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# The Buzz Aldrin's object. Part II. Apparent magnitude and first set of odds.

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posted on Nov, 15 2014 @ 12:24 PM
I was challenged to present the odds of the object observed during the Apollo 11 mission to be a SLA panel, travelling in the same direction that the CSM-LM on their way to the moon, and showing a close to constant apparent magnitude (not flashing).

If you didn't read part I, you can find it here.

The model below, is a simple but representative model for the light reflection of the panel. Calculus of luminous power, apparent magnitude and a first set of odds.

Postulates

1. Let's agree that the convex reflective surface has an albedo of 0.8 and the concave surface of 0.1.
2. Let's agree that there is no diffuse light in space. This means that the sun light is reflected by the object and strikes directly the eye of the observer, or the object will not be seen.
3. Let's agree that visible light (visible spectrum) corresponds to 40% of the total irradiance of the sun.
4. Lets agree that the position of the object in relation to the observer is a random event (the reflecting surfaces could be facing any direction).

Calculus

The luminous power of the Sun reflected by the object (visible spectrum), is given by the equation:
Pr = (h x w) x P x Al, where Pr is the luminous power reflected, h=6.4 m, w=4.7 m, P is the solar constant and, for the visible spectrum is equal to 1.361 kw/m2 x 0.4 = 0.544 kw/m2, and Al is the albedo of the surface.

For Al=0.8, Pr= 13,090 w, and for Al=0.1, Pr=1,636 w.

The luminous power flux (Pf), luminous power by surface unit, at a distance L between the object and the observer is given by the equation:
Pf=Pr / (h x (Pi x L/2)), where Pi= 3.1415.

For L= 22,000 m and Al=0.8, Pf=0.0592 w/m2. For Al=0.1, Pf=0.0074 w/m2.

For L= 10,000 m and Al=0.8, Pf=0.1304 w/m2.For Al=0.1, Pf=0.016 w/m2 .

For L= 500 m and Al=0.8, Pf=2.604 w/m2 . For Al=0.1, Pf=0.325 w/m2 .

For comparison, if you applied this same equation to the moon (Al=0.12, diameter 3474 km, distance Earth- Moon of 380,000 km), you'll find luminous power flux on Earth, during the full Moon, of 0.000683 w/m2. This tell us that, even in the worst scenario (Al=0.1 and L=22 km), the object would be so bright and the power flux so high that would be sufficient to do a precision handwork.

Through this analysis, we see that even in the worst scenario, if it was the panel, the brightness should be the first descriptive feature, and looking to it through the telescope would be a painful experience.

First set of odds

The probability associated to the observer being in the path of the reflected light.
The imaginary illuminated surface at a distance L between the observer and the object is given by the equation:
S=h x (Pi x L/2), where S is the area of the imaginary surface, Pi= 3.1415, and h=6.4m.

For L= 22 km, S=0,221 km2 .
For L= 10 km, S=0,101 km2 .
For L= 22 km, S=0,005 km2 .

The surface area of a sphere of radius R, with R=L is given by the equation:
A=4 x Pi x R2, where A is the area of the surface of the sphere. Because of the fact that we have 2 reflecting surfaces, pointed in opposite directions, we'll take in account only the surface area of half of the sphere:

Ah=2 x Pi x L2, where Ah is the surface area of half of the sphere, Pi=3.1415.

For L= 22 km, Ah= 3041.062 km2 .
For L= 10 km, Ah= 628.319 km2 .
For L= 0.5 km, Ah= 1.571 km2 .

The probability of the observer to be in the path of the reflected light at distances L= 22 km, L=10 km and L=0.5 km is given by the equation:
P=S/Ah, where P is the probability.
Thus,
P(22 km)=0.00007
P(10 km)=0.00016
P(0.5km)=0.00318

The analysis show that the probability that the "non flashing object" was not the panel, is an interval of P=99,682% and P=99,993%. If we associate this to the fact that the light would be extremily bright reassures my opinion that it was not the panel ( ... the further away you consider the object the higher the probability that it was not the panel).

In the making of this analysis I came up with a theory that would help the panel cause (but coming from me would be highly speculative): NASA would need just one word (and then a lot of math and justifications), and even then, the odds would still be low. Until then, if ever, the object, in my opinion, was not the panel.

Next topic? If necessary, the odds associated with the strange motion of the panel can be analysed.

posted on Nov, 15 2014 @ 12:34 PM
I'm just going to trust your math is right.

If so, then I agree.

posted on Nov, 15 2014 @ 04:35 PM
you make a good case for it not being the panel...

posted on Nov, 16 2014 @ 10:18 AM

You can't be wrong on that ...

posted on Nov, 16 2014 @ 10:19 AM

Yes, but I don't know what it was.

posted on Nov, 16 2014 @ 10:23 AM
Here's how we can calibrate your math, since your process has never been tested against a known object. Tell us how bright an SLA panel ought to be as observed telescopically from Earth, and then we can go look at actual photographs and visual reports and see how close you got. Otherwise it's just mumbo-math-jumbo to impress the innumerate.

Try it in the real world and see if it works.

No verification, no credibility.

posted on Nov, 16 2014 @ 10:23 AM

The panel would have received the same impulse as the CSM at TLI, and would therefore travel along the same trajectory. The issue is whether or not the panel were tumbling.

posted on Nov, 16 2014 @ 10:32 AM

originally posted by: network dude
I'm just going to trust your math is right.

If so, then I agree.

Trust no one, especially handwavers who have a preset answer to prove.

Ever hear of the inverse square law of brightness drop-off by distance?

Apparently, handwaver hasn't. Do you see that term anywhere in his calculations?

posted on Nov, 16 2014 @ 10:38 AM

originally posted by: 2timesOO

The probability associated to the observer being in the path of the reflected light..

Uh, aren't you assuming a total specular -- i.e., mirror-like -- reflection with no light reflected in any other directions?

Who else thinks this is realistic? Who thinks it's preposterous?

And wouldn't the silliness of this assumption destroy the model of all the sunlight reflected in a narrow concentrated beam?

Epic fail. QED.

posted on Nov, 16 2014 @ 10:41 AM

I never call math and physics a mumbo-jumbo. Remember, unbiased. If you want to prove me wrong, do it mathematically please (that's the only thing we really got).
If you want a cooperative effort, you can help to clear things out and provide the albedo of the exterior surface (this was certainly calculated and there are records somewhere ... they wouldn't want the astronauts to suffer from extreme heat). Not that this would change the odds.
Just one correction to the maths, where you read m2 and L2, etc you should read m^2, L^2 (m to the power of 2; L to the power of 2, etc.).

posted on Nov, 16 2014 @ 10:45 AM

We can do the maths on that. I already spent 8 hours on the analysis and posts ... but I can give it another go (but it will be a new topic).

posted on Nov, 16 2014 @ 10:47 AM

We can do the maths on that. I already spent 8 hours on the analysis and posts ... but I can give it another go (but it will be a new topic).

posted on Nov, 16 2014 @ 10:52 AM

Those were the postulates and model. If you agree and are serious about this (unbiased ... and please, don't attack the me as if I was a child, saying things like epic fail ...), we can change the model of the panel to account a deformation of the panel, to the extent you which. Please make a suggestion.

posted on Nov, 16 2014 @ 10:55 AM

originally posted by: JimOberg
Here's how we can calibrate your math, since your process has never been tested against a known object. Tell us how bright an SLA panel ought to be as observed telescopically from Earth, and then we can go look at actual photographs and visual reports and see how close you got. Otherwise it's just mumbo-math-jumbo to impress the innumerate.

Try it in the real world and see if it works.

No verification, no credibility.

Okay, I'll do the math and post it here.

posted on Nov, 16 2014 @ 11:06 AM

originally posted by: JimOberg

originally posted by: network dude
I'm just going to trust your math is right.

If so, then I agree.

Trust no one, especially handwavers who have a preset answer to prove.

Ever hear of the inverse square law of brightness drop-off by distance?

Apparently, handwaver hasn't. Do you see that term anywhere in his calculations?

Well you should read carefully, that was exactly what I did ... didn't you recognise the equation ...

posted on Nov, 16 2014 @ 11:22 AM

originally posted by: JimOberg
Here's how we can calibrate your math, since your process has never been tested against a known object. Tell us how bright an SLA panel ought to be as observed telescopically from Earth, and then we can go look at actual photographs and visual reports and see how close you got. Otherwise it's just mumbo-math-jumbo to impress the innumerate.

Try it in the real world and see if it works.

No verification, no credibility.

Please, in order to answer your question, provide me the distance of the spacecraft/ panel to Earth? I don't have the time to look around.

posted on Nov, 17 2014 @ 12:23 AM

I don't believe your calculated values are probabilities - they are a ratio of two numbers derived from fairly arbitrary values. Probability in statistics is derived from the comparison of observed results with expected and I don't believe that this is what you are showing.

Why is it so difficult to accept the word of the person who actually saw it?

posted on Nov, 17 2014 @ 12:52 AM

Why is it so difficult to accept the word of the person who actually saw it?

Because this?

edit on 11/17/2014 by Phage because: (no reason given)

posted on Nov, 17 2014 @ 12:35 PM

If you have a random event, probability is defined as the ratio between the number of an observed value divided by the size of the sample (the greater the size of the sample, the smaller is the error associated to that probability).
The postulate is that it's a random event. My next post on this will be to show that it's a random event and that the odds associated with that event are very, very, very, low.

posted on Nov, 17 2014 @ 01:01 PM

Well, a man of science must accept when an observation, or the representation of the observation through a sound physical model, don't fit its previous knowledge, and seek for a new set of tools that explains that observation. Well I believe he saw something, I don't know what, but with a pretty basic use of trigonometry and some very solid physical concepts, I can show what he didn't see (or the odds associated with what he claims he saw).
The funny thing, and I wouldn't expect this from people with any formation in the science fields (just joking ... people are people), that people with a science background can use science to discredit any observation of an unknown aerial phenomena, many times in a bias and ridiculous way, and feel offended when a scientific aproach is used to show that the main stream theory on a subject is completely bogus.

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