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IMAQ LineProfile with an angled line

Hi Folks,

 

I have a question regarding how LineProfile works, especially when the line is angled (i.e. not horizontal or vertical).  I've searched this and other boards and have not been able to find an answer.

 

Lets say I have a line that runs from the point (0,0) to (x, y), and lets assume for the moment that x=y so that the line is at a 45 degree angle.  Now I want to get the pixel intensity along that line and plot it.  My conventional wisdom tells me that the horiztonal axis of that plot should run from 0 to sqrt(x^2 + y^2), or in this cae sqrt(2)*x since x=y.  However, the plot runs from 0 to x instead.

 

So, my question is what is the horizontal axis of this plot actually showing?  Is it the x coordinate of the pixel, or is it something more complicated than this?

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Hi mwray1981,

 

It seems to be the coordinate of the longer plane. If the y-axis is longer on the line, then it will choose that instead. It seems to do this in order to properly give you the intensity of each pixel. So it gives you the intensity of each pixel across the line with the x-axis on the graph being the relative position on the longest axis.

Paolo F.
National Instruments
Applications Engineer
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