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3d graph and reverse iteration

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Message 11 of 14
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hi evan,
 
your vi sure works. but i didnt really understand the boolean logic you used. could you please shed some light on it? i have also included the another graph in my vi showing the path this takes but it doest look like the right path.
 
 
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Message 12 of 14
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As far as the difference between the two VIs I am not sure which works correctly for your application.  If you rotate our 2 graphs around you will notice that thet are taking different paths.  When mine hits the end of an x,y plan it simply increases z by 1 and reverses the x,y scan.  yours looks like it always starts the x,y scan at the opposite x.  So yours finishes the z=0 scan at x,y of (9,15), however it starts the z=1 scan at (0,15)

All my boolean logic does is decide whether I am counting up or down.  The 2 booleans are for whether I am counting up in x or in y.  So for z=0 the y boolean is true or count up.  In the mean time the x boolean switches every time I increase y.  So for y=0 its true and therefore I scan up in x.  However after that 1 line I toggle the boolean. now I know to count down in x.  So at y=1 I count down in x.  I continue this toggling for the entire plan.  Now when I reach the end of the scan I toggle the y boolean and the x boolean got toggled as well. 

So for the last scan I was increasing y and x.  I just toggles the 2 so i will now count down on x and y.  Therefore I follow the exact same path backwards.

Let me know if this clears it up any.

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Message 13 of 14
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hi evan,

your vi is the one which works correctly. that is how it should function!! thanks for all the advice. although i am just wondering can this boolean logic be converted to the old odd even logic? using boolean logic in my vi is quite troublesome. i mean the actual vi where this 3d scanning has to be implemented.!




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Message 14 of 14
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