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So that kind of beggs the question; "How much did you process that image before you posted your thumbnail?
I did nothing to the thumbnail image except convert the FITS file to a TIF so it could be uploaded to ATS.
I don't see anything odd. The wide dynamic range makes it necessary to adjust the levels. You might find this program more useful than Gimp. It's better suited to the wide dynamic range.
I think we have a problem. In the image you referenced above, and provided a link to sees a bit odd.
My apologies, when you said thumbnail I thought this is what you meant:
So ,,, you did not post "raw" data, you posted highly processed data.
The dynamic range of each pixel.
Dynamic range of what? The hardware out there will handle 256 levels of gray, that's it.
You're thinking wrong. As I said, background subtraction is processing which is performed on the FITS data. It is not performed by the CCD.
if it's what I'm thinking of, is an automatic process performed by the CCD to eliminate any substrate noise it may have.
There is no "added" data in the source but could very well be data loss in the processed images. As I said, I didn't know what you were going to do so I had no way of knowing what effect it would have on your plans. All I did was provide some information which you obviously did not have.
I also doubt that there will be enough "added" data to make a difference. So, in the end we will only have spent a lot of time producing what we already have.
Apparently you don't. You don't know what process of background subtraction is or involves.
Oh, but I do understand the nature of that processing. It brings up the impossibly weak "signals" so that we can work with them.
Cosmic rays can easily penetrate the entire thickness of the CCD. They produce a track in the photoactive layer no matter which side they originate from.
You will notice that my calculation is different from yours; cosmic rays must interact with the sensing portion of the device to be recorded, thus those entering from the back aren't "seen" as they do not reach the sensing areas
SOHO was launched in 1995.
60 microns!?! Seriously? Soho isn't that old. A modern device has an active layer more on the order of 0.2 to 0.3 nanometers, much less than you 60 microns (60,000nm).
That would seem to be an oxymoron. I suspect I'm missing your point.
I suspect your numbers are reasonably accurate, given the huge error.
Not according to experimental evidence (as provided). I read it to show the probability of a track of 32 to 55 pixels in length as being 0.008. That matches my figures.
However, your estimation of the probability is wholly inaccurate.
No, it doesn't. There are a range of values for elevation which will produce a track, not a just a single radian. This increases the probability that a track will be produced beyond 1/360 (or 1/180, if that's what you used). You have not produced the odds for the "correct" path, you have produced the odds for matching path, a path of the same length and azimuth.
The probability of a cosmic ray having the "correct" path remains the same 1:255,102.04
Only if the first track is there can we then calculate the probabilty that the second one will meet the required conditions. How can a single track be "special"? It's like calculating the odds of rolling snakeyes with one die. It's a null value. An again, the odds you calculated are for a matching path, not an intersecting path.
So yes we have two "tracks" each with the sae probability, which hasn't changed from it's original 1:255,102.04 (your wee song and dance was mostly deleted
Vector is incorrect, azimuth is not incorrect because the azimuth is the direction of the path. Yes we are working on a single plane. The vertical plane only determines the length of the track. You made no mention of the importance of the length of the track therefore we are only dealing with the azimuth of the track. The azimuth is the direction of the path in the horizontal plane so it is the correct term but path will also work. In order to intersect the only requirement is that the azimuths of the two paths are different.
Actually "vectors" and "azimuths" are both incorrect, we shall call them "paths". And no we are not working on a single plane.
Not really, it's actually a measure of the level of radiation. But flux levels will affect the odds of any particular track occuring. The more events, the more likely the event is to occur. The more times you roll the dice the better the odds of getting snake eyes.
You like the term "flus" which is really just anther way of quantizing the probability.
Interesting. I assume that you were counting paths rather than strikes. What was the minimum path length you used? This would relate to the question of the definition of "special". But if you used the browse images you would have missed (at the minimum) any strikes behind the occulter disk and outside the optical path. On the C2 images that amounts to about 10% of the image.
This value was found by "counting" cosmic ray strikes on 86 SOHO images
Yes. I am aware of that and that was a blunder on my part. The problem came from the way I applied the number of hits to the odds of an intersecting track occurring. The proper formula would be 1-(1-p)^n, where p is the probability and n is the number of "tries". This revises the probability for an intersecting track (assuming that there is a track for it to intersect) to 0.938. Close to certainty but not quite because; a) there must be at least two tracks of the required length (between 32 and 55 pixels) and b) the pair must not have the same azimuth.
You are aware that it is not possible to have probabilities greater than 1, right?
The proper formula would be 1-(1-p)^n, where p is the probability and n is the number of "tries".
The formula for the probability that a single event of probability p will occur given multiple "tries". Simplest case scenario:
And this is the formula for what exactly?
Originally posted by Phage
reply to post by AnthraAndromda
The formula for the probability that a single event of probability p will occur given multiple "tries". In the case of cosmic ray tracks we have a probability 0.0079 for an intersecting azimuth. We have 350 "tries" in a given frame. The probability of getting at least one intersecting azimuth:
And this is the formula for what exactly?
1-(1-0.0079)^350 = 0.938
The formula is correct but the variables you used are not. Again, you are calculating the probability for an exact duplication of the first track, this is not a requirement for your definition of an event. Your probability for the first track is also too restrictive since there is a range of values for the elevation of the vector.
Thus the joint probability becomes; P(A and B) = P(A) * P(B). Giving us the results I've already posted.
Originally posted by Phage
reply to post by AnthraAndromda
I understand the point. I look at it this way; If there is no "first" track at all there can be no relationship between two tracks. It is a null event. So calculating the odds doesn't really make sense in my point of view. But, if you like, we can include the odds for the first track occurring. It does reduce the odds but not to an extreme degree.
Thus the joint probability becomes; P(A and B) = P(A) * P(B). Giving us the results I've already posted.
The formula is correct but the variables you used are not. Again, you are calculating the probability for an exact duplication of the first track, this is not a requirement for your definition of an event. Your probability for the first track is also too restrictive since there is a range of values for the elevation of the vector.
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
I think I understand part of the reason for your incredulity. The trouble is that you have not rigidly defined the requirements of a "special" event. As I have pointed out these are the probabilities for two tracks of lengths between 32 and 55 pixels ...
So we should only see a cosmic ray track once in 160 days of images?
1. The probability of a event path is still: 0.00006241 or 1 in 16,023.073225444640282006088767826.
That translates to one in about 7,000 years of images. Yet intersecting tracks are easy to find.
3. The probability of two cosmic rays intersecting is: 0.0000000038950081 or 1 in 256,738,875.5879
You are attempting to apply your definition of "special event" prematurely. At this point in the discussion, we have not reach a point where any of that even matters. Yet you insist on applying it; incorrectly.
So what do you do rather than reanalyzing your approach? You conclude that they must not be cosmic rays. The problem is your math is wrong and your science is none.
You are attempting to apply your definition of "special event" prematurely. At this point in the discussion, we have not reach a point where any of that even matters. Yet you insist on applying it; incorrectly.
I had to do so since you haven't. I used the three images you provided (still no dates and times) to try to deduce what is is you call "special".
So we should only see a cosmic ray track once in 160 days of images?
1. The probability of a event path is still: 0.00006241 or 1 in 16,023.073225444640282006088767826.
No, Phage, my Math is correct, and my science is as well.
Originally posted by Phage
reply to post by AnthraAndromda
So we should only see a cosmic ray track once in 160 days of images?
1. The probability of a event path is still: 0.00006241 or 1 in 16,023.073225444640282006088767826.
No, Phage, my Math is correct, and my science is as well.
Sure it is. I think you should report your findings to Joseph Gurman.
I'm sure he would be fascinated to discover that what he has be thinking are cosmic rays can't be because your math proves it.
Originally posted by ScientificUAPer
AnthraAndromda, I think Phage does not need a teacher to argue his case convincingly, also in this thread.
Instead, the problem is that your stuff got pwned beyond recovery already on page 2, but you are seemingly incapable of understanding that, or psychologically incapable of admitting it.
You shan't be allowed to conclude that your case is as credible as Phage's when Phage has demonstrated he is correct, and you are not.
Let it go and move on.
edit on 14-8-2012 by ScientificUAPer because: typo
Originally posted by AnthraAndromda
..
But, I'll tell you what; get someone who is at least as expert as I, and we will see, what we will see.
Originally posted by ScientificUAPer
Originally posted by AnthraAndromda
..
But, I'll tell you what; get someone who is at least as expert as I, and we will see, what we will see.
Your stuff got busted, end of story. The fact that you claim to be an expert doesn't make it any better.
But, I'll tell you what; get someone who is at least as expert as I, and we will see, what we will see.
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?