I have 2 circles that collide in a certain collision point and under a certain collision angle which I calculate using this formula :
C1(x1,y1) C2(x2,y2)
and the angle between the line uniting their centre and the x axis is
X = arctg (|y2 - y1| / |x2 - x1|)
and what I want is to translate the circle on top under the same angle that collided with the other circle. I mean with the angle X and I don't know what translation coordinates should I give for a proper and a straight translation!
For what I think you mean, here's how to do it cleanly.
Think in vectors.
Suppose the centre of the bottom circle has coordinates (x1,y1), and the centre of the top circle has coordinates (x2,y2). Then define two vectors
support = (x1,y1)
direction = (x2,y2) - (x1,y1)
now, the line between the two centres is fully described by the parametric representation
line = support + k*direction
with k any value in (-inf,+inf). At the initial time, substituting k=1 in the equation above indeed give the coordinates of the top circle. On some later time t, the value of k will have increased, and substituting that new value of k in the equation will give the new coordinates of the centre of the top circle.
How much k increases at value t is equal to the speed of the circle, and I leave that entirely up to you :)
Doing it this way, you never need to mess around with any angles and/or coordinate transformations etc. It even works in 3D (provided you add in z-coordinates everywhere).
Related
I have some triangles in 3D space, which originate from 0,0,0 and extend towards two points p1= -x0, 0, z0 and p2= +x0, 0, z0. This is in Unity, such that +z is the forward axis (i.e. they lie flat). Each triangle is its own mesh, pivot is at 0,0,0.
Now, I would like to rotate these (using Quaternion.LookRotation) such that their ends form a continuous polygon, in case of three triangles a triangle, in case of four triangles a square, etc.
My approach is to calculate the incircle radius of the resulting polygon based on the length of each triangle (which is 2*x0). If I now calculate n points on this circle (where n is the number of triangles I have), I get x/y coordinates which I can directly use to set the "up" axis of each triangle correctly, i.e. Quaternion.LookRotation(Vector3.forward, new Vector3(x,y,0)). This orients the triangle correctly around the z axis, i.e. the center is still on 0,0,1.
However, and this has me stumped, I still need to change the forward axis of the triangles such that they tilt to form the final polygon. I tried using new Vector3(x,y,z0) which gives an almost correct result, but leads to an overlap at the edges. I suspect this is somehow due to the fact that rotation of the triangles effectively changes z0, but I am not sure how to proceed.
My question is, how to calculate the new forward axis such that the triangles align properly?
The problem is setting the forward axis to (x,y,z0), which is wrong since the length of the vector (x,y,z0) does not equal the original length (which is just z0). The z value thus needs to be adjusted such that new Vector(x,y,z1).magnitude == z0. This can be done by calculating
Mathf.Sqrt(Mathf.Pow(z0, 2) - Mathf.Pow(x, 2) - Mathf.Pow(y, 2))
Problem solved.
I have two rectangles where one is a clipping for the other one.
Now I want to rotate the bigger rectangle around the center of the clipping rectangle and adjust x/y values.
How can I calculate the new x/y values after rotation?
I actually just want to rotate the x/y of the bigger box around the center of the smaller box. So the x/y point of the bigger box is relative to the top/left point of the smaller box. I have the width and height of the smaller box so I can calculate x/y point of the big box relative to the center of small box. The angle to rotate is in degrees. The rotation can be any degree, for example 10.
You can do as follows:
determine the angle by which you want to rotate, make sure it suitable for the trigonometric functions (sin(), cos(), ...), i.e. right angle is usually Pi/2
in case of rotating counterclockwise, it is negative
determine the coordinates of c, as cx,cy
process each of the corners of the rectanlge, one by one, for a total of four
for each corner P, currently at coordinates px,py and to move to px2,py2
determine angle between current P and C, using atan2(py-cy, px-cx)
to get from degrees to radians (for use with trigonometry) calculate radians=(pi*degrees)/180.0
add the desired rotation angle to that current angle, to get newangle
determine the distance of current P to C, sqrt((px-cx)(px-cx) + (py-cy)(py-cy))
multiply the distance (which is not changing by rotation), with the appropriate trigonometric function
px2 = distance * cos(newangle)
py2 = distance * sin(newangle)
If you want to rotate a given point P around a point C, which are defined in the same coordinate system you can use a simple rotation matrix. Calculate the P coordinates with respect to C (subtraction), then apply rotation with the matrix and go back to original coordinates by adding C again.
All that matters is the coordinates of the rotation center and the angle.
The most compact formulation is by means of complex numbers (of which I hope you have some understanding; you actually don't need a complex data type, you can expand the formulas).
Let C be the center and α the angle. Then for any point P, the image Q is given by
Q = (P - C) cis(α) + C
where cis(α) = cos(α) + i sin(α).
The inverse rotation is simply given by
P = (Q - C) cis(-α) + C.
I have two points in 2D space, centred on origin (0,0). The first point represents the starting location and the second represents the end location. I need to calculate the angle of rotation between the two points, my problem being that the hypoteneuse from each point to (0,0) is not equal.
Could someone tell me how to work out the angle between the two points, bearing in mind that they could be anywhere relative to (0,0).
Many thanks,
Matt.
Let's say point 1 is (x1,y1) and point 2 is (x2,y2)
The tangent of the Angle from X axis to point 1, relative to (0,0) is y1/x1
The tangent of the Angle from X axis to point 2, relative to (0,0) is y2/x2
Take the arc tangent (is that the right term? Tan-1 on a calculator) to get the actual angle for each, then subtract to get the answer you're looking for
This is easily accomplished taking the arccosine of the normalized inner product of the two vectors. That is, given u = (ux, uy) and v = (vx, vy), the angle between the two is given by θ = acos(u·v/|u||v|), where u · v = uxvx + uyvy is the dot product of the two and the | | operator is the l2 normal given by |u| = sqrt(ux2 + uy2). This will result in the smallest rotation that can be applied to one of the vectors that will make them linear multiples of each other. Therefore, you may need to fiddle with the sign of θ to make sure you're going in the right direction if you have one you want to start from.
I'm using CML to manage the 3D math in an OpenGL-based interface project I'm making for work. I need to know the width of the viewing frustum at a given distance from the eye point, which is kept as a part of a 4x4 matrix that represents the camera. My goal is to position gui objects along the apparent edge of the viewport, but at some distance into the screen from the near clipping plane.
CML has a function to extract the planes of the frustum, giving them back in Ax + By + Cz + D = 0 form. This frustum is perpendicular to the camera, which isn't necessarily aligned with the z axis of the perspective projection.
I'd like to extract x and z coordinates so as to pin graphical elements to the sides of the screen at different distances from the camera. What is the best way to go about doing it?
Thanks!
This seems to be a duplicate of Finding side length of a cross-section of a pyramid frustum/truncated pyramid, if you already have a cross-section of known width a known distance from the apex. If you don't have that and you want to derive the answer yourself you can follow these steps.
Take two adjacent planes and find
their line of intersection L1. You
can use the steps here. Really
what you need is the direction
vector of the line.
Take two more planes, one the same
as in the previous step, and find
their line of intersection L2.
Note that all planes of the form Ax + By + Cz + D = 0 go through the origin, so you know that L1 and L2
intersect.
Draw yourself a picture of the
direction vectors for L1 and L2,
tails at the origin. These form an
angle; call it theta. Find theta
using the formula for the angle
between two vectors, e.g. here.
Draw a bisector of that angle. Draw
a perpendicular to the bisector at
the distance d you want from the
origin (this creates an isosceles
triangle, bisected into two
congruent right triangles). The
length of the perpendicular is your
desired frustum width w. Note that w is
twice the length of one of the bases
of the right triangles.
Let r be the length of the
hypotenuses of the right triangles.
Then rcos(theta/2)=d and
rsin(theta/2)=w/2, so
tan(theta/2)=(w/2)/d which implies
w=2d*tan(theta/2). Since you know d
and theta, you are done.
Note that we have found the length of one side of a cross-section of a frustrum. This will work with any perpendicular cross-section of any frustum. This can be extended to adapt it to a non-perpendicular cross-section.
I'm trying to make a triangle (isosceles triangle) to move around the screen and at the same time slightly rotate it when a user presses a directional key (like right or left).
I would like the nose (top point) of the triangle to lead the triangle at all times. (Like that old asteroids game).
My problem is with the maths behind this. At every X time interval, I want the triangle to move in "some direction", I need help finding this direction (x and y increments/decrements).
I can find the center point (Centroid) of the triangle, and I have the top most x an y points, so I have a line vector to work with, but not a clue as to "how" to work with it.
I think it has something to do with the old Sin and Cos methods and the amount (angle) that the triangle has been rotated, but I'm a bit rusty on that stuff.
Any help is greatly appreciated.
The arctangent (inverse tangent) of vy/vx, where vx and vy are the components of your (centroid->tip) vector, gives you the angle the vector is facing.
The classical arctangent gives you an angle normalized to -90° < r < +90° degrees, however, so you have to add or subtract 90 degrees from the result depending on the sign of the result and the sign of vx.
Luckily, your standard library should proive an atan2() function that takes vx and vy seperately as parameters, and returns you an angle between 0° and 360°, or -180° and +180° degrees. It will also deal with the special case where vx=0, which would result in a division by zero if you were not careful.
See http://www.arctangent.net/atan.html or just search for "arctangent".
Edit: I've used degrees in my post for clarity, but Java and many other languages/libraries work in radians where 180° = π.
You can also just add vx and vy to the triangle's points to make it move in the "forward" direction, but make sure that the vector is normalized (vx² + vy² = 1), else the speed will depend on your triangle's size.
#Mark:
I've tried writing a primer on vectors, coordinates, points and angles in this answer box twice, but changed my mind on both occasions because it would take too long and I'm sure there are many tutorials out there explaining stuff better than I ever can.
Your centroid and "tip" coordinates are not vectors; that is to say, there is nothing to be gained from thinking of them as vectors.
The vector you want, vForward = pTip - pCentroid, can be calculated by subtracting the coordinates of the "tip" corner from the centroid point. The atan2() of this vector, i.e. atan2(tipY-centY, tipX-centX), gives you the angle your triangle is "facing".
As for what it's relative to, it doesn't matter. Your library will probably use the convention that the increasing X axis (---> the right/east direction on presumably all the 2D graphs you've seen) is 0° or 0π. The increasing Y (top, north) direction will correspond to 90° or (1/2)π.
It seems to me that you need to store the rotation angle of the triangle and possibly it's current speed.
x' = x + speed * cos(angle)
y' = y + speed * sin(angle)
Note that angle is in radians, not degrees!
Radians = Degrees * RadiansInACircle / DegreesInACircle
RadiansInACircle = 2 * Pi
DegressInACircle = 360
For the locations of the vertices, each is located at a certain distance and angle from the center. Add the current rotation angle before doing this calculation. It's the same math as for figuring the movement.
Here's some more:
Vectors represent displacement. Displacement, translation, movement or whatever you want to call it, is meaningless without a starting point, that's why I referred to the "forward" vector above as "from the centroid," and that's why the "centroid vector," the vector with the x/y components of the centroid point doesn't make sense. Those components give you the displacement of the centroid point from the origin. In other words, pOrigin + vCentroid = pCentroid. If you start from the 0 point, then add a vector representing the centroid point's displacement, you get the centroid point.
Note that:
vector + vector = vector
(addition of two displacements gives you a third, different displacement)
point + vector = point
(moving/displacing a point gives you another point)
point + point = ???
(adding two points doesn't make sense; however:)
point - point = vector
(the difference of two points is the displacement between them)
Now, these displacements can be thought of in (at least) two different ways. The one you're already familiar with is the rectangular (x, y) system, where the two components of a vector represent the displacement in the x and y directions, respectively. However, you can also use polar coordinates, (r, Θ). Here, Θ represents the direction of the displacement (in angles relative to an arbitary zero angle) and r, the distance.
Take the (1, 1) vector, for example. It represents a movement one unit to the right and one unit upwards in the coordinate system we're all used to seeing. The polar equivalent of this vector would be (1.414, 45°); the same movement, but represented as a "displacement of 1.414 units in the 45°-angle direction. (Again, using a convenient polar coordinate system where the East direction is 0° and angles increase counter-clockwise.)
The relationship between polar and rectangular coordinates are:
Θ = atan2(y, x)
r = sqrt(x²+y²) (now do you see where the right triangle comes in?)
and conversely,
x = r * cos(Θ)
y = r * sin(Θ)
Now, since a line segment drawn from your triangle's centroid to the "tip" corner would represent the direction your triangle is "facing," if we were to obtain a vector parallel to that line (e.g. vForward = pTip - pCentroid), that vector's Θ-coordinate would correspond to the angle that your triangle is facing.
Take the (1, 1) vector again. If this was vForward, then that would have meant that your "tip" point's x and y coordinates were both 1 more than those of your centroid. Let's say the centroid is on (10, 10). That puts the "tip" corner over at (11, 11). (Remember, pTip = pCentroid + vForward by adding "+ pCentroid" to both sides of the previous equation.) Now in which direction is this triangle facing? 45°, right? That's the Θ-coordinate of our (1, 1) vector!
keep the centroid at the origin. use the vector from the centroid to the nose as the direction vector. http://en.wikipedia.org/wiki/Coordinate_rotation#Two_dimensions will rotate this vector. construct the other two points from this vector. translate the three points to where they are on the screen and draw.
double v; // velocity
double theta; // direction of travel (angle)
double dt; // time elapsed
// To compute increments
double dx = v*dt*cos(theta);
double dy = v*dt*sin(theta);
// To compute position of the top of the triangle
double size; // distance between centroid and top
double top_x = x + size*cos(theta);
double top_y = y + size*sin(theta);
I can see that I need to apply the common 2d rotation formulas to my triangle to get my result, Im just having a little bit of trouble with the relationships between the different components here.
aib, stated that:
The arctangent (inverse tangent) of
vy/vx, where vx and vy are the
components of your (centroid->tip)
vector, gives you the angle the vector
is facing.
Is vx and vy the x and y coords of the centriod or the tip? I think Im getting confused as to the terminology of a "vector" here. I was under the impression that a Vector was just a point in 2d (in this case) space that represented direction.
So in this case, how is the vector of the centroid->tip calculated? Is it just the centriod?
meyahoocomlorenpechtel stated:
It seems to me that you need to store
the rotation angle of the triangle and
possibly it's current speed.
What is the rotation angle relative to? The origin of the triangle, or the game window itself? Also, for future rotations, is the angle the angle from the last rotation or the original position of the triangle?
Thanks all for the help so far, I really appreciate it!
you will want the topmost vertex to be the centroid in order to achieve the desired effect.
First, I would start with the centroid rather than calculate it. You know the position of the centroid and the angle of rotation of the triangle, I would use this to calculate the locations of the verticies. (I apologize in advance for any syntax errors, I have just started to dabble in Java.)
//starting point
double tip_x = 10;
double tip_y = 10;
should be
double center_x = 10;
double center_y = 10;
//triangle details
int width = 6; //base
int height = 9;
should be an array of 3 angle, distance pairs.
angle = rotation_angle + vertex[1].angle;
dist = vertex[1].distance;
p1_x = center_x + math.cos(angle) * dist;
p1_y = center_y - math.sin(angle) * dist;
// and the same for the other two points
Note that I am subtracting the Y distance. You're being tripped up by the fact that screen space is inverted. In our minds Y increases as you go up--but screen coordinates don't work that way.
The math is a lot simpler if you track things as position and rotation angle rather than deriving the rotation angle.
Also, in your final piece of code you're modifying the location by the rotation angle. The result will be that your ship turns by the rotation angle every update cycle. I think the objective is something like Asteroids, not a cat chasing it's tail!