altitude?) This is different for detecting random entities in free space where most of the first n^2 trip really is lost. So it’s like you say but for different reasons??!?!

mcguire 5y ago

The altimeter does not want to know the average distance to the ground over a several square mile region. It wants to know the distance to the ground directly under the aircraft.

deleted 5y ago

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