http://www.k-wave.org/forum/topic/cartesian-co-ordinate-and-kgrid-points-1 Support for the k-Wave MATLAB toolbox en-US Wed, 13 May 2026 01:12:39 +0000 http://bbpress.org/?v=1.0.2 q http://www.k-wave.org/forum/search.php http://www.k-wave.org/forum/topic/cartesian-co-ordinate-and-kgrid-points-1#post-4893 Fri, 12 Dec 2014 12:25:34 +0000 Bradley Treeby 4893@http://www.k-wave.org/forum/ <p>Hi Zia,</p> <p>You can calculate this using some simple trigonometry. If the sensor is positioned with grid coordinates <code>[x, y]</code>, the position of the source (in grid points) will be <code>round([x + R*cos(theta), y + R*sin(theta)])</code> (or similar, depending on how you define your angles).</p> <p>Brad. </p> http://www.k-wave.org/forum/topic/cartesian-co-ordinate-and-kgrid-points-1#post-4880 Tue, 09 Dec 2014 14:46:05 +0000 zia.roghani 4880@http://www.k-wave.org/forum/ <p>Hi Bradley Treeby,<br /> Thank you so much for looking into the problem. Your help is always to the point.please look into the following,</p> <p>Nx=100;<br /> Ny=100;<br /> dx=1;<br /> dy=1;<br /> kgrid= makeGrid(Nx, dx, Ny, dy);</p> <p>for above are the following correct.orign (0,0) is at Nx=Ny=51;<br /> the coordinate of the first quardent Nx=1 upto 51, and Ny=51 upto 100;<br /> the coordinate of the second quardent Nx=1 upto 51, and Ny=1 upto 51;<br /> the coordinate of the third quardent Nx=51 upto 100, and Ny=1 upto 51;<br /> the coordinate of the forth quardent Nx=51 upto 100,and Ny=51 upto 100;<br /> '<br /> what actually i want to do is that i have an arry of 3-sensors along the x-axis or Y-axis.<br /> Let the middle sensor to be at the orign. now i want toplace the source at some distance say R,<br /> R is making some angle theta with middle sensor. How i can proceed...?</p> <p>regards. </p> http://www.k-wave.org/forum/topic/cartesian-co-ordinate-and-kgrid-points-1#post-4871 Fri, 05 Dec 2014 17:54:27 +0000 Bradley Treeby 4871@http://www.k-wave.org/forum/ <p>Hi Zia,</p> <p>I'm not sure I completely follow what you're asking? k-Wave indexes the grid points from the centre, where the grid index of the centre point is given by <code>floor(Nx/2) + 1</code>, i.e., </p> <p>for <code>Nx = 4</code>:<br /> <code>-2 -1 0 1</code></p> <p>for <code>Nx = 5</code>:<br /> <code>-2 -1 0 1 2</code></p> <p>To convert these to cartesian points, these are just multiplied by the grid point spacing (e.g., <code>kgrid.dx</code>).</p> <p>Hope that helps,</p> <p>Brad. </p> http://www.k-wave.org/forum/topic/cartesian-co-ordinate-and-kgrid-points-1#post-4869 Fri, 05 Dec 2014 17:27:57 +0000 zia.roghani 4869@http://www.k-wave.org/forum/ <p>Hi,,<br /> does any body can guide me how to developed a Cartesian coordinate type relation ships between kgrid points.<br /> let<br /> Nx = 100;<br /> Ny = 100;<br /> dx = 1;<br /> dy = 1;<br /> kgrid = makeGrid(Nx, dx, Ny, dy);<br /> I know that the distance of each kgrid point can be calculated by using the formula<br /> sqrt(kgrid.x.^2 + kgrid.y.^2);<br /> how can the (0,0) of the kgrid be points be related to cartesian cordiantes. i wants a general relationship ..<br /> any help will be higly appreciated. </p>