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Robert Clark
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Reged: 01/09/05
Posts: 16
Loc: Philadelphia, PA.
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Hello. I copied below a post to sci.astro about a possible means of communicating at high data rates from planetary missions using optical light transmissions. A couple of questions about the communication from the Moon case. Based on the example of the "Iridium flares", would a 4mx4m size mirror be sufficient size to see flashes from a lunar distance with the naked-eye? How large would be the size of the illuminated area on the Earth from the reflection of the Sun by the mirrors at the lunar distance? Would parabolic mirrors increase the area illuminated or the intensity of the light as viewed at the Earth? BTW, here's some info on how the intensity and the location of the "Iridium flares" can be predicted:
SeeSat-L Apr-98: Method for predicting flare. http://satobs.org/seesat/Apr-1998/0175.html
Bob Clark
======================================================= From: Robert Clark Date: Sat, 27 Jun 2009 23:54:14 -0700 (PDT) Subject: High data rate space transmissions through visible light communication. Newsgroups: sci.astro, rec.radio.amateur.space, rec.radio.amateur.antenna, sci.astro.seti, sci.physics
I had been thinking about methods of high data rate transmission in regards to getting *video* transmissions from Mars orbiter missions. I was irritated by the spotty coverage of the Mars surface at the best resolutions so I wanted to send real-time *continuous* imaging back to Earth receiving stations at the highest imaging resolutions. This would require very high transmission rates, much higher than what is currently used. The idea would be to use light transmissions but only of the on-off variety. You would use a large surface, many meters across, capable of being alternatively lit up and darkened. There are computer chips of course capable of operating at Ghz rates. This would determine if the large surface was lit up or not electrically, possibly by using a material whose reflective properties can be changed electrically. I was worried though about the twinkling seen in point sources, which this would appear to be, such as with stars due to atmospheric effects. So this might require the telescope(s) to be in Earth orbit. The question I had though was whether the atmospheric distortion would cause an "on" signal to appear "off" and vice versa? My understanding of atmospheric distortion is that it causes the point source to be constantly apparently undergoing small shifts in position. But this wouldn't be a problem if what you want to determine is whether it is on or off. If that is the case then ground based telescopes would work. In the large reflecting surface, I actually wanted to use separate, say, squares on the reflecting surface that could be put separately in the on-off position to increase the information transmission rate. But that would require being able to distinguish the squares from Earth millions of kilometers away. This is why I wanted to use light rather than radio for this since the larger wavelengths in radio would make the reflecting surface impractically large for diffraction limited resolution. Even with light you couldn't do this with a single telescope. They would have to be widely separated. Combining the signals from widely separated scopes is common in radio astronomy but is not nearly as successful in optical astronomy. That is because the light wavelengths are so much smaller and you would have to have nanoscale accuracy in positioning the widely separate mirrors in relationship to each other. However, in the case of just detecting an on-off signal this shouldn't be as big of a problem as you're not trying to form a usable image, but only trying to see if a particular location is on or off. You would need though highly accurate timing synchrony between the separate scopes, within nanoseconds, to be sure they are detecting the same on-off square. Note also here that the shifting in the image due to atmospheric distortion very definitely would be bad for using ground based scopes.
It occurred to me this might be a means of acquiring advertising support for a Google Lunar X Prize entrant. I had also been trying to come up with a method of having an illuminated image either on the Moon or in lunar orbit that would be visible to the naked eye on Earth. Such an idea was discussed here:
moon advertising. put a billboard on the moon. http://www.halfbakery.com/idea/moon_20advertising
I wouldn't be in favor of doing this in a way that would actually advertise a product. But I was thinking about it as a way of sending a message in favor of, for example, world peace. In this case you could still have advertisers who could say in TV commercials for example they supplied funding to support the mission and the message. BTW, I would be in favor of advertisers who could pay to have advertising signs set up at the rover landing site so that if anyone who wanted to log on to the the rover transmissions or who watched a TV program on the rover transmissions would see the ads. This to me is something different than an ad that someone would be forced to see just by looking up at the Moon. In any case you would need something large enough so that with naked eye resolution at the lunar distance it would still be distinguishable. This page gives the naked eye resolution at the lunar distance:
Purpose of Building Telescopes. http://www.astronomy.org/astronomy-survival/telepur.htm
According to this page the resolution of the human eye at the lunar distance would be about 22 miles. One single object clearly couldn't do this. However, if you had separate illuminated landers or orbiters at this large distance apart they could be used to send a message visible to the naked eye on Earth. It could work with orbiters by the example set of satellite formation flying by the Cluster mission:
Cluster mission. http://en.wikipedia.org/wiki/Cluster_mission
I also needed to find how large a brightly illuminated surface needed to be at the lunar surface to be visible by the naked eye on Earth. I thought of the example of the "Iridium flares":
Satellite flare. http://en.wikipedia.org/wiki/Satellite_flare
The Iridium satellites have 3 antennas that happen to be also reflective in visible light, totaling 4.8 m^2 in area. According to the Wikipedia page, the flares can be up to -8 in apparent magnitude, though typically at +6 magnitude, and are produced by an individual antenna, so by one of area 1.6 m^2. I'll assume the brightest flares are produced just by the orientation the antennas happen to be in so we could make our reflective surfaces be oriented with respect to the Sun to get the greatest brightness. For the same size surface, the brightness would be lessened by the greater distance to the Moon. The Iridium satellites are at about 780 km altitude so the Moon is about 500 times further. This would lower the brightness by a factor of 500^2 = 250,000. This page gives the apparent brightness commonly visible by the naked eye in urban areas as +3:
Apparent magnitude. http://en.wikipedia.org/wiki/Apparent_magnitude
The 250,000 times lesser brightness at the lunar distance for an Iridium sized reflective surface would give it a +13.5 higher apparent magnitude so up to +5.5 in apparent magnitude. To make our reflective surface be at +3 apparent magnitude we could make the area be 10 times larger, so at 16 m^2 area, or a square 4 meters across. We would need a method for a flat reflecting surface of unfolding it to this size. It might be easier instead to have the reflecting surface be a balloon inflated by stored gas. Since this would be in a vacuum, you wouldn't need much gas pressure or mass to accomplish this. Another consideration is that because of the brightness of the Moon it could swamp out our illuminated surface. For the orbiter, this could probably be alleviated by having the orbiter have a highly elliptical orbit, (this also would be beneficial in minimizing the required delta-v and fuel load) then it would be visible at the higher distances from the Moon in its orbit. For the landers it might work for them to land in the dark lunar maria.
To communicate the message though we would need a method to turn on and off the reflecting surface. One possibility would be to have the reflecting surface consist of very many small squares that could be rotated to reflect toward the Earth or away. Another possibility might be to have it covered with LCD's. Whichever method it would have to be both lightweight and low power. For our first attempts we probably would not want to send so many orbiter or landers at once to form a naked-eye visible image. We would first send just a single one to test it out. Note that this method with a single vehicle could still be used to send high definition video by having our single reflective surface be turned on and off at the required rate, about 256,000 times per sec with compression.
Bob Clark =======================================================
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Robert Clark
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Reged: 01/09/05
Posts: 16
Loc: Philadelphia, PA.
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We might want to test this out in Earth orbit first before sending to the Moon. There are low cost CubeSats only 10 cm on a side that can be launched to orbit at low cost:
CubeSat. http://en.wikipedia.org/wiki/CubeSat
Say we made the reflective surface on the CubeSat be a square 10 cm wide. This is .01 m^2 in area, so a factor of 6.25x10^(-3) smaller than the area of the Iridium antennas. This would result in the apparent magnitude being dimmed to -2.5. This is still quite bright and would be brighter than the star Sirius, the brightest star in the sky. Hmmm. I'm wondering about that. The sunlight at the Earth's distance amounts to about 1,000 watts/m^2. So even if this 10 cm wide reflector in space reflected all the light that fell on it to the Earth's surface it would still be only 10 watts. So this is saying a 10 watt light source at a distance of 780 km would be as bright as a -2.5 magnitude star. Is this right even if you put behind it say a parabolic reflector as with a flashlight so all the light was directed forward?
Bob Clark
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Charlie B
sage
Reged: 03/22/08
Posts: 241
Loc: Virginia
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You may want to look at this
Charlie B
-------------------- Meade SN-8, DS-90, AT-66
DSI Pro II (Schuler Filters), DSI-C, LPI, Canon XTi
AIP4WIN, Nebulosity, DSS, Registrax, GIMP
running on Dell 1420 with Vista
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Robert Clark
member
Reged: 01/09/05
Posts: 16
Loc: Philadelphia, PA.
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Quote:
You may want to look at this
Charlie B
Thanks for that. I'll read it in detail.
This should be a standard question in telescope optics if anyone knows the answer. You have a mirror of a certain size. How large is the reflected image according to distance to the imaging screen (the Earth in this case)? For a flat mirror? For a parabolic mirror? For a bright source, how bright can the image be at the imaging screen?
Bob Clark
Edited by Robert Clark (07/02/09 08:52 AM)
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coolrocketdude
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Reged: 08/26/08
Posts: 9
Loc: Alamogordo, NM
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Here is a NASA link for optical sat comm: http://lasers.jpl.nasa.gov/PAGES/about.html
Some current mission transmission rates: Hubble 1.024 Mbps (high rate) GLAST: 40 Mbps (highest rate)
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Jarad
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Reged: 04/28/03
Posts: 4330
Loc: Atlanta, GA
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Quote:
You have a mirror of a certain size. How large is the reflected image according to distance to the imaging screen (the Earth in this case)? For a flat mirror?
If the mirror is flat, it does not change the image scale. If the original light rays were parallel (like a laser), they stay parallel. If they weren't parallel, they continue to converge or diverge at the same rate as before the reflection.
Quote:
For a parabolic mirror?
It would depend on the FL of the mirror and the relative positions of the source and the earth.
Usually, the transmitter is placed at the focal point of the reflector, which results in the reflection producing parallel rays going in the direction of the reflector's axis (which is usually pointed towards earth). Of course, they won't be perfectly parallel, so the beam will spread out somewhat, and given the distances involved it will probably be wider than the earth by the time it gets here. That is an advantage in that requires less pointing accuracy for the transmitter, and a disadvantage in that it requires more power to be bright enough to detect. Even with some spreading, you still save a lot of power compared to just having the light go out in all directions.
Jarad
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Robert Clark
member
Reged: 01/09/05
Posts: 16
Loc: Philadelphia, PA.
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Quote:
If the mirror is flat, it does not change the image scale. If the original light rays were parallel (like a laser), they stay parallel. If they weren't parallel, they continue to converge or diverge at the same rate as before the reflection. ...
Jarad
Still puzzling about this. This page also says that about a flat mirror:
Image Size. http://www.darylscience.com/Demos/ImageSize.html
But it seems like you should be able to have a reflected image of the Sun on the Moon on the flat surface of a lake. But surely this image on the surface can't be the same *actual* size of the celestial body?
Bob Clark
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Jarad
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Reged: 04/28/03
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Loc: Atlanta, GA
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I didn't say a flat mirror makes an image the same size as the object, I said it doesn't change the image size. It is the same size as it would be if you were looking directly at the object with no reflection at the same total distance (your eye to mirror plus mirror to object). In other words, it's magnification is 1x.
Jarad
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Robert Clark
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Reged: 01/09/05
Posts: 16
Loc: Philadelphia, PA.
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O.K. I found this page with the equation for concave mirrors:
The Mirror Equation - Concave Mirrors. http://www.glenbrook.k12.il.us/gbssci/phys/class/refln/u13l3f.html
The mirror equation gives the relationship between the focal length, the object distance, and the image distance:
1/f = 1/do + 1/di
The object and image distances determine the magnification, which is the ratio of the image size to object size:
M = hi/ho = - di/do, where the minus sign indicates the image would be inverted.
The object distance is the distance to the Sun at about 150,000,000 km. The image distance would be the distance from the Moon to the Earth at 380,000 km. Because the solar distance is so much larger than the lunar distance, the Earth would be quite close to the focus. The size of the image is obtained from the equation hi/ho = - di/do, hi/1,400,000 = - 380,000/150,000,000, using 1,400,000 km for the size of the Sun. So hi = -3,547 km, with an inverted image. What I'm still puzzling about, assuming the idea that a 4 m mirror at lunar distance would have an apparent magnitude of +3 is correct, is would this mean all the observers in that 3,547 km wide area would view it as being of +3 magnitude?
Bob Clark
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Robert Clark
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Reged: 01/09/05
Posts: 16
Loc: Philadelphia, PA.
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Let's see how big the mirror would have to be at Mars to be visible with a telescope a (serious) amateur might have. The distance between Earth and Mars varies from around 60 million km to 400 million km. So we have to take into account the diminution with distance compared to the lunar distance of 380,000 km. However, the distance of Mars from the Sun is also larger than the Earth's so the reflected image would be dimmer because of that as well.
The furthest distance of Mars from the Sun is about 250 million km. This is about 1.6 times as far as the Earth distance so the mirror reflection would be dimmed by a factor of 1.6^2 = 2.56 to begin with. At the closest Earth-Mars distance of 60 million km, this is 158 times
further than the Earth-Moon distance so the reflection seen at the Earth would be dimmer by an additional factor of 158^2, about 25,000 times dimmer. At the furthest Earth-Mars distance of 400 million km, this is 1053 times further than the Moon, so it would be dimmed by a
factor of 1,100,000 times.
So the total diminution at Mars' closest approach would be (1/2.56)x (1/25,000) = 1/64,000. And at furthest distance, it would be (1/2.56)x(1/1,100,000) = 1/2,816,000. The closest distance dimming of 1/64,000 amounts to a change of apparent magnitude of 12, so to +15 for the
same 4 m sized mirror. The furthest distance dimming of 1/2,816,000 amounts to a change of apparent magnitude of 16 so to +19.
This article gives the apparent magnitude an 8-inch telescope might detect as +14 under good seeing conditions:
A practical guide to buying telescopes.
By Jeff Kanipe
Special to SPACE.com
posted: 04:05 pm ET
19 June 2000
http://www.space.com/scienceastronomy/astronomy/telescope_II.html
So at Mars closest approach you would need a larger reflecting surface to amount to a change in apparent magnitude of +1, so one of 2.51 times greater collecting area than the 4 meter one, so to 6.4 meters across. However, for the furthest distance magnitude of +19 you would need a larger reflecting surface than the 4 meter one to amount to a change in apparent magnitude of +5 in order to be visible by the 8-inch scope. So it would need to be larger by a factor of 100 in collecting area, so to 40 meters across.
Certainly a large reflective surface. But the Echo 2 satellite used a microwave reflective balloon surface of 41 meter diameter in the 1960's so it is technically feasible:
Echo satellite.
http://en.wikipedia.org/wiki/Echo_satellites
The image size as before would be calculated from the image and object distances, and the Sun's diameter. It would range from about 1/4 the Sun's size to about 1-1/2 times the Sun's size. This again raises the question: would all observers within the large observing region really see the reflecting surface as having those apparent magnitudes calculated?
Bob Clark
Edited by Robert Clark (07/14/09 08:41 AM)
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