# [geocentrism] Re: The resolution of Mars

• From: "Jack Lewis" <jack.lewis@xxxxxxxxxxxx>
• To: <geocentrism@xxxxxxxxxxxxx>
• Date: Tue, 27 Nov 2007 17:33:45 -0000

```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 will
```
be between 1e6 and 2e7 brighter than Mars, depending on the relative
```        orbital positions of Mars and Earth.
```
If you put the coin at between 970m and 4.7km away from you, Mars
```        and the coin would be the same brightness - provided you catch the
```
glint. The coin would now have an angular diameter between 0.88-4.25
```        arcseconds - well below the resolution limit of your eye.
```
Another point, of course, is that you wouldn't be able to see the
```        coin 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 to
```
be about 10cm in diameter, which means that planes more than 350m away will
```have 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-accepted
```astronomical data regarding the planet, Mars. For instance, on the
```
29.09.2003, the "Starwatch" column on page 22 of the UK’s Guardian newspaper
```
```
reported that Mars was 72,000,000 km away from us, and that it subtended an
```angle of 19.4 – 20 arcseconds.

If Mars has a radius 0.5327 times that of the World (Encyclopaedia
```
Britannica) 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 and
```
that their figure of 24’’ represents an absolute minimum. In fact, the
angular resolution of the human eye is usually taken as being about 1
```
arcminute (3 times bigger than the angular extent claimed for Mars‘ disc at
```the 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 more
```than the coin is.

Mars is an object. A big object, fair enough, but nevertheless still an
```
object. We see it because light is reflected off it, just the same as we see
```
```
the pound coin between our friend's finger and thumb. It is sunlight that is
```
```
reflected 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, like
```blades of grass, etc.

The smallest object that can be detected by an optical instrument of
```
circular entrance pupil is well-established and is given by something called
```
the 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 as
```
560-nm, corresponding to the peak spectral sensitivity of cones in the human
```
```
retina. 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 of
```
```
seeing 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 24
```arcseconds. This means that at the distance quoted to Mars, the claimed
```
diameter of Mars is (well) below this limit. If the distance of 72,000,000
```km 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 are
```wrong.

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 by
```
cloud). 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 the
```spreading 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 is
```
important only for long-exposure images in telescopes. The response time of
```the cones in the fovea of the human eye make this spreading almost
completely non detectable. (Through a telescope the only thing you will
```
notice 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 Mars
```would make it completely indistinct at best and invisible at worst to an
```
unaided 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 to
```adapt and dilate, was a distinct, sizeable orange disk.

```
In conclusion, based upon this data, Mars would have been invisible to the
```unaided eye, rather than the extremely brilliant object that we could all
```
very easily see (this was a few weeks after its closest approach). Since the
```
eye has been extensively studied in terrestrial laboratories, and since
```
Fraunhoffer 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.

```
----- 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
>>
>> Jack
>>
> You find out that you are measuring two different things
> - at least in this case.
>
>   Regards,
>
>      Regner

```
```

```
```

```