Need help in algorithm
Hello everyOne,
I am trying to make an algorithm which can detect the solar cells of solar module one by one, declare the active area by co-ordinate system like where is Xmin,Xmax, Ymin, Ymax are and then calculate the area covered by the solar module. For this I made a program and at the moment, I am stuck badly and need help.
What I am trying to do is, find the white pixel and convert it to gray. I managed to do it on cell however it perform only 3/4 of the cell area. For the last two days I am trying to find the error but could not locate it. Can someone please look at my code and guide me through, Please.
You can download all my Sub VIs on my other thread which is http://forums.ni.com/t5/Machine-Vision/Detect-the-rectangular/td-p/2370836.
Many Thanks
Lazer
Attachments:
Cell detection - main.vi 181 KB
convert 16Bits to 8Bits.vi 23 KB
Jiangs_SSC_test_V130308_SJ204_POS6_10m_25deg_34_6m.jpg 9 KB
Hey Lazer,
I'm not entirely sure what you are trying to do, I'm guessing it is the big while loop which is not giving the correct results? What exactly are you trying to do, e.g. what is going into the loop, and what is supposed to come out? Is the pixel array going in going to look like the jpg you attached? It sounds like you are trying to search for particular colours (which will be a certain value in U8) and look around them, or change some colours... maybe if we get a better idea of what is trying to be achieved, we could help a little.
In general, it's worth posting questions about vision in the machine vision forum, where there are a lot of other vision customers who will have experience. It may be easier to do this all with the vision functions, there are tools for colour matching among other things.
Also, if you have a lot of vi's, its easier to pack them all into a zipped folder so that people can download the whole lot at once.
There is lots of information about different functions in labview help, if you are stuck about what something does, or what to see if there is a function to do what you are looking for, you can have context help open while you look around the palette, and it will give information about each function.
Ian S
Applications Engineer CLD
National Instruments UK&Ireland
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Write an algorithm to compute the arrival time of a train, if you know the starting time and the distance to destination (in km). Assume that the average speed of the train is 60km/h. The start time is given as the hour (using 24h notation) and the minutes. For example, if the start time is 13:46, the user will input 13 and 46. Assume that the hour and minutes provided are valid values. The output will consist in two numbers, one for the hours and one for the minutes.
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THANKS, FROM A DESPERATE NOOBslvr99 wrote:
Assume that the average speed of the train is 60km/h.
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Now if you add the travel time of 210 minutes to the start time of 620 minutes you get the arrival time. It becomes 620+210 = 830 minutes.
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Still stuck with the same old producer consumer weight problem need help
Hello All,
This is the problem I am stuck with right now.
I have two array lists one producer array list and one consumer array list denoted by a and b
P1 P2 P3 P4 P5
5 6 7 8 9
C1 C2 C3 C4 C5
2 3 4 5 6
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We have the following sets
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(5,3)(6,3)(7,3),(8,3),(9,3)
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(5,5)(6,5)(7,5),(8,5),(9,5)
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Thanks.
Regards.
NP
What I have tried till now.
I have one more question could you help me with this.
I have an array list of this form.
package mypackage1;
import java.util.*;
public class DD
private int P;
private int C;
private int weight;
public void set_p(int P1)
P=P1;
public void set_c(int C1)
C=C1;
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weight=W1;
public int get_p()
return P;
public int get_c()
return C;
public int get_x()
return weight;
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ArrayList min_weights_int=new ArrayList();
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String v;
int o=0;
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[\code]In your example you selected the pair with the greatest
distance from the first set, and the pair with the least
distance from the second. I don't see how the distance
function was used.
Also it's not clear to me that there is always a solution,
and, if there is, whether consistently choosing the
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The most obvious approach is to systematically try
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are no possibilities left. This means backtracking whenever
a point is reached where you cannot continue. In this case
backtrack one step and try another possibility at this
step. After all possible choices of the previous step,
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This seems rather involved, and it probably is.
Interestingly, if you know Prolog, it is ridiculously
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In Java, you can implement the algorithm in much the same
way as Prolog implementations do it--keep a list of all the
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If you do know Prolog, you could generate lots of random
problems and see if there is always a solution. -
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the site is the following
http://www.uni-koeln.de/ew-fak/Mathe/Projekte/VisuPro/pythagoras/pythagoras.html
then double click on the screen to make it startPardon my ASCII art, but I've always liked the following, simple, geometric proof:
a b
---------------------------------------+
| | |
a| I | II |
| | |
---------------------------------------+
| | |
| | |
| | |
| | |
| | |
b| IV | III |
| | |
| | |
| | |
| | |
| | |
| | |
---------------------------------------+It almost goes without saying that I+II+III+IV == (a+b)^2, and II == IV == a*b,
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I hope the following sketch makes sense, stand back, ASCII art alert again: a b
---------------------------------------+
| . VI |
| . . |a
| V . |
| +
| |
| . |
b| . |
| |
| IX |
| . |
| . |b
| |
+ |
| . |
a| . . VII |
| VIII . |
---------------------------------------+
a bThe total area equals (a+b)^2 again and equals the sum of the smaller areas:
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Given this fundamental result, the others can easily be derived from this one,
or did I answer a question you didn't ask?
kind regards,
Jos -
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Message was edited by: B0OmB0Om wrote:
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Message was edited by: DarkxFlamexCaster
Message was edited by: DarkxFlamexCaster+it comes up when im almost done installing the newest/newer versions of itunes. it says+ +"the feature you are trying to use is on a network resource that is unavailable" "click ok to try again or enter an alternate path to a folder containing the installation package 'iTunes.msi' in the box below"+
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