I'm drawing squares along a circular path for an iOS application. However, at certain points along the circle, the squares start to go out of the circle's circumference. How do I make sure that the squares stay inside?
Here's an illustration I made. The green squares represent the positions I need the squares to actually be in. The red squares are where they actually appear given the following values for each square's upper-left corner:
x = origin.x + radius * cos(DEGREES_TO_RADIANS(angle));
y = origin.y + radius * sin(DEGREES_TO_RADIANS(angle));
Origin refers to the center of the circle. I have a loop that repeats this for every angle from 1 till 360 degrees.
EDIT: I've changed my design to position the centers of the squares along the circular path rather than their upper left corners.
why not just draw the centers of the squares along a smaller circle inside of the bigger one?
You could do the math to figure out exactly what the radius would have to be to ensure an exact fit, but you could probably trial and error your way there quickly too.
Doing it this way ensures that your objects would end up laid out in an actual circle too, which is not the case if you were merely making sure that one and only one corner of each square touched the larger bounding circle (that would create a slightly octagonal shape instead of a circle)
ryan cumley's answer made me realize how dumb I was all along. I just needed to change each square's anchor point to its center & that solved it. Now every calculated value for x & y would position every square's center exactly on the circular path.
Option 1) You could always find the diameter of the circle and then using Pythagorean Theorem, you could create a square that would fit perfectly within the circle. You could then loop through the square that was just made in the circle to create smaller squares, but I doubt this is what you are aiming for.
Option2) Find out what half of the length of one of the diagonals of the squares should be, and create a ring within the first ring. Then lay down squares at key points (like ever 30 degrees or 15 degrees, etc) along the inner path. Ex: http://i.imgur.com/1XYhoQ0.png
As you can see, the smaller (inner) circle is in the center of each green square, and that ensures that the corners of each square just touches the larger (outer) circle. Obviously my cheaply made picture in paint is not perfect, but mathematically it will work.
Related
Suppose that I want to find the 3D position of a cup with its rotation, with image input like this (this cup can be rotated to point in any direction):
Given that I have a bunch of 2D points specifying the top circle and bottom circle like the following image. (Let's assume that these points are given by a person drawing the lines around the cup, so it won't be very accurate. Ellipse fitting or SolvePnP might be needed to recover a good approximation. And the bottom circle is not a complete circle, it's just part of a circle. Sometimes the top part will be occluded as well so we cannot rely that there will be a complete circle)
I also know the physical radius of the top and bottom circle, and the distance between them by using a ruler to measure them beforehand.
I want to find the complete 2 circle like following image (I think I need to find the position of the cup and its up direction before I could project the complete circles):
Let's say that my ultimate goal is to be able to find the closest 2D top point and closest 2D bottom point, given a 2D point on the side of the cup, like the following image:
A point can also be inside of the cup, like so:
Let's define distance(a, b) as a function that find euclidean distance from point a and point b in pixel units.
From that I would be able to calculate the distance(side point, bottom point) / distance(top point, bottom point) which will be a scale number from 0 to 1, if I multiply this number to the physical height of the cup measured by the ruler, then I will know how high the point is from the bottom of the cup in metric unit.
What is the method I can use to find the corresponding top and bottom point given point on the side, so that I can finally find out the height of the point from the bottom of the cup?
I'm thinking of using PnP to solve this but my points do not have correct IDs associated with them. And I don't want to know the exact rotation of the cup, I only want to know the up direction of the cup.
I also think that fitting the ellipse might help somewhat, but maybe it's not the best because the circle is not complete.
If you have any suggestions, please tell me how to obtain the point height from the bottom of the cup.
Given the accuracy issues, I don't think it is worth performing a 3D reconstruction of the cone.
I would perform a "standard" ellipse fit on the top outline, which is the most accurate, then a constrained one on the bottom, knowing the position of the vertical axis. After reduction of the coordinates, the bottom ellipse can be written as
x²/a² + (y - h)²/b² = 1
which can be solved by least-squares.
Note that it could be advantageous to ask the user to point at the endpoints of the straight edges at the bottom, plus the lowest point, instead of the whole curve.
Solving for the closest top and bottom points is a pure 2D problem (draw the line through the given point and the intersection of the sides, and find the intersection points with the ellipse.
In OpenCV or object detection models, they represent bounding box as 4 numbers e.g. x,y,width,height or x1,y1,x2,y2.
These numbers seem to be ill-defined but it's fine when the resolution is big.
But it causes me to think when the image has very low resolution e.g. 8x8, the one-pixel error can cause things to go very wrong.
So I want to know, what exactly does it mean when you say that a bounding box has x1=0, x2=100?
Specifically, I want to clear these confusions when understood well:
Does the bounding box border occupy the 0th pixel or is it surrounding 0th pixel (its border is at x=-1)?
Where is the exact end of the bounding box? If the image have shape=(8,8), would the end be at 7 or 8?
If you want to represent a bounding box that occupy the entire image, what should be its values?
So I think the right question should be, how do I think about bounding box intuitively so that these are not confusing for me?
OK. After many days working with bounding boxes, I have my own intuition on how to think about bounding box coordinates now.
I divide coordinates in 2 categories: continuous and discrete. The mental problems usually arise when you try to convert between them.
Suppose the image have width=100, height=100 then you can have a continuous point with x,y that can have any real value in the range [0,100].
It means that points like (0,0), (0.5,7.1,39.83,99.9999) are valid points.
Now you can convert a continuous point to a discrete point on the image by taking the floor of the number. E.g. (5.5, 8.9) gets mapped to pixel number (5,8) on the image. It's very important to understand that you should not use the ceiling or rounding operation to convert it to the discrete version. Suppose you have a continuous point (0.9,0.9) this point lies in the (0,0) pixel so it's closest to (0,0) pixel, not (1,1) pixel.
From this foundation, let's try to answer my question:
So I want to know, what exactly does it mean when you say that a bounding box has x1=0, x2=100?
It means that the continuous point 1 has x value = 0, and continuous point 2, has x value = 100. Continuous point has zero size. It's not a pixel.
Does the bounding box border occupy the 0th pixel or is it surrounding 0th pixel (its border is at x=-1)?
In continuous-space, the bounding box border occupy zero space. The border is infinitesimally slim. But when we want to draw it onto an image, the border will have the size of at least 1 pixel thick. So if we have a continuous point (0,0), it will occupy 0th pixel of the image. But theoretically, it represents a slim border at the left side and top side of the 0th pixel.
Where is the exact end of the bounding box? If the image have shape=(8,8), would the end be at 7 or 8?
The biggest x,y value you can have is 7.999... but when converted to discrete version you will be left with 7 which represent the last pixel.
If you want to represent a bounding box that occupy the entire image, what should be its values?
You should represent bounding box coordinates in continuous space instead of discrete space because of the precision that you have. It means the largest bounding box starts at (0,0) and ends at (100,100). But if you want to draw this box, you need to convert it to discrete version and draws the bounding box at (0,0) and end at (99,99).
In OpenCv the bounding rectangle can be defined in many ways. One way is its top-left corner and bottom-right corner. In case of constructor Rect(int x1, int y1, int x2, int y2) it defines those two points. The rectangle starts exactly on that pixel and coordinate. For subpixel rectangles there are also variants holding the floating point coordinates.
So I want to know, what exactly does it mean when you say that a bounding box has x1=0, x2=100?
That means the top-left corner x-coordinate starts at 0 and bottom-right x-coordinate
starts at 100.
Does the bounding box border occupy the 0th pixel or is it surrounding 0th pixel (its border is at x=-1)?
The border starts exactly on the 0-th pixel. Meaning that rectangle with width and height of 1px when drawn is just a signle dot (1px)
Where is the exact end of the bounding box? If the image have shape=(8,8), would the end be at 7 or 8?
The end would be at 7, see below.
If you want to represent a bounding box that occupy the entire image, what should be its values?
Lets have an image size of 100,100. The around the image rectangle defined by two points would be Rect(Point(0,0), Point(99,99)) by starting point and size Rect(0, 0, 100, 100)
The basic is to know that image of size X,Y has a minimum top-left coordinate at (0,0) and maximum at bottom-right (X-1,Y-1)
I am making a reaction game, where you can destroy enemys and earn points, but if the enemy nodes intersect the players node, it reduces the lives by one.
Its a reaction game so I want to give some tolerance points in the intersection. So if they are intersecting, I don't want to see immediately a reduction.
Here is my code:
if (CGRectIntersectsRect(enemy.frame, player.frame)) {
lives--;
}
My goal is a tolerance threshold for the intersection.
I hope someone can help me. (code in objective c please)
From the comments on the question I'm guessing you are not understanding what "inset" means.
If you have a rectangle with origin (10,20), width 100 and height 50 then if you inset the rectangle by 10 on the x-axis and 5 on the y-axis the result is a rectangle with origin (20,25), width 80 and height 40.
Note that the inset rectangle is contained within the original rectangle. #luk2302 suggestion is instead of testing whether the frame rectangles of your two objects intersect you test whether two smaller rectangles intersect. You can generate these smaller rectangles by insetting the frame rectangles by some amount.
You can of course calculate an inset rectangle directly with a few lines of code, but there is an API call CGRectInset (and also NSInsetRect) which will do it for you in one line.
HTH
I am trying to find circles in images and warp them back to a canonical view (i.e. as if looking into the center). However, circles in general project to ellipses under perspective transformations. So I am first detecting ellipses, roughly doing the following (in OpenCV):
1. Find contours in the image
2. Estimate area of contour
3. Fitting a bounded box to contour and estimating area by width/2 * height/2 * PI (area of ellipse)
4. checking if area of contour and estimated area of ellipse is < a threhsold
Assuming I have found an ellipse by this method, how can I rectify it back to a circle such that I "undo" the perspective transform (although not in plane rotation as this cannot be done I guess). For example, if it was a rectangle I would just compute the homography from the 4 corners of an uprigh rectangle to the detected projected one.
I have no idea how to do this with an ellipse, any help is much appreciated.
Thanks
A circle is indeed transformed into an ellipse by a perspective transformation, however its axes are not the same as the axes of the initial circle, as shown in this illustration:
(source: brian-curtis.com)
You can refer to this link for a detailled demonstration. As a consequence, the bounding rectangle of the ellipse is not the image of the initial square by the perspective tranformation.
EDIT:
This means that the center and the axes of the ellipse you observe are not the images, by the perspective mapping, of the center and axes of the original circle. I tried to make a clearer illustration:
On this image, I drew in green the axes and center of the original circle, after perspective transformation, and the axes and center of the ellipse in red. On this specific example, the vertical axis is not deformed by the perspective mapping, but it would be deformed in the general case. Hence, deforming a circle by a perspective transformation gives an ellipse, but the axes and center that you see are not the axes and center of the original circle.
As a consequence, you cannot simply use the top, bottom, left and right points on the ellipse (the red points, which can easily be detected from the ellipse) to map these onto the top, bottom, left and right points of the circle because they do not correspond under the perspective mapping (the green points do, but they cannot be detected easily from the ellipse).
In the end, I don't think that it is at all possible to estimate the perspective mapping from a single detected ellipse.
This looks like an indeterminate problem.
The projection of a rectangle supplies 8 equations in 8 unknowns (homography coefficients).
With an ellipse, you can only retrieve the center coordinates (2 DOF), the axis (2 DOF) and the axis orientation (1 DOF).
How to draw a circle overlapping another circle in the moved phase of touch event,such that no gap is left out between the circles.The circles must be tightly packed to one another,so that even when user moves his hand on the screen faster or lightly,no gap must be present between the circles.
Just two circles? Or many circles? If just two, then detecting if they overlap is simply verifying that their centers are not closer than the sum of their radii. For example, if Circle1's raduis is 10 pixels, and Circle2's radius is 25 pixels, then they overlap if the center of Circle1 is less than 35 pixels from the center of Circle2.
So if you do your calculations in the "moved" phase and find that they're too close, you have to adjust the position of one of them. How you go about that will depend on the specifics of your application. You could:
Keep the y coordinate of the moving circle the same, and calculate the necessary x coordinate to maintain the required distance.
Same as above but swap x and y.
As above, but move the "unmoving" circle away from the "moving" circle.
Some other calculation that makes sense for your application.
NOTE: You should accept some of the answers you've been given.