second+quarter

To start off the fourth quarter, I am finishing up my model of the CMS detector with the hole, and then I am going to use the tools to find the angle of the center of the detector to the top of the side of the hole. Here are some pictures of my project:



I restarted the project, and made the CMS detector a circular shape which is much more realistic. I made some assumptions in making this model, like where exactly the CMS detector is under the hole. If I run into problems I can look at the picture to make another assumption.

Now that I have made the new detector in sketch-up, I am going to try and find out the angles at which the cosmic rays can come into the detector. Here is a picture of the length of the center of the detector all the way up to the top of the hole.

--> As shown in the picture above, the length from the middle of the detector to the top of the tunnel is around 45.39 meters. By looking at this, now all I need to do is find the length from the center of the CMS detector to the edge of the other side of the top of the tunnel. With this sketch-up tool, I can do this, and then use trigonometry to find the angle of where the cosmic rays are coming in. I now found the top length, and can use an inverse tangent idea to find the angle.

EQUATION--> x=tan-1(opposite/adjacent) x=tan-1(45.39/23.74)-->62.39 degrees. Now after doing this, I have found that the angle at which the cosmic rays are coming into the center of the detector is 62.39 degrees. This is the degree of elevation. The degree of depression would then be 27.61 degrees. This is the angle from the sky at which the cosmic rays are coming from the edge of the hole and going through the center. observations- When looking at the cosmic rays passing through the middle of the detector, I remember seeing them come from all sides, and even up and down. How is this possible if it is just coming from the tunnel?

Now that I have found the angles, I have started to look at some things to help me out with the angles. I am starting to look at pseudorapidity, and I am putting together what eta is compared to the angles of the particles coming out of the detector. To start off, I found a very useful chart of pseudorapidity versus degree, and this will be helpful when working through this next step for a while.

Now using a graph that I made at the beginning of last year, I am going to try to use the pseudorapidity versus degree graph to find out the maximum and minimum of the degrees at which the cosmic rays are coming into the detector. I am also going to be looking for a formula for which I can get the precise measurements.



These are the two equations I have found online that can be used to find the precise angle or pseudorapidity of the particles. When looking at the graph above, eta looks to be around .2 and -.2. This can give us a good idea with pseudorapidity, and so I can put it into the equation above to get the degrees.

After plugging the numbes into the equation, I found that 2tan-1(e^-.2) is equal to 78.61 degrees. With this number, I can now use the angle in google sketchup to make a cone shape coming from the very edge of the top of the tunnel. In doing so, I am finding the spray of cosmic rays that are coming down onto the detector from the opening at the top of the CMS. When doing a rough sketch of the design and then looking for the angle, I came up around 10 degrees short, with around 65 degrees. This is somewhat accurate, but I would like to make touch ups to the detector to make the error smaller.

46.74 meters long opposite* 37.65-(25/2 meters from half the detector) --> meters on top ajacent* =

tan-1(46.74/25.15)= 62 degrees

Now I am adding in a cylinder to see what th maximum and minimum angles for the cosmic rays coming through the detector will be. The outer limit seems to be 1 and -1 for n. 2tan-1(e^1) is equal to 140 degrees, and 2tan-1(e^-1)= 41.5 degrees. Using this, I can now find the radius of the end of the cylinder.

Because we know the angles, we can say that 41.5=tan-1(45.59/x) and also put in 140 degrees for 41.5 as well. 45.59 m comes from the height of the middle of the detector to the top of the hole, and x represents the distance horizontally. the horizontal distances are 51.53 meters and 54.33 meters. This is where any of the cosmic rays can be coming in.

With the n being smaller for the clump, there is a -.1 to .3 difference in eta. When changing the pseudorapidity into degrees, you get 58 degrees, along with 73 degrees. I then found the distance horizontally, which came out to be 28 meters and 17 meters away from the middle of the detector, which agrees with the 22 meters in the middle that we have found.

This is a graph of the final product. Here the purple cylinder is the bulk of the cosmic rays coming in, while the large cylinder is showing the areas at which all of the cosmic rays are coming from.

Below are some pictures of the cylinders being put into the right position.