I'm making an inventory screen for a game I'm working on, and I'd like to be able to draw a series of panels representing each item in the inventory. I want to be able to fit those panels on a circular path.
Here's a mock up I made of what I mean
http://fc02.deviantart.net/fs70/f/2010/280/7/2/khmp_magic_menu_concept_by_magna_alphamon-d30a7em.png
basically I'd like to be able to, give a radius, a center point, and the y co-ordinate to start drawing at, draw this series of panels so they align with the path of the circle like in the image.
Computing the y dimension is easy, its just the startposition y + panel height * panel index, but I'm unsure how to compute the x for a variable radius/center point circle.
Any help would be appreciated.
This is in C#, but something similar in C/C++ will be fine as long as I can convert it
Thanks in advance
EDIT: To calirify, y's position is relative to the top or bottom of the screen and is independent of the circle. If a given y does not map to a point on the circle, then I'll discard that point and not draw the panel.
While ideally I'd like to be able to use any elliptical shape (given two radii), a circle would be good too
Let cx, cy be the coordinates of the center point. Let r be the radius of the circle. Let y be the drawing y-coordinate and x, the x-coordinate. You observe that y = cy + panel height * panel index. By the magic of right triangles, this means that x^2 + y^2 = r^2. Solving for x, we get x = cx + sqrt(r^2 - (y-cy)^2).
EDIT: Converting to code:
#include <math>
float ordinate(float cx, float cy, float r, float y) {
// assumes cx and cy are in the same coordinate system as x and y
// assumes the coordinate origin is in the lower left corner.
return cx + sqrtf(powf(r,2) - powf(y-cy,2));
}
I'm dumb. After seeing Eric's answer, I remembered I can just rearrange and solve the equations of a circle or elipse as necessary.
Thanks
You can use a rotational matrix for this. Here is a simple algorithm that finds the next point {x, y} such that it is rotated theta radians around a circle. You can start with the first item at x=radius and y=radius (wherever really, just a point that you know will contain an item), and then just continue to increment theta as you loop through your items.
Point Rotate(int x, int y, float theta)
int x_p = (x * Math.Cos(theta)) - (y * Math.Sin(theta));
int y_p = (y * Math.Cos(theta)) + (x * Math.Sin(theta));
return new Point(x_p, y_p);
end
On a side note; I always preferred "Bolt1, Bolt2, Bolt3" to "Thunder, Thundara, Thundaga" =P
Related
I have some code which will detect the start and end point of a click-and-drag action, and will save it to 2 vector2 points. I then use this code to convert:
public Rectangle toRect(Vector2 a, Vector2 b)
{
return new Rectangle((int)a.X, (int)a.Y, (int)(b.X - a.X), (int)(b.Y - a.Y));
}
The code above does not work and googling, so far has come up inconclusive.
Could anyone please provide me with some code or a formula to properly convert this?
Note: a vector2 has an x and a y, and a rectangle has an x, a y, a width, and a height.
Any help is appreciated! Thanks
I think you need to have additional logic in there to decide which vector to use as the top left and which to use as the bottom right.
Try this:
public Rectangle toRect(Vector2 a, Vector2 b)
{
//we need to figure out the top left and bottom right coordinates
//we need to account for the fact that a and b could be any two opposite points of a rectangle, not always coming into this method as topleft and bottomright already.
int smallestX = (int)Math.Min(a.X, b.X); //Smallest X
int smallestY = (int)Math.Min(a.Y, b.Y); //Smallest Y
int largestX = (int)Math.Max(a.X, b.X); //Largest X
int largestY = (int)Math.Max(a.Y, b.Y); //Largest Y
//calc the width and height
int width = largestX - smallestX;
int height = largestY - smallestY;
//assuming Y is small at the top of screen
return new Rectangle(smallestX, smallestY, width, height);
}
I'm using C#, I have a list of Vector points and would like to try and approximate how close they look like a circle.
Any ideas how to implement this or if someone else has?
Based on the "gestures" tag I guess you not only want to know how close are these points to the smallest-circle (search for "Smallest-circle problem"), but you have to be concerned also about their order and spread:
I would start with the distance from the smallest-circle. If they are too far, you're done, it's not a circle.
If they are close enough to your configured threshold, compute the angle between the vector defined by the circle center, first point and each other point (picture bellow)
Check that each angle is greater than the previous.
Check that difference between any two angles next to each other is not over some configured threshold.
Check that the last point is close enough to the first one.
You will probably think of some other checks eventually, so make it simple to extend.
Another possibility:
Find the centroid
(http://en.wikipedia.org/wiki/Centroid#Of_a_finite_set_of_points) of
the points
Determine the distance (radius) of each point from the location of
the centroid
Calculate the distribution and determine if the points
are within acceptable tolerances (e.g. standard deviation < ±0.05 × mean radius or something like that)
Without knowing more about the source of the points, it's hard to suggest the best solution.
These might be of use: http://link.springer.com/article/10.1007%2FBF02276879#page-1 and http://www.dtcenter.org/met/users/docs/write_ups/circle_fit.pdf. Those methods will give you the best fitting circle through the points, but you are still going to need to determine whether your data points are close enough for your purposes.
UPDATE: based on the 'gesture' tag, somebody has already implemented it: http://depts.washington.edu/aimgroup/proj/dollar/
1) Pick any three points from that list, find the center of their appropriate circle
We can do this, using triangle circumcircle construction method, you find the medians of all three sides (two are sufficient) and their intersection is the center of the circle. Something like this:
public PointF findCenter(PointF a, PointF b, PointF c)
{
float k1 = (a.Y - b.Y) / (a.X - b.X) //Two-point slope equation
float k2 = (a.Y - c.Y) / (a.X - c.X) //Same for the (A,C) pair
PointF midAB = new PointF((a.X + b.X) / 2, (a.Y + b.Y) / 2) //Midpoint formula
PointF midAC = new PointF((a.X + c.X) / 2, (a.Y + c.Y) / 2) //Same for the (A,C) pair
k1 = -1*k1; //If two lines are perpendicular, then the product of their slopes is -1.
k2 = -1*k2; //Same for the other slope
float n1 = midAB.Y - k1*midAB.X; //Determining the n element
float n2 = midAC.Y - k2*midAC.Y; //Same for (A,C) pair
//Solve y1=y2 for y1=k1*x1 + n1 and y2=k2*x2 + n2
float x = (n2-n1) / (k1-k2);
float y = k1*x + n1;
return new PointF(x,y);
}
2) Check if the other points are equivalently distanced from this center, if yes, you have a circle, if no, you don't.
P.S. I haven't tested the code, so be prepared to debug. Ask if you need anything else
Take any three points from your point set.
If the points are co-linear then, your point set doesn't all lie on a circle.
Find the circumcircle of the triangle. The diameter is given by: d = (a*b*c)/2*area. The center of the circle is the point of intersection of the perpendicular bisectors of the three sides.
Now for every remaining point in the point set, if the distance from the center is not equal to the radius then the points are not on a circle. You can speed up the calculations by comparing the square of the radius against the square of the distance between the given point and the center.
Ok so I have searched and searched for a solution to my problem, but non seem to fix it.
I need to make a game with a rotating "cannon", my cannon is a simple rectangle placed in the middle of my panel that I can rotate with my keyboard. It rotates around one edge. I want to shoot out of the edge on the other side. I have found the starting point of where to shoot my bullets by using:
x = a + dia * (float)Math.Cos(angle);
y = b + dia * (float)Math.Sin(angle)
where "a, b" is the center coordinate I rotate it around and "dia" is the diagonal of the rectangle and "angle" is the angle of the one half of my rectangle.
public float rotate = 0.0f;
g.TranslateTransform(a , b);
g.RotateTransform(rotate);
I have a own class for my bullets that I put in a List.
So far so good. But when I rotate my cannon, the bullets don't come out from the tip anymore..they just start appearing far off where I want them to. it's because of this code:
x = (float)((x * Math.Cos(rotate)) - (y * Math.Sin(rotate)));
y = (float)((x * Math.Sin(rotate)) + (y * Math.Cos(rotate)));
that's supposed to update the x, y coordinates of the tip of the cannon.
If I delete it, it just fires from the same spot(no shit).
Can someone please explain to me what code I need to write to update the X, Y so they come out of my rectangle edge? It's driving me crazy..
Edit:
Found my answer staring at the screen in the early mornings. I had no need for any "find new x, y coordinates". I simply made a updater that updated the original angle with the float number it needed to move a little bit each time i rotated it.
hah! so simple, yet so hard to see.
First of all,
x = (float)((x * Math.Cos(rotate)) - (y * Math.Sin(rotate)));
y = (float)((x * Math.Sin(rotate)) + (y * Math.Cos(rotate)));
needs to be something like:
float oldx = x;
float oldy = y;
x = (float)((oldx * Math.Cos(rotate)) - (oldy * Math.Sin(rotate)));
y = (float)((oldx * Math.Sin(rotate)) + (oldy * Math.Cos(rotate)));
your new values need to be based purely off the old values..
If there's any other problem after fixing this, it may be related to how the rectangle is translated on the plane.
Edit: If this were a code review, I'd say the solution I just gave isn't quite the best solution either (it just doesn't suffer from the bug you introduced by using the new value of x to calculate the new value of y). See, Math.Cos and Math.Sin are generally expensive operations compared to multiplication and addition. If you had a bunch of points that need transformed the same way, best to calculate Math.Sin(rotate) and Math.Cos(rotate) once and use those values for every point. This might be a good place to use the Flyweight pattern and define a class where an instance would hold all your points for a given object/context so that operations can be done in aggregate.
Is there a formula to average all the x, y coordinates and find the location in the dead center of them.
I have 100x100 squares and inside them are large clumps of 1x1 red and black points, I want to determine out of the red points which one is in the middle.
I looked into line of best fit formulas but I am not sure if this is what I need.
Sometimes all the red will be on one side, or the other side. I want to essentially draw a line then find the center point of that line, or just find the center point of the red squares only. based on the 100x100 grid.
List<Point> dots = new List<Point>();
int totalX = 0, totalY = 0;
foreach (Point p in dots)
{
totalX += p.X;
totalY += p.Y;
}
int centerX = totalX / dots.Count;
int centerY = totalY / dots.Count;
Simply average separately the x coordinates and the y coordinates, the result will be the coordinates of the "center".
What if there are two or more subsets of red points ? Do you want the black point inside them?
Otherwis, if I understood your question, just give a weight of 1 to red points and 0 to blacks. Then do the weighted mean on X and Y coordinate
I am drawing an imaginary circle around the middle of a button.
The radius of the circle is the Height/2 if Height>Width or Width/2 if Width>Height.
Now i have to calculate which coordinates (in pixels) are in this circle.
The idea is that if e.g. the mouse cursor hovers over that circle, something happens.
Calculating each co-ordinate is overkill; just compare the distance to the center. For example:
int radius = 5; // whatever
int deltaX = originX - mouseX, deltaY = originY - mouseY;
// compare the square distance, to avoid an unnecessary square-root
if((deltaX * deltaX) + (deltaY * deltaY) <= (radius * radius)) {
// inside the circle, or on the edge
}
To avoid a little math, you could also do a quick bounding-box check, i.e. checking the rectangular region (just addition/subtraction). This can be used in combination, i.e.
check the bounding box
if it isn't in the bounding box it certainly isn't in the circle
if it is in the bounding box, do the math to compare the square-distance
You are inside the circle when this equation is satisfied:
Math.pow(mouse_pos_x-center_circle_x,2)+Math.pow(mouse_pos_y-center_circle_y,2)<Math.pow(radius,2)
The area of a a circle by definition is a group of points whose distance is equal to or less than the center.
All you must do to test if a point is within a circle is to calculate the distance between it and the center point. If this distance is smaller than the radius of the circle, the point is within the circle.
double Distance(Point p1, Point p2)
{
int x = p1.X - p2.X;
int y = p1.Y - p2.Y;
return Math.Sqrt(x * x + y * y);
}
You can use next condition:
x^2+y^2<R^2
Where R - radius,
All this points are in circle.