Detect collision with lines and limit movement
I'm making a game with libGDX in Java. I'm trying to make a collision detection. As you can see in the image, I have a line which is a wall and a player with specified radius. The desired position is the next location which the player is trying to be in. But because there is a wall, he's placed in the Actual Position which is on the Velocity vector, but more closer to the prev location. I'm trying to figure out how can I detect that closer position?
My attempt:
private void move(float deltaTime)
float step;
beginMovementAltitude();
if (playerComponent.isWalking())
step = handleAcceleration(playerComponent.getSpeed() + playerComponent.getAcceleration());
else step = handleDeacceleration(playerComponent.getSpeed(), playerComponent.getAcceleration());
playerComponent.setSpeed(step);
if (step == 0) return;
takeStep(deltaTime, step, 0);
private void takeStep(float deltaTime, float step, int rotate)
Vector3 position = playerComponent.getCamera().position;
float x = position.x;
float y = position.y;
int radius = playerComponent.getRadius();
auxEnvelope.init(x, x + radius, y, y + radius);
List<Line> nearbyLines = lines.query(auxEnvelope);
float theta;
int numberOfIntersections = 0;
float angleToMove = 0;
Gdx.app.log(step + "", "");
for (Line line : nearbyLines)
VertexElement src = line.getSrc();
VertexElement dst = line.getDst();
auxVector3.set(playerComponent.getCamera().direction);
auxVector3.rotate(Vector3.Z, rotate);
float nextX = x + (step * deltaTime) * (auxVector3.x);
float nextY = y + (step * deltaTime) * playerComponent.getCamera().direction.y;
float dis = Intersector.distanceLinePoint(src.getX(), src.getY(), dst.getX(), dst.getY(), nextX, nextY);
boolean bodyIntersection = dis <= 0.5f;
auxVector21.set(src.getX(), src.getY());
auxVector22.set(dst.getX(), dst.getY());
auxVector23.set(nextX, nextY);
if (bodyIntersection)
numberOfIntersections++;
if (numberOfIntersections > 1)
return;
theta = auxVector22.sub(auxVector21).nor().angle();
float angle = (float) (180.0 / MathUtils.PI * MathUtils.atan2(auxVector23.y - position.y, auxVector23.x - position.x));
if (angle < 0) angle += 360;
float diff = (theta > angle) ? theta - angle : angle - theta;
if (step < 0) step *=-1;
angleToMove = (diff > 90) ? theta + 180 : theta;
if (numberOfIntersections == 0)
moveCameraByWalking(deltaTime, step, rotate);
else
moveCameraInDirection(deltaTime, step, angleToMove);
java libgdx 2d collision-detection
|
show 1 more comment
I'm making a game with libGDX in Java. I'm trying to make a collision detection. As you can see in the image, I have a line which is a wall and a player with specified radius. The desired position is the next location which the player is trying to be in. But because there is a wall, he's placed in the Actual Position which is on the Velocity vector, but more closer to the prev location. I'm trying to figure out how can I detect that closer position?
My attempt:
private void move(float deltaTime)
float step;
beginMovementAltitude();
if (playerComponent.isWalking())
step = handleAcceleration(playerComponent.getSpeed() + playerComponent.getAcceleration());
else step = handleDeacceleration(playerComponent.getSpeed(), playerComponent.getAcceleration());
playerComponent.setSpeed(step);
if (step == 0) return;
takeStep(deltaTime, step, 0);
private void takeStep(float deltaTime, float step, int rotate)
Vector3 position = playerComponent.getCamera().position;
float x = position.x;
float y = position.y;
int radius = playerComponent.getRadius();
auxEnvelope.init(x, x + radius, y, y + radius);
List<Line> nearbyLines = lines.query(auxEnvelope);
float theta;
int numberOfIntersections = 0;
float angleToMove = 0;
Gdx.app.log(step + "", "");
for (Line line : nearbyLines)
VertexElement src = line.getSrc();
VertexElement dst = line.getDst();
auxVector3.set(playerComponent.getCamera().direction);
auxVector3.rotate(Vector3.Z, rotate);
float nextX = x + (step * deltaTime) * (auxVector3.x);
float nextY = y + (step * deltaTime) * playerComponent.getCamera().direction.y;
float dis = Intersector.distanceLinePoint(src.getX(), src.getY(), dst.getX(), dst.getY(), nextX, nextY);
boolean bodyIntersection = dis <= 0.5f;
auxVector21.set(src.getX(), src.getY());
auxVector22.set(dst.getX(), dst.getY());
auxVector23.set(nextX, nextY);
if (bodyIntersection)
numberOfIntersections++;
if (numberOfIntersections > 1)
return;
theta = auxVector22.sub(auxVector21).nor().angle();
float angle = (float) (180.0 / MathUtils.PI * MathUtils.atan2(auxVector23.y - position.y, auxVector23.x - position.x));
if (angle < 0) angle += 360;
float diff = (theta > angle) ? theta - angle : angle - theta;
if (step < 0) step *=-1;
angleToMove = (diff > 90) ? theta + 180 : theta;
if (numberOfIntersections == 0)
moveCameraByWalking(deltaTime, step, rotate);
else
moveCameraInDirection(deltaTime, step, angleToMove);
java libgdx 2d collision-detection
What did you try, what is your error?
– michaeak
Nov 13 '18 at 7:22
In the link you gave I didn't see relevant part to put the player in the "fixed" position. What I tried at the beginning is to check if the next position is in distance from wall smaller than radius. If true, I won't let the player move there. This behavior is fine if the speed of player is fixed. But in my game the speed is accelerating and decelerating according to input - so what happens is the player doesn't move if the speed is high but when it is low it does (making some "joggy" movement effect when decelerating near the wall)
– Gad Wissberg
Nov 13 '18 at 7:27
Can you please provide the code for this attempt you described in the question?
– michaeak
Nov 13 '18 at 7:28
Yes, posted to the question
– Gad Wissberg
Nov 13 '18 at 7:32
In the first method it seems a lot of lines are duplicated, is this the actual code? I guess not, as this would not compile (duplicate variable name). Same applies to the second method
– michaeak
Nov 13 '18 at 7:33
|
show 1 more comment
I'm making a game with libGDX in Java. I'm trying to make a collision detection. As you can see in the image, I have a line which is a wall and a player with specified radius. The desired position is the next location which the player is trying to be in. But because there is a wall, he's placed in the Actual Position which is on the Velocity vector, but more closer to the prev location. I'm trying to figure out how can I detect that closer position?
My attempt:
private void move(float deltaTime)
float step;
beginMovementAltitude();
if (playerComponent.isWalking())
step = handleAcceleration(playerComponent.getSpeed() + playerComponent.getAcceleration());
else step = handleDeacceleration(playerComponent.getSpeed(), playerComponent.getAcceleration());
playerComponent.setSpeed(step);
if (step == 0) return;
takeStep(deltaTime, step, 0);
private void takeStep(float deltaTime, float step, int rotate)
Vector3 position = playerComponent.getCamera().position;
float x = position.x;
float y = position.y;
int radius = playerComponent.getRadius();
auxEnvelope.init(x, x + radius, y, y + radius);
List<Line> nearbyLines = lines.query(auxEnvelope);
float theta;
int numberOfIntersections = 0;
float angleToMove = 0;
Gdx.app.log(step + "", "");
for (Line line : nearbyLines)
VertexElement src = line.getSrc();
VertexElement dst = line.getDst();
auxVector3.set(playerComponent.getCamera().direction);
auxVector3.rotate(Vector3.Z, rotate);
float nextX = x + (step * deltaTime) * (auxVector3.x);
float nextY = y + (step * deltaTime) * playerComponent.getCamera().direction.y;
float dis = Intersector.distanceLinePoint(src.getX(), src.getY(), dst.getX(), dst.getY(), nextX, nextY);
boolean bodyIntersection = dis <= 0.5f;
auxVector21.set(src.getX(), src.getY());
auxVector22.set(dst.getX(), dst.getY());
auxVector23.set(nextX, nextY);
if (bodyIntersection)
numberOfIntersections++;
if (numberOfIntersections > 1)
return;
theta = auxVector22.sub(auxVector21).nor().angle();
float angle = (float) (180.0 / MathUtils.PI * MathUtils.atan2(auxVector23.y - position.y, auxVector23.x - position.x));
if (angle < 0) angle += 360;
float diff = (theta > angle) ? theta - angle : angle - theta;
if (step < 0) step *=-1;
angleToMove = (diff > 90) ? theta + 180 : theta;
if (numberOfIntersections == 0)
moveCameraByWalking(deltaTime, step, rotate);
else
moveCameraInDirection(deltaTime, step, angleToMove);
java libgdx 2d collision-detection
I'm making a game with libGDX in Java. I'm trying to make a collision detection. As you can see in the image, I have a line which is a wall and a player with specified radius. The desired position is the next location which the player is trying to be in. But because there is a wall, he's placed in the Actual Position which is on the Velocity vector, but more closer to the prev location. I'm trying to figure out how can I detect that closer position?
My attempt:
private void move(float deltaTime)
float step;
beginMovementAltitude();
if (playerComponent.isWalking())
step = handleAcceleration(playerComponent.getSpeed() + playerComponent.getAcceleration());
else step = handleDeacceleration(playerComponent.getSpeed(), playerComponent.getAcceleration());
playerComponent.setSpeed(step);
if (step == 0) return;
takeStep(deltaTime, step, 0);
private void takeStep(float deltaTime, float step, int rotate)
Vector3 position = playerComponent.getCamera().position;
float x = position.x;
float y = position.y;
int radius = playerComponent.getRadius();
auxEnvelope.init(x, x + radius, y, y + radius);
List<Line> nearbyLines = lines.query(auxEnvelope);
float theta;
int numberOfIntersections = 0;
float angleToMove = 0;
Gdx.app.log(step + "", "");
for (Line line : nearbyLines)
VertexElement src = line.getSrc();
VertexElement dst = line.getDst();
auxVector3.set(playerComponent.getCamera().direction);
auxVector3.rotate(Vector3.Z, rotate);
float nextX = x + (step * deltaTime) * (auxVector3.x);
float nextY = y + (step * deltaTime) * playerComponent.getCamera().direction.y;
float dis = Intersector.distanceLinePoint(src.getX(), src.getY(), dst.getX(), dst.getY(), nextX, nextY);
boolean bodyIntersection = dis <= 0.5f;
auxVector21.set(src.getX(), src.getY());
auxVector22.set(dst.getX(), dst.getY());
auxVector23.set(nextX, nextY);
if (bodyIntersection)
numberOfIntersections++;
if (numberOfIntersections > 1)
return;
theta = auxVector22.sub(auxVector21).nor().angle();
float angle = (float) (180.0 / MathUtils.PI * MathUtils.atan2(auxVector23.y - position.y, auxVector23.x - position.x));
if (angle < 0) angle += 360;
float diff = (theta > angle) ? theta - angle : angle - theta;
if (step < 0) step *=-1;
angleToMove = (diff > 90) ? theta + 180 : theta;
if (numberOfIntersections == 0)
moveCameraByWalking(deltaTime, step, rotate);
else
moveCameraInDirection(deltaTime, step, angleToMove);
java libgdx 2d collision-detection
java libgdx 2d collision-detection
edited Nov 13 '18 at 7:45
Gad Wissberg
asked Nov 13 '18 at 7:02
Gad WissbergGad Wissberg
7118
7118
What did you try, what is your error?
– michaeak
Nov 13 '18 at 7:22
In the link you gave I didn't see relevant part to put the player in the "fixed" position. What I tried at the beginning is to check if the next position is in distance from wall smaller than radius. If true, I won't let the player move there. This behavior is fine if the speed of player is fixed. But in my game the speed is accelerating and decelerating according to input - so what happens is the player doesn't move if the speed is high but when it is low it does (making some "joggy" movement effect when decelerating near the wall)
– Gad Wissberg
Nov 13 '18 at 7:27
Can you please provide the code for this attempt you described in the question?
– michaeak
Nov 13 '18 at 7:28
Yes, posted to the question
– Gad Wissberg
Nov 13 '18 at 7:32
In the first method it seems a lot of lines are duplicated, is this the actual code? I guess not, as this would not compile (duplicate variable name). Same applies to the second method
– michaeak
Nov 13 '18 at 7:33
|
show 1 more comment
What did you try, what is your error?
– michaeak
Nov 13 '18 at 7:22
In the link you gave I didn't see relevant part to put the player in the "fixed" position. What I tried at the beginning is to check if the next position is in distance from wall smaller than radius. If true, I won't let the player move there. This behavior is fine if the speed of player is fixed. But in my game the speed is accelerating and decelerating according to input - so what happens is the player doesn't move if the speed is high but when it is low it does (making some "joggy" movement effect when decelerating near the wall)
– Gad Wissberg
Nov 13 '18 at 7:27
Can you please provide the code for this attempt you described in the question?
– michaeak
Nov 13 '18 at 7:28
Yes, posted to the question
– Gad Wissberg
Nov 13 '18 at 7:32
In the first method it seems a lot of lines are duplicated, is this the actual code? I guess not, as this would not compile (duplicate variable name). Same applies to the second method
– michaeak
Nov 13 '18 at 7:33
What did you try, what is your error?
– michaeak
Nov 13 '18 at 7:22
What did you try, what is your error?
– michaeak
Nov 13 '18 at 7:22
In the link you gave I didn't see relevant part to put the player in the "fixed" position. What I tried at the beginning is to check if the next position is in distance from wall smaller than radius. If true, I won't let the player move there. This behavior is fine if the speed of player is fixed. But in my game the speed is accelerating and decelerating according to input - so what happens is the player doesn't move if the speed is high but when it is low it does (making some "joggy" movement effect when decelerating near the wall)
– Gad Wissberg
Nov 13 '18 at 7:27
In the link you gave I didn't see relevant part to put the player in the "fixed" position. What I tried at the beginning is to check if the next position is in distance from wall smaller than radius. If true, I won't let the player move there. This behavior is fine if the speed of player is fixed. But in my game the speed is accelerating and decelerating according to input - so what happens is the player doesn't move if the speed is high but when it is low it does (making some "joggy" movement effect when decelerating near the wall)
– Gad Wissberg
Nov 13 '18 at 7:27
Can you please provide the code for this attempt you described in the question?
– michaeak
Nov 13 '18 at 7:28
Can you please provide the code for this attempt you described in the question?
– michaeak
Nov 13 '18 at 7:28
Yes, posted to the question
– Gad Wissberg
Nov 13 '18 at 7:32
Yes, posted to the question
– Gad Wissberg
Nov 13 '18 at 7:32
In the first method it seems a lot of lines are duplicated, is this the actual code? I guess not, as this would not compile (duplicate variable name). Same applies to the second method
– michaeak
Nov 13 '18 at 7:33
In the first method it seems a lot of lines are duplicated, is this the actual code? I guess not, as this would not compile (duplicate variable name). Same applies to the second method
– michaeak
Nov 13 '18 at 7:33
|
show 1 more comment
1 Answer
1
active
oldest
votes
The idea is to find intersection of path of object center and the line moved by radius of the circle, see that picture.
At first, you need to find a normal to the line. How to do it, depends on how the line is defined, if it's defined by two points, the formula is
nx = ay - by
ny = bx - ax
If the line is defined by canonical equation, then coefficients at x and y define normal, if I remembered correctly.
When normal is found, we need to normalize it - set length to 1 by dividing coordinates by vector length. Let it be n.
Then, we will project starting point, desired point and randomly chosen point on line to n, treating them as radius vectors.
Projection of vector a to vector b is
project (a, b) = scalar_product (a, b) / length (b)**2 * b
but since b is n which length equals 1, we will not apply division, and also we want to only find length of the result, we do not multiply by b. So we only compute scalar product with n for each of three aforementioned points, getting three numbers, let s be the result for starting point, d for desired point, l for chosen point on the line.
Then we should modify l by radius of the circle:
if (s < d) l -= r;
else if (s > d) l += r;
If s = d, your object moves in parallel along the line, so line can't obstruct its movement. It's highly improbable case but should be dealt with.
Also, that's important, if l was initially between s and d but after modifying is no longer between then, it's a special case you may want to handle (restrict object movement for example)
Ather that, you should compute (d - s) / (l - s).
If the result is greater or equals 1, the object will not reach the line.
If the result is between 0 and 1, the line obstructs movement and the result indicates part of the path the object will complete. 0.5 means that object will stop halfway.
If the result is negative, it means the line is behind the object and will not obstruct movement.
Note that when using floating point numbers the result will not be perfectly precise, that's why we handle that special case. If you want to prevent this from happening at all, organize loop and try approximations until needed precision is reached.
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The idea is to find intersection of path of object center and the line moved by radius of the circle, see that picture.
At first, you need to find a normal to the line. How to do it, depends on how the line is defined, if it's defined by two points, the formula is
nx = ay - by
ny = bx - ax
If the line is defined by canonical equation, then coefficients at x and y define normal, if I remembered correctly.
When normal is found, we need to normalize it - set length to 1 by dividing coordinates by vector length. Let it be n.
Then, we will project starting point, desired point and randomly chosen point on line to n, treating them as radius vectors.
Projection of vector a to vector b is
project (a, b) = scalar_product (a, b) / length (b)**2 * b
but since b is n which length equals 1, we will not apply division, and also we want to only find length of the result, we do not multiply by b. So we only compute scalar product with n for each of three aforementioned points, getting three numbers, let s be the result for starting point, d for desired point, l for chosen point on the line.
Then we should modify l by radius of the circle:
if (s < d) l -= r;
else if (s > d) l += r;
If s = d, your object moves in parallel along the line, so line can't obstruct its movement. It's highly improbable case but should be dealt with.
Also, that's important, if l was initially between s and d but after modifying is no longer between then, it's a special case you may want to handle (restrict object movement for example)
Ather that, you should compute (d - s) / (l - s).
If the result is greater or equals 1, the object will not reach the line.
If the result is between 0 and 1, the line obstructs movement and the result indicates part of the path the object will complete. 0.5 means that object will stop halfway.
If the result is negative, it means the line is behind the object and will not obstruct movement.
Note that when using floating point numbers the result will not be perfectly precise, that's why we handle that special case. If you want to prevent this from happening at all, organize loop and try approximations until needed precision is reached.
add a comment |
The idea is to find intersection of path of object center and the line moved by radius of the circle, see that picture.
At first, you need to find a normal to the line. How to do it, depends on how the line is defined, if it's defined by two points, the formula is
nx = ay - by
ny = bx - ax
If the line is defined by canonical equation, then coefficients at x and y define normal, if I remembered correctly.
When normal is found, we need to normalize it - set length to 1 by dividing coordinates by vector length. Let it be n.
Then, we will project starting point, desired point and randomly chosen point on line to n, treating them as radius vectors.
Projection of vector a to vector b is
project (a, b) = scalar_product (a, b) / length (b)**2 * b
but since b is n which length equals 1, we will not apply division, and also we want to only find length of the result, we do not multiply by b. So we only compute scalar product with n for each of three aforementioned points, getting three numbers, let s be the result for starting point, d for desired point, l for chosen point on the line.
Then we should modify l by radius of the circle:
if (s < d) l -= r;
else if (s > d) l += r;
If s = d, your object moves in parallel along the line, so line can't obstruct its movement. It's highly improbable case but should be dealt with.
Also, that's important, if l was initially between s and d but after modifying is no longer between then, it's a special case you may want to handle (restrict object movement for example)
Ather that, you should compute (d - s) / (l - s).
If the result is greater or equals 1, the object will not reach the line.
If the result is between 0 and 1, the line obstructs movement and the result indicates part of the path the object will complete. 0.5 means that object will stop halfway.
If the result is negative, it means the line is behind the object and will not obstruct movement.
Note that when using floating point numbers the result will not be perfectly precise, that's why we handle that special case. If you want to prevent this from happening at all, organize loop and try approximations until needed precision is reached.
add a comment |
The idea is to find intersection of path of object center and the line moved by radius of the circle, see that picture.
At first, you need to find a normal to the line. How to do it, depends on how the line is defined, if it's defined by two points, the formula is
nx = ay - by
ny = bx - ax
If the line is defined by canonical equation, then coefficients at x and y define normal, if I remembered correctly.
When normal is found, we need to normalize it - set length to 1 by dividing coordinates by vector length. Let it be n.
Then, we will project starting point, desired point and randomly chosen point on line to n, treating them as radius vectors.
Projection of vector a to vector b is
project (a, b) = scalar_product (a, b) / length (b)**2 * b
but since b is n which length equals 1, we will not apply division, and also we want to only find length of the result, we do not multiply by b. So we only compute scalar product with n for each of three aforementioned points, getting three numbers, let s be the result for starting point, d for desired point, l for chosen point on the line.
Then we should modify l by radius of the circle:
if (s < d) l -= r;
else if (s > d) l += r;
If s = d, your object moves in parallel along the line, so line can't obstruct its movement. It's highly improbable case but should be dealt with.
Also, that's important, if l was initially between s and d but after modifying is no longer between then, it's a special case you may want to handle (restrict object movement for example)
Ather that, you should compute (d - s) / (l - s).
If the result is greater or equals 1, the object will not reach the line.
If the result is between 0 and 1, the line obstructs movement and the result indicates part of the path the object will complete. 0.5 means that object will stop halfway.
If the result is negative, it means the line is behind the object and will not obstruct movement.
Note that when using floating point numbers the result will not be perfectly precise, that's why we handle that special case. If you want to prevent this from happening at all, organize loop and try approximations until needed precision is reached.
The idea is to find intersection of path of object center and the line moved by radius of the circle, see that picture.
At first, you need to find a normal to the line. How to do it, depends on how the line is defined, if it's defined by two points, the formula is
nx = ay - by
ny = bx - ax
If the line is defined by canonical equation, then coefficients at x and y define normal, if I remembered correctly.
When normal is found, we need to normalize it - set length to 1 by dividing coordinates by vector length. Let it be n.
Then, we will project starting point, desired point and randomly chosen point on line to n, treating them as radius vectors.
Projection of vector a to vector b is
project (a, b) = scalar_product (a, b) / length (b)**2 * b
but since b is n which length equals 1, we will not apply division, and also we want to only find length of the result, we do not multiply by b. So we only compute scalar product with n for each of three aforementioned points, getting three numbers, let s be the result for starting point, d for desired point, l for chosen point on the line.
Then we should modify l by radius of the circle:
if (s < d) l -= r;
else if (s > d) l += r;
If s = d, your object moves in parallel along the line, so line can't obstruct its movement. It's highly improbable case but should be dealt with.
Also, that's important, if l was initially between s and d but after modifying is no longer between then, it's a special case you may want to handle (restrict object movement for example)
Ather that, you should compute (d - s) / (l - s).
If the result is greater or equals 1, the object will not reach the line.
If the result is between 0 and 1, the line obstructs movement and the result indicates part of the path the object will complete. 0.5 means that object will stop halfway.
If the result is negative, it means the line is behind the object and will not obstruct movement.
Note that when using floating point numbers the result will not be perfectly precise, that's why we handle that special case. If you want to prevent this from happening at all, organize loop and try approximations until needed precision is reached.
answered Nov 20 '18 at 10:16
trollingchartrollingchar
3187
3187
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What did you try, what is your error?
– michaeak
Nov 13 '18 at 7:22
In the link you gave I didn't see relevant part to put the player in the "fixed" position. What I tried at the beginning is to check if the next position is in distance from wall smaller than radius. If true, I won't let the player move there. This behavior is fine if the speed of player is fixed. But in my game the speed is accelerating and decelerating according to input - so what happens is the player doesn't move if the speed is high but when it is low it does (making some "joggy" movement effect when decelerating near the wall)
– Gad Wissberg
Nov 13 '18 at 7:27
Can you please provide the code for this attempt you described in the question?
– michaeak
Nov 13 '18 at 7:28
Yes, posted to the question
– Gad Wissberg
Nov 13 '18 at 7:32
In the first method it seems a lot of lines are duplicated, is this the actual code? I guess not, as this would not compile (duplicate variable name). Same applies to the second method
– michaeak
Nov 13 '18 at 7:33