Dear Regner,As I understood Neville's paper, I thought that he was saying that the light from Mars would be to dim to be seen. Was he not saying simply that the photo receptors of the eye would not register the photons? I may have used the word 'resolution' incorrectly.
Jack----- Original Message ----- From: "Regner Trampedach" <art@xxxxxxxxxx>
To: <geocentrism@xxxxxxxxxxxxx> Sent: Tuesday, November 27, 2007 2:40 PM Subject: [geocentrism] Re: The resolution of Mars
Okay, Jack. My previous posts on this subject still stand. I'll comment on a couple of the statements of Neville's that you included. Neville: Mars is not emitting visible light any more than the coin is. Regner: True. Both just reflect sunlight. BUT: Mars' surface area is some 1e17 (= 10^17 = 1 followed by 17 zeroes) times larger than a coin. The surface area of Mars' which is presented to the Sun is 1e17 times larger than for a coin. Granted the coin have a larger reflectivity (albedo) close to 1, and Mars' is about 0.15 - so that gives us a factor of 10. Also, Mars is farther from the Sun, than Earth is, so the sunlight will be diluted by the square of that factor = 2.3. To be generous I'll call those two a factor of 100 resulting in the coin still being 1e15 times dimmer than Mars. If we placed a coin next to Mars, Mars would be 1e15 brighter than the coin. Light that is emitted or reflected in all directions, will get diluted by the square of the distance. This means that if you look at a coin a meter away, glinting in the sunlight, it willbe between 1e6 and 2e7 brighter than Mars, depending on the relativeorbital positions of Mars and Earth.If you put the coin at between 970m and 4.7km away from you, Marsand the coin would be the same brightness - provided you catch theglint. The coin would now have an angular diameter between 0.88-4.25arcseconds - well below the resolution limit of your eye.Another point, of course, is that you wouldn't be able to see thecoin because you would have to be on the dayside of Earth to get a reflection off the coin, which means the general daylight will completely outshine the coin seen at that distance - just as we can't see Mars in the daytime sky. I agree with everything else that Neville says, except his statement that you can't see things that are below the resolution limit. I would like to add, though, that his statement is correct in sun-lit scenarios, where the contrast between objects is much lower than between the black Universe and a star or a planet, e.g., you will be able to see (not necessarily resolve - they might just be little dots) black letters on a white background at much larger distance than yellow letters on a white background. My example with the airplane is good, since you know the size of the lights and approximate distance to the plane. I'll estimate the lights tobe about 10cm in diameter, which means that planes more than 350m away willhave lights that are below the resolution limit of your eye (I have used the 1 arcminute measure that Neville also prefers - the diffraction limit of 24" is too optimistic). I hope this helps. Regner- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -Quoting Jack Lewis <jack.lewis@xxxxxxxxxxxx>:Dear Regner, I do understand what you are saying but I'm not expressing myself sufficiently well. Try reading this form Neville's website. Jack Mars Dr. Neville Jones, Ph.D., M.Inst.P.--------------------------------------------------------------------------------There seems to be some fundamental flaw in the almost universally-acceptedastronomical data regarding the planet, Mars. For instance, on the29.09.2003, the "Starwatch" column on page 22 of the UK’s Guardian newspaperreported that Mars was 72,000,000 km away from us, and that it subtended anangle of 19.4 – 20 arcseconds. If Mars has a radius 0.5327 times that of the World (EncyclopaediaBritannica) and was really 72,000,000 km away, then it would indeed subtend the stated angle. However, the lower limit of human sight, corresponding to the wavelength for peak spectral sensitivity, 560 nm (Hecht and Williams, J.Gen. Physiol., 5, 1-34), is given by Born and Wolf as 24 arcseconds(Principles of Optics, sixth edition, p. 415). Now it should be noted that Born and Wolf’s tome is widely recognized as the foremost work on optics andthat their figure of 24’’ represents an absolute minimum. In fact, the angular resolution of the human eye is usually taken as being about 1arcminute (3 times bigger than the angular extent claimed for Mars‘ disc atthe time in question).If someone holds a pound coin between their finger and thumb and walks away from you, there will come a point when you can no longer see the coin. I.e.,its angular dimension is too small for the optical instrument you are using (your eye in this case) to resolve. Exactly the same is true of a football, a hot air balloon or a planet. Mars is not emitting visible light any morethan the coin is. Mars is an object. A big object, fair enough, but nevertheless still anobject. We see it because light is reflected off it, just the same as we seethe pound coin between our friend's finger and thumb. It is sunlight that isreflected off the coin and it is sunlight that is reflected off Mars. Thus Mars is simply an illuminated object, like the coin, like a football, likeblades of grass, etc. The smallest object that can be detected by an optical instrument ofcircular entrance pupil is well-established and is given by something calledthe Rayleigh resolution criterion. This, in turn, depends upon only two things: the diameter of the aperture and the wavelength of the light. Distance does not come into it, for what we are talking about. The eye is an optical instrument which obeys the Rayleigh resolution criterion. There is no question about this. Let's take the wavelength as560-nm, corresponding to the peak spectral sensitivity of cones in the humanretina. Let's take the maximum pupil aperture, which corresponds to a fully dilated (i.e., dark-adapted) pupil. This gives us the very maximum chance ofseeing the object. What is the smallest illuminated object that we can see under these conditions? The answer to this question has been shown to be 24arcseconds. This means that at the distance quoted to Mars, the claimeddiameter of Mars is (well) below this limit. If the distance of 72,000,000km is correct, then Mars' diameter must be bigger. Conversely, if the diameter is correct, then the distance to this planet must be much less.Possibly (almost certainly, in my opinion), both distance and diameter arewrong. This point is based upon standard diffraction theory. It is all the more convincing if one bears in mind the fact that you do not need a totally dark-adapted eye to see Mars. It is (relatively) very large. It does not necessarily need to be bright (and indeed can be partially obscured bycloud). The angular resolution of the human eye under these circumstances, and at this wavelength, is one arcminute. The point? Secular data does not add up. Either the distance to Mars is wrong, or the size of Mars is wrong,or both.The only effect that does tend to slightly increase the size of Mars is thespreading of its point spread function by the Earth's atmosphere. This is true only to a very small extent. It is a statistical effect and isimportant only for long-exposure images in telescopes. The response time ofthe cones in the fovea of the human eye make this spreading almost completely non detectable. (Through a telescope the only thing you willnotice because of this is a slight blurring of features on Mars.) I repeat, then, that the accepted distance to Mars and the accepted diameter of Marswould make it completely indistinct at best and invisible at worst to anunaided human eye. On the contrary, it was a very sizeable nighttime object.Something is wrong. That something is the worldly data regarding Mars, for what we actually observed during the latter part of 2003, even if going out from a brightly-lit room and looking at Mars without waiting for our eye toadapt and dilate, was a distinct, sizeable orange disk.In conclusion, based upon this data, Mars would have been invisible to theunaided eye, rather than the extremely brilliant object that we could allvery easily see (this was a few weeks after its closest approach). Since theeye has been extensively studied in terrestrial laboratories, and sinceFraunhoffer diffraction theory is very well established, it seems obvious,from the simple observation of any one of us, that earthly data regarding Mars is significantly incorrect. Copyright © 2003-2005 Dr. N.T. Jones. All rights reserved.----- Original Message ----- From: "Regner Trampedach" <art@xxxxxxxxxx>To: <geocentrism@xxxxxxxxxxxxx> Sent: Tuesday, November 27, 2007 10:26 AM Subject: [geocentrism] Re: The resolution of Mars > Quoting Jack Lewis <jack.lewis@xxxxxxxxxxxx>: > >> Dear Regner,>> The point I was making, or rather Neville was, is that for its >> distance>> and >> size it should be invisible. >> > Exactly - and that was the point I was answering. I told you: > * It is still bright enough to be visible > * But it is too small to be resolved so you see it as a point, > as opposed to an extended object, like a planetary disk. > We see the stars - they have far smaller angular diameters than > Mars - I hope even GC folks will agree to that. > The stars are unresolved. > No matter how large a conventional telescope you use, you will > still only see a point of light - not a round star like our Sun. > With interferometers you can actually see the size of a few of > the closest stars, but that is irrelevant for this point. > > You need to distinguish between unresolved and invisible. > Unresolved means that you cannot distinguish it from a point, > but you can still see it if it is bright enough. > >> So what do you do when you measure something in >> one way and the measure it in another way and get completely different >> answers? >> >> Jack >> > You find out that you are measuring two different things > - at least in this case. > > Regards, > > Regner