k-Wave

1. zia.roghani
   

   Inactive
   Joined: Sep '14
   Posts: 12

   Hi,,
   does any body can guide me how to developed a Cartesian coordinate type relation ships between kgrid points.
   let
   Nx = 100;
   Ny = 100;
   dx = 1;
   dy = 1;
   kgrid = makeGrid(Nx, dx, Ny, dy);
   I know that the distance of each kgrid point can be calculated by using the formula
   sqrt(kgrid.x.^2 + kgrid.y.^2);
   how can the (0,0) of the kgrid be points be related to cartesian cordiantes. i wants a general relationship ..
   any help will be higly appreciated.

   Posted 9 years ago #

2. Bradley Treeby
   

   Developer
   Joined: Mar '10
   Posts: 1,643

   Hi Zia,

   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 floor(Nx/2) + 1, i.e.,

   for Nx = 4:
   -2 -1 0 1

   for Nx = 5:
   -2 -1 0 1 2

   To convert these to cartesian points, these are just multiplied by the grid point spacing (e.g., kgrid.dx).

   Hope that helps,

   Brad.

   Posted 9 years ago #

3. zia.roghani
   

   Inactive
   Joined: Sep '14
   Posts: 12

   Hi Bradley Treeby,
   Thank you so much for looking into the problem. Your help is always to the point.please look into the following,

   Nx=100;
   Ny=100;
   dx=1;
   dy=1;
   kgrid= makeGrid(Nx, dx, Ny, dy);

   for above are the following correct.orign (0,0) is at Nx=Ny=51;
   the coordinate of the first quardent Nx=1 upto 51, and Ny=51 upto 100;
   the coordinate of the second quardent Nx=1 upto 51, and Ny=1 upto 51;
   the coordinate of the third quardent Nx=51 upto 100, and Ny=1 upto 51;
   the coordinate of the forth quardent Nx=51 upto 100,and Ny=51 upto 100;
   '
   what actually i want to do is that i have an arry of 3-sensors along the x-axis or Y-axis.
   Let the middle sensor to be at the orign. now i want toplace the source at some distance say R,
   R is making some angle theta with middle sensor. How i can proceed...?

   regards.

   Posted 9 years ago #

4. Bradley Treeby
   

   Developer
   Joined: Mar '10
   Posts: 1,643

   Hi Zia,

   You can calculate this using some simple trigonometry. If the sensor is positioned with grid coordinates [x, y], the position of the source (in grid points) will be round([x + R*cos(theta), y + R*sin(theta)]) (or similar, depending on how you define your angles).

   Brad.

   Posted 9 years ago #

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