Of Cosmic Rays, UFO, and SOHO

reply to post by AnthraAndromda

I have shown, with proper mathematics, the reality of this. You may disagree if you like, however, be sure of yourself first. Get another opinion.

Well I know this wasn't a reply to me,but I did get another opinion and he said you are wrong..

I asked Phage.
(not really I just couldn't resist)

Seriously though it looks as though you have a need to be right and those who don't agree with your conclusions are wrong, why is that?

Originally posted by tsurfer2000h
reply to post by AnthraAndromda

 

Don't take this the wrong way, but what are you an expert in?

I am a hardware / software engineer (msee, mscs) with 38+ years expirence.

Most of Phage's argument can only work in a single event model, and we are working with a multi=event model. So tell me; just how does this equate to being right?

And how does this equate to being wrong?

A single event model does not apply to a multievent system.

So what your saying is because he is using this as a single event model he is wrong,but have you actually looked at this as a single event because if not you cannot say he wrong.

Yes, of course. However it is easy to "see". We are discussing the probability that of intersecting "paths"..For this to occur there must be two events , representing each "path". I'm sure you can see how the equations for the single event model can't work here (they are incapable of predicting or handling the second event).

Originally posted by tsurfer2000h
reply to post by AnthraAndromda

 

I have shown, with proper mathematics, the reality of this. You may disagree if you like, however, be sure of yourself first. Get another opinion.

Well I know this wasn't a reply to me,but I did get another opinion and he said you are wrong..

I asked Phage.

(not really I just couldn't resist)

Seriously though it looks as though you have a need to be right and those who don't agree with your conclusions are wrong, why is that?

Oh? Can you show us all just where I was wrong?

reply to post by AnthraAndromda

Oh? Can you show us all just where I was wrong?

I never said you were wrong,but when you call people wrong mainly because they proved a point that wasn't exactly what you think they should be, that is where you went wrong.

Still haven't answered the question of whether you did your equasions for a single event that would then prove one wrong,because that would be the only way that you could actually show him where he was wrong.

reply to post by AnthraAndromda

For this to occur there must be two events , representing each "path".

Yes, in order for it to occur there must be two tracks otherwise there is no possibility for the intersection of two tracks. Without 2 tracks it is null. But you didn't seem to notice that I changed my calculations to include the probability of that first track being produced. So, if you want to include images with less than two tracks you can.

If we do include that null case in which the first track does not exist this is how it is calculated:

1) Probability of the first track (32-55 pixels in length) existing: 0.008
2) Probability of a second track of the same length range existing: 0.008^2 = 0.000064
3) Probability of that track having an intersecting azimuth: 0.000064 x 0.99 = 0.000063
4) Probability of that pair of tracks appearing in any frame: 1-(1-0.000063)^350 = 0.0219

www.abovetopsecret.com...

Originally posted by Phage
reply to post by AnthraAndromda

 

For this to occur there must be two events , representing each "path".

Yes, in order for it to occur there must be two tracks otherwise there is no possibility for the intersection of two tracks. Without 2 tracks it is null. But you didn't seem to notice that I changed my calculations to include the probability of that first track being produced. So, if you want to include images with less than two tracks you can.

If we do include that null case in which the first track does not exist this is how it is calculated:

1) Probability of the first track (32-55 pixels in length) existing: 0.008
2) Probability of a second track of the same length range existing: 0.008^2 = 0.000064
3) Probability of that track having an intersecting azimuth: 0.000064 x 0.99 = 0.000063
4) Probability of that pair of tracks appearing in any frame: 1-(1-0.000063)^350 = 0.0219

And this is the probability that a path will occur at a specific elevation angle. Not rhe probability that an intersecting path. I know you think you have, however, the length of "track" isn't important.

" Probability of the first track (32-55 pixels in length) existing: 0.008" How is this possible since the probability of the path that creats the track is 0.0000078?

" Probability of that track having an intersecting azimuth: 0.000064 x 0.99 ..." Just where did the "0.99" come from?

" 1-(1-0.000063)^350 " and just where did you get this? P(not a) = 1 P(a). To raise it to a power indicates "tries".
To apply a "1 - P(not a)" is to create a double negative. Which iI think in this case is meaningless. The roblem here is that ou are still attempting to apply a single event model to a multi-event system. By the way "P(not a)" is the probability the event won't happen. So there is a 0.02 probability that your event that won't happen won't happen?

So far yo have written a lot of words, given us many numbers, but, haven't yet shown us anything real. You "flavor" of probability smacks of high school

reply to post by AnthraAndromda

We've been all through this.

And this is the probability that a path will occur at a specific elevation angle. Not rhe probability that an intersecting path. I know you think you have, however, the length of "track" isn't important.

No. It is the probability that the path will occur over a range of elevation values. The length of the track is a function of the elevation of the path. With an elevation of 90º the track will be one pixel long. With an elevation of 2º the track will be 80 pixels long. This is why you need to specify a track length (or range of lengths). The length of the track affects the probabilities.

" Probability of the first track (32-55 pixels in length) existing: 0.008" How is this possible since the probability of the path that creats the track is 0.0000078?

There are two component to the path, the vertical and the horizontal. To calculate the probabilities of a path creating a track only the vertical component is considered. The azimuth comes into play only if those two tracks exist. Without those two tracks there can be no intersection.

" Probability of that track having an intersecting azimuth: 0.000064 x 0.99 ..." Just where did the "0.99" come from?

In order for the "extended" tracks to intersect (as you did in your example) the only requirement is that the azimuths are different (otherwise the tracks are parallel). This means there are 359 degrees available for that intersection to occur. 359/360 = 0.99 (or 179/180).

To raise it to a power indicates "tries".

Yes. Due to flux levels and exposure times we get about 350 "tries" on each image.

The roblem here is that ou are still attempting to apply a single event model to a multi-event system.

No. Please note that item #2 calculates the probability of two tracks existing.

So there is a 0.02 probability that your event that won't happen won't happen?

Yes. Which means there is a 0.02 probability that it will happen. In order to calculate the probability that an event will occur over multiple tries it is necessary to consider the probability it will not occur. We then "not" that result to produce the probability that it will. wow.joystiq.com...

Originally posted by Phage
reply to post by AnthraAndromda

 

We've been all through this.

Yes we have! And, yu still don't understand.

And this is the probability that a path will occur at a specific elevation angle. Not rhe probability that an intersecting path. I know you think you have, however, the length of "track" isn't important.

No. It is the probability that the path will occur over a range of elevation values. The length of the track is a function of the elevation of the path. With an elevation of 90º the track will be one pixel long. With an elevation of 2º the track will be 80 pixels long. This is why you need to specify a track length (or range of lengths). The length of the track affects the probabilities.

Yes, I'm sorry,over your narrow range. Elevation is not an issue here Phage. It quite simply does not matter!

" Probability of the first track (32-55 pixels in length) existing: 0.008" How is this possible since the probability of the path that creats the track is 0.0000078?

There are two component to the path, the vertical and the horizontal. To calculate the probabilities of a path creating a track only the vertical component is considered. The azimuth comes into play only if those two tracks exist. Without those two tracks there can be no intersection.

You still have not accounted to the original probability! If you insist on your 0.008, then we have to multiply it by the original probability: 0.0000078 * 0.008 = 0.0000000624. 1:16,025,641.02 The second path will have the same probability! (this has just degraded your hypothesis by three orders of magnetude

[quoteIn order for the "extended" tracks to intersect (as you did in your example) the only requirement is that the azimuths are different (otherwise the tracks are parallel). This means there are 359 degrees available for that intersection to occur. 359/360 = 0.99 (or 179/180).

It appears that you may be thinking that the detector is at the center; its not. Non-intersecting tracks are permitted. This wholly invalidates your "0.99".

No. Please note that item #2 calculates the probability of two tracks existing.

Yes, but incorrectly!

So there is a 0.02 probability that your event that won't happen won't happen?

Yes. Which means there is a 0.02 probability that it will happen. In order to calculate the probability that an event will occur over multiple tries it is necessary to consider the probability it will not occur. We then "not" that result to produce the probability that it will. wow.joystiq.com...


And here you revert to the single event model.


It is clear you don't have a handle on this. You continue to attempt to obfucate the model, Inject "innovative" math, enter "objects" (in the prograing sense), and method that have no application in this model. And when you state an authority for your math, it is a World of Warcraft drop formula, also of "innovative" ath. I've played WOW, that formula is absolutely useless, at least in game. While this equation may apply in "gaming theory", it does not apply here.

Oh well, Phage, you seem to have your fans, I'm sure they will accept anything you say, and reject the reality here. And, y'all think I need help!

reply to post by AnthraAndromda


After the dust settles from the probability debate, your argument still seems very similar to invalidating a winning lottery ticket based on the odds against winning.

reply to post by AnthraAndromda

If you insist on your 0.008, then we have to multiply it by the original probability

We are talking about the probability of a track being produced at all. We must start with that. The only factor which determines if a track is produced is elevation. Your "original probability" includes azimuth and elevation. We are only concerned with elevation. The probability of one track being produced is 0.008.Therefore the probability of two tracks being produced is 0.008^2 = 0.000064.

It appears that you may be thinking that the detector is at the center

No.

Non-intersecting tracks are permitted

But according to you it is the intersecting tracks which are of interest. From the OP:

That is; in all three of the second set of images there are two cosmic rays with intersecting vectors. This makes the whole proposition much less tenable.

It is far more probable that tracks will be intersecting than not intersecting because there is only one azimuth which will not produce an intersect, the azimuth which is identical.

And here you revert to the single event model.

You are doing the same thing (using incorrect values). You are attempting to calculate the single probability for two events occurring. I guess you don't realize that in doing so you are, in effect, calculating the probability of a single event. That event being the occurrance of two intersecting tracks.

While this equation may apply in "gaming theory", it does not apply here.

It has nothing to do with game theory but it has everything to do with probabilities just as rolling dice has everything to do with probabilities. Would an explanation from a programming point of view help? wpquestions.com...

Originally posted by Phage
reply to post by AnthraAndromda

 

We are talking about the probability of a track being produced at all. We must start with that. The only factor which determines if a track is produced is elevation. Your "original probability" includes azimuth and elevation. We are only concerned with elevation. The probability of one track being produced is 0.008.Therefore the probability of two tracks being produced is 0.008^2 = 0.000064.

No Phage, we are talking about "paths", ya know the "thing" that "makes the track". With out the path there is no trace. Elevation be damned And, the probability of a path is: 0.0000078. and it never gets any better.

But according to you it is the intersecting tracks which are of interest. From the OP:

This is the kind of activety that obfuscates this.

Yes, Phage it is the intersecting paths we are interested in. However, the probability of an intersection can not be greater than it components. Thus the only thing you do with this sort of obfiscation attempt is make the probabilities orders of magnitude less, and probably less than they should be. You simply can not leave out critical data in your equations, doing so will lead to failure.

It is far more probable that tracks will be intersecting than not intersecting because there is only one azimuth which will not produce an intersect, the azimuth which is identical.

No so! It is possible to draw many traces in that area where no traces are parallel, nor intersecting.

You are doing the same thing (using incorrect values). You are attempting to calculate the single probability for two events occurring. I guess you don't realize that in doing so you are, in effect, calculating the probability of a single event. That event being the occurrance of two intersecting tracks.

No Phage, you are mistaken yet again. What I am computing is the probability of two such events during an observation period. You lack of engineering is showing.

It has nothing to do with game theory but it has everything to do with probabilities just as rolling dice has everything to do with probabilities. Would an explanation from a programming point of view help? wpquestions.com...

Sigh, wrong! It doesn't really matter much about gaming theory stuff anyway. Yes it is a method to compute certain probabilities. But, as I said, it is an equation that applies to single event models, not multi-event.

I threw PHP away 12 years ago. It is a fairly nice scripting language, as far as scripting languages go. But, it IS a scripting language. Scripting languages have always been cumbersom, slow, insecure, and inadiquate. Hence I don't use them unless absolutely required. Though it matters little in this discussion. If ya want the same in "C" (any dialect) I'd be happy to give it ya.

You might also want to get a bit better authority for your math. World of Warcraft, and other, shall we say "lesser" authorities are beneith you. Try any of the universities that have this content on-line.

Oh wait, I've been using them for a long time, and I don't remember any of them supporting your "hypothesis", nor your equation.

reply to post by AnthraAndromda

The elevation determines the probability of a track being produced. An elevation between 71º and 90º will produce of "track" of 1 pixel. The probability of getting a "track" of one pixel is higher than the probability of getting a track of any other length. Elevation cannot "be damned".

You are correct about the probability of any particular path. If I were to guess that a path was going to have an elevation of 30º and an azimuth of 60º the odds are are indeed 1:128,205 of getting that right. But that azimuth only matters for the second track, the azimuth for the first track is irrelevant because we are trying to "match" it with the second one.

Let's try simplifying the problem using dice. What we are trying to do is the equivalent of rolling doubles. It doesn't matter what the first die is, as long as we match it with the second one. What are the odds? They are 1/6.

But what you are doing is the equivalent of calculating the odds of rolling snake eyes. What are the odds for that? (1/6)*(1/6) = 1/36. But we don't have to roll snake eyes. Snake eyes would represent and exact match of tracks. That is not what you are asking for.

It is possible to draw many traces in that area where no traces are parallel, nor intersecting.

Not if you are going to arbitrarily "extend" them as you have done. They will either be parallel or intersect.

reply to post by Phage


And, you are failing to understand that the probability for all paths is the same (0.0000078).

You state that the paths are either parallel or intersecting; but you fail to account for two paths of the same origin on diverging paths (I've seen one, posted it too).

I am not "trying to roll snake-eyes", I'm simply producing another path. The probability for all paths is the same, as I have already said. And the probability for a path does not change with elevation; it remains the same. If we were to take this discussion into the "gaming world" for just a moment; your attempts at changing the probabilities might be considered "cheating". Good thing this isn't a "gaming" discussion, ay?

The probability of any given path is fixed by the mechanics of the problem, yet you seem to think that because of these mechanics, the probabilities might be variable, and, of course, they can't be.

Now, a slight change of subject. I've been thinking about this "cosmic Ray thing, and the "flus", plus a few other things. It occured to me that we might be able to use software to turn this SOHO instrument into a comsic ray detector.

Would you agree that the occulting disk will block all light (photons)? If so, we may have an area of the image within which we can determing cosmic ray flux.

reply to post by AnthraAndromda

And, you are failing to understand that the probability for all paths is the same (0.0000078)

No. That is the probability of each path. The probability of all paths is 1.0.

I am not "trying to roll snake-eyes", I'm simply producing another path.

In effect you are. You are using the probability of producing one particular path. That's what the "1" in 1/360 represents, one degree out of 360. Just like the "1" in 1/6 represents one particular number on the die. The odds of rolling doubles is (1/1) x (1/6) = 1/6. The first probability is 1.0 because no matter what you do, you will get a number. The second probability is 0.167 because you have 1 chance out of 6 to get a number which matchs that first number. If you want to roll snake eyes (or any other specified double, the odds are (1/6) x (1/6) = 1/36. The point is, for doubles all you have to do is roll the first die. The combined odds do not depend on what that number is.

With the paths it's the same thing, all you have to do is have that first path. You say elevation doesn't matter. Ok, that's fine. That means the path can an elevation of any of 360 degrees (just like the number on the first die doesn't matter) and the odds of that are 360/360. If the elevation of the first path doesn't matter, neither does the azimuth (just like the number on the first die. So, just like with dice the probability of that first path is 1, there will absolutely, positively be a first path.

Now, whether or not that first path will produce a track of any specified length is a matter of different probabilities. If you want to consider a length of between 1 and 1024 pixels, those odds are still 360/360. What about the azimuth of the first path? Well if that path produces a track it will have an azimuth (if the length is greater than 1 pixel) but the azimuth of the first track doesn't matter, just like the number of the first die doesn't matter. So what are the odds of that track with a length that doesn't matter and an azimuth that doesn't matter? The odds are 1:1, just like the odds of that first die of the pair. If you're going for doubles, it doesn't matter what the first die comes up. If you're going for some relationship between tracks, it doesn't matter what the path was which produced the track. According to your definition, the odds of that first track are not (1/360) x (1/360), they are (360/360) x (360/360) = 1.

Would you agree that the occulting disk will block all light (photons)? If so, we may have an area of the image within which we can determing cosmic ray flux.

That would make sense. However sometimes a part of the compression routine is employed to delete the area of the disk from the data for bandwidth considerations. It only seems to be there in older images though. But it's easy to determine when it's in place because the pixels have a value of 0. BTW, if you're interested, in the FITS files the image data begins at 21C0. Each pixel has two bytes, making each pixel row 2048 bytes long.

Originally posted by Phage
reply to post by AnthraAndromda

 

And, you are failing to understand that the probability for all paths is the same (0.0000078)

No. That is the probability of one path. The probability of all paths is 1.0.

Perhaps you misunderstood. The probability for any path is 0.0000078. Although the probability for "all" paths is most certainly not 1.

In effect you are. You are using the probability of producing one particular path. That's what the "1" in 1/360 represents, one degree out of 360. Just like the "1" in 1/6 represents one particular number on the die. The odds of rolling doubles is (1/1) x (1/6) = 1/6. The first probability is 1.0 because no matter what you do, you will get a number. The second probability is 0.167 because you have 1 chance out of 6 to get a number which matchs that first number. If you want to roll snake eyes (or any other specified double, the odds are (1/6) x (1/6) = 1/36. The point is, for doubles all you have to do is roll the first die. The combined odds do not depend on what that number is.

No, again you have missed the mark. the probability for snake-eyes is 1:36, as you have shown. Unfortunately only a few words eariler you said it was 1:6 You are contradicting yourself.

Your understanding of probability is on par with a high schol student who has had his first day of instruction. I'm sorry Phage, but you do not understand probability, and have amply demonstrated that fact.

You have failed to construct an accurate model of the "problem", and now seem to be "searching" for something that will overturn my argument. Problem with that is, you won't find anything.

The real math speaks for itself: not all of those "anomalies" are cosmic rays. This conclusion is inescapable.

But, then again, not all those "anomalies" really exist either; but, that is perhaps better saved for another discussion, as, probability may have nothing to do with it.

reply to post by AnthraAndromda

The probability for any path is 0.0000078. Although the probability for "all" paths is most certainly not 1.

The probability for each particular path is 0.0000078. The probability for a cosmic ray to follow any of those 12,960 paths is 1.0. It must follow one of them, just as when you roll a die you must get a 1, 2, 3, 4, 5, or 6. The odds of getting any number on a die are 6/6. The odds of a cosmic ray following any path are (360/360) x (360/360). The probability is 1.0

No, again you have missed the mark. the probability for snake-eyes is 1:36, as you have shown. Unfortunately only a few words eariler you said it was 1:6 You are contradicting yourself.

No. That is not what I said. I said the odds of rolling doubles is 1:6. Doubles is any pair. Snake eyes is a particular pair. A particular pair. The odds of rolling the first ace is 1:6, the odds of rolling the second ace is 1:6. The odds of rolling both aces is 1:36. The odds of rolling a non-specific pair is 1:6 because the odds of rolling any number with the first die are 6/6. This is what you seem to have a problem grasping.

You have failed to construct an accurate model of the "problem", and now seem to be "searching" for something that will overturn my argument. Problem with that is, you won't find anything.

No. You don't understand the problem. I have been searching for a way to get you to understand the mistake you are making. You are calculating the odds for "snake eyes", not the odds for "doubles". You should be calculating the odds for "doubles". As a starting point.

reply to post by Phage


Phage, you are misunderstanding!

What you "think" I am calculating is not the probability for "snake-eyes", nor is it "doubles". I am calculating the probabilities of a multi-event model.

By the way: the probability for rolling doubles is still 1:36 whether you predefine the "doubles" you want or not. The probability is still 1:36 The first element has an equal probability.

reply to post by AnthraAndromda

What you "think" I am calculating is not the probability for "snake-eyes", nor is it "doubles". I am calculating the probabilities of a multi-event model.

Rolling dice is a multi-event model. Each die is an independent event. Each path is an independent event.

The first element has an equal probability.

That's what I said. There is only one definition for doubles; a pair of matching numbers. How is an equal probability expressed? In the case of dice it would be 6/6 which reduces to 1/1. And the odds of rolling any doubles is 1/1 * 1/6 = 1/6. Snake eyes (or any other particular) pair is a special case of doubles with reduced odds because you do not have equal probabilities of getting an ace with the first die.
www.thinkingbettor.com...

And there is an equal probability that a cosmic ray will follow any path. Remember? You said it yourself:

Did I somewhere imply that any given vector had a different probability than any other? I don't believe I did, but in the off chance I was interpreted that way; all vectors have equal probability.

www.abovetopsecret.com...

Originally posted by Phage
reply to post by AnthraAndromda

 

By the way: the probability for rolling doubles is still 1:36 whether you predefine the "doubles" you want or not. The probability is still 1:36 The first element has an equal probability.

That's what I said. There is only one definition for doubles; a pair of any two numbers. And the odds of rolling doubles is 1/1 * 1/6 = 1/6. Snake eyes is a special case of doubles with reduced odds because you do not have equal probabilities of getting an ace with the first die.

So which is it? 1:6 or 1:36?

This is an example of your broken logic, Phage. This sort of thing can not exist; it is either 1 or the other. You should take it to a math teacher, perhaps when he tells you the same thing as I, you will begin to listen.

And there is an equal probability that a cosmic ray will follow any path. Remember? You said it yourself:

Did I somewhere imply that any given vector had a different probability than any other? I don't believe I did, but in the off chance I was interpreted that way; all vectors have equal probability.

www.abovetopsecret.com...

edit on 8/15/2012 by Phage because: (no reason given)

Yes, I did. And the probability for any path (1, 2, 3, ..., n) remains the same. 0.0000078. And yes yu did state that some paths have greater probability. (it might have been closer to an implicatin, but, it seemed rather "overt" to me)

So; all that I have said still stands. the probabilities have become wholly untenable. And, not all of these things are cosmic rays. Despite what you may want, the Universe doesn't care if you are comfortable. And, reality remains reality. Your concepts of this, the logic you are using, quite simply, don't work.

You remind me of myself; 40 years ago. Thankfuly, I've learned more since then.

reply to post by AnthraAndromda

So which is it? 1:6 or 1:36?

Why do i have to repeat myself?
The odds for getting snake eyes is 1:36.
The odds for getting any doubles at all is 1:6.

For some reason, a lot of people have trouble grasping that concept. The chances of rolling doubles with a single toss of a pair of dice is 1 in 6. People want to believe it’s 1 in 36, but that’s only if you specify which pair of doubles must be thrown.

www.thinkingbettor.com...

And yes yu did state that some paths have greater probability. (it might have been closer to an implicatin, but, it seemed rather "overt" to me)

No one will accuse you of having an underactive imagination. I neither said nor implied any such thing.

Your concepts of this, the logic you are using, quite simply, don't work.

You are quite simply, wrong.