Anylogic - line of sight











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Is there a way to check if there is a line of sight between two agents assuming some buildings and presentation markup?
(Meaning a function that would check if two agents can see each other assuming buildings and walls)










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    Is there a way to check if there is a line of sight between two agents assuming some buildings and presentation markup?
    (Meaning a function that would check if two agents can see each other assuming buildings and walls)










    share|improve this question
























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      favorite









      up vote
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      Is there a way to check if there is a line of sight between two agents assuming some buildings and presentation markup?
      (Meaning a function that would check if two agents can see each other assuming buildings and walls)










      share|improve this question













      Is there a way to check if there is a line of sight between two agents assuming some buildings and presentation markup?
      (Meaning a function that would check if two agents can see each other assuming buildings and walls)







      anylogic






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      asked Nov 12 at 18:10









      chana ross

      51




      51
























          2 Answers
          2






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          This is how I did it once in the past. The only problem is that it might be slow if you need to do that calculation thousands of times per second. And it's only in 2D. If you need 3D, the idea is not so different.



          1) add all your building nodes and everything that might be an obstacle between the two agents to a collection. You may want to put a rectangular node around your buildings to keep everything in one collection (assuming you are using space markup with nodes)



          2) generate a delta distance delta (equal to 1 for example) and find the angle of the line that passes through both agents.



          3) Make a loop from the position of agent1 till the position of agent2. It will be something like this:



          L=delta;
          while(L<LThatReachesSecondAgent){
          x1 = agent1.getX() + L*cos(angle);
          y1 = agent1.getY() + L*sin(angle);
          for(Node n : yourCollectionOfNodes){
          If(n.contains(x1,y1))
          return false
          }
          /*This can also work maybe faster
          //int numNodesInTheWay=count(yourCollectionOfNodes,n->n.contains(x1,y1));
          //if(numNodesInTheWay>0) return false
          */
          L+=delta;
          }
          return true





          share|improve this answer























          • interesting, and two points; 1- you can do this without calculating the angle of the line that passes through both agents. 2- in order to make it faster, you can use a zoned environment, and do the above for loop only for those nodes in possible zones, not all of them. something like this: i.stack.imgur.com/kOL1L.png
            – Mohammad Hasan N.
            Nov 13 at 8:29




















          up vote
          0
          down vote













          welcome to SOF.



          No, there is no build-in function, afaik. You would have to code something manually, which is possible but not straight forward. If you want help with that, you need to provide more details on the specifics.






          share|improve this answer





















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            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes








            up vote
            0
            down vote



            accepted










            This is how I did it once in the past. The only problem is that it might be slow if you need to do that calculation thousands of times per second. And it's only in 2D. If you need 3D, the idea is not so different.



            1) add all your building nodes and everything that might be an obstacle between the two agents to a collection. You may want to put a rectangular node around your buildings to keep everything in one collection (assuming you are using space markup with nodes)



            2) generate a delta distance delta (equal to 1 for example) and find the angle of the line that passes through both agents.



            3) Make a loop from the position of agent1 till the position of agent2. It will be something like this:



            L=delta;
            while(L<LThatReachesSecondAgent){
            x1 = agent1.getX() + L*cos(angle);
            y1 = agent1.getY() + L*sin(angle);
            for(Node n : yourCollectionOfNodes){
            If(n.contains(x1,y1))
            return false
            }
            /*This can also work maybe faster
            //int numNodesInTheWay=count(yourCollectionOfNodes,n->n.contains(x1,y1));
            //if(numNodesInTheWay>0) return false
            */
            L+=delta;
            }
            return true





            share|improve this answer























            • interesting, and two points; 1- you can do this without calculating the angle of the line that passes through both agents. 2- in order to make it faster, you can use a zoned environment, and do the above for loop only for those nodes in possible zones, not all of them. something like this: i.stack.imgur.com/kOL1L.png
              – Mohammad Hasan N.
              Nov 13 at 8:29

















            up vote
            0
            down vote



            accepted










            This is how I did it once in the past. The only problem is that it might be slow if you need to do that calculation thousands of times per second. And it's only in 2D. If you need 3D, the idea is not so different.



            1) add all your building nodes and everything that might be an obstacle between the two agents to a collection. You may want to put a rectangular node around your buildings to keep everything in one collection (assuming you are using space markup with nodes)



            2) generate a delta distance delta (equal to 1 for example) and find the angle of the line that passes through both agents.



            3) Make a loop from the position of agent1 till the position of agent2. It will be something like this:



            L=delta;
            while(L<LThatReachesSecondAgent){
            x1 = agent1.getX() + L*cos(angle);
            y1 = agent1.getY() + L*sin(angle);
            for(Node n : yourCollectionOfNodes){
            If(n.contains(x1,y1))
            return false
            }
            /*This can also work maybe faster
            //int numNodesInTheWay=count(yourCollectionOfNodes,n->n.contains(x1,y1));
            //if(numNodesInTheWay>0) return false
            */
            L+=delta;
            }
            return true





            share|improve this answer























            • interesting, and two points; 1- you can do this without calculating the angle of the line that passes through both agents. 2- in order to make it faster, you can use a zoned environment, and do the above for loop only for those nodes in possible zones, not all of them. something like this: i.stack.imgur.com/kOL1L.png
              – Mohammad Hasan N.
              Nov 13 at 8:29















            up vote
            0
            down vote



            accepted







            up vote
            0
            down vote



            accepted






            This is how I did it once in the past. The only problem is that it might be slow if you need to do that calculation thousands of times per second. And it's only in 2D. If you need 3D, the idea is not so different.



            1) add all your building nodes and everything that might be an obstacle between the two agents to a collection. You may want to put a rectangular node around your buildings to keep everything in one collection (assuming you are using space markup with nodes)



            2) generate a delta distance delta (equal to 1 for example) and find the angle of the line that passes through both agents.



            3) Make a loop from the position of agent1 till the position of agent2. It will be something like this:



            L=delta;
            while(L<LThatReachesSecondAgent){
            x1 = agent1.getX() + L*cos(angle);
            y1 = agent1.getY() + L*sin(angle);
            for(Node n : yourCollectionOfNodes){
            If(n.contains(x1,y1))
            return false
            }
            /*This can also work maybe faster
            //int numNodesInTheWay=count(yourCollectionOfNodes,n->n.contains(x1,y1));
            //if(numNodesInTheWay>0) return false
            */
            L+=delta;
            }
            return true





            share|improve this answer














            This is how I did it once in the past. The only problem is that it might be slow if you need to do that calculation thousands of times per second. And it's only in 2D. If you need 3D, the idea is not so different.



            1) add all your building nodes and everything that might be an obstacle between the two agents to a collection. You may want to put a rectangular node around your buildings to keep everything in one collection (assuming you are using space markup with nodes)



            2) generate a delta distance delta (equal to 1 for example) and find the angle of the line that passes through both agents.



            3) Make a loop from the position of agent1 till the position of agent2. It will be something like this:



            L=delta;
            while(L<LThatReachesSecondAgent){
            x1 = agent1.getX() + L*cos(angle);
            y1 = agent1.getY() + L*sin(angle);
            for(Node n : yourCollectionOfNodes){
            If(n.contains(x1,y1))
            return false
            }
            /*This can also work maybe faster
            //int numNodesInTheWay=count(yourCollectionOfNodes,n->n.contains(x1,y1));
            //if(numNodesInTheWay>0) return false
            */
            L+=delta;
            }
            return true






            share|improve this answer














            share|improve this answer



            share|improve this answer








            edited Nov 12 at 20:14

























            answered Nov 12 at 19:24









            Felipe

            1,8052721




            1,8052721












            • interesting, and two points; 1- you can do this without calculating the angle of the line that passes through both agents. 2- in order to make it faster, you can use a zoned environment, and do the above for loop only for those nodes in possible zones, not all of them. something like this: i.stack.imgur.com/kOL1L.png
              – Mohammad Hasan N.
              Nov 13 at 8:29




















            • interesting, and two points; 1- you can do this without calculating the angle of the line that passes through both agents. 2- in order to make it faster, you can use a zoned environment, and do the above for loop only for those nodes in possible zones, not all of them. something like this: i.stack.imgur.com/kOL1L.png
              – Mohammad Hasan N.
              Nov 13 at 8:29


















            interesting, and two points; 1- you can do this without calculating the angle of the line that passes through both agents. 2- in order to make it faster, you can use a zoned environment, and do the above for loop only for those nodes in possible zones, not all of them. something like this: i.stack.imgur.com/kOL1L.png
            – Mohammad Hasan N.
            Nov 13 at 8:29






            interesting, and two points; 1- you can do this without calculating the angle of the line that passes through both agents. 2- in order to make it faster, you can use a zoned environment, and do the above for loop only for those nodes in possible zones, not all of them. something like this: i.stack.imgur.com/kOL1L.png
            – Mohammad Hasan N.
            Nov 13 at 8:29














            up vote
            0
            down vote













            welcome to SOF.



            No, there is no build-in function, afaik. You would have to code something manually, which is possible but not straight forward. If you want help with that, you need to provide more details on the specifics.






            share|improve this answer

























              up vote
              0
              down vote













              welcome to SOF.



              No, there is no build-in function, afaik. You would have to code something manually, which is possible but not straight forward. If you want help with that, you need to provide more details on the specifics.






              share|improve this answer























                up vote
                0
                down vote










                up vote
                0
                down vote









                welcome to SOF.



                No, there is no build-in function, afaik. You would have to code something manually, which is possible but not straight forward. If you want help with that, you need to provide more details on the specifics.






                share|improve this answer












                welcome to SOF.



                No, there is no build-in function, afaik. You would have to code something manually, which is possible but not straight forward. If you want help with that, you need to provide more details on the specifics.







                share|improve this answer












                share|improve this answer



                share|improve this answer










                answered Nov 12 at 18:48









                Benjamin

                9301513




                9301513






























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