The answer to Direct way of computing clockwise angle between 2 vectors was helpful in getting the clockwise angle between two vectors, but what I need to know is how to get the angle from two vectors going in the clockwise direction from a specific vector.
Here's a visual explanation of what I am trying to say. When going in the clockwise direction from the red vector, the clockwise angle is this: (Excuse the bad drawing)
But when going in the clockwise direction from the black vector, the clockwise angle is this:
EDIT: I'll rephrase my question. This is what I am trying to achieve. I have been following this old but useful link on generating triangle meshes: http://www.geom.uiuc.edu/~samuelp/del_project.html#algorithms
With the above images, I always want the angle going in a clockwise direction from a specific vector. I have been using the code from the linked SO answer to get the angle, but its incorrect in my situation
float dotProd = vx*ux + vy*uy;
float det = vx*uy - vy*ux;
float angle = MathUtils.atan2(det, dotProd);//in radians
if(angle < 0) angle += MathUtils.PI;
The angle calculation incorrectly allows unnecessary edges to be connected.
How can I fix this?
I'm sure there is a shorter way to do it, but if you know where you 0,0 is
first do a check in which quarter your vector is and then perform the calculation accordingly.
it will solve your solution but mathematically I think you can get there faster.
I did these experiments:
int[] v1={5, 2}, v2={3, 4}, v3={5, 0};
double d1=v1[0]*v2[0]+v1[1]*v2[1], d2=v1[0]*v2[1]-v1[1]*v2[0];
System.out.println(Math.toDegrees(Math.atan2(d2, d1)));
d2=v2[0]*v1[1]-v2[1]*v1[0];
System.out.println(Math.toDegrees(Math.atan2(d2, d1)));
d1=v3[0]*v1[0]+v3[1]*v1[1]; d2=v3[0]*v1[1]-v3[1]*v1[0];
System.out.println(Math.toDegrees(Math.atan2(d2, d1)));
d1=v1[0]*v3[0]+v1[1]*v3[1]; d2=v1[0]*v3[1]-v1[1]*v3[0];
System.out.println(Math.toDegrees(Math.atan2(d2, d1)));
d1=v2[0]*v3[0]+v2[1]*v3[1]; d2=v3[0]*v2[1]-v3[1]*v2[0];
System.out.println(Math.toDegrees(Math.atan2(d2, d1)));
It seems the angle computed is the immediate angle: if it's positive the direction is counter clockwise (from v3 to v2 for example), if it is negative then it is clockwise.
I used standard java atan2 I dont know what you are using.
Therefore to get the clock wise always you have to check the result: if it is positive then subtract 360 degrees.
You have to do your own experiments to confirm this negative/positive.
Related
So I'm using box2d to handle physics and collisions in libGDX. I'm trying to figure out the angle of the collision so that I can apply certain effects at a given angle.
Doing some research, I found that atan2 should be the math function I need to get the correct angle, but it's returning curious results. For a contact that should be in the 2nd quadrant, I'm getting angles of less than 90. For contacts that are in the 3rd and 4th quadrants, I'm getting positive numbers for the contact angle. Does anyone have any insight as to what I could be doing wrong here?
float other_x = contact.getWorldManifold().getPoints()[0].x * GameScreen.BOX_TO_WORLD;
float other_y = contact.getWorldManifold().getPoints()[0].y * GameScreen.BOX_TO_WORLD;
float playerXpos = this.getX();
float playerYpos = this.getY();
float damageAngle = (float)Math.atan2((other_y - playerYpos), (other_x - playerXpos));
damageAngle = (float) (damageAngle * (180d/Math.PI));
And yes, that index of the world manifold does have coordinate information in it. I just can't seem to figure out why atan2 is returning those values.
edit So I'm thinking that the problem isn't with atan2, but rather with the contact information that's being returned by the WorldManifold. So I guess my question is, what could be the problem, such that contact information would be off like this?
I am studying the following code.
boolean convex(double x1, double y1, double x2, double y2,
double x3, double y3)
{
if (area(x1, y1, x2, y2, x3, y3) < 0)
return true;
else
return false;
}
/* area: determines area of triangle formed by three points
*/
double area(double x1, double y1, double x2, double y2,
double x3, double y3)
{
double areaSum = 0;
areaSum += x1 * (y3 - y2);
areaSum += x2 * (y1 - y3);
areaSum += x3 * (y2 - y1);
/* for actual area, we need to multiple areaSum * 0.5, but we are
* only interested in the sign of the area (+/-)
*/
return areaSum;
}
I do not understand the concept that area being negative.
Shouldn't area be always positive? maybe I am lacking some understanding of terms here.
I tried to contact the original writer but this code is about 8 years old and I have no way to contact the original writer.
This method of determining if the given vertex x2y2 is convex seems really mobile. I really want to understand it.
Any direction or reference to help me understand this piece of code will be appreciated greatly.
Source code : http://cgm.cs.mcgill.ca/~godfried/teaching/cg-projects/97/Ian/applets/BruteForceEarCut.java
The algorithm use a very simple formula with which you can compute twice the area of a triangle.
This formula has two advantages:
it doesn't require any division
it returns a negative area if the point are in the counterclockwise order.
In the code sample, the actual value of the area doesn't matter, only the sign of the result is needed.
The formula can also be used to check if three points are colinear.
You can find more information about this formula on this site : http://www.mathopenref.com/coordtrianglearea.html
This algorithm is basically using the dot product of two vectors and interpreting the results. This is the core of the Gift Wrapping Algorithm used to find convex hulls.
Since a dot b is also equal to |a|*|b|*cos(theta) then if the result is positive, cos of theta must be positive and thus convex. Per a wiki article on cross products...
Because the magnitude of the cross product goes by the sine of the
angle between its arguments, the cross product can be thought of as a
measure of ‘perpendicularity’ in the same way that the dot product is
a measure of ‘parallelism’. Given two unit vectors, their cross
product has a magnitude of 1 if the two are perpendicular and a
magnitude of zero if the two are parallel. The opposite is true for
the dot product of two unit vectors.
The use of "area" is slightly misleading on part of the original coder in my opinion.
You know about how integrals work, right? One way to think of integrals is in terms of the area under the integrated curve. For functions that are strictly positive, that definition works great, but when the function becomes negative at some point, there is a problem because then you have to take the absolute value, right?
That is not always so, actually, and it can be quite useful in some contexts to leave the curve negative. Think back to what was said earlier: the area under the curve. All that space between negative infinity and our function. Clearly, that is absurd, right? A better way to think of it is as the difference between the area under the curve, and the area under the x axis. That way, when the function is positive, our curve is gaining more area, and when it is negative, it is gaining less than the x axis.
The same thing applies to plane figures that are not strict functions. In order to really determine this, we have to define which direction our edge is going as it travels around the figure. We can define it so that all the area on the right of our curve is inside the region, and all the area to the left is outside (or we can define it the other way around, but I will use the first way).
So our figure includes all the area from there to the edge at infinity of the plane that is directly to our right. Regions enclosed clockwise really include their conventional interior twice. Regions enclosed counterclockwise don't include their conventional interior at all. The area, then, is the difference between our region and the whole plane.
The application of this to concavity is fairly simple, if you understand what it actually means to be concave or convex. The triangle you are given is concave if it is cutting an area out from the plane, and it is convex if you are adding extra area to it. That is the exact same thing we were doing to determine the our area, so positive area corresponds to a convex shape, and negative area corresponds to a concave shape.
You can also do other weird things with this conceptual model. For instance, you can turn a region 'inside out' by reversing the edge direction.
I'm sorry if this has been a little hard to follow, but this is the actual way I understand negative area.
I'm creating a very very simple game for fun. Realizing I needed the trajectory of an object given an angle and a velocity, it seemed logical to use this parametric equation:
x = (v*cos(ø))t and y = (v*sin(ø)t - 16t^2
I know that this equation works for a trajectory, but it isn't working with most ø values I use.
Do java angles work any differently from normal angle calculation?
My goal is for the object to start from bottom left of the window and follow an arc that is determined by the speed and angle given. However it tends to go strange directions.
The value of ø should be horizontal at 0 degrees and vertical at 90, and in the equation it refers to the angle at which the arc begins.
This is my first ever question post on this site, so if I'm missing anything in that regard please let me know.
Here is the calculating part of my code
not shown is the void time() that counts for each 5ms
also I should mention that the parX and parY are used to refer to the x and y coordinates in an unrounded form, as graphics coordinates require integer values.
Any help is much appreciated, and thank you in advance!
public void parametric()
{
parX = (float) ((speed*cos(-ø))*time);
gravity = (time*time)*(16);
parY = (float) ((float) ((speed*sin(-ø))*time)+gravity)+500;
xCoord = round(parX);
yCoord = round(parY);
}
Do java angles work any differently from normal angle calculation?
You only need to read the docs
public static double cos(double a)
Parameters:
a - an angle, in radians.
I guess you are using degrees instead of radians?
what I want to do is the following: I have an object (blue point) and I want to point it to other object no matter where it is located around it (green point). So I need to know the angle between these two objects to do what I want right?
http://s13.postimage.org/6jeuphcdj/android_angle.jpg
The problem is, I don't know what to do to achieve this. I've already used atan, math.tan and so many other functions but without any good results.
Could you help me? Thanks in advance.
Calculate a dot product of object vectors. Use Math.acos on the value you get. That will give you an angle in radians.
So, say your blue dot is at vec1 = (50, 100) and green one at vec2 = (100, 400).
A tuple (x, y) as a two dimensional vector describes object's position and distance from (0, 0) on your screen. To find the angle between these two vectors, you do a standard, binary dot product operation on them. This will get you a scalar (a value, cos(Theta)), but you want the inverse of it (acos) which is the angle you're looking for.
You can get a better understanding on the matter here
Suppose the coordinates of the blue and green points are (xblue, yblue) and (xgreen, ygreen) respectively.
The angle at which the blue point sees the green point is:
double angleRadians = Math.atan2(ygreen-yblue, xgreen-xblue);
If you want the angle in degrees:
double angleDegrees = Math.toDegrees(angleRadians);
I'm trying to write a 2D game in Java that uses the Separating Axis Theorem for collision detection. In order to resolve collisions between two polygons, I need to know the Minimum Translation Vector of the collision, and I need to know which direction it points relative to the polygons (so that I can give one polygon a penalty force along that direction and the other a penalty force in the opposite direction). For reference, I'm trying to implement the algorithm here.
I'd like to guarantee that if I call my collision detection function collide(Polygon polygon1, Polygon polygon2) and it detects a collision, the returned MTV will always point away from polygon1, toward polygon2. In order to do this, I need to guarantee that the separating axes that I generate, which are the normals of the polygon edges, always point away from the polygon that generated them. (That way, I know to negate any axis from polygon2 before using it as the MTV).
Unfortunately, it seems that whether or not the normal I generate for a polygon edge points towards the interior of the polygon or the exterior depends on whether the polygon's points are declared in clockwise or counterclockwise order. I'm using the algorithm described here to generate normals, and assuming that I pick (x, y) => (y, -x) for the "perpendicular" method, the resulting normals will only point away from the polygon if I iterate over the vertices in clockwise order.
Given that I can't force the client to declare the points of the polygon in clockwise order (I'm using java.awt.Polygon, which just exposes two arrays for x and y coordinates), is there a mathematical way to guarantee that the direction of the normal vectors I generate is toward the exterior of the polygon? I'm not very good at vector math, so there may be an obvious solution to this that I'm missing. Most Internet resources about the SAT just assume that you can always iterate over the vertices of a polygon in clockwise order.
You can just calculate which direction each polygon is ordered, using, for example, the answer to this question, and then multiply your normal by -1 if the two polygons have different orders.
You could also check each polygon passed to your algorithm to see if it is ordered incorrectly, again using the algorithm above, and reverse the vertex order if necessary.
Note that when calculating the vertex order, some algorithms will work for all polygons and some just for convex polygons.
I finally figured it out, but the one answer posted was not the complete solution so I'm not going to accept it. I was able to determine the ordering of the polygon using the basic algorithm described in this SO answer (also described less clearly in David Norman's link), which is:
for each edge in polygon:
sum += (x2 - x1) * (y2 + y1)
However, there's an important caveat which none of these answers mention. Normally, you can decide that the polygon's vertices are clockwise if this sum is positive, and counterclockwise if the sum is negative. But the comparison is inverted in Java's 2D graphics system, and in fact in many graphics systems, because the positive y axis points downward. So in a normal, mathematical coordinate system, you can say
if sum > 0 then polygon is clockwise
but in a graphics coordinate system with an inverted y-axis, it's actually
if sum < 0 then polygon is clockwise
My actual code, using Java's Polygon, looked something like this:
//First, find the normals as if the polygon was clockwise
int sum = 0;
for(int i = 0; i < polygon.npoints; i++) {
int nextI = (i + 1 == polygon.npoints ? 0 : i + 1);
sum += (polygon.xpoints[nextI] - polygon.xpoints[i]) *
(polygon.ypoints[nextI] + polygon.ypoints[i]);
}
if(sum > 0) {
//reverse all the normals (multiply them by -1)
}