20
$\begingroup$

The experiment

This experiment is documented in a documentary called Convex Earth. The exact location the following information is taken from starts at 14:25.

High frequency directional antennas are set up 14 km apart, 1.5m [I recall them saying 1m on the video, but the experiment note, location and height, I've added below says 1.5m] from water level. Both are on points along the on the coast of a large body of water, with sufficient coastal curvature for there to be only water between the two points. Thus, there are no objects or land masses obstructing them.

According to the experimenters, the curvature of the Earth over that distance, for an antenna 1 m above ground, would present an obstacle of 3.84 m. This should be sufficient to prevent the antennas from remaining in radio contact.

Illustration of curvature

The above image illustrates what is described above. (The house is in the image as an example of an object 3.84 m high.)

Research

I've read online that radio waves, especially small ones, would be almost entirely unaffected by gravity. Really large radio waves can bend ever so slightly around the Earth's curvature, beyond line of sight, but it's nominal. This would seem to be in contradiction to what this experiment has demonstrated.

One example of such information, and another.

Coordinate Info [Added Later]

On their web site (which I've looked up since posting this question, to get more exacting details), they state the locations of the two antenna as:

Team A: São Lourenço do Sul, RS 31 ° 22'42.37 "S 51 ° 57'40.79" W 
Team B: São Lourenço do Sul, RS 31 ° 30'0.91 "S 52 ° 0'26.88" W

I've checked, and the distance between those points is 14.24 km. I used this tool to check. Here is a screen shot of the result.

Screen shot of antenna locations

Radio equipment [added June 9]

The equipment used in the experiment was two sets of the following :

1 Radio Ubiquiti Bullet M5 HP Approved by Anatel to 400mW operating in the band 5800 Mhz

1 satellite dish Aquarius Approved by the FCC with 24 dB gain and 4 degrees of openness

1 UHF Radio HT

HT VHF Radio 1 with 2 dB omni directional antenna

1 radio HT VHF / UHF dual band

All other info on the experiment is detailed here.

location and height

In the video cited, I got the impression one antenna was on land, and the other antenna was out on a boat on water. But looking at the locations given in experiment notes, it looks to me that both are at points along the coastline. I was not sure why that was. I've since seen this note on the experiment notes:

Note: Team B coordinate the currently appears on the water, but when in 2011 the experiment was conducted had a cove where the equipment was installed. The equipment of both teams were positioned at 1.5 meters the water level height.

Question

What is the scientific explanation, using accepted laws of physics, to explain how these high frequency radio waves can make contact with the opposite antenna over a distance of 14 km?

Additional related info

An similar experiment was conducted by the same researchers, using a laser beam. It was transmitted across a distance of 33.78 km, at 1.5 m above water level. It too was successfully transmitted between the two points with that distance between them.

$\endgroup$
  • 5
    $\begingroup$ Do you know the frequency of the radio waves used? The diffraction properties of the radio signal depend pretty heavily on its frequency. $\endgroup$ – probably_someone Jun 6 '18 at 4:50
  • 2
    $\begingroup$ Here (en.wikipedia.org/wiki/High_frequency) it's defined as "3 to 30 MHz" with bending around Earth only possible when the wave is bounced off the ionosphere, which in this experiment is not the case. $\endgroup$ – inspirednz Jun 6 '18 at 4:59
  • 2
    $\begingroup$ Here's the exact specifications of the antenna used at both ends: Radio Ubiquiti Bullet M5 HP Approved by Anatel with 400mW operating in the range of 5800 Mhz, and the distance was 14.2 KM. $\endgroup$ – inspirednz Jun 6 '18 at 5:02
  • 2
    $\begingroup$ Also note shortwave signal will bounce in the ionosphere en.wikipedia.org/wiki/Skywave $\endgroup$ – jean Jun 6 '18 at 11:05
  • 1
    $\begingroup$ I'm not sure how much it plays into specific experiment this but a lot of these 'proofs' of the flat earth or other such nonsense such as the Bedford Level experiment are actually demonstrations of atmospheric refraction. $\endgroup$ – JimmyJames Jun 6 '18 at 13:38
59
$\begingroup$

EDIT: In the interest of avoiding spreading misleading information, I have removed the portions of this answer that have been disputed or refuted in the comments and edits on this question. Specifically, the parts about the ACK/Distance shown on the screen at 42:47 and the calculation of the curvature have been removed. The rest of this answer, however, still stands.

TL;DR: They erroneously believed that radio antennae were lasers. The antennae should still be able to connect even on a curved Earth.

The video pretends that the signal leaving the radio antennae is like a laser beam, focused in the line that emanates from transmitter to receiver without diverging. In reality, this isn't even close to true, even for directional radio antennae. Both the transmitted signal and the receiver acceptance get wider farther from the respective antennae, purely due to the diffractive properties of waves. This means that the signal actually propagates in a large ellipsoidal region between the antennae called the Fresnel zone**. The rule of thumb that is used in engineering systems is that as long as at least 60 percent of the Fresnel zone is unobstructed, signal reception should be possible.

The maximum radius $F$ of the Fresnel zone is given in the same Wikipedia article by

$$F=\frac{1}{2}\sqrt{\frac{cD}{f}}\,,$$

where $c=3\times {10}^8 \frac{\mathrm{m}}{\mathrm{s}}$ is the speed of light, $D$ is the propagation distance and $f$ is the frequency. Using $D=14 \, \mathrm{km}$ and $f=5.880 \, \mathrm{GHz},$ we see that $F=13.69 \, \mathrm{m}.$ As you can see, the beam expands massively over such a distance. If you cut out the lower $3.84 \, \mathrm{m}$ of that circle, you would find that the fraction of the beam that is obstructed for obstruction height $h$ from the formula for the area of the cut-out portion given here:

$$\frac{A_{\text{obstructed}}}{A_{\text{whole beam}}}=\frac{F^2\cos^{-1}\left(\frac{F-h}{F}\right)-(F-h)\sqrt{2Fh-h^2}}{\pi F^2}\,.$$

Evaluating this expression for $F=13.69 \, \mathrm{m}$ and $h=3.84 \, \mathrm{m}$ gives you an obstruction fraction of $\frac{A_{\text{obstructed}}}{A_{\text{whole beam}}}=0.085.$

So, even on a curved earth, only 8.5 percent of the beam would be obstructed. This is well within the rule of thumb (which required less than 40 percent obstruction), so the antennae should still be able to connect on a curved Earth.

**In reality, propagation of radio waves between two antennae is complicated, and I'm necessarily skipping over a lot of details here, or else this post would become a textbook. What I refer to as the "Fresnel zone" here is technically the first Fresnel zone, but the distinction is not necessary here.

$\endgroup$
  • 1
    $\begingroup$ Thanks very much for this. I've since looked up what coordinates they provide for the two antenna. I've added this into the question. So I suppose we have to assume they either lied about those numbers, or something else (such as the matter of the Fresnel Zone effect) means 14.2 Km is still doable. Although, I am wondering why Wikisource states "For an antenna 10 feet above the ground, 8 kilometers (5 miles) is the maximum line-of-sight distance." in relation to high frequency radio antenna / transmission. (other sources of info I've looked at make similar statements). $\endgroup$ – inspirednz Jun 6 '18 at 8:17
  • 2
    $\begingroup$ @inspirednz "For an antenna 10 feet above the ground, 8 kilometers (5 miles) is the maximum line-of-sight distance."* sounds like the distance to the horizon at 10 feet high. The antenna will be visible from 8km even if the observer is lying on the ground. But that doesn't mean that 8km is the max. communications distance. $\endgroup$ – LLlAMnYP Jun 6 '18 at 8:49
  • 1
    $\begingroup$ I think your third paragraph's (omnicalculator) calculations do not match the graphics the OP posted. You are calulating how much of a ship 14 km away would be obscured given an eyesight level, but the graphic is about the earth's (alleged ;-)) bulge's highest point in the middle between two objects at 0 height. Put in 7km distance (to the midpoint) and eyesight level 0 into the calculator and voilà...3.84555m obscured height. $\endgroup$ – arne.b Jun 6 '18 at 13:30
  • 5
    $\begingroup$ Note that if the distance actually was approximately 1.4Km or less that the people would have been able to visibly see the other team with their own eyes, let alone the antenna experiment. $\endgroup$ – Aaron Jun 6 '18 at 14:26
  • 2
    $\begingroup$ Refraction in the atmosphere also helps. Radio wave refraction effects mean that the radio horizon is farther than the optical horizon by an amount that can be approximated as if the earth's radius were 4/3 larger. It might be worth mentioning this in the answer. en.wikipedia.org/wiki/… $\endgroup$ – Doug Lipinski Jun 7 '18 at 12:45
2
$\begingroup$

Since the existing answer has a few errors (see my comments on that otherwise excellent answer), I wanted to offer another take. There are three key phenomena at play here, refraction, line of sight, and diffraction. I will tackle each in turn.

Refraction

Since the atmosphere decreases in density as you go up in altitude it acts to refract radio waves. This has the same root cause as light bending when in passes between mediums with different refractive indices (e.g. a prism or the bending of light as look down into a swimming pool). This means that the optical or radio horizon is actually farther away than the geometric horizon. Assuming standard atmospheric conditions, this can be accounted for by calculating the horizon distance as if the earth's radius was bigger by a factor of 4/3 (Wikipedia link). The distance to the horizon can be computed as $$d_{\rm horizon} = \sqrt{2kRh+h^2}$$ where $R$ is the earth's radius (about 6371 km), $k$ is the multiplicative factor (4/3 for radio waves), and $h$ is the height above ground. Based on the video, let's assume the first antenna is about 1.5 m above the ground, that gives a horizon distance of 5.048 km.

It is also possible that some ducting effects are at play. That would actually lead to a very good transmission and fully explain the successful link by itself.

Line of sight

So we know how far the horizon is based on the correct refraction of radio waves, but how far away can you see the other antenna? Some simple trigonometry reveals that to see another antenna distance $d_{\rm total}$ away, that antenna must have height $$ h_2 = \sqrt{(kR)^2+(d_{\rm total}-d_{\rm horizon})^2} - kR $$ Plugging in the numbers from the video ($d_{\rm total} = 14.24~{\rm km}$) we find that the second antenna must be at least 4.97 m above the water to have line of sight for radio waves. It's hard to tell from the video, but this might be a bit too high so we need something other than plain line of sight.

Diffraction

Electromagnetic waves diffract as they pass near objects. This phenomenon causes the waves to essentially bend around corners by some amount so it is not actually necessary to have line of sight to the EM source in order to receive a signal from that source. The effect of diffraction in this case is probably best captured by estimating the relative strength of the diffracted signal compared to the strength it would have if there were line of sight. Assuming it is actually out of line of sight, we can use nomograms in this publication1 to estimate the attenuation. The attenuation is very dependent on frequency and height of the transmitter and receiver, but using a 5.8 GHz frequency (as stated in the comments on the original post), $k=4/3$ as above, and assuming antenna heights of around 2-4 m above the water gives somewhere around 25-30 dB of attenuation. While this is a large attenuation factor, it is certainly believable that the transmission could still be received. It is the equivalent of moving the antennas from about 14.24 km apart (if they had line of sight) to 250+ km apart. After all, satellite communication satellites work and those transmitters are typically in geostationary orbit at about 36,000 km high.

Conclusions

This is a complicated radio propagation problem which is impossible to model fully without knowing the transmit power levels, transmit and receive antenna gain patterns, receiver noise characteristics, waveform properties, and signal processing details. It appears that the receive dish is likely (just) out of line of sight from the transmit dish, even when refraction is considered. However, the combination refraction and diffraction means that some amount of the transmit signal will reach the receiver. The attenuation due to non line of sight is large, but it is still possible that the receiver gets enough power to detect the transmission anyway.


[1] "Propagation by Diffraction," International Telecommunication Union, Recommendation ITU-R PN.526–7, 2001.

$\endgroup$
  • $\begingroup$ Welcome to Physics SE. Keep up the well-structured answers with citations! $\endgroup$ – user191954 Jun 8 '18 at 13:51
  • $\begingroup$ Thanks very much. Very informative. The details of the experiment state both antenna are 1m above water level, so is suggest it's probably most useful if we factor that into any relevant calculations. I've edited the question to add in the exact details of the equipment used, in case that helps us to be even more precise in our calculations and conclusions. $\endgroup$ – inspirednz Jun 8 '18 at 21:43
  • $\begingroup$ Correction to above comment: I've now noted in the question, the experiment done by boat shown in the video states a height of 1m, but the experiment written up on their website states a height of 1.5m, and both points are on the coast. So using 1.5m in our calculations is fine. $\endgroup$ – inspirednz Jun 8 '18 at 22:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.