As the title describes, I want to make a tiny circle that circulates on the edge of the sector of the another big circle. I have implemented sector of the circle, now only issue here is how to make small circle circulate on the edge of this sector. I have tried various ways, however, none of them was proved to be successful, therefore I plead you to give me some tips of how to implement it.
Thanks in advance.
You just have to consider that, for a circle of radius 1 centered on the origin, every point on the circle can be described as:
P = [sin(alpha); cos(alpha)]
With 0<=alpha<2*pi
Now, if you change the radius and the center you will have:
P = [(radius * sin(alpha))+x_center; (radius*cos(alpha))+y_center]
So, just have a loop for alpha going from 0 to 2*pi (or whatever section of circle you need) and use the above equation to calculate the position of the center of the small circle.
I presume you have a a function that can draw a circle at a given position in cartesian co-ordinates and radius.
Use polar co-ordinates (angle / radius), set the radius to the radius of the big circle minus the small circle. Set the angle to wherever you want to start the circle. Then set a loop up to increment the angle by a given amount. After each increment, clear the screen, draw the big circle. Then convert the polar co-orindates into cartesian, add on the centre of the big circle and draw the small circle. Hold for as long as you want.
Related
I have a sphere in my program and I intend to draw some rectangles over at a distance x from the centre of this sphere. The figure looks something below:
The rectangles are drawn at (x,y,z) points that I have already have in a vector of 3d points.
Let's say the distance x from centre is 10. Notice the orientation of these rectangles and these are tangential to an imaginary sphere of radius 10 (perpendicular to an imaginary line from the centre of sphere to the centre of rectangle)
Currently, I do something like the following:
For n points vector<vec3f> pointsInSpace where the rectnagles have to be plotted
for(int i=0;i<pointsInSpace.size();++i){
//draw rectnagle at (x,y,z)
}
which does not have this kind of tangential orientation that I am looking for.
It looked to me of applying roll,pitch,yaw rotations for each of these rectangles and using quaternions somehow to make them tangential as to what I am looking for.
However, it looked a bit complex to me and I wanted to ask about some better method to do this.
Also, the rectangle in future might change to some other shape, so a kind of generic solution would be appreciated.
I think you essentially want the same transformation as would be accomplished with a LookAt() function (you want the rectangle to 'look at' the sphere, along a vector from the rectangle's center, to the sphere's origin).
If your rectangle is formed of the points:
(-1, -1, 0)
(-1, 1, 0)
( 1, -1, 0)
( 1, 1, 0)
Then the rectangle's normal will be pointing along Z. This axis needs to be oriented towards the sphere.
So the normalised vector from your point to the center of the sphere is the Z-axis.
Then you need to define a distinct 'up' vector - (0,1,0) is typical, but you will need to choose a different one in cases where the Z-axis is pointing in the same direction.
The cross of the 'up' and 'z' axes gives the x axis, and then the cross of the 'x' and 'z' axes gives the 'y' axis.
These three axes (x,y,z) directly form a rotation matrix.
This resulting transformation matrix will orient the rectangle appropriately. Either use GL's fixed function pipeline (yuk), in which case you can just use gluLookAt(), or build and use the matrix above in whatever fashion is appropriate in your own code.
Personally I think the answer of JasonD is enough. But here is some info of the calculation involved.
Mathematically speaking this is a rather simple problem, What you have is a 2 known vectors. You know the position vector and the spheres normal vector. Since the square can be rotated arbitrarily along around the vector from center of your sphere you need to define one more vector, the up vector. Without defining up vector it becomes a impossible solution.
Once you define a up vector vector, the problem becomes simple. Assuming your square is on the XY-plane as JasonD suggest above. Then your matrix becomes:
up_dot_n_dot_n.X up_dot_n_dot_n.Y up_dot_n_dot_n.Z 0
n.X n.y n.z 0
up_dot_n.x up_dot_n.x up_dot_n.z 0
p.x p.y p.z 1
Where n is the normal unit vector of p - center of sphere (which is trivial if sphere is in the center of the coordinate system), up is a arbitrary unit vector vector. The p follows form definition and is the position.
The solution has a bit of a singularity at the up direction of the sphere. An alternate solution is to rotate first 360 around up, the 180 around rotated axis dot up. Produces same thing different approach no singularity problem.
I have black and white image after binarization. After that I get one object with irregular shape. Link to this image is below.
How can I inscribe this object to circle?? or How can I find "center" of this object??
You can find the center of gravity of the pixels using a simple formula which is the sum of the x coordinates divided by the number of points and the sum of the y coordinates divided by the number of points (I mean white points).
Then you can draw a circle centered in the center of gravity with radious half of the maximum distance between points.
Here you have a graphic explanation for this.
This sounds like a smallest circle problem on the set of white pixels. It can be found in linear time in the number of pixels. This is the best you will ever get it your input is just an array of binary pixels.
well, you could scan from top down for the top-most white pixel, then from the bottom up for the bottom-most white pixel, same for left and right. that gives you a rectangle. finding the center of the rectangle is easy (e.g. left + ( right - left ) / 2), and that's your circle center. then find the distance to a corner (any will do), and that's your circle radius.
I think, that center of the object can be easily found as an arithmetic mean of x and y coordinate. I you want to replace it by a circle, I'd say that the diameter is a double of the mean distance of all points to the center.
I am looking for a way to get all the vertices of a rectangle whose center, normal, length and height I know. I am a little weak in maths so please help me.
Edit : the plane is in 3D space.
You can easily calculate the x and y coordinates of the vertices of a rectangle in 2D space given the center, width and height by subtracting/adding half the width/height from the x/y position of the center point.
If you need this in 3D space, this becomes a little more tricky and relies on a bit of trigonometry, but still follows the same principle. You'll need one extra piece of information. You need some way of fixing the orientation of the square in some direction; ie, which direction is the rectangle 'facing'. The normal will allow you to work out what plane the rectangle is on, but without some orientation on that plane, the best you can do is work out a set of possible values in a circle around the center for each of the vertices.
I am trying to implement a gradient brush from scratch in C++ with GDI. I don't want to use GDI+ or any other graphics framework. I want the gradient to be of any direction (arbitrary angle).
My algorithm in pseudocode:
For each pixel in x dirrection
For each pixel in the y direction
current position = current pixel - centre //translate origin
rotate this pixel according to the given angle
scalingFactor =( rotated pixel + centre ) / extentDistance //translate origin back
rgbColor = startColor + scalingFactor(endColor - startColor)
extentDistance is the length of the line passing from the centre of the rectangle and has gradient equal to the angle of the gradient
Ok so far so good. I can draw this and it looks nice. BUT unfortunately because of the rotation bit the rectangle corners have the wrong color. The result is perfect only for angle which are multiples of 90 degrees. The problem appears to be that the scaling factor doesn't scale over the entire size of the rectangle.
I am not sure if you got my point cz it's really hard to explain my problem without a visualisation of it.
If anyone can help or redirect me to some helpful material I'd be grateful.
Ok guys fixed it. Apparently the problem was that when I was rotating the gradient fill (not the rectangle) I wasn't calculating the scaling factor correctly. The distance over which the gradient is scaled changes according to the gradient direction. What must be done is to find where the edge points of the rect end up after the rotation and based on that you can find the distance over which the gradient should be scaled. So basically what needs to be corrected in my algorithm is the extentDistance.
How to do it:
•Transform the coordinates of all four corners
•Find the smallest of all four x's as minX
•Find the largest of all four x's and call it maxX
•Do the same for y's.
•The distance between these two point (max and min) is the extentDistance
Heyo,
I'm currently working on a project where I need to place the camera such that the full motion of a character would be viewable without moving the camera. I have the position where the character starts, as well as the maximum distance that the character will travel in all three directions (X,Y, & Z). I also have the field of view (which is 90 degrees).
Is there an equation that'll figure out where I need to place the camera so it won't have to move to see the full motion?
Note: this is using OpenGL.
Clarification: The camera should be "in front" of the character that's in the motion, not above.
It'll also be moving along a ground plane.
If you make a bounding sphere of the points, all you need to do is keep the camera at a distance greater than or equal to the radius of the bounding sphere / sin(FOV/2).
For example, if you have a bounding sphere with radius Radius, and a specified Field of View FOV, your camera just needs to be at a point "Dist" away, pointing towards the center of the bounding sphere.
The equation for calculating the distance is:
Dist = Radius / sin( FOV/2 );
This will work in 3D, for a camera at any orientation.
Simply having the maximum range of (X, Y, Z) is not on its own sufficient, because the viewing port is essentially pyramid shaped, with the apex of the pyramid being at the eye position.
For the sake of argument, let's assume that all movement is in the (X, Z) plane (i.e. the ground), and the eye is directly above the origin 10m along the Y axis.
Assuming a square viewport, with your 90˚ field of view you'd be able to see from ±10m along both the X and Z axis, but only for objects who are on the ground (Y = 0). As soon as they come off the ground your view is reduced. If it's 1m of the ground then your (X, Z) extent is only ±9m.
Clearly a real camera could be placed anyway in the scene, facing any direction. Even the "roll" angle of the camera could change how much is visible. There are actually infinitely many such camera points, so you will need to constrain your criteria somewhat.
Take the line segment from the startpoint to the endpoint. Construct a plane orthogonal to this line segment through the midpoint of the line segment. Then position the camera somewhere in this plane at an distance of more than the following from the intersection point of plane and line looking at the intersection point. The up vector of the camera must be in the plane and the horizontal field of view must be 90 degrees.
distance = sqrt(dx^2 + dy^2 + dz^2) / 2
This camera positions will all have the startpoint and the endpoint on the left or right border of the view port and verticaly centered.
Another solution might be to write a function that takes the startpoint, the endpoint, and the desired position of both points on the screen. Then just solve the projection equation for the camera transformation.
It depends, for example, if the object is gonna move in a plane, you can just place the camera outside a ball circumscribed its movement area (this depends on the fact that FOV is 90, which is a fortunate angle).
If the object is gonna move in 3D, it's much more difficult. It would help if you'd specify the region where the object moves (cube vs. ball...) and the direction you want to see it from.