Anylogic - line of sight
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
asked Nov 12 '18 at 18:10
chana ross
51
add a comment |
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
asked Nov 12 '18 at 18:10
chana ross
51
add a comment |
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
asked Nov 12 '18 at 18:10
chana ross
51
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
anylogic
asked Nov 12 '18 at 18:10
chana ross
51
asked Nov 12 '18 at 18:10
chana ross
51
asked Nov 12 '18 at 18:10
chana ross
51
asked Nov 12 '18 at 18:10
chana ross
51
asked Nov 12 '18 at 18:10
chana ross
51
51
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
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
answered Nov 12 '18 at 19:24
Felipe
1,8452721
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 aboveforloop 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 '18 at 8:29
add a comment |
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.
answered Nov 12 '18 at 18:48
Benjamin
9301513
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
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active
oldest
votes
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
answered Nov 12 '18 at 19:24
Felipe
1,8452721
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 aboveforloop 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 '18 at 8:29
add a comment |
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
answered Nov 12 '18 at 19:24
Felipe
1,8452721
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 aboveforloop 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 '18 at 8:29
add a comment |
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
answered Nov 12 '18 at 19:24
Felipe
1,8452721
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
answered Nov 12 '18 at 19:24
Felipe
1,8452721
edited Nov 12 '18 at 20:14
answered Nov 12 '18 at 19:24
Felipe
1,8452721
answered Nov 12 '18 at 19:24
Felipe
1,8452721
answered Nov 12 '18 at 19:24
Felipe
1,8452721
1,8452721
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 aboveforloop 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 '18 at 8:29
add a comment |
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 aboveforloop 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 '18 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 '18 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 '18 at 8:29
add a comment |
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.
answered Nov 12 '18 at 18:48
Benjamin
9301513
add a comment |
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.
answered Nov 12 '18 at 18:48
Benjamin
9301513
add a comment |
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.
answered Nov 12 '18 at 18:48
Benjamin
9301513
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.
answered Nov 12 '18 at 18:48
Benjamin
9301513
answered Nov 12 '18 at 18:48
Benjamin
9301513
answered Nov 12 '18 at 18:48
Benjamin
9301513
answered Nov 12 '18 at 18:48
Benjamin
9301513
9301513
add a comment |
add a comment |
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