Comments inserted below:
At 06:21 PM 3/20/2015, Roger Rehr W3SZ wrote:
Hi JJ and Ed et al,
I need to correct at least part of my statement
about the effects of the antenna beamwidth on
the depth and width of the notch in sun noise during an eclipse.
My thinking was faulty. Detector beamwidth
[antenna or eye] will affect the apparent
diameter of an object that is captured in
transit mode, either by allowing the object to
transit across the beamwidth of the detector, or
by sweeping the detector's beamwidth across the object;
If tracked exactly thru the center of the object
the result will be the same curve. Drift curves
are favored because antenna mechanical tracking
variations are eliminated (antenna fixed in position)(assumed no wind effects).
just consider the two extremes: A tiny object
will be detected by a sufficiently sensitive
detector when it enters the beamwidth at one
edge of the detector's field of view and
continue to be detected until it leaves the
beamwidth at the other edge. So its apparent
angular diameter will be on the order of the
beamwidth of the detector, as long as it is
well centered in the detector's field of view.
Yes. Usual example is the infinitely small object: point source.
On the other hand, the angular diameter of an
object much larger than the detector beamwidth
can be accurately measured as long as it is
well centered in the detector's field of view.
If angular size is much larger than that of the
antenna. Typically the reality is somewhere between these two extremes.
But for the eclipse experiments, the detector
would be locked onto the sun, with the sun centered well within its beamwidth.
e.g. the antenna is tracking the center of the sun.
So the larger beamwidth of the RF array vs the
optical array does not directly cause a widening
of the notch nor a decrease in the depth of the
notch for the RF experiment vs the optical
experiment by way of the mechanism described above.
What the larger beamwidth does do is to
contribute to a reduction in the signal to noise
ratio of the RF system as compared to a system
where the entire beamwidth is filled by the sun
[as it would be in an optical system with
appropriately chosen region of interest].
Yes. You see this with more significance when
tracking the moon. If antenna bw is much larger
than apparent angular size of the Moon antenna
noise is the sum of the moon thermal noise and
background sky noise. The lunar noise "bump" is smaller in dB.
For this case we have:
Diameter of sun and moon 30 minutes, or 0.5 degree.
That is the optical size of the sun. The radio
sun is larger in apparent angular size (per Flagg
about 2-degrees). Of course this depends on observing wavelength.
Optical resolution of the human eye 1 minute, or
1/60 degree. So each pixel will be much smaller than the sun.
Optical resolution of a telescope is even
better, ~ wavelength/Objective diameter. So
again each pixel will be much smaller than the sun.
Beamwidth of the antenna is 4.4 degrees. So each
pixel is much larger than the sun [and yes,
there is only one pixel]. The sun fills only
(0.5/4.4)**2 or 1.2% of the pixel.
So with the human eye or telescope the sun can
be very accurately represented and a region of
interest accurately drawn. Relative detector sensitivity will be 100%.
With a 4.4 degree beamwidth dish the relative
detector sensitivity based solely on this geometric argument will be 1.2%.
For this reason and others the RF system will
have poorer signal to noise ratio than the optical system.
You have assumed brilliance is the same at radio
as optical wavelengths - not true.
I believe any effect of the larger beamwidth of
the antenna vs the optical systems will be due
to the poorer statistics of the RF system, due
to the reduced signal to noise ratio of the RF
system vs the optical system. This would be
expected to produce a shallower null, but
instead of broadening the null as I first
stated, this statistical effect would narrow the
null. This latter effect would because with
poorer signal to noise ratio, the small
difference in signal level caused by a very
slight occultation of the sun at the very
beginning and end of the eclipse would be hidden
due to the poorer signal to noise ratio of the RF system.
You cannot make this argument as the sun's output
is different at optical vs RF. In fact the
angular size of the earth's shadow is what
determines the duration of the eclipse and amount
of variation in brightness is determined by the
amount of blockage of the earth's disc over the
apparent size of the sun. At visible light the
earth almost matches the sun in apparent size; at
mw frequencies the sun is larger than the earth
in apparent size. Thus the effect on radio
brightness will likely be less deep a drop.
JJ had mentioned the possibility of an RF halo
[my words] affecting the results. I had also
done some thinking along those lines, but then I
remembered that when I use moon noise to keep my
0.79 degree beamwidth dish peaked on the moon on
10 GHz, the moon seems like a tiny object and it
does not seem to subtend a greater angle than my beamwidth.
The Moon is a rock. It emits thermal noise from
its surface therefore visual size and radio size
are nearly equal. Truer at mw wavelengths where
one can discern a radio disc; at 2m the Moon is a
blob in the sky. The Moon has no corona
emission. The sun does. You have seen photos of
the sun corona extending farther out from the optical disc.
That observation suggests that there is not a
significant RF halo, at least at 10 GHz. I
checked on the net, and found an NRAO page that
says that at 4.6 GHz the edge of the RF limb of
the sun is only about 20,000 km above the
optical edge of the sun; a tiny difference
given that the diameter of the sun is 1,392,000
km. Given these results on 10 GHz and 1.2 GHz,
I suspect that on 1-1.5 GHz the result is
similar. No halo sufficiently large or intense to affect the results directly.
OK, that is contrary to what I have read. The
apparent radio sun varies in brilliance and size
depending on wavelength of observation. I'll see
if I can find the paper I cited earlier. Maybe my memory is faulty.
On an image, the RF limb of the sun looks like
the thickness of an eggshell above the optical
limb. There is a nice [and brief] discussion of
this below a picture showing the RF limb superimposed on the optical limb at
Scroll down to the second image, the one showing
a blue green disk that is the sun and read the
caption for more info on this. The rest of the
page is very interesting too, as you might guess
since it is titled "A Tour of the Radio Universe".
Hope the above is of interest to some folks out there!
Maybe our radioastronomers in the group will
tell us what is really going on :)
On 3/20/2015 6:55 PM, JJ Mx wrote:
The beamwidth of my dish is 4.4° (3.3m @ 1.4 GHz)
Find below the geometry of the sun/moon eclipse for my antenna location.
I did the comparison with the optical eclipse
time for the beginning and end of contact between moon and sun.
Maximum was around 9:22 UTC.
[mailto:email@example.com] De la part de Roger Rehr W3SZ
Envoyé : vendredi 20 mars 2015 22:00
À : <mailto:firstname.lastname@example.org>email@example.com
Objet : Re: [Moon-Net] Partial eclipse tomorrow
The differences you observed RF vs optical [both depth and width of null]:
What was the beamwidth of your dish?
I would think that the wider RF beamwidth could
explain both observations, depending upon just
how wide it was compared to the optical
beamwidth, which would of course be quite
narrow. One would need to know the beamwidth
and the geometric details of the occultation to do the calculation.
On 3/20/2015 2:13 PM, JJ Mx wrote:
Moreover the radio eclipse started before the
optical eclipse and finished later than the
forecast. The sun seemed to be bigger at this
frequency (probably coming from its hot
chromosphere I am not an expert and I have to find more info).
Moon-Net posting and subscription instructions
are at http://www.nlsa.com/nets/moon-net-help.html
73, Ed - KL7UW
"Kits made by KL7UW"
Dubus Mag business: