I have two 3d points, as Eigen::Vector3d.
I need to rotate the second point by a quaternion, around the first point'
I usually use this to rotate a Vector:
Eigen::Vector3d point; //point to rotate
Eigen::Quaterniond quat, p2; //quat to rotate by, temp
p2.w() = 0;
p2.vec() = point;
Eigen::Quaterniond rotatedP1 = quat * p2 * quat.inverse();
Eigen::Vector3d rotatedPoint = rotatedP1.vec();
This works, but it rotates the point around zero.
How can I rotate the point around another point?
As if the first point is rotating, and the second point is 'parented' to it.
In the image below, I am getting a rotation around the green axis, and i want it around the red axis, by passing point2 into the function somehow.
To rotate a vector point around the origin by quat, you just write quat * point (because the * operator is overloaded accordingly).
To rotate around a different point, you need to shift the point then rotate and shift it back.
quat * (point - point2) + point2
If quat and point2 are known before you can also calculate
quat * point + (point2 - quat * point2)
where the part in parentheses can be precalculated.
Related
I am trying to make an image move towards my mouse pointer. Basically, I get the angle between the points, and move along the x axis by the cosine of the angle, and move along the y axis the sine of the angle.
However, I don't have a good way of calculating the angle. I get the difference in x and the difference in y, and use arctangent of Δy/Δx. The resulting angle in quadrant 1 is correct, but the other three quadrants are wrong. Quadrant 2 ranges from -1 to -90 degrees. Quadrant 3 is always equal to quadrant 1, and quadrant 4 always equals quadrant 4. Is there an equation that I can use to find the angle between the two points from 1-360 degrees?
Note: I cannot use atan2(), and I do not know what a vector is.
// This working code is for Windows HDC mouse coordinates gives the angle back that is used in Windows. It assumes point 1 is your origin point
// Tested and working on Visual Studio 2017 using two mouse coordinates in HDC.
//
// Code to call our function.
float angler = get_angle_2points(Point1X, Point1Y, Point2X, Point2Y);
// Takes two window coordinates (points), turns them into vectors using the origin and calculates the angle around the x-axis between them.
// This function can be used for any HDC window. I.e., two mouse points.
float get_angle_2points(int p1x, int p1y, int p2x,int p2y)
{
// Make point1 the origin, and make point2 relative to the origin so we do point1 - point1, and point2-point1,
// Since we don’t need point1 for the equation to work, the equation works correctly with the origin 0,0.
int deltaY = p2y - p1y;
int deltaX = p2x - p1x; // Vector 2 is now relative to origin, the angle is the same, we have just transformed it to use the origin.
float angleInDegrees = atan2(deltaY, deltaX) * 180 / 3.141;
angleInDegrees *= -1; // Y axis is inverted in computer windows, Y goes down, so invert the angle.
//Angle returned as:
// 90
// 135 45
//
// 180 Origin 0
//
//
// -135 -45
//
// -90
// The returned angle can now be used in the C++ window function used in text angle alignment. I.e., plf->lfEscapement = angle*10;
return angleInDegrees;
}
The answers regarding atan2 are correct. For reference, here is atan2 in Scratch block form:
If you're unable to use atan2() directly, you could implement its internal calculations on your own:
atan2(y,x) = atan(y/x) if x>0
atan(y/x) + π if x<0 and y>0
atan(y/x) - π if x<0 and y<0
This is the code I use, and it seems to work perfectly fine.
atan(x/y) + (180*(y<0))
where X is the difference between the Xs of the points (x2 - x1), and Y is the difference between the Ys (y2 - y1).
atan((x2-x1)/(y1-y2)) + (180*((y1-y2)<0))
I would like to draw a textured circle in Direct3D which looks like a real 3D sphere. For this purpose, I took a texture of a billard ball and tried to write a pixel shader in HLSL, which maps it onto a simple pre-transformed quad in such a way that it looks like a 3-dimensional sphere (apart from the lighting, of course).
This is what I've got so far:
struct PS_INPUT
{
float2 Texture : TEXCOORD0;
};
struct PS_OUTPUT
{
float4 Color : COLOR0;
};
sampler2D Tex0;
// main function
PS_OUTPUT ps_main( PS_INPUT In )
{
// default color for points outside the sphere (alpha=0, i.e. invisible)
PS_OUTPUT Out;
Out.Color = float4(0, 0, 0, 0);
float pi = acos(-1);
// map texel coordinates to [-1, 1]
float x = 2.0 * (In.Texture.x - 0.5);
float y = 2.0 * (In.Texture.y - 0.5);
float r = sqrt(x * x + y * y);
// if the texel is not inside the sphere
if(r > 1.0f)
return Out;
// 3D position on the front half of the sphere
float p[3] = {x, y, sqrt(1 - x*x + y*y)};
// calculate UV mapping
float u = 0.5 + atan2(p[2], p[0]) / (2.0*pi);
float v = 0.5 - asin(p[1]) / pi;
// do some simple antialiasing
float alpha = saturate((1-r) * 32); // scale by half quad width
Out.Color = tex2D(Tex0, float2(u, v));
Out.Color.a = alpha;
return Out;
}
The texture coordinates of my quad range from 0 to 1, so I first map them to [-1, 1]. After that I followed the formula in this article to calculate the correct texture coordinates for the current point.
At first, the outcome looked ok, but I'd like to be able to rotate this illusion of a sphere arbitrarily. So I gradually increased u in the hope of rotating the sphere around the vertical axis. This is the result:
As you can see, the imprint of the ball looks unnaturally deformed when it reaches the edge. Can anyone see any reason for this? And additionally, how could I implement rotations around an arbitrary axis?
Thanks in advance!
I finally found the mistake by myself: The calculation of the z value which corresponds to the current point (x, y) on the front half of the sphere was wrong. It must of course be:
That's all, it works as exspected now. Furthermore, I figured out how to rotate the sphere. You just have to rotate the point p before calculating u and v by multiplying it with a 3D rotation matrix like this one for example.
The result looks like the following:
If anyone has any advice as to how I could smooth the texture a litte bit, please leave a comment.
I want to calculate points of elliptical arc. I knwo coordinates of start point, end point, center of ellipse, radian for x and y, i know angle of x rotate, i know angles beetwen end point and center, and start point and center.
I have problem with calculate points and rotate them.
I use euqation:
x = rx * cos(alfa)
y = ry * sin(alfa)
[resultx, resulty] = ([x,y] * rotatex) + [cx,cy] - rotate and translate with cx,cy (center of ellipse)
The main problem is, that start point and end point is rotated too, but these points should be static (without rotate). I don't know how to inlcude this case.
Now i get points for rotated ellpitical arc, but my start point and end point are also rotated.
Your question is unclear, but I think this is what you want:
x = rx * cos(alfa-beta)
y = ry * sin(alfa-beta)
[resultx, resulty] = ([x,y] * rotatex(beta)) + [cx,cy]
i have an object that is doing a circular motion in a 3d space, the center or the circle is at x:0,y:0,z:0 the radius is a variable. i know where the object is on the circle (by its angle [lets call that ar] or by the distance it has moved). the circle can be tilted in all 3 directions, so i got three variables for angles, lets call them ax,ay and az. now i need to calculate where exactly the object is in space. i need its x,y and z coordinates.
float radius = someValue;
float ax = someValue;
float ay = someValue;
float az = someValue;
float ar = someValue; //this is representing the angle of the object on circle
//what i need to know
object.x = ?;
object.y = ?;
object.z = ?;
You need to provide more information to get the exact formula. The answer depends on which order you apply your rotations, which direction you are rotating in, and what the starting orientation of your circle is. Also, it will be much easier to calculate the position of the object considering one rotation at a time.
So, where is your object if all rotations are 0?
Let's assume it's at (r,0,0).
The pseudo-code will be something like:
pos0 = (r,0,0)
pos1 = pos0, rotated around Z-axis by ar (may not be Z-axis!)
pos2 = pos1, rotated around Z-axis by az
pos3 = pos2, rotated around Y-axis by ay
pos4 = pos3, rotated around X-axis by ax
pos4 will be the position of your object, if everything is set up right. If you have trouble setting it up, try keeping ax=ay=az=0 and worry about just ar, until your get that right. Then, start setting the other angles one at a time and updating your formula.
Each rotation can be performed with
x' = x * cos(angle) - y * sin(angle)
y' = y * cos(angle) + x * sin(angle)
This is rotation on the Z-axis. To rotate on the Y-axis, use z and x instead of x and y, etc. Also, note that angle is in radians here. You may need to make angle negative for some of the rotations (depending which direction ar, ax, ay, az are).
You can also accomplish this rotation with matrix multiplication, like Marcelo said, but that may be overkill for your project.
Use a rotation matrix. Make sure you use a unit vector.
I'm writing a script where icons rotate around a given pivot (or origin). I've been able to make this work for rotating the icons around an ellipse but I also want to have them move around the perimeter of a rectangle of a certain width, height and origin.
I'm doing it this way because my current code stores all the coords in an array with each angle integer as the key, and reusing this code would be much easier to work with.
If someone could give me an example of a 100x150 rectangle, that would be great.
EDIT: to clarify, by rotating around I mean moving around the perimeter (or orbiting) of a shape.
You know the size of the rectangle and you need to split up the whole angle interval into four different, so you know if a ray from the center of the rectangle intersects right, top, left or bottom of the rectangle.
If the angle is: -atan(d/w) < alfa < atan(d/w) the ray intersects the right side of the rectangle. Then since you know that the x-displacement from the center of the rectangle to the right side is d/2, the displacement dy divided by d/2 is tan(alfa), so
dy = d/2 * tan(alfa)
You would handle this similarily with the other three angle intervals.
Ok, here goes. You have a rect with width w and depth d. In the middle you have the center point, cp. I assume you want to calculate P, for different values of the angle alfa.
I divided the rectangle in four different areas, or angle intervals (1 to 4). The interval I mentioned above is the first one to the right. I hope this makes sense to you.
First you need to calculate the angle intervals, these are determined completely by w and d. Depending on what value alfa has, calculate P accordingly, i.e. if the "ray" from CP to P intersects the upper, lower, right or left sides of the rectangle.
Cheers
This was made for and verified to work on the Pebble smartwatch, but modified to be pseudocode:
struct GPoint {
int x;
int y;
}
// Return point on rectangle edge. Rectangle is centered on (0,0) and has a width of w and height of h
GPoint getPointOnRect(int angle, int w, int h) {
var sine = sin(angle), cosine = cos(angle); // Calculate once and store, to make quicker and cleaner
var dy = sin>0 ? h/2 : h/-2; // Distance to top or bottom edge (from center)
var dx = cos>0 ? w/2 : w/-2; // Distance to left or right edge (from center)
if(abs(dx*sine) < abs(dy*cosine)) { // if (distance to vertical line) < (distance to horizontal line)
dy = (dx * sine) / cosine; // calculate distance to vertical line
} else { // else: (distance to top or bottom edge) < (distance to left or right edge)
dx = (dy * cosine) / sine; // move to top or bottom line
}
return GPoint(dx, dy); // Return point on rectangle edge
}
Use:
rectangle_width = 100;
rectangle_height = 150;
rectangle_center_x = 300;
rectangle_center_y = 300;
draw_rect(rectangle_center_x - (rectangle_width/2), rectangle_center_y - (rectangle_center_h/2), rectangle_width, rectangle_height);
GPoint point = getPointOnRect(angle, rectangle_width, rectangle_height);
point.x += rectangle_center_x;
point.y += rectangle_center_y;
draw_line(rectangle_center_x, rectangle_center_y, point.x, point.y);
One simple way to do this using an angle as a parameter is to simply clip the X and Y values using the bounds of the rectangle. In other words, calculate position as though the icon will rotate around a circular or elliptical path, then apply this:
(Assuming axis-aligned rectangle centered at (0,0), with X-axis length of XAxis and Y-axis length of YAxis):
if (X > XAxis/2)
X = XAxis/2;
if (X < 0 - XAxis/2)
X = 0 - XAxis/2;
if (Y > YAxis/2)
Y = YAxis/2;
if (Y < 0 - YAxis/2)
Y = 0 - YAxis/2;
The problem with this approach is that the angle will not be entirely accurate and the speed along the perimeter of the rectangle will not be constant. Modelling an ellipse that osculates the rectangle at its corners can minimize the effect, but if you are looking for a smooth, constant-speed "orbit," this method will not be adequate.
If think you mean rotate like the earth rotates around the sun (not the self-rotation... so your question is about how to slide along the edges of a rectangle?)
If so, you can give this a try:
# pseudo coode
for i = 0 to 499
if i < 100: x++
else if i < 250: y--
else if i < 350: x--
else y++
drawTheIcon(x, y)
Update: (please see comment below)
to use an angle, one line will be
y / x = tan(th) # th is the angle
the other lines are simple since they are just horizontal or vertical. so for example, it is x = 50 and you can put that into the line above to get the y. do that for the intersection of the horizontal line and vertical line (for example, angle is 60 degree and it shoot "NorthEast"... now you have two points. Then the point that is closest to the origin is the one that hits the rectangle first).
Use a 2D transformation matrix. Many languages (e.g. Java) support this natively (look up AffineTransformation); otherwise, write out a routine to do rotation yourself, once, debug it well, and use it forever. I must have five of them written in different languages.
Once you can do the rotation simply, find the location on the rectangle by doing line-line intersection. Find the center of the orbited icon by intersecting two lines:
A ray from your center of rotation at the angle you desire
One of the four sides, bounded by what angle you want (the four quadrants).
Draw yourself a sketch on a piece of paper with a rectangle and a centre of rotation. First translate the rectangle to centre at the origin of your coordinate system (remember the translation parameters, you'll need to reverse the translation later). Rotate the rectangle so that its sides are parallel to the coordinate axes (same reason).
Now you have a triangle with known angle at the origin, the opposite side is of known length (half of the length of one side of the rectangle), and you can now:
-- solve the triangle
-- undo the rotation
-- undo the translation