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Post by gillianren on Aug 27, 2010 19:51:41 GMT -4
If there is, they haven't sent it to me for proofreading yet.
Yes, it is "who are." I should stop trying to do so many things at once.
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Post by ka9q on Aug 27, 2010 22:38:39 GMT -4
The beamwidth of a parabolic dish antenna depends on its diameter and the wavelength of the radio signal.
The wavelength is equal to the speed of light divided by the radio frequency. The higher the frequency, the shorter the wavelength. Apollo used "S-band" radio frequencies just below 2300 megahertz (millions of cycles per second). The wavelength is thus around 13 centimeters.
The wider the dish, the narrower the beam. The shorter the wavelength, the narrower the beam.
As you move off the antenna boresight (beam center) the signal decreases; it doesn't stay at 100% strength and then suddenly drop to zero. Any beamwidth measurement has to state the relative signal strength at the beam edge. Typically this is 3 decibels, a power drop of 50%.
For a parabolic dish, the 3dB beamwidth in degrees is about 70 * w/D, where w is the wavelength and D is the diameter in the same units. For example, on the 13 cm S-band a 13 meter diameter dish (100 wavelengths) has a 3 dB beamwidth of about 0.70 degrees.
0.7 degrees is a little bigger than the apparent size of the moon as seen from the earth, so by pointing this dish at the center of the moon it could pick up a spacecraft anywhere on the moon's near side plus anything orbiting the moon up to an altitude of about 700 km.
The 64m monsters at Goldstone and Parkes have a correspondingly narrower beamwidth of about 0.14 degrees. Since this is only about a quarter of the moon's apparent diameter, it no longer suffices to simply point the antenna at the center of the moon to pick up anything on its near side or in orbit. Because of this it was sometimes necessary to use separate antennas to track the CSM and LM during those portions of an Apollo mission when they were apart.
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Post by PhantomWolf on Aug 28, 2010 9:32:23 GMT -4
Just as an aside, some people simply don't understand, or even realise that 1/3 is smaller than 1/2, after all, 3 is bigger than 2.
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Post by fireballs on Aug 28, 2010 14:26:23 GMT -4
Thanks for the help guys. I understand it better now I'll ask another question if I get confused
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Post by fireballs on Aug 31, 2010 20:52:09 GMT -4
At the S-band wavelength, a radio telescope beam width is about 1/3 degree. For comparison, the diameter of the Moon as viewed from Earth is about 1/2 degree. So you can see that the antenna beam width is narrower than the size of the Moon. If the signal was not coming from the Moon it would be immediately obvious. I was thinking about this more and here's a question: Could I take a protractor and measure 1/2 degree? Would that be equivalent to the 1/2 degree that the moon is viewed as? See what I'm getting at? Because if that is equivalent, then you guys are right when you say that transmission from the moon are extremely narrow.
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Post by Ginnie on Aug 31, 2010 21:25:52 GMT -4
For comparison, the diameter of the Moon as viewed from Earth is about 1/2 degree. So you can see that the antenna beam width is narrower than the size of the Moon. If the signal was not coming from the Moon it would be immediately obvious. I was thinking about this more and here's a question: Could I take a protractor and measure 1/2 degree? Would that be equivalent to the 1/2 degree that the moon is viewed as? See what I'm getting at? Because if that is equivalent, then you guys are right when you say that transmission from the moon are extremely narrow. I don't think the protractor would work unless you had a really, really big one. Remember, fireballs - we're not measuring the angle or where the moon is in the sky - we're measuring its size. A very rough way to see how big one degree of the sky is - hold up you hand at arms length and stick up your little finger. Its width would be about 1 degree. Look at it like this - from directly overhead to the horizon is 90 0. So, the width of the moon is 1/180 of that. It moves across the sky at about 15 0 per hour. Let's say, you were facing North... The moon seems to be bigger than that when you look at it, doesn't it? But this is the size it would be when looking at the whole sky from the eastern horizon to the western horizon. I think the diagram illustrates just how small the Moon actually is in the sky. You need accurate instruments to send and receive signals to and from the moon.
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Bob B.
Bob the Excel Guru?
Posts: 3,072
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Post by Bob B. on Aug 31, 2010 22:15:51 GMT -4
For comparison, the diameter of the Moon as viewed from Earth is about 1/2 degree. So you can see that the antenna beam width is narrower than the size of the Moon. If the signal was not coming from the Moon it would be immediately obvious. I was thinking about this more and here's a question: Could I take a protractor and measure 1/2 degree? Would that be equivalent to the 1/2 degree that the moon is viewed as? See what I'm getting at? Because if that is equivalent, then you guys are right when you say that transmission from the moon are extremely narrow. I'll refer you to Ginnie's previous post, which I think answers the question. Another way to think about the apparent size of the Moon is that it is equivalent to viewing a quarter from nine feet away. It doesn't seem that small when you look at it in the sky but it really is.
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Post by Ginnie on Aug 31, 2010 22:28:06 GMT -4
I was thinking about this more and here's a question: Could I take a protractor and measure 1/2 degree? Would that be equivalent to the 1/2 degree that the moon is viewed as? See what I'm getting at? Because if that is equivalent, then you guys are right when you say that transmission from the moon are extremely narrow. I'll refer you to Ginnie's previous post, which I think answers the question. Another way to think about the apparent size of the Moon is t hat it is equivalent to viewing a quarter from nine feet away. It doesn't seem that small when you look at it in the sky but it really is. Okay, now that just seems a bit too small. I can't even see a quarter from nine feet away.
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Post by Data Cable on Sept 1, 2010 1:21:19 GMT -4
Another way to think about the apparent size of the Moon is t hat it is equivalent to viewing a quarter from nine feet away.Okay, now that just seems a bit too small. I can't even see a quarter from nine feet away. Seems so, but the math works out: A standard American quarter dollar coin selected randomly from my own pocket change measures ~0.955 in. in diameter. The tangent of 0.25° (approximate half-angle subtended by the moon as viewed from Earth) is .0043633508. 0.4775 (the radius of a quarter) divided by that value yields a distance of 109.4 in., or 9 ft. 1.4 in.
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Post by PhantomWolf on Sept 1, 2010 2:40:48 GMT -4
I'll refer you to Ginnie's previous post, which I think answers the question. Another way to think about the apparent size of the Moon is t hat it is equivalent to viewing a quarter from nine feet away. It doesn't seem that small when you look at it in the sky but it really is. Okay, now that just seems a bit too small. I can't even see a quarter from nine feet away. You must have bad eye sight, 9 feet is under 3 metres away.
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Post by randombloke on Sept 1, 2010 5:19:45 GMT -4
Try shining a searchlight on it to get the full "full moon" effect.
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Post by Mr Gorsky on Sept 1, 2010 6:44:51 GMT -4
What's a quarter ... ?
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Post by dwight on Sept 1, 2010 7:49:45 GMT -4
.25 of the whole.
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Post by Ginnie on Sept 1, 2010 16:22:23 GMT -4
Okay, now that just seems a bit too small. I can't even see a quarter from nine feet away. You must have bad eye sight, 9 feet is under 3 metres away. I guess if it was as bright as the actual moon I'd have no problem seeing the quarter.
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Post by svector on Sept 1, 2010 18:17:18 GMT -4
Okay, now that just seems a bit too small. I can't even see a quarter from nine feet away. Then try two bits from three yards. .
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