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      "body": "@@ -1,8 +1,397 @@\n+%3Ccenter%3E!%5Btangent circles%5D(https://steemitimages.com/DQmc6bvhHbejzhgqg848oNqGybCjkBPLpHx1AQNsWJvdvhA/triplet.png)%0A*An initial sphere triplet located on the vertices of a triangle with side lengths constructed from sphere radii: %3Cimg src=%22https://latex.codecogs.com/png.latex?a=r_B+r_C,&space;b=r_A+r_C,&space;c=r_A+r_B%22 title=%22a=r_B+r_C, b=r_A+r_C, c=r_A+r_B%22 /%3E*%3C/center%3E%0A%0A\n So you w\n@@ -925,398 +925,8 @@\n e.%0A%0A\n-%3Ccenter%3E!%5Btangent circles%5D(https://steemitimages.com/DQmc6bvhHbejzhgqg848oNqGybCjkBPLpHx1AQNsWJvdvhA/triplet.png)%0A*An initial sphere triplet located on the vertices of a triangle with side lengths constructed from sphere radii: %3Cimg src=%22https://latex.codecogs.com/png.latex?a=r_B+r_C,&space;b=r_A+r_C,&space;c=r_A+r_B%22 title=%22a=r_B+r_C, b=r_A+r_C, c=r_A+r_B%22 /%3E*%3C/center%3E%0A%0A%0A\n Ther\n",
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{
  "trx_id": "cce85815b6e7c65f20af85754a0179f9dd387363",
  "block": 20808408,
  "trx_in_block": 24,
  "op_in_trx": 0,
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  "timestamp": "2018-03-19T09:36:03",
  "op": [
    "comment",
    {
      "parent_author": "",
      "parent_permlink": "math",
      "author": "libbum",
      "permlink": "clustering-tangent-spheres",
      "title": "Clustering tangent spheres",
      "body": "So you want to do some sphere packing to make your inflatable [space habitats](https://github.com/Libbum/space-habitats) for deployment on the Moon and Mars snug and cozy. Perfectly straightforward you think to yourself; since one can start with a triplet of spheres tangent pairwise by placing one at some origin on a 2D plane, extending the second out in the <img src=\"https://latex.codecogs.com/png.latex?x\" title=\"x\" /> direction and completing the triangle by identifying the third sphere's position based on the radii of all three.\n\n<center>![tangent circles](https://steemitimages.com/DQmc6bvhHbejzhgqg848oNqGybCjkBPLpHx1AQNsWJvdvhA/triplet.png)\n*An initial sphere triplet located on the vertices of a triangle with side lengths constructed from sphere radii: <img src=\"https://latex.codecogs.com/png.latex?a=r_B&plus;r_C,&space;b=r_A&plus;r_C,&space;c=r_A&plus;r_B\" title=\"a=r_B+r_C, b=r_A+r_C, c=r_A+r_B\" />*</center>\n\n\nThere's not even time to take a mirco-break to refresh your favourite time sink website before you've busted out the coordinates of <img src=\"https://latex.codecogs.com/png.latex?C\" title=\"C\" />\n\n<center><img src=\"https://latex.codecogs.com/png.latex?e&space;=&space;\\frac{b^2&plus;c^2-a^2}{2c},&space;\\quad&space;f=&space;\\sqrt{b^2-e^2}.\" title=\"e = \\frac{b^2+c^2-a^2}{2c}, \\quad f= \\sqrt{b^2-e^2}.\" /></center>\n\nOn a roll, you extend the coordinates of <img src=\"https://latex.codecogs.com/png.latex?A\" title=\"A\" />, <img src=\"https://latex.codecogs.com/png.latex?B\" title=\"B\" />and <img src=\"https://latex.codecogs.com/png.latex?C\" title=\"C\" /> into three dimensions setting their <img src=\"https://latex.codecogs.com/png.latex?z\" title=\"z\" /> component to 0 and construct a tetrahedron from this base triangle and three new distances <img src=\"https://latex.codecogs.com/png.latex?m&space;=&space;r_A&plus;r_D\" title=\"m = r_A+r_D\" />, <img src=\"https://latex.codecogs.com/png.latex?n&space;=&space;r_B&plus;r_D\" title=\"n = r_B+r_D\" />, <img src=\"https://latex.codecogs.com/png.latex?p=r_C&plus;r_D\" title=\"p=r_C+r_D\" />; finding the tetrahedron's fourth vertex\n\n<center><img src=\"https://latex.codecogs.com/png.latex?\\inline&space;x_D&space;=&space;\\frac{m^2&plus;c^2-n^2}{2c},&space;\\quad&space;y_D&space;=&space;\\frac{f^2c&plus;m^2c-m^2e&plus;n^2e-p^2c-c^2e&plus;e^2c}{2fc},&space;\\quad&space;z_D&space;=&space;\\pm\\sqrt{m^2-x_D^2-y_D^2}.\" title=\"x_D = \\frac{m^2+c^2-n^2}{2c}, \\quad y_D = \\frac{f^2c+m^2c-m^2e+n^2e-p^2c-c^2e+e^2c}{2fc}, \\quad z_D = \\pm\\sqrt{m^2-x_D^2-y_D^2}.\" /></center>\n\nPausing for just a moment, you realise you've been on autopilot for a second&mdash;since such trivial calculations are self evident. Reflecting on a time that feels so long ago now, when you learnt that a minimal set of four tangent spheres in three dimensions can be described by a tetrahedron. The locus of each sphere's center point is one of the tetrahedron's vertices, and the lengths of each edge of the tetrahedron are defined via the radii of the pair of spheres for which the edge connects their centers. Using the already-known properties of the base triangle; introducing a fourth sphere <img src=\"https://latex.codecogs.com/png.latex?D\" title=\"D\" /> with known radius <img src=\"https://latex.codecogs.com/png.latex?r_D\" title=\"r_D\" /> and finding its coordinates, is now just some basic geometry&mdash;extending the system of equations used to solve the locus of <img src=\"https://latex.codecogs.com/png.latex?C\" title=\"C\" />\n\n<center><img src=\"https://latex.codecogs.com/png.latex?\\begin{aligned}&space;e^2&plus;f^2&=b^2\\\\&space;(e-c)^2&plus;f^2&=a^2,\\\\&space;\\end{aligned}\" title=\"\\begin{aligned} e^2+f^2&=b^2\\\\ (e-c)^2+f^2&=a^2,\\\\ \\end{aligned}\" /></center>\n\nto the intersection represented by\n<center><img src=\"https://latex.codecogs.com/png.latex?\\begin{aligned}&space;x_D^2&plus;y_D^2&plus;z_D^2&=m^2\\\\&space;(x_D-c)^2&plus;y_D^2&plus;z_D^2&=n^2\\\\&space;(x_D-e)^2&plus;(y_D-f)^2&plus;z_D^2&=p^2.\\\\&space;\\end{aligned}\" title=\"\\begin{aligned} x_D^2+y_D^2+z_D^2&=m^2\\\\ (x_D-c)^2+y_D^2+z_D^2&=n^2\\\\ (x_D-e)^2+(y_D-f)^2+z_D^2&=p^2.\\\\ \\end{aligned}\" /></center>\n\nBack from your daydream&mdash;that time sink website is calling. Since you've pretty much wrapped this problem up, why not take a peek? After a few sensible chuckles, it dawns on you that there's still work to do. Your sphere packing algorithm is going to feed you positions and radii of arbitrary sets of pairwise tangent triplets for you to calculate the fourth position. This means that if you are solving the tetrahedron in this simple basis, translation and rotation of <img src=\"https://latex.codecogs.com/png.latex?D\" title=\"D\" />'s position must occur.\n\nMoreover, this is going to have to happen in a number of steps.\n\n1. Remove <img src=\"https://latex.codecogs.com/png.latex?\\triangle&space;ABC\" title=\"\\triangle ABC\" />'s translation component, aligning <img src=\"https://latex.codecogs.com/png.latex?A\" title=\"A\" /> to the origin.\n2. Recalculate <img src=\"https://latex.codecogs.com/png.latex?c\" title=\"c\" />, <img src=\"https://latex.codecogs.com/png.latex?e\" title=\"e\" />, <img src=\"https://latex.codecogs.com/png.latex?f\" title=\"f\" />in the simple basis for this triplet to solve for the fourth locus.\n3. Denote the solution as <img src=\"https://latex.codecogs.com/png.latex?H\" title=\"H\" /> in tetrahedron <img src=\"https://latex.codecogs.com/png.latex?EFGH\" title=\"EFGH\" />, which now must be translated to tetrahedron <img src=\"https://latex.codecogs.com/png.latex?ABCD\" title=\"ABCD\" />'s basis to find <img src=\"https://latex.codecogs.com/png.latex?D\" title=\"D\" />.\n4. Rotate <img src=\"https://latex.codecogs.com/png.latex?\\overrightarrow{EF}\" title=\"\\overrightarrow{EF}\" /> onto <img src=\"https://latex.codecogs.com/png.latex?\\overrightarrow{AB}\" title=\"\\overrightarrow{AB}\" />.\n5. Rotate <img src=\"https://latex.codecogs.com/png.latex?G\" title=\"G\" /> to <img src=\"https://latex.codecogs.com/png.latex?C\" title=\"C\" /> by rotating the face normals of the triangle identified in 4 and <img src=\"https://latex.codecogs.com/png.latex?\\triangle&space;ABC\" title=\"\\triangle ABC\" />.\n6. Translate this triangle to the <img src=\"https://latex.codecogs.com/png.latex?ABC\" title=\"ABC\" /> basis by returning the translation component removed in 1.\n\nNone of this is too difficult in of its own, but without proper forethought&mdash;you're gonna have a bad time.\nBasically you're going to end up catching edge cases just to the left of center like a tennis ball to the nuts of a daydreaming ball boy at the US Open.\n\nWhat happens when you're sent a triplet that has coordinates (in the translated basis) <img src=\"https://latex.codecogs.com/png.latex?(0,0,0)\" title=\"(0,0,0)\" />, <img src=\"https://latex.codecogs.com/png.latex?(e,f,0)\" title=\"(e,f,0)\" />, <img src=\"https://latex.codecogs.com/png.latex?(c,0,0)\" title=\"(c,0,0)\" /> or in other words: when <img src=\"https://latex.codecogs.com/png.latex?B=G\" title=\"B=G\" /> and <img src=\"https://latex.codecogs.com/png.latex?C=F\" title=\"C=F\" />? You must check for this mirror symmetry to make sure that the first rotation doesn't place your <img src=\"https://latex.codecogs.com/png.latex?F\" title=\"F\" /> position closer to Alpha Centauri than it does to it's wanted position at <img src=\"https://latex.codecogs.com/png.latex?B\" title=\"B\" />.\n\nThe list goes on, but it suffices to say that you start juggling possibilities like some guy trying to wash a car in an infomercial.\n\nFret not! There is a better way!\n\n---\n\nConsider that we have a pairwise tangent triplet with known radii and arbitrary positions in <img src=\"https://latex.codecogs.com/png.latex?\\mathbb{R}^3\" title=\"\\mathbb{R}^3\" />.\nAdditionally, we know the radius of a fourth sphere we wish to add to the cluster.\n\nTo not confuse the skimming reader&mdash;I will change notations here, and define a tetrahedron with vertices <img src=\"https://latex.codecogs.com/png.latex?S_1\" title=\"S_1\" />, <img src=\"https://latex.codecogs.com/png.latex?S_2\" title=\"S_2\" /> and <img src=\"https://latex.codecogs.com/png.latex?S_3\" title=\"S_3\" /> residing at distances <img src=\"https://latex.codecogs.com/png.latex?d_1\" title=\"d_1\" />, <img src=\"https://latex.codecogs.com/png.latex?d_2\" title=\"d_2\" />, <img src=\"https://latex.codecogs.com/png.latex?d_3\" title=\"d_3\" /> from a fourth vertex <img src=\"https://latex.codecogs.com/png.latex?S_0\" title=\"S_0\" />, whose coordinates <img src=\"https://latex.codecogs.com/png.latex?(x,y,z)\" title=\"(x,y,z)\" /> must be found.\n\nThe distance values here are straightforward, since they are defined as the sum of the radii along a given edge: <img src=\"https://latex.codecogs.com/png.latex?d_i=r_i&plus;r_0\" title=\"d_i=r_i+r_0\" />. Since <img src=\"https://latex.codecogs.com/png.latex?S_0\" title=\"S_0\" /> is at the intersection of three spheres centered at <img src=\"https://latex.codecogs.com/png.latex?S_1\" title=\"S_1\" />, <img src=\"https://latex.codecogs.com/png.latex?S_2\" title=\"S_2\" /> and <img src=\"https://latex.codecogs.com/png.latex?S_3\" title=\"S_3\" />  with radii <img src=\"https://latex.codecogs.com/png.latex?d_1\" title=\"d_1\" />, <img src=\"https://latex.codecogs.com/png.latex?d_2\" title=\"d_2\" />, <img src=\"https://latex.codecogs.com/png.latex?d_3\" title=\"d_3\" /> , we know its coordinates satisfy the equations\n\n<center><img src=\"https://latex.codecogs.com/png.latex?\\begin{aligned}&space;(x-x_1)^2&plus;(y-y_1)^2&plus;(z-z_1)^2&=d_1^2\\qquad(1)\\\\&space;(x-x_2)^2&plus;(y-y_2)^2&plus;(z-z_2)^2&=d_2^2\\qquad(2)\\\\&space;(x-x_3)^2&plus;(y-y_3)^2&plus;(z-z_3)^2&=d_3^2\\qquad(3)\\\\&space;\\end{aligned}\" title=\"\\begin{aligned} (x-x_1)^2+(y-y_1)^2+(z-z_1)^2&=d_1^2\\qquad(1)\\\\ (x-x_2)^2+(y-y_2)^2+(z-z_2)^2&=d_2^2\\qquad(2)\\\\ (x-x_3)^2+(y-y_3)^2+(z-z_3)^2&=d_3^2\\qquad(3)\\\\ \\end{aligned}\" /></center>\n\nSubtract (1) from (2) & (3) to obtain the equations for two planes, and couple this set with (1) to yield the quadratic system:\n\n<center><img src=\"https://latex.codecogs.com/png.latex?\\begin{aligned}&space;2(x_1-x_2)x&plus;2(y_1-y_2)y&plus;2(z_1-z_2)z&=d_2^2-d_1^2&plus;x_1^2-x_2^2&plus;y_1^2-y_2^2&plus;z_1^2-z_2^2\\\\&space;2(x_1-x_3)x&plus;2(y_1-y_3)y&plus;2(z_1-z_3)z&=d_3^2-d_1^2&plus;x_1^2-x_3^2&plus;y_1^2-y_3^2&plus;z_1^2-z_3^2\\\\&space;x^2&plus;y^2&plus;z^2-2&space;x_1&space;x-2&space;y_1&space;y-2&space;z_1&space;z&space;&=&space;d_1^2-x_1^2-y_1^2-z_1^2.\\\\&space;\\end{aligned}\" title=\"\\begin{aligned} 2(x_1-x_2)x+2(y_1-y_2)y+2(z_1-z_2)z&=d_2^2-d_1^2+x_1^2-x_2^2+y_1^2-y_2^2+z_1^2-z_2^2\\\\ 2(x_1-x_3)x+2(y_1-y_3)y+2(z_1-z_3)z&=d_3^2-d_1^2+x_1^2-x_3^2+y_1^2-y_3^2+z_1^2-z_3^2\\\\ x^2+y^2+z^2-2 x_1 x-2 y_1 y-2 z_1 z &= d_1^2-x_1^2-y_1^2-z_1^2.\\\\ \\end{aligned}\" /></center>\n\nThe closed form solution to this system is not the nicest of expressions, so we can consider the equivalent set below in vector format\n\n<center><img src=\"https://latex.codecogs.com/png.latex?\\begin{aligned}&space;\\vec&space;u\\cdot\\vec&space;r=a\\\\&space;\\vec&space;v\\cdot\\vec&space;r=b\\\\&space;\\vec&space;r\\cdot\\vec&space;r&plus;\\vec&space;w\\cdot\\vec&space;r=c\\\\&space;\\end{aligned}\" title=\"\\begin{aligned} \\vec u\\cdot\\vec r=a\\\\ \\vec v\\cdot\\vec r=b\\\\ \\vec r\\cdot\\vec r+\\vec w\\cdot\\vec r=c\\\\ \\end{aligned}\" /></center>\n\nwhere <img src=\"https://latex.codecogs.com/png.latex?\\vec&space;r=(x,y,z)\" title=\"\\vec r=(x,y,z)\" /> is our unknown vector. Following from this, we obtain values of most of the components directly,\n\n<center><img src=\"https://latex.codecogs.com/png.latex?\\begin{aligned}&space;\\vec&space;u&=\\frac{S_1-S_2}{\\left\\|S_1-S_2\\right\\|}\\\\&space;\\vec&space;v&=\\frac{S_1-S_3}{\\left\\|S_1-S_3\\right\\|}\\\\&space;\\vec&space;w&=-2S_1\\\\&space;a&=\\frac{d_2^2-d_1^2&plus;x_1^2-x_2^2&plus;y_1^2-y_2^2&plus;z_1^2-z_2^2}{2\\left\\|S_1-S_2\\right\\|}\\\\&space;b&=\\frac{d_3^2-d_1^2&plus;x_1^2-x_3^2&plus;y_1^2-y_3^2&plus;z_1^2-z_3^2}{2\\left\\|S_1-S_3\\right\\|}\\\\&space;c&=d_1^2-x_1^2-y_1^2-z_1^2\\\\&space;\\end{aligned}\" title=\"\\begin{aligned} \\vec u&=\\frac{S_1-S_2}{\\left\\|S_1-S_2\\right\\|}\\\\ \\vec v&=\\frac{S_1-S_3}{\\left\\|S_1-S_3\\right\\|}\\\\ \\vec w&=-2S_1\\\\ a&=\\frac{d_2^2-d_1^2+x_1^2-x_2^2+y_1^2-y_2^2+z_1^2-z_2^2}{2\\left\\|S_1-S_2\\right\\|}\\\\ b&=\\frac{d_3^2-d_1^2+x_1^2-x_3^2+y_1^2-y_3^2+z_1^2-z_3^2}{2\\left\\|S_1-S_3\\right\\|}\\\\ c&=d_1^2-x_1^2-y_1^2-z_1^2\\\\ \\end{aligned}\" /></center>\n\nNotice that <img src=\"https://latex.codecogs.com/png.latex?\\vec&space;u\" title=\"\\vec u\" /> and <img src=\"https://latex.codecogs.com/png.latex?\\vec&space;v\" title=\"\\vec v\" /> are unit vectors, so we can form a basis with a third vector:\n\n<center><img src=\"https://latex.codecogs.com/png.latex?\\vec&space;t=\\frac{\\vec&space;u\\times\\vec&space;v}{\\left\\|\\vec&space;u\\times\\vec&space;v\\right\\|}.\" title=\"\\vec t=\\frac{\\vec u\\times\\vec v}{\\left\\|\\vec u\\times\\vec v\\right\\|}.\" /></center>\n\nA solution to <img src=\"https://latex.codecogs.com/png.latex?\\vec&space;r\" title=\"\\vec r\" /> can then be obtained in the form <img src=\"https://latex.codecogs.com/png.latex?\\vec&space;r=\\alpha\\vec&space;u&plus;\\beta\\vec&space;v&plus;\\gamma\\vec&space;t\" title=\"\\vec r=\\alpha\\vec u+\\beta\\vec v+\\gamma\\vec t\" /> once <img src=\"https://latex.codecogs.com/png.latex?\\alpha\" title=\"\\alpha\" />, <img src=\"https://latex.codecogs.com/png.latex?\\beta\" title=\"\\beta\" /> and <img src=\"https://latex.codecogs.com/png.latex?\\gamma\" title=\"\\gamma\" /> are identified. Plugging this expression for <img src=\"https://latex.codecogs.com/png.latex?\\vec&space;r\" title=\"\\vec r\" /> into the vector equations for <img src=\"https://latex.codecogs.com/png.latex?a\" title=\"a\" /> and <img src=\"https://latex.codecogs.com/png.latex?b\" title=\"b\" /> above immediately yields\n\n<center><img src=\"https://latex.codecogs.com/png.latex?\\alpha=\\frac{a-b(\\vec&space;u\\cdot\\vec&space;v)}{1-(\\vec&space;u\\cdot\\vec&space;v)^2},\\quad\\beta=\\frac{b-a(\\vec&space;u\\cdot\\vec&space;v)}{1-(\\vec&space;u\\cdot\\vec&space;v)^2},\" title=\"\\alpha=\\frac{a-b(\\vec u\\cdot\\vec v)}{1-(\\vec u\\cdot\\vec v)^2},\\quad\\beta=\\frac{b-a(\\vec u\\cdot\\vec v)}{1-(\\vec u\\cdot\\vec v)^2},\" /></center>\n\nwhile the third equation becomes <img src=\"https://latex.codecogs.com/png.latex?\\gamma^2&plus;(\\vec&space;w\\cdot\\vec&space;t)\\gamma&plus;d=0\" title=\"\\gamma^2+(\\vec w\\cdot\\vec t)\\gamma+d=0\" /> where <img src=\"https://latex.codecogs.com/png.latex?d=\\alpha^2&plus;\\beta^2&plus;2\\alpha\\beta(\\vec&space;u\\cdot\\vec&space;v)&plus;\\alpha(\\vec&space;u\\cdot\\vec&space;w)&plus;\\beta(\\vec&space;v\\cdot\\vec&space;w)-c\" title=\"d=\\alpha^2+\\beta^2+2\\alpha\\beta(\\vec u\\cdot\\vec v)+\\alpha(\\vec u\\cdot\\vec w)+\\beta(\\vec v\\cdot\\vec w)-c\" />.\n\nThis last equation can finally be solved for <img src=\"https://latex.codecogs.com/png.latex?\\gamma\" title=\"\\gamma\" />:\n\n<center><img src=\"https://latex.codecogs.com/png.latex?\\gamma=\\frac{1}{2}\\left(-(\\vec&space;w\\cdot\\vec&space;t)\\pm\\sqrt{(\\vec&space;w\\cdot\\vec&space;t)^2-4d}\\right).\" title=\"\\gamma=\\frac{1}{2}\\left(-(\\vec w\\cdot\\vec t)\\pm\\sqrt{(\\vec w\\cdot\\vec t)^2-4d}\\right).\" /></center>\n\nWith these values you can now solve the irregular tetrahedron in 3D directly! This means no edge cases, no recalculation of previously known values or complex `if elseif else` blocks based on various input conditions.\n\n<center>![sphere cluster](https://steemitimages.com/DQmZcNdZUDdThUwpifaHUTPb4i1jb4QgNSDfsvs6Rv3tpB5/spherecluster.png)\n*The fruits of your labour made manifest.*</center>",
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      "title": "Introduceyourself: Physicist at the Stockholm Resilience Centre",
      "body": "Hi everyone! I'll try to keep this intro short and sweet. They say it's best to leave your audience before your audience leaves you.\n\nI'd classify myself as a computational physicist (for the moment), meaning I primarily develop theoretical models or simulation tools which help to deeper understanding of our universe.\n\n![Aluminium thin foil interacting with an intense laser pulse](https://steemitimages.com/DQmZD5hWjAcUQRDZ2rH62iDRBWMYQo5HNexQ4L9BN78Pebi/fancy.png)\n*Simulatation (using my adaptive mesh relativistic Vlasov-Maxwell solver [Veritas](http://ft.nephy.chalmers.se/veritas/)) of a 3 µm thick aluminium foil getting hit by an ultra-intense, ultra-short laser pulse.*\n\nThe above image for example is from a project when I was a Postdoctoral Researcher at the [Plasma Theory Research Group](http://ft.nephy.chalmers.se/) at [Chalmers University of Technology](https://www.chalmers.se) in Gothenburg, Sweden. \n\n![Tim](https://steemitimages.com/DQmVQcA6irwT39EpqQBxxsVZnRZBjLVQb5w6QdrPDvLBJcb/1011602_10152489753627266_4312764894393305237_n.jpg)\n*This is me. There is no science being done here, therefore no smiling.*\n\n## A quick academic history\n\n- Started out in Computer Science and Electronic Engineering, then took some sidesteps into undergraduate physics. Went off the rails a bit as well, getting qualifications in Ancient Languages (Attic Greek & Sahidic Coptic) and Philosophy.\n- PhD from [RMIT University](https://www.rmit.edu.au/) in Melbourne, Australia with the [Chemical and Quantum Physics Group](http://tcqp.science). That research focused on microscopic models of defects and quantum transport in condensed matter systems, phase qubits and large scale computer simulation using *ab initio* techniques.\n- Conducted research with the [Defence Science and Technology Organisation](https://www.dst.defence.gov.au/); specialising in statistical properties of Lagrangian particles in application to the areas of turbulent dispersion, release characterisation and blast dynamics.\n- Postdoctoral Researcher at the [Plasma Theory Research Group](http://ft.nephy.chalmers.se/) at [Chalmers University of Technology](https://www.chalmers.se) as stated above:  attempting to generate high acceleration laser-plasma based ion sources for near future use in applications such as hadron therapy for cancers, cosmic radiation studies on living cells and spacecraft, short-life medical isotope production, proton imaging and many other important applications.\n- Recently started a second Postdoctoral position at the [Stockholm Resilience Centre](http://www.stockholmresilience.org/) in Stockholm, Sweden. Here, I'll be applying my skills to advancing global social-ecological modelling analysis of systemic change and \"tipping points\" in terrestrial ecosystems.\n\n## What to expect from my feed\n\nIf any of the things above seemed interesting but also sounded mostly like white noise with no actual meaningful content: I hope that in the future I'll be able to delve into some of my work and interests at a popular science level - with a bit of a technical edge. Think more [3Blue1Brown](https://www.youtube.com/channel/UCYO_jab_esuFRV4b17AJtAw) rather than IFLScience.\n\nI'll also delve into some side projects I've been working on and why I think they're cool. Why the maths is fun and what sort of applications we can apply our ideas to. A teaser: [spherical-cow](https://github.com/Libbum/spherical-cow) is a sphere packing library I've built so you can densely pack spheres of random sizes into arbitrarily shaped containers.\n\n![Spherical Cow](https://steemitimages.com/DQmUUMMDFPJ1kd8ws4d6Vmp8hsX6JvtwxDD4FHWVTjLBDTF/cow_output.jpg)\n*A spherical cow in vacuum.*\n\nNot only can you reference old physics jokes with this tool, but you can also design [space habitats](http://digitalstructures.mit.edu/page/research#form-finding-of-deep-exploration-surface-habitats) for future Moon and Mars colonies.\n\n---\n\nFeel free to introduce yourself in the comments below, and let me know what specifically you'd like to know more about!",
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