![\"\"](\"https://i.ibb.co/QjFs65D/Representation-of-azimuth-and-zenith-angles.png\")
I have some questions/clarifications about this topic as I struggle to visualize things like this.
\nLet me give it a go though and hopefully, someone can help.
\n
The simple diagram above makes sense to me.
\n\n\nx = cos(azimuth) * sin(zenith)\ny = sin(azimuth) * sin(zenith)\nz = cos(zenith)\nPlease let me know if anything above is incorrect. But assuming it's all good, I can see that to be able to determine the Azimuth angle, we are required to know the relative position of the Cardinal North direction.
\nTake this example:
\nFor this competition, it's unclear to me where the IceCube detector is located and if that's important in developing an understanding here. FYI, I know I'm probably over-complicating this. I checked it out on Google Earth and looks to be slightly North by North-East of the South Pole. Please see the photo below:
\n![\"\"](\"https://i.ibb.co/6HF1YYr/Screenshot-2023-01-26-at-8-41-47-PM.png\")
Is my understanding correct? Obviously then, the detector itself has it's own relative vector pointing North, but every sensor has its own independent relative vector pointing North. At this point, I am now thoroughly confused, haha.
\nThe other thing that throws me a bit when I think about all this is that the local zenith is coincident (but in opposite directions) to the Axis of the Earth and therefore the North Pole. I don't think it matters though because for Zenith we only care about elevation and for Azimuth we only care about the relative direction of the Neutrino and the North Cardinal Direction.
\nOut of curiosity, would this mean that a Zenith of 180 degrees and an Azimuth of 0 degrees would be a Neutrino that travels a path that takes it straight through the North Pole, along the Earth's axis, and straight out the South Pole?
\nSorry if this is all incoherent, as I said above, I can struggle to visualize things properly so I may have completely messed something up. Any help/support would be greatly appreciated.
\nps. I realize my image uses the Sun, but obviously, in this competition instead of the sun, it would be the origin of the Neutrino that the vector points to.
\n[EDIT:] Additional image for helpful context:
\n![\"\"](\"https://i.ibb.co/vcLHzcB/Screenshot-2023-01-27-at-10-16-21-AM.png\")
I have some questions/clarifications about this topic as I struggle to visualize things like this.
\nLet me give it a go though and hopefully, someone can help.
\n
The simple diagram above makes sense to me.
\n\n\nx = cos(azimuth) * sin(zenith)\ny = sin(azimuth) * sin(zenith)\nz = cos(zenith)\nPlease let me know if anything above is incorrect. But assuming it's all good, I can see that to be able to determine the Azimuth angle, we are required to know the relative position of the Cardinal North direction.
\nTake this example:
\nFor this competition, it's unclear to me where the IceCube detector is located and if that's important in developing an understanding here. FYI, I know I'm probably over-complicating this. I checked it out on Google Earth and looks to be slightly North by North-East of the South Pole. Please see the photo below:
\n
Is my understanding correct? Obviously then, the detector itself has it's own relative vector pointing North, but every sensor has its own independent relative vector pointing North. At this point, I am now thoroughly confused, haha.
\nThe other thing that throws me a bit when I think about all this is that the local zenith is coincident (but in opposite directions) to the Axis of the Earth and therefore the North Pole. I don't think it matters though because for Zenith we only care about elevation and for Azimuth we only care about the relative direction of the Neutrino and the North Cardinal Direction.
\nOut of curiosity, would this mean that a Zenith of 180 degrees and an Azimuth of 0 degrees would be a Neutrino that travels a path that takes it straight through the North Pole, along the Earth's axis, and straight out the South Pole?
\nSorry if this is all incoherent, as I said above, I can struggle to visualize things properly so I may have completely messed something up. Any help/support would be greatly appreciated.
\nps. I realize my image uses the Sun, but obviously, in this competition instead of the sun, it would be the origin of the Neutrino that the vector points to.
\n[EDIT:] Additional image for helpful context:
\n
For the data we give you, the two angles are simply spherical coordinates defined in the same coordinate system as the sensor geometry is defined in. So this means it is entirely independent of the location of IceCube.
\nIn order to do physics and then map these values to actual sky coordinates, or for example galactic coordinates, we have existing functions in icecube, so no need from your side to worry about that part. :)
", "rawMarkdown": "For the data we give you, the two angles are simply spherical coordinates defined in the same coordinate system as the sensor geometry is defined in. So this means it is entirely independent of the location of IceCube.\n\nIn order to do physics and then map these values to actual sky coordinates, or for example galactic coordinates, we have existing functions in icecube, so no need from your side to worry about that part. :)", "votes": null }, { "id": "2117854", "postDate": "01/27/2023 15:31:17", "content": "Oh! I see! Thanks for the response @pellerphys.
\nSo the center of the sensor array (IceCube) is the origin of the Azimuth and Zenith used in this competition? Then the 'local zenith' is always the z-axis (Positive coincident with South-Pole pointing to space)?
\nAnd the XY plane is located at the halfway depth w.r.t. the sensor array area, ie. 1950m down(not the overall area – considering the 1KM w/ no sensors, 1225m down)? Or would the XY plane be at the surface?
\nDoes the picture below make sense (it assumes roughly the XY plane is at the halfway point w.r.t. Sensor Array volume)? Also, I THINK the deep-core sensors are in the center… so my Z vector may be in the wrong location and should be centred within the deep-core sensor area.
\n![\"\"](\"https://i.ibb.co/28Pg6Cn/icecube-drawio.png\")
Yes, that looks about right to me 👍
\nThe origin [0,0,0] is roughly in the middle of all sensors. But keep in mind that direction vectors are invariant under translations anyway. These vectors really only encode the direction, and not the position of the neutrino.
", "rawMarkdown": "Yes, that looks about right to me 👍\n\nThe origin [0,0,0] is roughly in the middle of all sensors. But keep in mind that direction vectors are invariant under translations anyway. These vectors really only encode the direction, and not the position of the neutrino.", "votes": null }, { "id": "2120616", "postDate": "01/29/2023 18:02:02", "content": "Thx for thoses explanations. I was also completely lost with the coordinates system. But still 2 questions:
\nYes, all parallel paths will have identical zenith and azimuth values, there is no positional information!
", "rawMarkdown": "Yes, all parallel paths will have identical zenith and azimuth values, there is **no** positional information!", "votes": null }, { "id": "2129877", "postDate": "02/05/2023 01:08:43", "content": "Adding as a reference a a similar question asked before this one which also has some relevant responses…
\nhttps://www.kaggle.com/competitions/icecube-neutrinos-in-deep-ice/discussion/379857
I truly do not want to confuse the situation with my possible ignorance in this area and this is my first post on Kaggle.
\nIf a Zenith angle is defined from 0 to 90 degrees and the Azimuth angle is from 0 to 360 degrees then any vector is fully defined that passes through the origin.
\nFor example, at sunrise, it could be Zenith 90 degrees and Azimuth 90 degrees and at sunset, it could be Zenith 90 degrees and Azimuth 270 degrees (need a 180-degree difference between sunrise to sunset Azimuth). High noon Zenith is 0.
\nIt appears from the links below the above is one accepted standard.
\n\nhttps://en.wikipedia.org/wiki/Solar_azimuth_angle
", "rawMarkdown": "I truly do not want to confuse the situation with my possible ignorance in this area and this is my first post on Kaggle.\n\nIf a Zenith angle is defined from 0 to 90 degrees and the Azimuth angle is from 0 to 360 degrees then any vector is fully defined that passes through the origin. \n\nFor example, at sunrise, it could be Zenith 90 degrees and Azimuth 90 degrees and at sunset, it could be Zenith 90 degrees and Azimuth 270 degrees (need a 180-degree difference between sunrise to sunset Azimuth). High noon Zenith is 0.\n\nIt appears from the links below the above is one accepted standard.\n\nhttps://sinovoltaics.com/learning-center/basics/elevation-angle/#:~:text=Sunrise%2FSunset,as%20refracted%20by%20the%20atmosphere.\n\nhttps://en.wikipedia.org/wiki/Solar_azimuth_angle", "votes": null }, { "id": "2139907", "postDate": "02/11/2023 09:21:37", "content": "Neutrinos can also come from below the horizon, and therefore zenith needs to go from 0 all the way to 180 to cover the full sphere.
", "rawMarkdown": "Neutrinos can also come from below the horizon, and therefore zenith needs to go from 0 all the way to 180 to cover the full sphere.", "votes": null } ], "comments": [ { "id": 2117820, "author_name": "pellerphys", "author_url": "", "post_date": "01/27/2023 15:11:40", "content": "For the data we give you, the two angles are simply spherical coordinates defined in the same coordinate system as the sensor geometry is defined in. So this means it is entirely independent of the location of IceCube.
\nIn order to do physics and then map these values to actual sky coordinates, or for example galactic coordinates, we have existing functions in icecube, so no need from your side to worry about that part. :)
", "votes": null, "replies": [ { "id": 2117854, "author_name": "dschettler8845", "author_url": "", "post_date": "01/27/2023 15:31:17", "content": "Oh! I see! Thanks for the response @pellerphys.
\nSo the center of the sensor array (IceCube) is the origin of the Azimuth and Zenith used in this competition? Then the 'local zenith' is always the z-axis (Positive coincident with South-Pole pointing to space)?
\nAnd the XY plane is located at the halfway depth w.r.t. the sensor array area, ie. 1950m down(not the overall area – considering the 1KM w/ no sensors, 1225m down)? Or would the XY plane be at the surface?
\nDoes the picture below make sense (it assumes roughly the XY plane is at the halfway point w.r.t. Sensor Array volume)? Also, I THINK the deep-core sensors are in the center… so my Z vector may be in the wrong location and should be centred within the deep-core sensor area.
\n
Yes, that looks about right to me 👍
\nThe origin [0,0,0] is roughly in the middle of all sensors. But keep in mind that direction vectors are invariant under translations anyway. These vectors really only encode the direction, and not the position of the neutrino.
", "votes": null, "replies": [ { "id": 2120616, "author_name": "aristhene", "author_url": "", "post_date": "01/29/2023 18:02:02", "content": "Thx for thoses explanations. I was also completely lost with the coordinates system. But still 2 questions:
\nYes, all parallel paths will have identical zenith and azimuth values, there is no positional information!
", "votes": null, "replies": [] } ] } ] } ] } ] }, { "id": 2129877, "author_name": "jirkaborovec", "author_url": "", "post_date": "02/05/2023 01:08:43", "content": "Adding as a reference a a similar question asked before this one which also has some relevant responses…
\nhttps://www.kaggle.com/competitions/icecube-neutrinos-in-deep-ice/discussion/379857
I truly do not want to confuse the situation with my possible ignorance in this area and this is my first post on Kaggle.
\nIf a Zenith angle is defined from 0 to 90 degrees and the Azimuth angle is from 0 to 360 degrees then any vector is fully defined that passes through the origin.
\nFor example, at sunrise, it could be Zenith 90 degrees and Azimuth 90 degrees and at sunset, it could be Zenith 90 degrees and Azimuth 270 degrees (need a 180-degree difference between sunrise to sunset Azimuth). High noon Zenith is 0.
\nIt appears from the links below the above is one accepted standard.
\n\nhttps://en.wikipedia.org/wiki/Solar_azimuth_angle
", "votes": null, "replies": [ { "id": 2139907, "author_name": "pellerphys", "author_url": "", "post_date": "02/11/2023 09:21:37", "content": "Neutrinos can also come from below the horizon, and therefore zenith needs to go from 0 all the way to 180 to cover the full sphere.
", "votes": null, "replies": [] } ] } ], "raw_markdown_by_id": { "2117798": "I have some questions/clarifications about this topic as I struggle to visualize things like this. \n\nLet me give it a go though and hopefully, someone can help.\n\n\n\nThe simple diagram above makes sense to me.\n* **Azimuth** in this context is:\n * The angle, measured clockwise from the North cardinal direction, of the projection of the path of the Neutrino onto the Earth's surface. \n * This can be 0-360 degrees (in radians that's 0 to 2pi). This is because any value around the compass is valid and possible (like spinning a pencil on a table it can land pointing in any *direction*).\n* **Zenith** in this context is:\n * The angle of inclination ranging from the local zenith (a line perpendicular to the Earth's surface) to the path of the Neutrino\n * This value ranges from 0-180 degrees (in radians that's 0 to pi). Straight up (equal to the local zenith) to straight down.\n * Note that **Elevation/Altitude** are often used as the term(s) that is/are the ***complement*** of Zenith (i.e. 90-Zenith=Elevation)\n\n```python\n# It is easy to see why this formula makes sense with basic trig\n# --> however, for this to make sense, the normalized sphere radius must be equal to 1\nx = cos(azimuth) * sin(zenith)\ny = sin(azimuth) * sin(zenith)\nz = cos(zenith)\n```\n\n**Please let me know if anything above is incorrect.** But assuming it's all good, I can see that to be able to determine the Azimuth angle, we are required to know the relative position of the Cardinal North direction.\n\nTake this example:\n* If I am standing due south of the North Pole and point to an object due East in the sky, the Azimuth is 90 degrees\n* If I walk until I'm standing due north of the North Pole and want to point to the same object as before I now have to point due West, meaning the Azimuth is now 270 degrees.\n* This example is with examples showing Cardinal directions to simplify but it obviously shows that the relative position of the person pointing (or the detector) w.r.t. Cardinal North is important in determining the correct Azimuth angle.\n\nFor this competition, it's unclear to me where the IceCube detector is located and if that's important in developing an understanding here. FYI, I know I'm probably over-complicating this. I checked it out on Google Earth and looks to be slightly North by North-East of the South Pole. Please see the photo below:\n\n\n\nIs my understanding correct? Obviously then, the detector itself has it's own relative vector pointing North, but every sensor has its own independent relative vector pointing North. At this point, I am now thoroughly confused, haha.\n\n---\n\nThe other thing that throws me a bit when I think about all this is that the **local zenith** is coincident (but in opposite directions) to the Axis of the Earth and therefore the North Pole. I don't think it matters though because for Zenith we only care about elevation and for Azimuth we only care about the relative direction of the Neutrino and the North Cardinal Direction.\n\nOut of curiosity, would this mean that a Zenith of 180 degrees and an Azimuth of 0 degrees would be a Neutrino that travels a path that takes it straight through the North Pole, along the Earth's axis, and straight out the South Pole?\n\n---\n\nSorry if this is all incoherent, as I said above, I can struggle to visualize things properly so I may have completely messed something up. Any help/support would be greatly appreciated.\n\nps. I realize my image uses the Sun, but obviously, in this competition instead of the sun, it would be the origin of the Neutrino that the vector points to.\n\n---\n\n[EDIT:] Additional image for helpful context:\n", "2117820": "For the data we give you, the two angles are simply spherical coordinates defined in the same coordinate system as the sensor geometry is defined in. So this means it is entirely independent of the location of IceCube.\n\nIn order to do physics and then map these values to actual sky coordinates, or for example galactic coordinates, we have existing functions in icecube, so no need from your side to worry about that part. :)", "2117854": "Oh! I see! Thanks for the response @pellerphys. \n\nSo the center of the sensor array (IceCube) is the origin of the Azimuth and Zenith used in this competition? Then the 'local zenith' is always the z-axis (Positive coincident with South-Pole pointing to space)?\n\nAnd the XY plane is located at the halfway depth w.r.t. the sensor array area, ie. 1950m down(not the overall area – considering the 1KM w/ no sensors, 1225m down)? Or would the XY plane be at the surface?\n\nDoes the picture below make sense (it assumes roughly the XY plane is at the halfway point w.r.t. Sensor Array volume)? Also, I *THINK* the deep-core sensors are in the center... so my Z vector may be in the wrong location and should be centred within the deep-core sensor area.\n\n", "2117923": "Yes, that looks about right to me 👍\n\nThe origin [0,0,0] is roughly in the middle of all sensors. But keep in mind that direction vectors are invariant under translations anyway. These vectors really only encode the direction, and not the position of the neutrino.", "2120616": "Thx for thoses explanations. I was also completely lost with the coordinates system. But still 2 questions:\n1. Does your answer mean that all parallel impact paths will have the same azimuth/zenith? I guess there is no reason for all observed impact paths to run through the origin. Are the azimuth/zenith train Data normalized to origin?\n2. Does an impact path have an entry and exit point on the ice cube? If yes the azimuth/zenith peer should relate to entry or to exit?", "2121561": "Yes, all parallel paths will have identical zenith and azimuth values, there is **no** positional information!", "2129877": "Adding as a reference a a similar question asked before this one which also has some relevant responses...\nhttps://www.kaggle.com/competitions/icecube-neutrinos-in-deep-ice/discussion/379857", "2139652": "I truly do not want to confuse the situation with my possible ignorance in this area and this is my first post on Kaggle.\n\nIf a Zenith angle is defined from 0 to 90 degrees and the Azimuth angle is from 0 to 360 degrees then any vector is fully defined that passes through the origin. \n\nFor example, at sunrise, it could be Zenith 90 degrees and Azimuth 90 degrees and at sunset, it could be Zenith 90 degrees and Azimuth 270 degrees (need a 180-degree difference between sunrise to sunset Azimuth). High noon Zenith is 0.\n\nIt appears from the links below the above is one accepted standard.\n\nhttps://sinovoltaics.com/learning-center/basics/elevation-angle/#:~:text=Sunrise%2FSunset,as%20refracted%20by%20the%20atmosphere.\n\nhttps://en.wikipedia.org/wiki/Solar_azimuth_angle", "2139907": "Neutrinos can also come from below the horizon, and therefore zenith needs to go from 0 all the way to 180 to cover the full sphere." }, "source": "meta" }