Let’s say you are staying in a tacky motel room. You lie down it the bed and, looking up, you
see that the ceiling is covered with a mirror. You reach into your bag and pull out a long
cylindrical object. You turn it on.
You shine your flashlight straight up and the spot of light is reflected off the mirror straight
back down, illuminating a spot surrounding the flashlight. You then slowly deflect the
flashlight beam from straight up and see that the reflected spot moves down the bed, passes
your toes, lights up the footboard, then moves across the floor and eventually the reflected
spots falls on the end table near the door. You are looking for your cellphone but the reflected
spot of light is now too dim to adequately illuminate the end table all the way across the room
so you now aim the flashlight directly at the end table and there it is, the missing cellphone.
This is an analogy to explain the TIGHAR “donut hole.”
see:
http://tighar.org/Publications/TTracks/2008Vol_24/donut.pdfRadio waves at the frequencies being used by Earhart behaved exactly like the flashlight beam.
They traveled up into the ionosphere about 160 nautical miles (300 kilometers ) to the “E
layer” which then acts like a mirror and reflects them back down to earth, just like the
flashlight beam. The signals that went straight up came straight back down and the part of the
signal that went up at a slight angle came down near the location of the transmitter. At
progressively lower angles, the reflected signal moved progressively further away from the location of
the transmitter. This process has the technical name of “near vertical incidence skywave
propagation.” It should be obvious that there is no “hole” in the pattern, it is continuous from
the location of the transmitter outward, but growing progressively weaker further from the
transmitter since the signals have further to travel. However, most of the path length is the
distance up 160 nautical miles and then back down the same 160 NM so even at the location of
the transmitter, the received signals have traveled 320 NM. Using trig, or the Pythagorean
theorem, you can calculate how far the signals would travel to arrive at various distances from
the transmitter. (We can ignore the curvature of the earth in this calculation.) At 80 NM from
the transmitter, the signals would have traveled a total of 330 NM, at 210 NM from the
transmitter the signals would have traveled 380 NM. You will note, that because most of the
travel distance is due to traveling up and back to the “E layer,” that the total distance
doesn’t change very much even though the distance away from the transmitter increases a lot
more. This extra travel distance caused the signal strength to get weaker but not very much.
Using the analogy, you wouldn’t notice much, if any, difference in the brightness of the
reflected flashlight beam when it was shining on your toes compared to when it came straight
back down to shine on the location of the flashlight.
So you can see that the signal should actually be a little bit stronger inside the “TIGHAR donut
hole” than in the donut itself, so how does TIGHAR come up with the hole?
Brandenburg and Varney analyzed the transmission pattern of the antenna on Earhart’s plane
and calculated that it put out a very weak signal straight up, and nearly straight up, but got
much stronger at lower angles. So there would only be a very weak signal to start
with going straight up, or nearly straight up, that would be available to be
reflected back down near the airplane. So TIGHAR claims, that due to the weak signal that
would be reflected back within 80 NM of Earhart’s transmitter, that Itasca couldn’t hear her
when she was within that distance. But as she got further away, the much stronger signal
transmitted at lower angles by the antenna, even when combined with the slightly longer path distance and
path losses, resulted in a strong signal at Itasca. So the reflected signal from the ionosphere would be strong
everywhere within the 80 NM proposed "donut hole" except just very near the transmitter so there would only
be a much smaller hole there, certainly less than 38 NM.
To use our analogy, when you first shine the flashlight straight up it is very dim because the
batteries are almost dead and when you deflect it slightly from straight up, the reflected spot of
light is not bright enough to illuminate your toes. So you replace the batteries and now the light
is much brighter and aiming the light at the mirrored ceiling at the lower angle, you can now
see your toes, the footboard and the end table across the room, but dimmer over by the end
table since the light had to travel up to the ceiling and then all the way across the room to the
end table.
There is nothing wrong with this analysis of the skywave propagation of Earhart’s radio.
But there is something wrong with the “donut hole” theory. Brandenburg ignored the “direct
wave” propagation of Earhart’s radio which is in addition to the skywave propagation.
Brandenburg admitted that he did not consider this part of the propagation:
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“It's possible that there was direct path propagation at short distances, due to excitation of the
airframe, but ICEPAC only calculates path loss for an ionospheric path. However, at 1,000
feet altitude (where Earhart said she was flying then), the horizon distance is about 38 miles.
So outside about 40 miles, there wouldn't be any direct path, and skywave would govern. “
See:
https://tighar.org/smf/index.php/topic,285.msg2537.html#msg2537------------------------------------------------------------------------------
Direct wave propagation requires that the transmitter and the receiver be able to see each other,
that there be “line of sight” between the stations. This distance varies with altitude of the plane
and also with the height of the receiving antenna. Brandenburg calculates this as being 38
nautical miles but he then ignores this direct wave propagation! (Using 1,000 feet for the plane
and 60 feet for the Itasca antenna actually makes the distance 47 NM.) Varney calculated
that the strongest signal was sent horizontally which is contrary to
Brandenburg’s speculation of only a “possibility” of a direct wave. Look at my analogy of
shining the flashlight directly on the end table to see the missing cellphone. The direct light is
much brighter than any reflected light off the ceiling and the direct signal from Earhart to the
Itasca would have been much stronger than any reflected skywave signal since it had to travel
only 38 NM (using Brandenburg’s number) instead of 322 NM using skywave to arrive at the
Itasca if it was 38 NM away. So even using Brandenburg’s number, there would be no “hole”
within 38 NM!
So what about beyond that 38 NM and the “donut” at 80 NM? First, it would be filled in by skywave, only
slightly diminished in strength.
Second, the TIGHAR theory, of course, places the plane south of the Itasca where the air was clear enabling
Earhart to climb to higher altitude to facilitate a better visual search for Howland. To reach 80
NM with a direct wave required the plane to climb to only 3300 feet from which a direct wave
would reach the Itasca much stronger than a skywave covering the same radius since the direct
wave would only have traveled 80 NM while the skywave would have gone 330 NM. (And this
ignores the fact that the horizontal direct wave signal was transmitted much stronger than the
skywave signals (according to Varney's table of antenna signal pattern) so the direct wave signal
would have been much stronger as received by the Itasca than the skywave “donut” signal.) So
both of these types of signals would fill in the donut hole out to 80 nm and then the skywave continues
to supply a strong signal beyond that distance.
So my advice about “donut holes,” just buy them at your local bakery.
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As long as we are talking about skywave propagation, this is a good time to get rid of the idea
that there was a “skip zone” for Earhart’s signals that prevented nearby stations from hearing
her while stations further away could. At the frequencies Earhart was using there was no skip
zone because even signals sent straight up would be reflected back down to earth. Skip
involves higher frequency transmissions that would pass up through the ionosphere and not
return to earth if the signals were sent up at a high angle. At some lower angle, however, these
higher frequency signals are reflected back to earth where they can be heard some distance
away but this leaves a gap near the transmitter, this gap is the “skip zone.”
Return to our analogy, only this time paint a black circle on the mirror centered directly above
the flashlight on the bed. Shining the light straight up, or nearly so, will create no reflected
beam since the light is absorbed by the black paint, the same as the blackness of outer space.
Now deflect the beam slowly lower, like you did before, and you eventually reach a point
where the beam hits the mirror outside the black circle and the mirror then
reflects the beam back down, say the beam now lights up the
footboard but it didn’t light up your toes even though your toes were nearer to the flashlight.
The space between the flashlight and the footboard (in this example) is the “skip zone” and no
reflected light beam could be seen in this area. From the footboard outward the flashlight beam
can be reflected and seen.
gl