So I have one point in 3D space, and I have location and rotation of the Camera in 3D space.
So basically there is Vector3 on the object.
Camera Vector3 and Quaternion.
I need to get how to look at that point.
I want to tell user how to move to that point.
Should user direct camera left or right or behind?
One way to do this is to calculate the direction the camera is currently facing as a yaw angle (like a compass heading), and calculate the direction it needs to face in order to be looking at the point.
Subtract one from the other and adjust the result so that it is in the range of -180 to 180 degrees (or -pi to pi radians) and then tell the user to turn left or right based on the sign. If the absolute value is more than 120 degrees (or some configurable value) then tell them it is behind them.
To find the camera's current heading, transform the vector (0, 0, 1) by the quaternion to get the forward vector, then calculate the heading using atan2(forward.z, forward.x).
To calculate the heading required to look at the point, subtract the current camera position from the point to get a desired forward vector and then pass to atan:
Vector3 desired_forward = point - camera_pos;
float desired_heading = atan2(desired_forward.z, desired_forward.x);
Then find the rotation needed:
float rotation_needed = desired_heading - heading;
if(rotation_needed > Math.PI)
rotation_needed -= 2 * Math.PI;
if(rotation_needed < -Math.PI)
rotation_needed += 2 * Math.PI;
Now tell the user to rotate left or right based on the sign of the rotation needed.
If you want to do it for look up/down, you can calculate a pitch angle by first calculating the length of the forward vector in the XZ plane and then using atan2 again:
float xzLength = sqrt(forward.x * forward.x + forward.z * forward.z);
float pitch_angle = atan2(forward.y, xzLength);
Do the same for the desired forward vector and subtract the current from the desired. Check the sign to tell the user whether to look up or down.
There's a few likely complications. For example, depending on whether the camera quaternion specifies the transform from world space to camera space or vice versa, you might need to negate the calculated camera heading.
Related
Currently I'm working on a orbit system for a game. I've got it so an object will move along a circle based on a progress value that'll be between 0.0 and 1.0 (0.5 being half way around the circle). I calculate this like this:
float angle = Mathf.Deg2Rad * 360 * t;
float xPos = Mathf.Sin(angle) * xAxis;
float yPos = Mathf.Cos(angle) * yAxis;
return new Vector3(xPos, yPos, 0.0f);
With t simply being deltatime and the xAxis/yAxis variables being the radius of the circle.
What I'm a little stuck on currently though is how I could possibly get the progress around the circle based on a poisition. So if I have an object that hits the bottom of the circle, how do I calculate that to be a progress of 0.5?
First step: Find out the angle of your given position with the y-axis.
Second step: Calculate the fraction of a full circle (360 degs) that your angle has.
First step involves a bit of trigonometry, and there you have to make sure to get the right type of angle based on what quadrant you're in. Second step should be trivial then.
You can check out the atan2 function that's available in many programming languages: https://en.wikipedia.org/wiki/Atan2
It gives the angle between a point (x, y) and the positive x-axis. So then in your case, depending on where your circle starts, you'd then shift that by 90 degrees to get the angle with the positive y-axis. Other than that it should work fine though.
A similar question was asked before, unfortunately I cannot comment Samgaks answer so I open up a new post with this one. Here is the link to the old question:
How to calculate ray in real-world coordinate system from image using projection matrix?
My goal is to map from image coordinates to world coordinates. In fact I am trying to do this with the Camera Intrinsics Parameters of the HoloLens Camera.
Of course this mapping will only give me a ray connecting the Camera Optical Centre and all points, which can lie on that ray. For the mapping from image coordinates to world coordinates we can use the inverse camera matrix which is:
K^-1 = [1/fx 0 -cx/fx; 0 1/fy -cy/fy; 0 0 1]
Pcam = K^-1 * Ppix;
Pcam_x = P_pix_x/fx - cx/fx;
Pcam_y = P_pix_y/fy - cy/fy;
Pcam_z = 1
Orientation of Camera Coordinate System and Image Plane
In this specific case the image plane is probably at Z = -1 (However, I am a bit uncertain about this). The Section Pixel to Application-specified Coordinate System on page HoloLens CameraProjectionTransform describes how to go form pixel coordinates to world coordinates. To what I understand two signs in the K^-1 are flipped s.t. we calculate the coordinates as follows:
Pcam_x = (Ppix_x/fx) - (cx*(-1)/fx) = P_pix_x/fx + cx/fx;
Pcam_y = (Ppix_y/fy) - (cy*(-1)/fy) = P_pix_y/fy + cy/fy;
Pcam_z = -1
Pcam = (Pcam_x, Pcam_y, -1)
CameraOpticalCentre = (0,0,0)
Ray = Pcam - CameraOpticalCentre
I do not understand how to create the Camera Intrinsics for the case of the image plane being at a negative Z-coordinate. And I would like to have a mathematical explanation or intuitive understanding of why we have the sign flip (P_pix_x/fx + cx/fx instead of P_pix_x/fx - cx/fx).
Edit: I read in another post that the thirst column of the camera matrix has to be negated for the case that the camera is facing down the negative z-direction. This would explain the sign flip. However, why do we need to change the sign of the third column. I would like to have a intuitive understanding of this.
Here the link to the post Negation of third column
Thanks a lot in advance,
Lisa
why do we need to change the sign of the third column
To understand why we need to negate the third column of K (i.e. negate the principal points of the intrinsic matrix) let's first understand how to get the pixel coordinates of a 3D point already in the camera coordinates frame. After that, it is easier to understand why -z requires negating things.
let's imagine a Camera c, and one point B in the space (w.r.t. the camera coordinate frame), let's put the camera sensor (i.e. image) at E' as in the image below. Therefore f (in red) will be the focal length and ? (in blue) will be the x coordinate in pixels of B (from the center of the image). To simplify things let's place B at the corner of the field of view (i.e. in the corner of the image)
We need to calculate the coordinates of B projected into the sensor d (which is the same as the 2d image). Because the triangles AEB and AE'B' are similar triangles then ?/f = X/Z therefore ? = X*f/Z. X*f is the first operation of the K matrix is. We can multiply K*B (with B as a column vector) to check.
This will give us coordinates in pixels w.r.t. the center of the image. Let's imagine the image is size 480x480. Therefore B' will look like this in the image below. Keep in mind that in image coordinates, the y-axis increases going down and the x-axis increases going right.
In images, the pixel at coordinates 0,0 is in the top left corner, therefore we need to add half of the width of the image to the point we have. then px = X*f/Z + cx. Where cx is the principal point in the x-axis, usually W/2. px = X*f/Z + cx is exactly as doing K * B / Z. So X*f/Z was -240, if we add cx (W/2 = 480/2 = 240) and therefore X*f/Z + cx = 0, same with the Y. The final pixel coordinates in the image are 0,0 (i.e. top left corner)
Now in the case where we use z as negative, when we divide X and Y by Z, because Z is negative, it will change the sign of X and Y, therefore it will be projected to B'' at the opposite quadrant as in the image below.
Now the second image will instead be:
Because of this, instead of adding the principal point, we need to subtract it. That is the same as negating the last column of K.
So we have 240 - 240 = 0 (where the second 240 is the principal point in x, cx) and the same for Y. The pixel coordinates are 0,0 as in the example when z was positive. If we do not negate the last column we will end up with 480,480 instead of 0,0.
Hope this helped a little bit
I'm about to project image into cylindrical panorama. But first I need to get the pixel (or color from pixel) I'm going to draw, then then do some Math in shaders with polar coordinates to get new position of pixel and then finally draw pixel.
Using this way I'll be able to change shape of image from polygon shape to whatever I want.
But I cannot find anything about this method (get pixel first, then do the Math and get new position for pixel).
Is there something like this, please?
OpenGL historically doesn't work that way around; it forward renders — from geometry to pixels — rather than backwards — from pixel to geometry.
The most natural way to achieve what you want to do is to calculate texture coordinates based on geometry, then render as usual. For a cylindrical mapping:
establish a mapping from cylindrical coordinates to texture coordinates;
with your actual geometry, imagine it placed within the cylinder, then from each vertex proceed along the normal until you intersect the cylinder. Use that location to determine the texture coordinate for the original vertex.
The latter is most easily and conveniently done within your geometry shader; it's a simple ray intersection test, with attributes therefore being only vertex location and vertex normal, and texture location being a varying that is calculated purely from the location and normal.
Extemporaneously, something like:
// get intersection as if ray hits the circular region of the cylinder,
// i.e. where |(position + n*normal).xy| = 1
float planarLengthOfPosition = length(position.xy);
float planarLengthOfNormal = length(normal.xy);
float planarDistanceToPerimeter = 1.0 - planarLengthOfNormal;
vec3 circularIntersection = position +
(planarDistanceToPerimeter/planarLengthOfNormal)*normal;
// get intersection as if ray hits the bottom or top of the cylinder,
// i.e. where |(position + n*normal).z| = 1
float linearLengthOfPosition = abs(position.z);
float linearLengthOfNormal = abs(normal.z);
float linearDistanceToEdge = 1.0 - linearLengthOfPosition;
vec3 endIntersection = position +
(linearDistanceToEdge/linearLengthOfNormal)*normal;
// pick whichever of those was lesser
vec3 cylindricalIntersection = mix(circularIntersection,
endIntersection,
step(linearDistanceToEdge,
planarDistanceToPerimeter));
// ... do something to map cylindrical intersection to texture coordinates ...
textureCoordinateVarying =
coordinateFromCylindricalPosition(cylindricalIntersection);
With a common implementation of coordinateFromCylindricalPosition possibly being simply return vec2(atan(cylindricalIntersection.y, cylindricalIntersection.x) / 6.28318530717959, cylindricalIntersection.z * 0.5);.
I'm trying to figure out if I have points that make for example a square:
* *
* *
and let's say I know the center of this square.
I want a formula that will make it for eample twice its size but from the center
* *
* *
* *
* *
Therefore the new shape is twice as large and from the center of the polygon. It has to work for any shape not just squares.
I'm looking more for the theory behind it more than the implementation.
If you know the center point cp and a point v in the polygon you would like to scale by scale, then:
v2 = v - cp; // get a vector to v relative to the centerpoint
v2_scaled = v2 * scale; // scale the cp-relative-vector
v1_scaled = v2_scaled + cp; // translate the scaled vector back
This translate-scale-translate pattern can be performed on vectors of any dimension.
If you want the shape twice as large, scale the distance of the coordinates to be sqrt(2) times further from the center.
In other words, let's say your point is at (x, y) and the center is (xcent, ycent). Your new point should be at
(xcent + sqrt(2)*(x - xcent), ycent + sqrt(2)*(y - ycent))
This will scale the distances from the new 'origin', (xcent, ycent) in such a way that the area doubles. (Because sqrt(2)*sqrt(2) == 2).
I'm not sure there's a clean way to do this for all types of objects. For relatively simple ones, you should be able to find the "center" as the average of all the X and Y values of the individual points. To double the size, you find the length and angle of a vector from the center to the point. Double the length of the vector, and retain the same angle to get the new point.
Edit: of course, "twice the size" is open to several interpretations (e.g., doubling the perimeter vs. doubling the area) These would change the multiplier used above, but the basic algorithm would remain essentially the same.
To do what you want you need to perform three operations: translate the square so that its centroid coincides with the origin of the coordinate system, scale the resulting square, translate it back.
I am writing a drawing program, Whyteboard -- http://code.google.com/p/whyteboard/
I have implemented image rotating functionality, except that its behaviour is a little odd. I can't figure out the proper logic to make rotating the image in relation to the mouse position
My code is something similar to this:
(these are called from a mouse event handler)
def resize(self, x, y, direction=None):
"""Rotate the image"""
self.angle += 1
if self.angle > 360:
self.angle = 0
self.rotate()
def rotate(self, angle=None):
"""Rotate the image (in radians), turn it back into a bitmap"""
rad = (2 * math.pi * self.angle) / 360
if angle:
rad = (2 * math.pi * angle) / 360
img = self.img.Rotate(rad, (0, 0))
So, basically the angle to rotate the image keeps getting increased when the user moves the mouse. However, this sometimes means you have to "circle" the mouse many times to rotate an image 90 degrees, let alone 360.
But, I need it similar to other programs - how the image is rotated in relation to your mouse's position to the image.
This is the bit I'm having trouble with. I've left the question language-independent, although using Python and wxPython it could be applicable to any language
I'm assuming resize() is called for every mouse movement update. Your problem seems to be the self.angle += 1, which makes you update your angle by 1 degree on each mouse event.
A solution to your problem would be: pick the point on the image where the rotation will be centered (on this case, it's your (0,0) point on self.img.Rotate(), but usually it is the center of the image). The rotation angle should be the angle formed by the line that goes from this point to the mouse cursor minus the angle formed by the line that goes from this point to the mouse position when the user clicked.
To calculate the angle between two points, use math.atan2(y2-y1, x2-x1) which will give you the angle in radians. (you may have to change the order of the subtractions depending on your mouse position axis).
fserb's solution is the way I would go about the rotation too, but something additional to consider is your use of:
img = self.img.Rotate(rad, (0, 0))
If you are performing a bitmap image rotation in response to every mouse drag event, you are going to get a lot of data loss from the combined effect of all the interpolation required for the rotation. For example, rotating by 1 degree 360 times will give you a much blurrier image than the original.
Try having a rotation system something like this:
display_img = self.img.Rotate(rad, pos)
then use the display_img image while you are in rotation mode. When you end rotation mode (onMouseUp maybe), img = display_img.
This type of strategy is good whenever you have a lossy operation with a user preview.
Here's the solution in the end,
def rotate(self, position, origin):
""" position: mouse x/y position, origin: x/y to rotate around"""
origin_angle = self.find_angle(origin, self.center)
mouse_angle = self.find_angle(position, self.center)
angle = mouse_angle - origin_angle
# do the rotation here
def find_angle(self, a, b):
try:
answer = math.atan2((a[0] - b[0]) , (a[1] - b[1]))
except:
answer = 0
return answer