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circle in three dimensional coordiantes [closed]
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I want to draw a circle in three dimensional coordiantes, i'm given a vector, the angle where vector's intersects with the circle is 90 degrees, the intersection point is the centre of the circle. The radius can be parametrized. EDIT: I am programming a server plugin for minecraft. At this point I have made a sword that can be thrown. I want to add some decor. I want that after the sword there was a trace in the form of a circle. But I don't understand how to draw a circle in 3D coordinates so that the angle of intersection of the sword throw vector with the center of the circle is 90 degrees. The radius can be arbitrary, and the vector can enter the center of the circle. I thought I could just rotate the throw vector on 3 axes and get a circle, but nothing worked. I need an equation with which I can draw a given circle.
You have center C, normal vector N, radius R. Seems you want to get points at the circumference.
At first get some base vector in the circle plane.
Possible way:
Reveal normal component with the largest magnitude and with the second magnitude. For example, abs(N.X) is the largest, abs(N.Z) has the second magnitude, and abs(N.Y) is the smallest. Make the smallest component zero, exchange two larger ones, and negate the largest. For this example base vector will be:
A = (N.Z, 0, -N.X)
It is perpendicular to normal, hence lies in the circle plane.
Then get the next basis vector using vector product (B will be perpendicular both to A and to N, it lies in the plane too)
B = N x A
Now normalize vectors A and B (make them unit length)
A = A / len(A)
B = B / len(B)
and you can get any point at the circumeference with parametric equation where t changes in the range 0..2*Pi
P(t) = C + R * A * Cos(t) + R * B * Sin(t)
or in components:
P.X = C.X + R * A.X * Cos(t) + R * B.X * Sin(t)
and so on
Related
I want to find the blue point
I know the red point b
and the angle between 2 radii
image
We know that the red point has coordinates (B.x, B.y). We know the origin point (A.x, A.y). We know the angle between the points theta. We want to find (C.x, C.y). I will be assuming 2d geometry, as this is what your reference shows.
Using a rotation matrix, we can subtract B by A to place A at the origin, then applying the matrix with the angle given.
Thus, given
B_new = (B.x - A.x, B.y - A.y),
C = (B_new.x * cos(theta) - B_new.y * sin(theta), B_new.x * sin(theta) + B_new.y * cos(theta)) + (A.x, A.y)
I am building a minecraft plugin.
Basically my world limit is set by radius off the spawn location, instead of having 4 corners to determine the limited region of the world.
So my world limit is a round radius for example of distance of 1000 blocks.
Now I want to place a small 50x50 round area of blocks inside of that area of 1000 blocks radius, so I need to find a random coordinate which will be the center point of the round area I am going to make.
Question
How to get a random coordinate in a radius based area, is there an equation for that? I can't just do spawnX + rand(-1000,1000) and for spawnZ, because the area is round.
How can I make sure that the whole round area I want to put in will fit in the big radius area?
To pick a random spot in a circle would be a random angle and a random number within the radius:
random_angle = random(2π)
random_dist = random(radius)
random_x = cos(random_angle) * random_dist
random_y = sin(random_angle) * random_dist
To ensure a round area fits into another round area you calculate the distance from the middle for the small round + its radius and as long as that's equal to or less than the bigger round then it fits in.
To find the distance is a² + b² = c². So subsitute that for x and y to get:
distance = √(x_diff² + y_diff²)
This all assumes your co-ords start at (0, 0) in the middle of the map circle. If not you can calculate that by adding half the width and height to all co-ords.
If you want each point of your circle to have an equal chance of being selected, you can generate a set of polar coordinates instead, then convert it to cartesian coordinates.
Generate these random numbers:
r: between 0 and 1000
theta: between 0 and 2π
Assuming the centre of the circle is (0,0), the cartesian coordinates are:
x = r * cos(theta)
y = r * sin(theta)
To ensure that the smaller circle with radius 50 is inside the big circle, just limit r to be between 0 and 950 instead.
I'm not sure if I understand your problem correctly, but I think you can use circle formula for your coordinates. In a circle, we always have:
X^2+Y^2=R^2
so if you select x = spawnX + rand(-1000,1000), then you can calculate possible y range with aforementioned formula.
Purpose
I'm implementing a polygon rotation with Java AWT.
I'm already able to draw polygons on the screen, and I'd like to apply a rotation matrix manually upon my polygons coordinates (rotation is done around the lookAt point of the user).
What I've already done
In order to rotate the world, the user first clicks on the screen and then drags the mouse around to perform the rotation.
Let's note the first click point as S, the following point from the drag event as L, and the center of the screen as C.
In order to calculate the rotation angle, when first clicking the screen, I keep a vector from C to S: C-S.
Then, when a drag event occurs, I calculate the vector from C to L: C-L.
I then calculate the angle in radians between C-S to C-L, and that's what I apply on my world.
This works well, and the polygon is indeed rotation around the lookAt point.
My problem
The problem occurs when the user finishes a rotation of PI, and then the polygon is rotating backward.
e.g. When the user starts rotating, the angle starts from 0.1.... 0.2... 1.. 2.. 3.. and in value ~3.1 (I assume PI), the values are starting to go down: 3... 2.. 1.. until 0, and vice versa.
This makes sense since the radians range is [0, PI].
I assume the base vector C-S lies on the right side of X axis, and when the rotation goes down below the X axis the polygon is rotating backwards.
However, I have no idea how to keep the polygon rotating in the same direction all the time (when the user performs a full rotation around the polygon).
Edit
Angle function is:
public final double angle(Vector2D v1)
{
double vDot = this.dot(v1) / ( this.length()*v1.length() );
if( vDot < -1.0) vDot = -1.0;
if( vDot > 1.0) vDot = 1.0;
return ((double) (Math.acos( vDot )));
}
This is a problem of the arcus cosine, acos(cos(x)) is a periodic hat function moving up and down in the range of 0 to pi.
In higher dimensions that can not be avoided, as there is no preferred frame of reference, so there is no way to say that phi should really be -phi. In 2 dimensions there is a prefered orientation of the plane so that one can say what is the first and what the second vector and define a unique angle in positive orientation. Rotate the situation so that the first vector comes to lay on the positive real half axis to get the angle and correct quadrant from the coordinates of the rotated second vector.
Easiest to reconstruct is the complex picture, to compute the angle from a=a.x+i*a.y to b=b.x+i*b.y rotate b back by multiplying with the conjugate of a to get an angle from the zero angle resp. the positive real axis,
arg((a.x-i*a.y)*(b.x+i*b.y))
=arg((a.x*b.x+a.y*b.y)+i*(a.x*b.y-a.y*b.x))
=atan2( a.x*b.y-a.y*b.x , a.x*b.x+a.y*b.y )
Note that screen coordinates use the opposite orientation to the cartesian/complex plane, thus change atan2(y,x) to atan2(-y,x) to get an angle in the usual direction.
public Point rotate(Point original, Point vertex, double angle){
Point translated = new Point(original.x - vertex.x, original.y - vertex.y);
int x = (int)Math.round(translated.x * Math.cos(angle) - translated.y * Math.sin(angle));
int y = (int)Math.round(translated.x * Math.sin(angle) + translated.y * Math.cos(angle));
return new Point(vertex.x+x,vertex.y+y);
}
This is a simple rotation method that you can use to rotate a point around a given vertex.
Assume I have several bounding boxes with 4 coordinates pair (long/lat only) each representing the 4 corners of a square box. How can I check if 2 of those boxes intersects?
I know I could use java.awt.Rectangle to check if 2 rectangles intersects, but the problem is it is calculated using X/Y/Width/Height instead of coordinates.
Can someone please give me some directions on how can I do this calculations?
Thanks.
EDIT
What I am trying to accomplish is the same represented by this library.
Basically it calculates a square bounding box around a given point and check if the (imaginary) squares intersects with each other, like in this image:
(source: google.com)
So far I've been able to calculate the corners for each marker and now I need to somehow check if they intersect with each other. How can I do this intersection calculation?
EDIT 2
This is how I am calculating the corners:
private static double getLatitude(double distance, double lat, double angle) {
return toDegrees(asin(sin(toRadians(lat)) * cos(distance / RADIUS) + cos(toRadians(lat)) * sin(distance / RADIUS) * cos(toRadians(angle))));
}
private static double getLongitude(double distance, double lat, double lng, double angle) {
double newLat = getLatitude(distance, lat, angle);
return toDegrees(toRadians(lng) + atan2(sin(toRadians(angle)) * sin(distance / RADIUS) * cos(toRadians(lat)), cos(distance / RADIUS) - sin(toRadians(lat)) * sin(toRadians(newLat))));
}
Where RADIUS = 6378.1 and angle = 45/135/225/315 (top right, bottom right, bottom left and top left).
Example output
I'm assuming that in your "lat/long bounding box' each side follows the lines of constant longitude and latitude - in other words that the top side follows the line of constant latitude, and the left side the line of constant longitude.
While this is not actually a rectangle in real life, it can actually be treated as one for our purposes. Mathematically you can think of this as transforming the bounding box into a "lat/long' space, where the shape is in fact a rectangle. If that doesn't make sense you may have to take my word for it. In any case it is possible to show that the curved shapes in real space intersect if and only if the rectangles intersect in curved space.
The short version of this is: if you do a standard test for intersection of rectangles (using the Java Rectangle class code, and using latitude and longitude as the rectangle bounds) you will get the right result.
EXAMPLE
You have two areas, defined as:
The area between 50 and 52 degrees N and 75 and 77 degrees E
The area between 51 and 53 degrees N and 76 and 79 degrees E
You can correctly test for their intersection by doing:
Rectangle r1 = new Rectangle(75,50,2,2);
Rectangle r2 = new Rectangle(76,51,2,3);
boolean intersects = r1.insersects(r2);
It doesn't matter that the rectangles are not rectangular in Euclidean space.
P.S. This will not work if one of your rectangles actually contains either the north or south pole. In that case you will need to split each rectangle into two, one on each side of the pole. You need to normalize everything to +/- 90 latitude and +/- 180 longitude. You will need to do something clever if one or more of the rectangles overlaps the +/-180 longitude line.
I am trying to make a circle (actually a flat cylinder) rotate so that the edge crosses two points in world position. These two points can be anywhere on a sphere. The sphere has the same radius and position as the cylinder. the origin of both is [0,0,0].
It's a little bit hard to explain, so I included three pictures that I hope illustrates what I am trying to accomplish.
Here you see what I am trying to accomplish. The yellow circle represents one of the points along the sphere, while the red circle represents the other point. The blue line is actually a flat cylinder going through the sphere, and is rotated so that it goes through both points.
Here is another similar picture, but with the points at different locations.
In this picture one can see the cylinder in full, as the sphere has been hidden.
Now, I am really terrible at math, so I would really love an answer made up of pseudo code or a programming language. And if I should be so lucky, java.
The circles rotation can be represented with either a quaternion or a matrix
So far, what I have tried, is rotating the cylinder with an up vector towards one of the points, and a forward vector towards the other point. But I can't seem to make it work. I have also tried other solutions, most of them involving two rotations (one for each point), but I end up having trouble when combining the rotations.
Here is my current non-working code:
This code makes the circle go through the first point, and then rotates it with an "up vector" towards the same point, this second rotation varies depending on the first point position, and is kind off all over the place.
//calculate direction vector between the two points
point1point2dir.set(point1Pos);
//subtract point two position
point1point2dir.sub(point2Pos);
//normalize
point1point2dir.nor();
//make two quaternions for rotation
Quaternion rot1=new Quaternion();
Quaternion rot2=new Quaternion();
//set first rotation two a rotation between X-axis and point1 position. Vector3.X = (1,0,0)
rot1.set(m.quatUtils.getRot(Vector3.X, point1Pos));
//crossmuliply direction vector between the two points with X-axis
point1point2dir.crs(Vector3.X);
//set the second rotation to a rotation between Z-Axis and the crossmultiplied direction vector
rot2.set(m.quatUtils.getRot(Vector3.Z, point1point2dir));
//multiply the two rotations
rot1.mul(rot2);
//apply the rotation to the cylinders matrix
cylinderMatrix.rotate(rot1);
//the function that gets the quaternion rotation between two vectors
Quaternion getRot(Vector3 pStart, Vector3 pDest) {
start.set(pStart);
dest.set(pDest);
start.nor();
dest.nor();
cosTheta = Vector3.dot(start.x, start.y, start.z, dest.x, dest.y,
dest.z);
rotationAxis.set(0.0f, 0.0f, 0.0f);
if (cosTheta < -1.0f + 0.001f) {
rotationAxis.set(Z_AXIS);
rotationAxis.crs(start);
if (rotationAxis.len2() < 0.01f) {
rotationAxis.set(X_AXIS);
rotationAxis.crs(start);
}
rotationAxis.nor();
resultQuat.set(rotationAxis, 180.0f);
return resultQuat;
}
rotationAxis.set(start);
rotationAxis.crs(dest);
s = (float) Math.sqrt((1 + cosTheta) * 2);
invs = 1.0f / s;
resultQuat.set(rotationAxis.x * invs, rotationAxis.y * invs,
rotationAxis.z * invs, s * 0.5f);
return resultQuat;
}
I would suggest this solution:
Calculate v1 and v2 as the vectors from the center of the sphere to each point that you want the cylinder to pass trough.
Cross product v1 and v2 to get the vector up of the cylinder, let's call it n.
Position the center of the cylinder in the center of the sphere.
Rotate the cylinder using n as vector up.
I figured out the solution! It was actually really simple. I don't know how I managed to bungle the math as much as I did earlier. I actually did spend alot of time on this >:)
Sorry if I wasted anybodys time!
The solution:
find direction vector from point1 (A) to point2 (B).
crossmultiply direction vector with point2 to get (C)
Find the quaternion which represents the rotation from Z-axis to the crossmultiplied direction vector (C), function for doing this included in the code attached to the question.
apply rotation.
Here is the working code (yay):
//the rotation
Quaternion rot=new Quaternion();
//the direction from point1 to point 2 (the point positions are in this case also the direction vectors from center)
point1point2dir.set(point1Pos);
point1point2dir.sub(point2Pos);
point1point2dir.nor();
//crossmultiplied with point2
point1point2dir.crs(point2Pos);
//set the rotation to the rotation between Z-axis and the crossmultiplied direction between point 1 and 2
rot.set(m.quatUtils.getRot(Vector3.Z, point1point2dir));
//apply rotation
ekvatorMatrix.rotate(rot);
And here is the code for the function that returns the quaternion between two vectors:
Quaternion getRot(Vector3 pStart, Vector3 pDest) {
start.set(pStart);
dest.set(pDest);
start.nor();
dest.nor();
cosTheta = Vector3.dot(start.x, start.y, start.z, dest.x, dest.y,
dest.z);
rotationAxis.set(0.0f, 0.0f, 0.0f);
if (cosTheta < -1.0f + 0.001f) {
rotationAxis.set(Z_AXIS);
rotationAxis.crs(start);
if (rotationAxis.len2() < 0.01f) {
rotationAxis.set(X_AXIS);
rotationAxis.crs(start);
}
rotationAxis.nor();
resultQuat.set(rotationAxis, 180.0f);
return resultQuat;
}
rotationAxis.set(start);
rotationAxis.crs(dest);
s = (float) Math.sqrt((1 + cosTheta) * 2);
invs = 1.0f / s;
resultQuat.set(rotationAxis.x * invs, rotationAxis.y * invs,
rotationAxis.z * invs, s * 0.5f);
return resultQuat;
}
Assuming that the initial cylinder is axis aligned with the "circle" ends in positive and negative X direction, and assuming cylinder and sphere is initially unit size (radius=1.0) I would do the following:
Convert the world coordinate representation of the Red and "Yellow" points (let's just for fun call them A and B shall we) to normalized vectors pointing from centre [0,0,0] (from now on called C)
Calculate the angle between CA and CB (which is really just between A and B). Let's call this angle W
Calculate the vector perpendicular to both A and B by doing a cross product. Lets call this new vector D.
Find the rotation matrix that rotates from [0,0,1] to B. Lets call this M1. This can be done in the same way as in point 3 (create a perpendicular vector and rotate identity matrix around it with the angle between the normalized vectors).
Find the rotation matrix that rotates W around D. Let's call this M2
Combine M1 + M2 into M3
You result is M3
This was not tested and so I don't know if it works.