I have a circle being drawn at a certain position. I can move it just fine with speed set to 10f but when it starts to circle it becomes extremely fast. Its obviously not moving at (units/second) I'm not sure whats going on. I thought that the archSpeed needed to be in radians or something, that slowed it down - still not right though.
Here's the Circle Arc Equation I'm basing off of:
s = r * theta
Here are the functions I'm using:
private void moveOut(double deltaTime)
{
SetPosition(x += direction * speed * deltaTime, y, 0);
if (x - (direction * GetWidth() / 2f) >= centerX + radius + GetWidth() / 2f)
{
//onOutside = true;
}
Log.d(TAG, "moving out");
}
private void circleCenter(double deltaTime)
{
float angleSpeed = (float) (radius * (speed * Math.PI / 180) * deltaTime);
currentAngle += angleSpeed;
if (currentAngle >= 2 * Math.PI)
{
currentAngle = (float) (2 * Math.PI - currentAngle);
}
SetPosition(centerX + radius * FloatMath.cos(currentAngle), centerY + radius * FloatMath.sin(currentAngle), 0);
}
Your angleSpeed formula looks wrong.
I'd work it out first by saying What is the distance I travel in that time. The answer as you already know is speed*deltaTime. Now you have a distance you can work out the angle by using the arc forumla that says arclength = radius*angle. So angle = arclength/radius.
Put these two together to get
angle = speed*deltaTime/radius
This will be in radians of course.
Essentially this boils down to the fact you were multiplying by radius instead of dividing by it (looking at it in terms of units would have helped spot this but that is outside the scope of a programming forum).
Related
I currently have a sprite following a path...all is working well from a steering perspective, however I am trying to make it so the center of the sprite tracks along the path, rather than the corner (0,0) tracing along the path. Essentially, I would like the center of the fish to follow the line.
Below is an image of what I have implemented, and what I would like to achieve.
The core implementation of this mechanic lies within my update() method; as follows;
private void update(float deltaTime) {
float angle = (float) Math.atan2(path.get(waypoint).y - getY(), path.get(waypoint).x - getX());
velocity.set((float) Math.cos(angle) * speed, (float) Math.sin(angle) * speed);
velocity_normal = new Vector2(-velocity.y, velocity.x).nor();
setPosition(
getX() + (velocity.x * deltaTime),
getY() + (velocity.y * deltaTime)
);
setRotation(angle * MathUtils.radiansToDegrees);
if(isWayPointReached()){
setPosition(path.get(waypoint).x, path.get(waypoint).y);
if(waypoint + 1 >= path.size){
waypoint = 0;
} else {
waypoint++;
}
}
}
In particular the setPosition call. My initial thoughts were to calculate the normal vector to the velocity vector, normalize, and multiply the x and y components respectively by the fish height... to my mind this would offset the fish by its height (150px). The code attempted is as follows;
velocity_normal = new Vector2(-velocity.y, velocity.x).nor();
setPosition(
getX() + (velocity.x * deltaTime) + velocity_normal.x * getHeight() * deltaTime,
getY() + (velocity.y * deltaTime) + velocity_normal.y * getHeight() * deltaTime
);
The results in some odd behavior, the fish gets progressively further from the line, it seems the vector is getting compounded and added each frame.
I have also tried to update the normal vector once each way-point has been reached, however this does not work either.
I think the above logic is correct, however have I made a fundamental error in my vector maths?
Your assistance would be greatly appreciated.
EDIT:
Added to constructor:
setPosition(
path.get(waypoint).x - 0.5f * getWidth() ,
path.get(waypoint).y - 0.5f * getHeight()
);
Amended update() method;
private void update(float deltaTime) {
float angle = (float) Math.atan2(path.get(waypoint).y - getY(), path.get(waypoint).x - getX());
velocity.set((float) Math.cos(angle) * speed, (float) Math.sin(angle) * speed);
Vector2 velocity_normal = new Vector2();
velocity_normal.set(velocity).nor().scl( speed * deltaTime ); // multiply the speed to scale the unit vector up
translate( velocity_normal.x, velocity_normal.y );
setRotation(angle * MathUtils.radiansToDegrees);
if(isWayPointReached()){
setPosition(path.get(waypoint).x, path.get(waypoint).y);
if(waypoint + 1 >= path.size){
waypoint = 0;
} else {
waypoint++;
}
}
}
Note the omission of the setPosition call, and the replacement with;
Vector2 velocity_normal = new Vector2();
velocity_normal.set(velocity).nor().scl( speed * deltaTime );
translate( velocity_normal.x, velocity_normal.y );
How would I influence the pointA / pointB as mentioned below?
Thanks.
The Sprite position is from the left bottom corner and the origin (where the sprite rotates around) is already set to the center of the sprite. So only the local offset is wrong. You have to substract half the size from the position and then the sprite can move relatively from that offset.
Where you set the sprite:
setPosition(
path.get(waypoint).x - 0.5f * getWidth(),
path.get(waypoint).y - 0.5f * getHeight()
);
In the update method. Because you are adding the velocity every frame you can translate the sprite.
void update(float deltaTime){
// directional global vector
Vector2 velocity = tmp.set(path.get(waypoint)).sub(path.get(waypoint - 1)).nor().scl(speed * deltaTime);
// reference angle is relative to the right vector(1, 0)
float angle = velocity.angle();
setRotation(angle);
translate(velocity.x, velocity.y);
if (isWayPointReached()){
setPosition(
path.get(waypoint).x - 0.5f * getWidth(),
path.get(waypoint).y - 0.5f * getHeight()
);
if(waypoint + 1 >= path.size){
waypoint = 1;
} else {
waypoint++;
}
}
}
There is setCenter() method in Sprite class:
https://libgdx.badlogicgames.com/nightlies/docs/api/com/badlogic/gdx/graphics/g2d/Sprite.html#setCenter-float-float-
Try setting center like that instead of calculating it additionally.
I'm trying to add some distance (e.g. 10px) between a segment (arc) of the pie chart and it's center without success, here's what i've tried so far:
int value = 20; // example
double arcAngle = (value * 360 / 100);
double angle = 360 - (arcAngle / 2); // direction to add the distance to (center of arc)
double newX = pieCenterX + Math.cos(angle * Math.PI / 180.0) * 10;
double newY = pieCenterY + Math.sin(angle * Math.PI / 180.0) * 10;
// then drawing the arc with new x and y
g.fill(new Arc2D.Double(newX, newY, bounds.getWidth(), bounds.getHeight(), startAngle, arcAngle, Arc2D.PIE));
Ideally i should end up with something like that:
I don't know much on how to approach this, so my code was taken from examples i found elsewhere.
Usually zero angle is OX direction (right). So you have to make correction by 90 degrees (if your coordinate system is counterclockwise)
double angle = 90 + 360 - (arcAngle / 2);
I have a custom View, IndicatorView, which is essentially a triangle that orients itself according to a specified angle of a circle with a radius equal to the triangle's length. The angle the triangle points to is frequently updated and I would like to animate between these two positions similar to how a hand on a clock moves. Below is an illustration of my custom view (not drawn proportionally or to scale; drawn according to the Android View coordinate plane):
In the IndicatorView class, I draw the triangle using a Path object and three PointF objects:
#Override
protected void onDraw(Canvas canvas){
path = new Path();
path.setFillType(Path.FillType.EVEN_ODD);
//a, b, and c are PointF objects
path.moveTo(a.x, a.y);
path.lineTo(b.x, b.y);
path.lineTo(c.x, c.y);
path.close();
canvas.drawPath(path, paint);
}
To calculate the different points, given the angle, I use parametric equations:
public void showAngle(){
//x = centerX + radius * cos(angle)
//y = centerY + radius * sin(angle)
//TODO sloppy; tidy up / optimize once finished
//centerX, centerY, length, and bottomWidth are all values
//calculated in onSizeChanged
a = new PointF((float) (centerX + (length * Math.cos(angle))), (float) (centerY + (length * Math.sin(angle))));
//perpendicular bilateral radius
double pRadius = bottomWidth / 2;
//perpendicular angle plus or minus 90 degrees depending on point
float pAngle = angle - 90;
pAngle = (pAngle < 0) ? 360 - Math.abs(pAngle) : pAngle;
pAngle = (pAngle > 360) ? pAngle % 360 : pAngle;
b = new PointF((float) (centerX + (pRadius * Math.cos(pAngle))), (float) (centerY + (pRadius * Math.sin(pAngle))));
pAngle = angle + 90;
pAngle = (pAngle < 0) ? 360 - Math.abs(pAngle) : pAngle;
pAngle = (pAngle > 360) ? pAngle % 360 : pAngle;
c = new PointF((float) (centerX + (pRadius * Math.cos(pAngle))), (float) (centerY + pRadius * Math.sin(pAngle)));
invalidate();
}
When I have a new angle, I use an ObjectAnimator to animate between the two angles. I place an AnimatorUpdateListener on the ObjectAnimator and call my showAngle() method in my IndicatorView using the intermediate values specified from the Animator:
public void updateAngle(float newAngle){
//don't animate to an angle if the previous angle is the same
if(view.getAngle() != newAngle){
if(anim != null && anim.isRunning()){
anim.cancel();
}
anim = ObjectAnimator.ofFloat(view, "angle", view.getAngle(), newAngle);
anim.setDuration(duration);
anim.addUpdateListener(new ValueAnimator.AnimatorUpdateListener() {
#Override
public void onAnimationUpdate(ValueAnimator animation) {
if(view != null){
view.showAngle();
}
}
});
}
}
However, this code produces some strange and unexpected behavior:
The width size of the triangle changes somewhat drastically. This could be due to casting between different types but it shouldn't be that dramatic.
The point of the triangle never stops at the specified angle. Instead it just keeps moving in a circle.
The angle seems to dictate the animations speed rather than where the triangle should stop.
Sometimes it seems as though there are numerous triangles on the screen. This could be due to the speed, perhaps it's moving very fast.
Obviously, somewhere along the line my calculations must be incorrect, though, I'm struggling to find out where I went wrong. Question(s): Is there a more efficient way of getting my custom view to animate rotation to a given angle? If I am approaching this correctly, where am I going wrong?
So, the solution to my problem was rather simple but simple enough to be overlooked. The angle field that was being used for the calculations was in degrees and it just had to be converted to radians in order for it to work with the sin and cos methods.
Change all PointF instantiations, for instance:
a = new PointF((float) (centerX + (length * Math.cos(angle))), (float) (centerY + (length * Math.sin(angle))));
to use the angle in radians:
a = new PointF((float) (centerX + (length * Math.cos(Math.toRadians(angle))),
(float) (centerY + (length * Math.sin(Math.toRadians(angle)))));
Also, part of the problem was due to sound constantly being analyzed and the View being updated before the previous animation had time to render a few frames. This led to the IndicatorView hardly moving when the angle was being updated often and when it was not it would quickly move to its destination. This happens because the previous animation is canceled before another animation is set (which is necessary to prevent a delay). This is a tricky problem to fix but one optimization I found was to avoid starting a new animation if the current angle and the previous angle were relatively close to each other.
Hopefully this will be useful for someone stuck with a similar problem. This was all part of a guitar tuner project I was working on and the source can be found on GitHub.
I have a method in my android app that looks like this:
//get point after rotation
public static PointF getRotatedPoint(PointF pt,PointF center, float degrees)
{
double angleInRadians = degrees * (Math.PI / 180);
pt.x = (float) (Math.cos(angleInRadians) * (pt.x-center.x) - Math.sin(angleInRadians) * (pt.y-center.y) + center.x);
pt.y = (float) (Math.sin(angleInRadians) * (pt.x-center.x) + Math.cos(angleInRadians) * (pt.y-center.y) + center.y);
return pt;
}
I have a rectangle that I rotate by 45 degrees. I can touch any point on the rotated rectangle and it gives me the touched point I want to get the coordinates of the point if the rectangle wasn't rotated. So I pass in -45 in the degrees argument. Here is how I call it:
getRotatedPoint(touchedPoint, centerOfRectangle,-45);
When I draw the point on the rectangle before it gets rotated, it gives me a result close to the position I touched on the rotated rectangle but off by a pretty big difference.
Here is a picture to explain my problem:
I think this might be a problem with my math so any answers are greatly appreciated.
You are mixing initial and final values in the calculations. You re-assign pt.x:
pt.x = (float) (Math.cos(angleInRadians) * (pt.x-center.x) - Math.sin(angleInRadians) * (pt.y-center.y) + center.x);
which doesn't immediately pose any problems. But the calculation for pt.y relies on the original value of pt.x, not the rotated value:
pt.y = (float) (Math.sin(angleInRadians) * (pt.x-center.x) + Math.cos(angleInRadians) * (pt.y-center.y) + center.y);
Thus just use some temporary variables to hold the initial values.
public static PointF getRotatedPoint(PointF pt,PointF center, float degrees)
{
double x0 = pt.x;
double y0 = pt.y;
double angleInRadians = degrees * (Math.PI / 180);
pt.x = (float) (Math.cos(angleInRadians) * (x0-center.x) - Math.sin(angleInRadians) * (y0-center.y) + center.x);
pt.y = (float) (Math.sin(angleInRadians) * (x0-center.x) + Math.cos(angleInRadians) * (y0-center.y) + center.y);
return pt;
}
Since in the digital world a real collision almost never happens, we will always have a situation where the "colliding" circle overlaps the rectangle.
How to put back the circle in the situation where it collides perfectly with the rectangle without overlap?
Suppose that the rectangle is stopped (null velocity) and axis-aligned.
I would solve this problem with a posteriori approach (in two dimensions).
In short I have to solve this equation for t:
Where:
is a number that answers to the question: how many frames ago did the
collision happen perfectly?
is the radius of the circle.
is the center of the circle
is its velocity.
and are functions that return the x and y coordinates of
the point where the circle and the rectangle collide (when the circle is
at position, that is in the position in which perfectly collide with the rectangle).
Recently I solved a similar problem for collisions between circles, but now I don't know the law of the functions A and B.
After years of staring at this problem, and never coming up with a perfect solution, I've finally done it!
It's pretty much a straight forward algorithm, no need for looping and approximations.
This is how it works at a higher level:
Calculate intersection times with each side's plane IF the path from current point to future point crosses that plane.
Check each side's quadrant for single-side intersection, return the intersection.
Determine the corner that the circle is colliding with.
Solve the triangle between the current point, the corner, and the intersecting center (radius away from the corner).
Calculate time, normal, and intersection center.
And now to the gory details!
The input to the function is bounds (which has a left, top, right, bottom) and a current point (start) and a future point (end).
The output is a class called Intersection which has x, y, time, nx, and ny.
{x, y} is the center of the circle at intersection time.
time is a value from 0 to 1 where 0 is at start and 1 is at end
{nx, ny} is the normal, used for reflecting the velocity to determine the new velocity of the circle
We start off with caching variables we use often:
float L = bounds.left;
float T = bounds.top;
float R = bounds.right;
float B = bounds.bottom;
float dx = end.x - start.x;
float dy = end.y - start.y;
And calculating intersection times with each side's plane (if the vector between start and end pass over that plane):
float ltime = Float.MAX_VALUE;
float rtime = Float.MAX_VALUE;
float ttime = Float.MAX_VALUE;
float btime = Float.MAX_VALUE;
if (start.x - radius < L && end.x + radius > L) {
ltime = ((L - radius) - start.x) / dx;
}
if (start.x + radius > R && end.x - radius < R) {
rtime = (start.x - (R + radius)) / -dx;
}
if (start.y - radius < T && end.y + radius > T) {
ttime = ((T - radius) - start.y) / dy;
}
if (start.y + radius > B && end.y - radius < B) {
btime = (start.y - (B + radius)) / -dy;
}
Now we try to see if it's strictly a side intersection (and not corner). If the point of collision lies on the side then return the intersection:
if (ltime >= 0.0f && ltime <= 1.0f) {
float ly = dy * ltime + start.y;
if (ly >= T && ly <= B) {
return new Intersection( dx * ltime + start.x, ly, ltime, -1, 0 );
}
}
else if (rtime >= 0.0f && rtime <= 1.0f) {
float ry = dy * rtime + start.y;
if (ry >= T && ry <= B) {
return new Intersection( dx * rtime + start.x, ry, rtime, 1, 0 );
}
}
if (ttime >= 0.0f && ttime <= 1.0f) {
float tx = dx * ttime + start.x;
if (tx >= L && tx <= R) {
return new Intersection( tx, dy * ttime + start.y, ttime, 0, -1 );
}
}
else if (btime >= 0.0f && btime <= 1.0f) {
float bx = dx * btime + start.x;
if (bx >= L && bx <= R) {
return new Intersection( bx, dy * btime + start.y, btime, 0, 1 );
}
}
We've gotten this far so we know either there's no intersection, or it's collided with a corner. We need to determine the corner:
float cornerX = Float.MAX_VALUE;
float cornerY = Float.MAX_VALUE;
if (ltime != Float.MAX_VALUE) {
cornerX = L;
} else if (rtime != Float.MAX_VALUE) {
cornerX = R;
}
if (ttime != Float.MAX_VALUE) {
cornerY = T;
} else if (btime != Float.MAX_VALUE) {
cornerY = B;
}
// Account for the times where we don't pass over a side but we do hit it's corner
if (cornerX != Float.MAX_VALUE && cornerY == Float.MAX_VALUE) {
cornerY = (dy > 0.0f ? B : T);
}
if (cornerY != Float.MAX_VALUE && cornerX == Float.MAX_VALUE) {
cornerX = (dx > 0.0f ? R : L);
}
Now we have enough information to solve for the triangle. This uses the distance formula, finding the angle between two vectors, and the law of sines (twice):
double inverseRadius = 1.0 / radius;
double lineLength = Math.sqrt( dx * dx + dy * dy );
double cornerdx = cornerX - start.x;
double cornerdy = cornerY - start.y;
double cornerdist = Math.sqrt( cornerdx * cornerdx + cornerdy * cornerdy );
double innerAngle = Math.acos( (cornerdx * dx + cornerdy * dy) / (lineLength * cornerdist) );
double innerAngleSin = Math.sin( innerAngle );
double angle1Sin = innerAngleSin * cornerdist * inverseRadius;
// The angle is too large, there cannot be an intersection
if (Math.abs( angle1Sin ) > 1.0f) {
return null;
}
double angle1 = Math.PI - Math.asin( angle1Sin );
double angle2 = Math.PI - innerAngle - angle1;
double intersectionDistance = radius * Math.sin( angle2 ) / innerAngleSin;
Now that we solved for all sides and angles, we can determine time and everything else:
// Solve for time
float time = (float)(intersectionDistance / lineLength);
// If time is outside the boundaries, return null. This algorithm can
// return a negative time which indicates the previous intersection.
if (time > 1.0f || time < 0.0f) {
return null;
}
// Solve the intersection and normal
float ix = time * dx + start.x;
float iy = time * dy + start.y;
float nx = (float)((ix - cornerX) * inverseRadius);
float ny = (float)((iy - cornerY) * inverseRadius);
return new Intersection( ix, iy, time, nx, ny );
Woo! That was fun... this has plenty of room for improvements as far as efficiency goes. You could reorder the side intersection checking to escape as early as possible while making as few calculations as possible.
I was hoping there would be a way to do it without trigonometric functions, but I had to give in!
Here's an example of me calling it and using it to calculate the new position of the circle using the normal to reflect and the intersection time to calculate the magnitude of reflection:
Intersection inter = handleIntersection( bounds, start, end, radius );
if (inter != null)
{
// Project Future Position
float remainingTime = 1.0f - inter.time;
float dx = end.x - start.x;
float dy = end.y - start.y;
float dot = dx * inter.nx + dy * inter.ny;
float ndx = dx - 2 * dot * inter.nx;
float ndy = dy - 2 * dot * inter.ny;
float newx = inter.x + ndx * remainingTime;
float newy = inter.y + ndy * remainingTime;
// new circle position = {newx, newy}
}
And I've posted the full code on pastebin with a completely interactive example where you can plot the starting and ending points and it shows you the time and resulting bounce off of the rectangle.
If you want to get it running right away you'll have to download code from my blog, otherwise stick it in your own Java2D application.
EDIT:
I've modified the code in pastebin to also include the collision point, and also made some speed improvements.
EDIT:
You can modify this for a rotating rectangle by using that rectangle's angle to un-rotate the rectangle with the circle start and end points. You'll perform the intersection check and then rotate the resulting points and normals.
EDIT:
I modified the code on pastebin to exit early if the bounding volume of the path of the circle does not intersect with the rectangle.
Finding the moment of contact isn't too hard:
You need the position of the circle and rectangle at the timestep before the collision (B) and the timestep after (A). Calculate the distance from the center of the circle to the line of the rectangle it collides with at times A and B (ie, a common formula for a distance from a point to a line), and then the time of collision is:
tC = dt*(dB-R)/(dA+dB),
where tC is the time of collision, dt is the timestep, dB is the distance to line before the collision, dA is the distance after the collision, and R is the radius of the circle.
This assumes everything is locally linear, that is, that your timesteps are reasonably small, and so that the velocity, etc, don't change much in the timestep where you calculate the collision. This is, after all, the point of timesteps: in that with a small enough timestep, non-linear problems are locally linear. In the equation above I take advantage of that: dB-R is the distance from the circle to the line, and dA+dB is the total distance moved, so this question just equates the distance ratio to the time ratio assuming everything is approximately linear within the timestep. (Of course, at the moment of collision the linear approximation isn't its best, but to find the moment of collision, the question is whether it's linear within a timestep up to to moment of collision.)
It's a non-linear problem, right?
You take a time step and move the ball by its displacement calculated using velocity at the start of the step. If you find overlap, reduce the step size and recalculate til convergence.
Are you assuming that the balls and rectangles are both rigid, no deformation? Frictionless contact? How will you handle the motion of the ball after contact is made? Are you transforming to a coordinate system of the contact (normal + tangential), calculating, then transforming back?
It's not a trivial problem.
Maybe you should look into a physics engine, like Box2D, rather than coding it yourself.