Jesterface23 Posted December 22, 2015 Share Posted December 22, 2015 Here is the math I have on CME arrival times. The one thing that all of the full halo CMEs have in common is this, 'y=(1/(40x+28500))(x+1765)^2-100' (or at least really really close). First, you need to get three times, 1st the time when the CME exits the sun in a SDO AIA 131 image, 2nd is the time of when the CME edge first hits the outer circle in the LASCO C3 image, and the 3rd from that point where the CME first hit the outer edge of the circle in the C3 image, go around the outer circle 180 degrees and get the time of when the CME hits that point. Then the math, get the difference of time #2 and #3 and write that down as time #4, get the time that is between time #2 and #3 and write that down as time #5, then get the difference of time #1 and time #5 and turn that into all minutes and write it down as number #6. Next, turn time #4 into minutes and put that number as x into the equation y=(1/(40x+28500))(x+1765)^2-100 and you now have #7. Now multiply #7 by #6 and you have the estimated travel time in minutes, just change that to days hours and minutes and add it to time #1. The issues are mainly if the image isn't clear like CME #1 a week ago, or there is a big chunk or missing images, then you are left to guessing the time of when it gets to the edge of the C3 circle edge and the time will likely be way off. Currently what I have is for CMEs where you need to guess one of either time #2 or #3 is likely +-3 hours off the arrival time and perfectly clear CMEs are at +-1.5 hours off the arrival time. For the equation, there is a lack of data before x=80 and after x=580, so I just went with the pattern and guessed and will need to modify the equation once there are some clear events to get that data from. The next thing I want to work on is get the arrival time from inside of the C3 image circle and not need to get times #2 and #3 from faint CMEs all the way at the edge of the C3 circle. Link to comment Share on other sites More sharing options...
Marcel de Bont Posted December 23, 2015 Share Posted December 23, 2015 I assume your math if flawless and invite you to predict the next CME impact. The problem is however, between the outer edge of LASCO C3 and ACE at L1... how much will the CME slow down? Still a lot of variables we can't know. It's not easy. Link to comment Share on other sites More sharing options...
Jesterface23 Posted December 23, 2015 Author Share Posted December 23, 2015 (edited) Well, currently the equation was made off of the average of 18 data points, so it should be fairly close like with the arrival time like with CME #2 a week ago at 1 hour and 20 minutes off. You did give me an idea though, I need to go back and through those CMEs again and write down any other possible info around the event then see if there are any other patterns with it. Edit: Started work on version 3. One thing I completely forgot about in making this version and the first is that the earth is a moving target, then I'll need to see if there is a pattern in the density once I get all of the data. Edited December 26, 2015 by Jesterface23 Link to comment Share on other sites More sharing options...
Jesterface23 Posted December 28, 2015 Author Share Posted December 28, 2015 (edited) Looks like a quick one, it looks fairly close to the last CMEs direction, so I'm not making any estimated changes. Arrival time looks like somewhere around December 30th at 3:36z and it looks like the possible error time is just over +-3 hours. Then an update on version 3, I believe I have the math done to get the CME direction, now I just need to figure out how to make a chart off of that to get the CME speed, then make another chart for impact times at the earths orbit. Edited December 28, 2015 by Jesterface23 Link to comment Share on other sites More sharing options...
Marcel de Bont Posted December 28, 2015 Share Posted December 28, 2015 I haven't attempted yet to determine an impact time but that sounds incredibly quick. How did you come to this conclusion if you don't have a CME speed? Link to comment Share on other sites More sharing options...
Jesterface23 Posted December 28, 2015 Author Share Posted December 28, 2015 Its basically cross multiplying fractions Link to comment Share on other sites More sharing options...
Guest Engelsman Finland Posted December 28, 2015 Share Posted December 28, 2015 Is that UT? Or GMT? Link to comment Share on other sites More sharing options...
Jesterface23 Posted December 28, 2015 Author Share Posted December 28, 2015 (edited) It would be GMT or UTC+-00:00. And one of the points I am going off of is in the area of the very faint halo in C3 near the top side to the left side, if that actually does count as a halo. Edit: I am going to check back through the C3 images and see if there were any big changes recently. Another Edit: Well, either it is fast or I am looking at something that looks like the halo but isn't the halo. Guess time will tell lol. Edited December 28, 2015 by Jesterface23 Link to comment Share on other sites More sharing options...
Waldo Hazeleger Posted December 28, 2015 Share Posted December 28, 2015 The NASA model thinks it is fast. Impact between 30/1800 and 31/0000 Link to comment Share on other sites More sharing options...
Guest Engelsman Finland Posted December 28, 2015 Share Posted December 28, 2015 Looks like a happy new year... Link to comment Share on other sites More sharing options...
Waldo Hazeleger Posted December 28, 2015 Share Posted December 28, 2015 And Noaa thinks about the same. Link to comment Share on other sites More sharing options...
Guest Stephane Mabille Posted December 29, 2015 Share Posted December 29, 2015 Link to comment Share on other sites More sharing options...
Marcel de Bont Posted December 29, 2015 Share Posted December 29, 2015 Jesterface23 might be onto something. NOAA goes for 11:00 UTC tomorrow on their ENLIL model. Will be interesting to see who will be closest with their predictions. Stephane: that model shows the polarity of the IMF. That has nothing to do with the Bz. Link to comment Share on other sites More sharing options...
Jesterface23 Posted December 30, 2015 Author Share Posted December 30, 2015 Welp, there goes my time. But now going off of NOAA's ENLIL model, if it did add around 5 hours from the CME plowing through the denser plasma, it would be closer to their time at around 8:36z +-3 hours. Link to comment Share on other sites More sharing options...
Marcel de Bont Posted December 30, 2015 Share Posted December 30, 2015 I just wonder what kind of speed you used for your calculations. It seems NOAA used a similar speed as you did as your and NOAA's arrival time were pretty similar. I am curious to what speed you got with your methods. Link to comment Share on other sites More sharing options...
Jesterface23 Posted December 30, 2015 Author Share Posted December 30, 2015 (edited) What I was going off of is one point hits the edge of the image at around 16:30, then a very faint halo hits the other end at around 18:06. Now looking back the real 2nd point may have been somewhere around 3:42 when the halo was more clear instead of 18:06. So off of that it might be a far side hit instead of a near direct hit. A new arrival time could be somewhere around the 31st mid day UTC but is very unclear from lack of data. Edited December 30, 2015 by Jesterface23 Link to comment Share on other sites More sharing options...
Guest James Noyes Posted December 30, 2015 Share Posted December 30, 2015 Just out of interest Jesterface have you considered that the CHS from CH707 is it may be pushing the CME along in your speed calculations? I saw no one had mentioned to I thought i'd ask. Link to comment Share on other sites More sharing options...
Jesterface23 Posted December 30, 2015 Author Share Posted December 30, 2015 The velocity and density in the area are two things I am going to look at when making the next version. Link to comment Share on other sites More sharing options...
Jesterface23 Posted January 24, 2016 Author Share Posted January 24, 2016 I think I found one of the big issues. There are no big changes to the C3 camera, there is just the change to keep the field of view 32 solar radii, so that is ruled out for there being a big difference between one line of data points over a second line of data points. What looks to be the thing that is making the big difference is the top line of data points are a direct hit of a CME and the line below is a side hit. So I guess now I need to go back, get more data, and find out more about the lines in the CME lol Link to comment Share on other sites More sharing options...
Jesterface23 Posted February 12, 2016 Author Share Posted February 12, 2016 (edited) First real test run of the new second equation. The second point in C3 seems to fall somewhere near the data gap, but I'm going with an Earth arrival time at around 9:30UTC on the 15th. Not sure of the +/- on it, I just want to see how far off it ends up. Edited February 12, 2016 by Jesterface23 1 Link to comment Share on other sites More sharing options...
Guest Engelsman Finland Posted February 13, 2016 Share Posted February 13, 2016 Does any one have the speed of it yet? It's great being able to do the math. But when I am watching the data I look to match or look for a spike in the wind speed. 483 kms. Link to comment Share on other sites More sharing options...
Marcel de Bont Posted February 13, 2016 Share Posted February 13, 2016 It should be pretty obvious when the CME arrives. The EPAM protons should spike and you will likely see a distinct increase in the solar wind speed and the IMF Bt. Link to comment Share on other sites More sharing options...
Jesterface23 Posted September 20, 2017 Author Share Posted September 20, 2017 That latest crazy sunspot region pushed me back into space weather a bit more. I ended up redoing the formula below to try and make it work at any angle or distance (not perfect yet) and I am hoping someone might be able to answer a question. In the formula, x is difference in hours from when the CME first hits the 30 solar radii edge of the C3 imagery and the second point when the back of the CME hits the edge. The m is the multiplier and this is where things get tricky. A multiplier 1 is most common before the first trough in the formula and 0 is most common after the first trough. But that CME on the 10th made it into a rare category with a multiplier of 2. I haven't had much time to do more research, but I've only had 3 other data points in that multiplier. This is where I would have my question, would anyone possibly know what might be the cause of those multiplier and how to spot it? I thought it might have been a direct hit vs side hit, but going back though some CMEs I don't think that adds up. Thanks y = cos(120x/213) - (((120x/213)/4.9358)^(pi/2)) + (((sqrt(pi/2)^m)*0.68*213)/30) + (pi/2) Link to comment Share on other sites More sharing options...
Jesterface23 Posted October 4, 2017 Author Share Posted October 4, 2017 I have around 100 useable CMEs with clear enough C2 and C3 imagery archived now, so hopefully that will be enough to finish up this project haha. I have another guess on that might be causing those 3 different multipliers, and there might actually be a 4th lower multiplier. Guess I'll find out once I get the time to go through everything. 1 Link to comment Share on other sites More sharing options...
Jesterface23 Posted November 29, 2017 Author Share Posted November 29, 2017 I did a bit of thinking today and I think I know where that number sqrt(pi/2) comes in to play with looking at some side views of CMEs. In the image the outer lines is the Lasco C3 field of view and the yellow dot is the sun. Now this would be a perfect world CME, the full ring would be the CME shock wave (if there is a name for it, I don't know it) from launch but not much of the shock wave would make it to the other side with the sun in the way. That leaves the part that is the main part of the CME. In the imagery the distance traveled of the shock wave times sqrt(pi/2) looks really close to the distance traveled by the leading edge of some CMEs. That is where the multiplier from the formula comes from, but I'll still need to go through all the data to figure out what to look for to get one of the multipliers. Link to comment Share on other sites More sharing options...
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