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# reflectivity2

## reflectivity

(OP)
What is the difference in the reflectivity of the sea for horizontally as opposed to vertically polarised em waves?

### RE: reflectivity

Depends upon the angle.  For low angles, think about how polarized sun glasses work.

Of course, defining H and V for waves with normal incidence (straight in at 90 degrees) is tricky...

### RE: reflectivity

(OP)
could you elaborate on the former please

### RE: reflectivity

When discussing reflectivity, it's normal to define the electric field vector components as s and p.  The p-component is in the plane of incidence, while the s-component is normal to the plan of incidence.

The p-component usually has an increasingly higher reflectivity than the s-component as the angle of incidence increases.

TTFN

### RE: reflectivity

(OP)
I am trying to determine why H pol radars, over a sea path, have a more loby structure than V pol radars. Is it due to different reflectivities of the sea for each plane of polarisation?

### RE: reflectivity

As you probably already know, light is the same thing as RF except that the frequency is much higher.

H-polarization predominates in reflections.  That is why the 'crystals' (?) within polarized sunglasses are oriented vertically.

For your radar, presumably the H-polarization pattern is more affected by the in-and-out of phase reflections and therefore shows a greater variation in the pattern amplitude as you vary the angle.  If everything is as it should be, then the maximum effect (largest amplitude variation) will be at about 37 degrees.

At radar wavelengths, waves (sea state, direction) would have a great effect.

### RE: reflectivity

I'd have thought you would use perpendicular and parallel polarization as you would for radome calculations.  You define a plane that includes the incidence vector and the normal to the surface and your signal is perpendicular or parallel to this plane. Although with the movement of the sea this is going to be all over the place and you probably might as well use vertical and horizontal.

### RE: reflectivity

2
The two polarizations transition into the ocean with impedances based on the angles.
For a wave propagating into the ocean there is a mismatch upon entering the ocean where the impedance of air 377 ohms transitions to the impedance of water 377/(square root of water dielectric = approx. 80)= 42 ohms.
This effective impedance of 42 ohms of water changes with angle as mentioned before. Here is a simple formula that gives some hint of effective impedance over a certain range of angles.
Vertical pole Impedance of water=42 Ohms x Sin(angle)
Horizontal pole Impedance of water= 42 Ohms/Sin(angle)
Where skimming the water is zero degree angle, straight down from directly above is 90 degrees.

Example: Radar Energy hits the water at 6.4 degrees, impedance is 42/sine(6.4)= 42/0.1115= 377 degrees and all the energy for vertical polarization hits the ocean and is absorbed by it since the air and the waters effective impedance is the same. For the same angle 6.4 degrees, the effective horizontal pole impedance of the ocean is 42 x 0.1115 = 4.7 ohms and this huge mismatch to 377 ohms reflects the power. Vertical still reflects off the ocean outside this angle, but it's always less than horizontal, though sometimes just barely less.

Why this occurs relates to the boundary conditions for the two polarizations and  I wish I had a more intuitive feeling for the explanation, but I don't.  This answer applies only for a certain range of angles, probably from 2 to 50 degrees or so. Energy going from 90 degrees angle, or straight down reflects the same for V and H pole. RF energy that "reflects" off something actually re-radiates, it creates current on the surface of that object and that current re-radiates. It may be that energy on either side of the magic 6.4 degrees re-radiates with a phase difference where the energy at lower angles than 6.4 compared to reflected energy at higher angles than 6.4 degrees cancel due to the phase of their respective reflections (higher versus lower impedances give a sign difference of 180 degrees in the standard reflection formula).
Kevin.

### RE: reflectivity

Higgler,
Thats the best explanation I've seen of this. Thanks.

### RE: reflectivity

(OP)
Many thanks Higgler this is just what I was after. Do you have a reference for the V and H pol Z equations?

### RE: reflectivity

Formulas thanks to Rutgers University and the Internet;

http://www.ece.rutgers.edu/~orfanidi/ewa/ch06.pdf this one has the information versus angle as we are discussing here, Snells law and formulas 6.2.5, 6.2.12 and 6.2.13 in this reference shows the Impedance dependent on the cosine function, which is the same as my description with the sine function only the angle is referenced from vertical instead of horizontal.

I printed out these two, just to refresh my old memory. I don't fully understand the material.

For a more general discussion, try this one.
http://www.ece.rutgers.edu/~orfanidi/ewa/ch04.pdf has a good tutuorial on electromagnetic wave reflection, this stuff is a bit complicated to just pick up quickly.

Kevin
One other curious note about reflections off the water, the phase of the vertical and horizontal reflections are opposite. So if you are on "smooth water" and transmitting Vertical or Horizontal (or slant 45 linear) from a ship to an aircraft, if there is a dip in the Vertical polarization due to the water reflection, then that will be a peak for Horizontal Polarization. Vertical is still the one to choose for reflection. I think Horizontal is common on ships if the energy has to go through a vertical mast, horizontal polarization will have a slight dip in the pattern, vertical will have a large suckout due to the mast blockage.

### RE: reflectivity

(OP)
Thanks Higgler - What happens to the plane of polarisation (V and H) after its has been reflected by the sea?

### RE: reflectivity

greebling;
The phrase "plane" of the polarization is a bit confusing to me. I'll guess that the question is "does the ocean create some H pole when you radiate V pole into it, and vice versa". A polarization of a pure V or H will not change if the water is perfectly flat, but of course there are ocean waves, so if you think of the RF energy inducing current on the water and re-radiating, this reradiation will follow the contour of the water shape. It's equivalent to having an antenna on the water with the polarization that you've sent to it. In general I would think that a very pure vertically polarized wave would reflect with some horizontal component(-10 to -30 dBish with respect to the incoming polarization) , as would a horizontal wave reflect some vertical.

kch
PS:A Microwave Oven Note;
RF loss in water is about 10 dB per inch at Microwave oven frequencies (@2.54 Ghz), so you never really heat the center of a large cup of coffee/tea directly. Plus, from an antenna perspective you can leave a spoon in your coffee when you microwave it. Even though current is induced on it, the energy mostly is absorbed into the coffee/tea loading and the Radar Cross Section of a small spoon with coffee/tea load is fairly small producing minimal reflections. If you tell someone you just want it to heat quicker, it is likely to be true. Use a small spoon too, I don't have a spare oven to send you to quash my guilt due to this note of interest.

### RE: reflectivity

(OP)
Thanks Higgler. Refering back to a previous point of V pol hitting the surface of the sea at an angle of 6.4 degrees where its Z is 377 Ohms, over what distance is the z become 42 Ohms thereby not having a boundry condition, impedance miss match and energy reflected? Or is the energy just absorbrd as discribed above?

### RE: reflectivity

the 377 ohms is at 6.4 degrees incident angle (as in your transmitter is just above the water and radiating towards the horizon), this is for vertical polarization entering the water and the impedance changes to 42 ohms when the energy enters the water from directly above it. The 377 ohms value changes with angle, so closer to the transmit antenna you would have a steeper angle and lower impedance, farther away from the transmit antenna you have a shallower angle and higher impedance,  based on that cosine angular function. It would be something to plot on Excel just to show how quickly things change at the lower angles.
Straight downward into the water don't forget the impedance is 377/(square root of water dielectric 80) = 42 ohms, the sine of 90 degrees or cosine of zero degrees being = 1.
kch.
This same thing happens on normal ground, but it's much tougher to predict since you don't know the soil dielectric properties very well.

### RE: reflectivity

(OP)
Deep inside the water I understand the impeadance is 42 ohms. V pol energy enters the water at 6.4 degrees with an impedance of 377 ohms as the energy penetrates deeper. There must be a transition from 377 to 42 ohms. Any miss match in impedance will cause energy to be reflected. So is the change in impedance gradual, is energy reflected or is there something else I have miss understood?

### RE: reflectivity

"Deep inside the water..."

So far as I'm aware, radio waves (radar) do not propagate very well under water.

### RE: reflectivity

True, the loss is high in fresh water (10 dB per inch at 2.5 ghz), and very high in salt water. Magnetic Resonance Imaging (MRI) machines commonly use 63 Mhz to penetrate humans who are ballpark similar in loss to salt water, but they only need to go a 10 inches maximum usually.
kch

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