Using Sprite Kit I am trying to set an SKPhysicsBody moving according to a given angle, so for example if you wanted the sprite to travel to the right you would specify 1.571 radians. To turn the specified angle into a velocity I am using the method below to convert radians to a CGVector. The ORIGINAL version that I implemented from memory has the strange effect of offsetting all the angles by 90degrees. (i.e. if 0 degrees is used the sprite moves right (just like it would if you specified 90degrees)
Question:
I have fixed this in the NEW version by swapping the dx and dy assignments. My question is why does this happen, do I have it wrong in the original (there do seem to be others doing it that way on the web) or is there some reason based on the particular coordinate system being used.
// ORIGINAL
- (CGVector)convertAngleToVector:(CGFloat)radians {
CGVector vector;
vector.dx = cos(radians) * 10;
vector.dy = sin(radians) * 10;
NSLog(#"DX: %0.2f DY: %0.2f", vector.dx, vector.dy);
return vector;
}
// NEW, SWAPPED DX & DY
- (CGVector)convertAngleToVector:(CGFloat)radians {
CGVector vector;
vector.dy = cos(radians) * 10;
vector.dx = sin(radians) * 10;
NSLog(#"DX: %0.2f DY: %0.2f", vector.dx, vector.dy);
return vector;
}
NOTE: also in Sprite Kit clockwise rotations are negative, so far convertAngleToVector is doing positive clockwise rotations (i.e. 1.571 radians is right, where it should be left) I could just do cos(radians*-1) and sin(radians*-1) but there might be some underlying reason for this based on me swapping dx and dy.
Sprite Kit (SKView Coordinates):
Yeah, SpriteKit defaults to the right. The Physics Collision sample project solves this by implementing this method:
- (CGFloat)shipOrientation
{
// The ship art is oriented so that it faces the top of the scene, but Sprite Kit's rotation default is to the right.
// This method calculates the ship orientation for use in other calculations.
return self.zRotation + M_PI_2;
}
You can then just get the existing orientation by calling something like:
CGFloat shipDirection = [self shipOrientation];
And then adjust the zRotation property from there.
From the Sprite Kit Programming Guide (emphasis added):
Sprite Kit also has a standard rotation convention. Figure 4-2 shows the polar coordinate convention. An angle of 0 radians specifies the positive x axis. A positive angle is in the counterclockwise direction.
In this coordinate system, an angle of zero radians pointing to the right is correct. If you want to use a system in which a zero angle is straight up (along positive y axis) and increase clockwise, you'll want to transform your angles before converting them to vectors.
Related
This is a question for Unity people or Math geniuses.
I'm making a game where I have a circle object that I can move, but I don't want it to intersect or go into other (static) circles in the world (Physics system isn't good enough in Unity to simply use that, btw).
It's in 3D world, but the circles only ever move on 2 axis.
I was able to get this working perfectly if circle hits only 1 other circle, but not 2 or more.
FYI: All circles are the same size.
Here's my working formula for 1 circle to move it to the edge of the colliding circle if intersecting:
newPosition = PositionOfStaticCircleThatWasJustIntersected + ((positionCircleWasMovedTo - PositionOfStaticCircleThatWasJustIntersected).normalized * circleSize);
But I can't figure out a formula if the moving circle hits 2 (or more) static circles at the same time.
One of the things that confuse me the most is the direction issue depending on how all the circles are positioned and what direction the moving circle is coming from.
Here's an example image of what I'm trying to do.
Since we're operating in a 2D space, let's approach this with some geometry. Taking a close look at your desired outcome, a particular shape become apparent:
There's a triangle here! And since all circles are the same radius, we know even more: this is an isosceles triangle, where two sides are the same length. With that information in hand, the problem basically boils down to:
We know what d is, since it's the distance between the two circles being collided with. And we know what a is, since it's the radius of all the circles. With that information, we can figure out where to place the moved circle. We need to move it d/2 between the two circles (since the point will be equidistant between them), and h away from them.
Calculating the height h is straightforward, since this is a right-angle triangle. According to the Pythagorean theorem:
// a^2 + b^2 = c^2, or rewritten as:
// a = root(c^2 - b^2)
float h = Mathf.Sqrt(Mathf.Pow(2 * a, 2) - Mathf.Pow(d / 2, 2))
Now need to turn these scalar quantities into vectors within our game space. For the vector between the two circles, that's easy:
Vector3 betweenVector = circle2Position - circle1Position
But what about the height vector along the h direction? Well, since all movement is on 2D space, find a direction that your circles don't move along and use it to get the cross product (the perpendicular vector) with the betweenVector using Vector3.Cross(). For
example, if the circles only move laterally:
Vector3 heightVector = Vector3.Cross(betweenVector, Vector3.up)
Bringing this all together, you might have a method like:
Vector3 GetNewPosition(Vector3 movingCirclePosition, Vector3 circle1Position,
Vector3 circle2Position, float radius)
{
float halfDistance = Vector3.Distance(circle1Position, circle2Position) / 2;
float height = Mathf.Sqrt(Mathf.Pow(2 * radius, 2) - Mathf.Pow(halfDistance, 2));
Vector3 betweenVector = circle2Position - circle1Position;
Vector3 heightVector = Vector3.Cross(betweenVector, Vector3.up);
// Two possible positions, on either side of betweenVector
Vector3 candidatePosition1 = circle1Position
+ betweenVector.normalized * halfDistance
+ heightVector.normalized * height;
Vector3 candidatePosition2 = circle1Position
+ betweenVector.normalized * halfDistance
- heightVector.normalized * height;
// Absent any other information, the closer position will be assumed as correct
float distToCandidate1 = Vector3.Distance(movingCirclePosition, candidatePosition1);
float distToCandidate2 = Vector3.Distance(movingCirclePosition, candidatePosition2);
if (distToCandidate1 < distToCandidate2){
return candidatePosition1;
}
else{
return candidatePosition2;
}
}
I'm trying to constrain an object's rotation, so that it behaves like a joystick (meaning it can only rotate up to some maximum angle from center).
I tried to constrain rotation on each individual axis, but they behaved really weirdly when rotating (the angle values didn't just grow linearly). The input I'm using to supply this rotation is the rotation of physical controller. How can I do this?
This is how it works now and how I want it to work:
Images are 2D but it applies to all axes.
It sounds to me like there are two parts to the problem:
Determine whether supplied rotation exceeds maximum allowable rotation of object
If the supplied rotation is too large, reducing that rotation so it fits within the rotational constraints, then applying it
In my code examples, I'll be assuming that the initial rotation of the virtual joystick is zero across the board (Quaternion.identity), and that your supplied rotation is called newRotation.
For the first part, Quaternion.Angle() comes to mind. This gives the angle between two given rotations, and can be used like so:
if (Quaternion.Angle(Quaternion.identity, newRotation) < 30){
// Angle from initial to new rotation is under 30 degrees
}
For the second part, you'll need some way of reducing the supplied rotation so it is within the allowable angle from the initial rotation. For that, Quaternion.Slerp() is useful. This allows you to interpolate between two rotations, returning a Quaternion that is a combination of the two supplied. For example, this gives back half of newRotation:
Quaternion.Slerp(Quaternion.identity, newRotation, 0.5f);
Putting these two methods together, you can write a clamping method which ensures that the supplied rotation never exceeds a certain angle from the initial rotation. Here's an example usage:
// Maximum angle joystick can tilt at
public float tiltAngle;
// If this isn't your initial rotation, set it in Awake() or Start()
Quaternion initialRotation = Quaternion.identity;
void Update(){
Quaternion newRotation = MethodToGetSuppliedRotation();
Quaternion clampedRotation = ClampRotation(initialRotation, newRotation, tiltAngle);
transform.localRotation = clampedRotation;
}
// Clamps "b" such that it never exceeds "maxAngle" degrees from "a"
Quaternion ClampRotation(Quaternion a, Quaternion b, float maxAngle){
float newAngle = Quaternion.Angle(a, b);
if (newAngle <= maxAngle){
// Rotation within allowable constraint
return b;
}
else{
// This is the proportion of the new rotation that is within the constraint
float angleRatio = maxAngle / newAngle;
return Quaternion.Slerp(a, b, angleRatio);
}
}
Hope this helps! Let me know if you have any questions.
I am using the following code to handle rotating my player model to the position of my mouse.
void Update() {
// Generate a plane that intersects the transform's position with an upwards normal.
Plane playerPlane = new Plane(Vector3.up, transform.position);
// Generate a ray from the cursor position
Ray ray = Camera.main.ScreenPointToRay(Input.mousePosition);
// Determine the point where the cursor ray intersects the plane.
// This will be the point that the object must look towards to be looking at the mouse.
// Raycasting to a Plane object only gives us a distance, so we'll have to take the distance,
// then find the point along that ray that meets that distance. This will be the point
// to look at.
float hitdist = 0f;
// If the ray is parallel to the plane, Raycast will return false.
if (playerPlane.Raycast(ray, out hitdist)) {
// Get the point along the ray that hits the calculated distance.
var targetPoint = ray.GetPoint(hitdist);
// Determine the target rotation. This is the rotation if the transform looks at the target point.
Quaternion targetRotation = Quaternion.LookRotation(targetPoint - transform.position);
// Smoothly rotate towards the target point.
transform.rotation = Quaternion.Slerp(transform.rotation, targetRotation, speed * Time.deltaTime); // WITH SPEED
//transform.rotation = Quaternion.Slerp(transform.rotation, targetRotation, 1); // WITHOUT SPEED!!!
}
I would like to be able to determine if the rotation is clockwise or counter-clockwise for animation purposes. What would be the best way of handling this? I'm fairly unfamiliar with quaternions so I'm not really sure how to approach this.
Angles between quaternions are unsigned. You will always get the shortest distance, and there's no way of defining "counter-clockwise" or "clockwise" unless you actively specify an axis (a point of view).
What you CAN do, however, is to take the axis that you're interested in (I assume it's the normal to your base plane.. perhaps the vertical of your world?) and take the flat 2D components of your quaternions, map them there and compute a simple 2D angle between those.
Quaternion A; //first Quaternion - this is your desired rotation
Quaternion B; //second Quaternion - this is your current rotation
// define an axis, usually just up
Vector3 axis = new Vector3(0.0f, 1.0f, 0.0f);
// mock rotate the axis with each quaternion
Vector3 vecA = A * axis;
Vector3 vecB = B * axis;
// now we need to compute the actual 2D rotation projections on the base plane
float angleA = Mathf.Atan2(vecA.x, vecA.z) * Mathf.Rad2Deg;
float angleB = Mathf.Atan2(vecB.x, vecB.z) * Mathf.Rad2Deg;
// get the signed difference in these angles
var angleDiff = Mathf.DeltaAngle( angleA, angleB );
This should be it. I never had to do it myself and the code above is not tested. Similar to: http://answers.unity3d.com/questions/26783/how-to-get-the-signed-angle-between-two-quaternion.html
This should work even if A or B are not Quaternions, but one of them is an euler-angle rotation.
Two dimensional quaternions (complex numbers) have a signed angle. But, the more correct way to think about complex numbers is with an unsigned angle which is relative to either the XY oriented plane or the YX oriented plane. I.E. a combination of an unsigned angle an an oriented plane of rotation.
In 2D there are only two oriented planes of rotation so the idea of a "signed angle" is really just a trick to get both the unsigned angle and the oriented plane of rotation packed into a single number.
For a quaternion the "signed angle" trick cannot be used because in 3D you have an infinite number of oriented planes you can rotate in, so a single signed angle cannot encode all the rotation information like it can in the 2D case.
The only way for a signed angle to make sense in 3D is with reference to a particular oriented plane, such as the XY oriented plane.
-- UPDATE --
This is pretty easy to solve as a method on a quaternion class. If all you want to know is "is this counter clockwise", then since we know the rotation angle is from 0 to 180, a positive dot product between the quat's axis of rotation and the surface normal should indicate that we're rotating counter clockwise from the perspective of that surface. And a negative dot product indicates the opposite. Ignoring the zero case, this should do the trick with much less work:
public bool IsCounterClockwise( in Vector3 normal ) => I*normal.X + J*normal.Y + K*normal.Z >= 0;
I'm trying figure out how to get an OpenGL specified object to be displayed correctly according to the device orientation (ie. according to the gravity vector from the accelerometer, and heading from compass).
The GLGravity sample project has an example which is almost like this (despite ignoring heading), but it has some glitches. For example, the teapot jumps 180deg as the device viewing angle crosses the horizon, and it also rotates spuriously if you tilt the device from portrait into landscape. This is fine for the context of this app, as it just shows off an object and it doesn't matter that it does these things. But it means that the code just doesn't work when you attempt to emulate real life viewing of an OpenGL object according to the device's orientation. What happens is that it almost works, but the heading rotation you apply from the compass gets "corrupted" by the spurious additional rotations seen in the GLGravity example project.
Can anyone provide sample code that shows how to adjust correctly for the device orientation (ie. gravity vector), or to fix the GLGravity example so that it doesn't include spurious heading changes?
//Clear matrix to be used to rotate from the current referential to one based on the gravity vector
bzero(matrix, sizeof(matrix));
matrix[3][3] = 1.0;
//Setup first matrix column as gravity vector
matrix[0][0] = accel[0] / length;
matrix[0][1] = accel[1] / length;
matrix[0][2] = accel[2] / length;
//Setup second matrix column as an arbitrary vector in the plane perpendicular to the gravity vector {Gx, Gy, Gz} defined by by the equation "Gx * x + Gy * y + Gz * z = 0" in which we arbitrarily set x=0 and y=1
matrix[1][0] = 0.0;
matrix[1][1] = 1.0;
matrix[1][2] = -accel[1] / accel[2];
length = sqrtf(matrix[1][0] * matrix[1][0] + matrix[1][1] * matrix[1][1] + matrix[1][2] * matrix[1][2]);
matrix[1][0] /= length;
matrix[1][1] /= length;
matrix[1][2] /= length;
//Setup third matrix column as the cross product of the first two
matrix[2][0] = matrix[0][1] * matrix[1][2] - matrix[0][2] * matrix[1][1];
matrix[2][1] = matrix[1][0] * matrix[0][2] - matrix[1][2] * matrix[0][0];
matrix[2][2] = matrix[0][0] * matrix[1][1] - matrix[0][1] * matrix[1][0];
//Finally load matrix
glMultMatrixf((GLfloat*)matrix);
Here's a clarification showing how to get the elevation and tilt that are needed for gluLookAt solution as shown in my last answer:
// elevation comes from z component (0 = facing horizon)
elevationRadians = asin(gravityVector.z / Vector3DMagnitude(gravityVector));
// tilt is how far screen is from vertical, looking along z axis
tiltRadians = atan2(-gravityVector.y, -gravityVector.x) - M_PI_2;
Following up on Chris's suggestion: I'm not sure if I've got this all correct due to differing conventions of row/column order and heading cw or ccw. However the following code is what I came up with:
Vector3D forward = Vector3DMake(0.0f, 0.0f, -1.0f);
// Multiply it by current rotation matrix to get teapot direction
Vector3D direction;
direction.x = matrix[0][0] * forward.x + matrix[1][0] * forward.y + matrix[2][0] * forward.z;
direction.y = matrix[0][1] * forward.x + matrix[1][1] * forward.y + matrix[2][1] * forward.z;
direction.z = matrix[0][2] * forward.x + matrix[1][2] * forward.y + matrix[2][2] * forward.z;
heading = atan2(direction.z, direction.x) * 180 / M_PI;
// Use this heading to adjust the teapot direction back to keep it fixed
// Rotate about vertical axis (Y), as it is a heading adjustment
glRotatef(heading, 0.0, 1.0, 0.0);
When I run this code, the teapot behaviour has apparently "improved" eg. heading no longer flips 180deg when device screen (in portrait view) is pitched forward/back through upright. However, it still makes major jumps in heading when device (in landscape view) is pitched forward/back. So something's not right. It suggests that the above calculation of the actual heading is incorrect...
I finally found a solution that works. :-)
I dropped the rotation matrix approach, and instead adopted gluLookAt. To make this work you need to know the device "elevation" (viewing angle relative to horizon ie. 0 on horizon, +90 overhead), and the camera's "tilt" (how far the device is from vertical its x/y plane ie. 0 when vertical/portrait, +/-90 when horizontal/landscape), both of which are obtained from the device gravity vector components.
Vector3D eye, scene, up;
CGFloat distanceFromScene = 0.8;
// Adjust eye position for elevation (y/z)
eye.x = 0;
eye.y = distanceFromScene * -sin(elevationRadians); // eye position goes down as elevation angle goes up
eye.z = distanceFromScene * cos(elevationRadians); // z position is maximum when elevation is zero
// Lookat point is origin
scene = Vector3DMake(0, 0, 0); // Scene is at origin
// Camera tilt - involves x/y plane only - arbitrary vector length
up.x = sin(tiltRadians);
up.y = cos(tiltRadians);
up.z = 0;
Then you just apply the gluLookAt transformation, and also rotate the scene according to the device heading.
// Adjust view for device orientation
gluLookAt(eye.x, eye.y, eye.z, scene.x, scene.y, scene.z, up.x, up.y, up.z);
// Apply device heading to scene
glRotatef(currentHeadingDegrees, 0.0, 1.0, 0.0);
Try rotating the object depending upon iphone acceleration values.
float angle = -atan2(accelX, accelY);
glPushMatrix();
glTranslatef(centerPoint.x, centerPoint.y, 0);
glRotatef(angle, 0, 0, 1);
glTranslatef(-centerPoint.x, -centerPoint.y, 0);
glPopMatrix();
Where centerPoint is the middle point the object.
oo, nice.
GLGravity seems to get everything right except for the yaw. Here's what I would try. Do everything GLGravity does, and then this:
Project a vector in the direction you want the teapot to face, using the compass or whatever you so choose. Then multiply a "forward" vector by the teapot's current rotation matrix, which will give you the direction the teapot is facing. Flatten the two vectors to the horizontal plane and take the angle between them.
This angle is your corrective yaw. Then just glRotatef by it.
Whether or not the 3GS's compass is reliable and robust enough for this to work is another thing. Normal compasses don't work when the north vector is perpendicular to their face. But I just tried the Maps app on my workmate's 3GS and it seems to cope, so maybe they have got a mechanical solution in there. Knowing what the device is actually doing will help interpret the results it gives.
Make sure to test your app at the north and south poles once you're done. :-)
Getting a much more stable gravity-based reference, can now be done using CMMotionManager.
When starting motion updates with startDeviceMotionUpdates(), you can specify a reference frame.
This fuses the accelerometer, gyroscope and optionally (depending on chose reference frame) magnetometer data. Accelerometer data is pretty noisy and bouncy (any sideways motion of the device temporarily tilts the gravity vector by any device acceleration) and alone doesn't make a good reference.
I've been low-pass filtering the accelerometer data, which helps a bit but makes the system slow.
I would like to use Cocos2d on the iPhone to draw a 2D car and make it steer from left to right in a natural way.
Here is what I tried:
Calculate the angle of the wheels and just move it to the destination point where the wheels point to. But this creates a very unnatural feel. The car drifts half the time
After that I started some research on how to get a turning circle from a car, which meant that I needed a couple of constants like wheelbase and the width of the car.
After a lot of research, I created the following code:
float steerAngle = 30; // in degrees
float speed = 20;
float carWidth = 1.8f; // as in 1.8 meters
float wheelBase = 3.5f; // as in 3.5 meters
float x = (wheelBase / abs(tan(steerAngle)) + carWidth/ 2);
float wheelBaseHalf = wheelBase / 2;
float r = (float) sqrt(x * x + wheelBaseHalf * wheelBaseHalf);
float theta = speed * 1 / r;
if (steerAngle < 0.0f)
theta = theta * -1;
drawCircle(CGPointMake(carPosition.x - r, carPosition.y),
r, CC_DEGREES_TO_RADIANS(180), 50, NO);
The first couple of lines are my constants. carPosition is of the type CGPoint. After that I try to draw a circle which shows the turning circle of my car, but the circle it draws is far too small. I can just make my constants bigger, to make the circle bigger, but then I would still need to know how to move my sprite on this circle.
I tried following a .NET tutorial I found on the subject, but I can't really completely convert it because it uses Matrixes, which aren't supported by Cocoa.
Can someone give me a couple of pointers on how to start this? I have been looking for example code, but I can't find any.
EDIT After the comments given below
I corrected my constants, my wheelBase is now 50 (the sprite is 50px high), my carWidth is 30 (the sprite is 30px in width).
But now I have the problem, that when my car does it's first 'tick', the rotation is correct (and also the placement), but after that the calculations seem wrong.
The middle of the turning circle is moved instead of kept at it's original position. What I need (I think) is that at each angle of the car I need to recalculate the original centre of the turning circle. I would think this is easy, because I have the radius and the turning angle, but I can't seem to figure out how to keep the car moving in a nice circle.
Any more pointers?
You have the right idea. The constants are the problem in this case. You need to specify wheelBase and carWidth in units that match your view size. For example, if the image of your car on the screen has a wheel base of 30 pixels, you would use 30 for the WheelBase variable.
This explains why your on-screen circles are too small. Cocoa is trying to draw circles for a tiny little car which is only 1.8 pixels wide!
Now, for the matter of moving your car along the circle:
The theta variable you calculate in the code above is a rotational speed, which is what you would use to move the car around the center point of that circle:
Let's assume that your speed variable is in pixels per second, to make the calculations easier. With that assumption in place, you would simply execute the following code once every second:
// calculate the new position of the car
newCarPosition.x = (carPosition.x - r) + r*cos(theta);
newCarPosition.y = carPosition.y + r*sin(theta);
// rotate the car appropriately (pseudo-code)
[car rotateByAngle:theta];
Note: I'm not sure what the correct method is to rotate your car's image, so I just used rotateByAngle: to get the point across. I hope it helps!
update (after comments):
I hadn't thought about the center of the turning circle moving with the car. The original code doesn't take into account the angle that the car is already rotated to. I would change it as follows:
...
if (steerAngle < 0.0f)
theta = theta * -1;
// calculate the center of the turning circle,
// taking int account the rotation of the car
circleCenter.x = carPosition.x - r*cos(carAngle);
circleCenter.y = carPosition.y + r*sin(carAngle);
// draw the turning circle
drawCircle(circleCenter, r, CC_DEGREES_TO_RADIANS(180), 50, NO);
// calculate the new position of the car
newCarPosition.x = circleCenter.x + r*cos(theta);
newCarPosition.y = circleCenter.y + r*sin(theta);
// rotate the car appropriately (pseudo-code)
[car rotateByAngle:theta];
carAngle = carAngle + theta;
This should keep the center of the turning circle at the appropriate point, even if the car has been rotated.