My problem basically is to calculate the x and y coordinates of the second element in the following situation.
Is for a tool in unity3d, using c#
So I'm guessing you have the coordinates of A and B. Find the angle of the second line with:
float angle = atan2(B.y-A.y, B.x-A.x)
This only works if your situation is axis-aligned like in your diagram (i.e. if the original configuration is lined up along the x axis). If not, you can solve the formula |U x V| = |U| |V| sin(angle) for angle (you will need an arcsin -- the inverse of sin), where U and V are the old and new AB vectors.
Then rotate your point of interest (call it P) around A. You do this by first subtracting A's coordinates from P so the axis of rotation is at the origin. Then rotate P by multiplying with the rotation matrix:
[ cos(angle) -sin(angle) ] [ P.x ]
[ sin(angle) cos(angle) ] [ P.y ]
Which gives
x = cos(angle) * P.x - sin(angle) * P.y
y = sin(angle) * P.x + cos(angle) * P.y
After you have these, add A's coordinates back in.
In summary:
P_new = A + rotate(P_old - A)
The actual code will be more involved than this, but this is the abstract picture. I'll leave the coding to you.
If you know the angle between the x-axis and this box shaped object you can use some basic trig.
Lets assume the box shaped object has a length of "len". ( It would really help to better label what is going on, I don't fully understand.) and lets say you have an angle of θ.
I'm assuming the centre of the circle has coordinates (0,0).
The then the vertical distance from the centre of the circle to the edge of the circle where the box touches is y = len * sin(θ) and for the horizontal distance is len * cos(θ).
If you are only going about 2/5ths of the way up the box thing you would use len / 5 instead of len.
This is just the math behind it. In c# you want to use the Math class. it has all the functions you need. Be careful with radians and degrees.
Related
I am trying to draw an arc between two points that represents a projectile's path. The angle that the projectile leaves point A at is known, and the X/Y coordinates of both points are known.
I'm trying to figure out the math behind this, and how to draw it up in c#.
Here is my failed attempt, based on some path examples I found
var g = new StreamGeometry();
var xDistance = Math.Abs(pointA.X - pointB.X);
var yDistance = Math.Abs(pointA.Y - pointB.Y);
var angle = 60;
var radiusX = xDistance / angle;
var radiusY = yDistance / angle;
using (var gc = g.Open())
{
gc.BeginFigure(
startPoint: pointA,
isFilled: false,
isClosed: false);
gc.ArcTo(
point: pointB,
size: new Size(radiusX, radiusY),
rotationAngle: 0d,
isLargeArc: false,
sweepDirection: SweepDirection.Clockwise,
isStroked: true,
isSmoothJoin: false);
}
Any help would be greatly appreciated!
Edit #2 (added clarity): For this problem assume that physics play no role (no gravity, velocity, or any outside forces). The projectile is guaranteed to land at point B and move along a parabolic path. The vertex will be halfway between point A and point B on the horizontal axis. The angle that it launches at is the angle up from the ground (horizontal).
So Point A (Ax, Ay) is known.
Point B (Bx, By) is known.
The angle of departure is known.
The X half of the vertex is known (Vx = Abs(Ax - Bx)).
Does this really boil down to needing to figure out the Y coordinate of the vertex?
Following on from the comments, we need a quadratic Bezier curve. This is defined by 3 points, the start, end, and a control point:
It is defined by the following equation:
We therefore need to find P1 using the given conditions (note that the gravity strength is determined implicitly). For a 2D coordinate we need two constraints / boundary conditions. They are given by:
The tangent vector at P0:
We need to match the angle to the horizontal:
The apex of the curve must be directly below the control point P1:
Therefore the vertical coordinate is given by:
[Please let me know if you would like some example code for the above]
Now for how to add a quadratic Bezier; thankfully, once you have done the above, it is not too difficult
The following method creates the parabolic geometry for the simple symmetric case. The angle is measured in degrees counterclockwise from the horizontal.
public Geometry CreateParabola(double x1, double x2, double y, double angle)
{
var startPoint = new Point(x1, y);
var endPoint = new Point(x2, y);
var controlPoint = new Point(
0.5 * (x1 + x2),
y - 0.5 * (x2 - x1) * Math.Tan(angle * Math.PI / 180));
var geometry = new StreamGeometry();
using (var context = geometry.Open())
{
context.BeginFigure(startPoint, false, false);
context.QuadraticBezierTo(controlPoint, endPoint, true, false);
}
return geometry;
}
A body movement subject only to the force of gravity (air resistance is ignored) can be evaluated with the following equations:
DistanceX(t) = dx0 + Vx0·t
DistanceY(t) = dy0 + Vy0·t - g/2·t^2
Where
g : gravity acceleration (9.8 m/s^2)
dx0 : initial position in the X axis
dy0 : initial position in the Y axis
Vy0 : initial X velocity component (muzzle speed)
Vy0 : initial Y velocity component (muzzle speed)
Well that doesn't seem very helpful, but lets investigate further. Your cannon has a muzzle speed V we can consider constant, so Vx0 and Vy0 can be written as:
Vx0 = V·cos(X)
Vy0 = V·sin(X)
Where X is the angle at which you are shooting. Ok, that seems interesting, we finally have an input that is useful to whoever is shooting the cannon: X. Lets go back to our equations and rewrite them:
DistanceX(t) = dx0 + V·cos(X)·t
DistanceY(t) = dy0 + V·sin(X)·t - g/2·t^2
And now, lets think through what we are trying to do. We want to figure out a way to hit a specific point P. Lets give it coordinates: (A, B). And in order to do that, the projectile has to reach that point in both projections at the same time. We'll call that time T. Ok, lets rewrite our equations again:
A = dx0 + V·cos(X)·T
B = dy0 + V·sin(X)·T - g/2·T^2
Lets get ourselves rid of some unnecessary constants here; if our cannon is located at (0, 0) our equations are now:
A = V·cos(X)·T [1]
B = V·sin(X)·T - g/2·T^2 [2]
From [1] we know that: T = A/(V·cos(X)), so we use that in [2]:
B = V·sin(X)·A/(V·cos(X)) - g/2·A^2/(V^2·cos^2(X))
Or
B = A·tan(X) - g/2·A^2/(V^2*cos^2(X))
And now some trigonometry will tell you that 1/cos^2 = 1 + tan^2 so:
B = A·tan(X) - g/2·A^2/V^2·(1+tan^2(X)) [3]
And now you have quadratic equation in tan(X) you can solve.
DISCLAIMER: typing math is kind of hard, I might have an error in there somewhere, but you should get the idea.
UPDATE The previous approach would allow you to solve the angle X that hits a target P given a muzzle speed V. Based on your comments, the angle X is given, so what you need to figure out is the muzzle speed that will make the projectile hit the target with the specified cannon angle. If it makes you more comfortable, don't think of V as muzzle speed, think of it as a form factor of the parabola you are trying to find.
Solve Vin [3]:
B = A·tan(X) - g/2·A^2/V^2·(1+tan^2(X))
This is a trivial quadratic equation, simply isolate V and take the square root. Obviously the negative root has no physical meaning but it will also do, you can take any of the two solutions. If there is no real number solution for V, it would mean that there is simply no possible shot (or parabola) that reaches P(angle X is too big; imagine you shoot straight up, you'll hit yourself, nothing else).
Now simply eliminate t in the parametrized equations:
x = V·cos(X)·t [4]
y = V·sin(X)·t - g/2·t^2 [5]
From [4] you have t = x/(V·cos(X)). Substitute in [5]:
y = tan(X)·x - g·x^2 /(2·V^2*cos^2(X))
And there is your parabola equation. Draw it and see your shot hit the mark.
I've given it a physical interpretation because I find it easier to follow, but you could change all the names I've written here to purely mathematical terms, it doesn't really matter, at the end of the day its all maths and the parabola is the same, any which way you want to think about it.
I am trying to code for a game I am working on a specific curve with a specific rotation. I am not a great mathematician... At all... Tried searching for solutions for a few hours, but I'm affraid I do not find any solution.
So, a small picture to illustrate first:
This is an eighth of a circle, radius of 9, beggining is (0,0)
The end is now at about 6.364, -2.636. But I need this same curve, with a 45° direction at the end, but ending at aexactly 6.0,-3.0.
Could any of you show me how to do this? I need to be able to calculate precisly any point on this curve & its exact length. I would suppose using some kind of eliptical math could be a solution? I admit my math class are reaaaly far now and have now good clue for now...
Thank for any possible help
I think I found a quadratic curve which sastisfies your requirement:
f(x) = -1/12 x^2 + 9
Copy the following into https://www.desmos.com/calculator to see it:
-\frac{1}{12}x^2+9
f'(x) would be -1/6x, so when x=6, the derivative would be -1, which corresponds to a -45° inclination. There are probably infinite curves that satisfy your requirement but if my calculus isn't too rusty this is one of them.
I tried to fit an ellipse with foci starting at y=6 here and starting at y=9 here to your points but the slope doesn't look like 45°.
Also starting at any height k, here doesn't seem to work.
I don't think you've fully understood the question I asked in the comments about the "inclination" angle. So I will give a general case solution, where you have an explicit tangent vector for the end of the curve. (You can calculate this using the inclination angle; if we clarify what you mean by it then I will be happy to edit with a formula to calculate the tangent vector if necessary)
Let's draw a diagram of how the setup can look:
(Not 100% accurate)
A and B are your fixed points. T is the unit tangent vector. r and C are the radius and center of the arc we need to calculate.
The angle θ is given by the angle between BA and T minus π/2 radians (90 degrees). We can calculate it using the dot product:
The (signed) distance from the center of AB to C is given by:
Note that this is negative for the case on the right, and positive for the left. The radius is given by:
(You can simplify by substituting and using a cosine addition rule, but I prefer to keep things in terms of variables in the diagram). To obtain the point C, we need the perpendicular vector to AB (call it n):
Now that we have the radius and center of the circular arc, we still need to determine which direction we are moving in, i.e. whether we are moving clockwise or anti-clockwise when going from A to B. This is a simple test, using the cross-product:
If this is negative, then T is as in the diagram, and we need to move clockwise, and vice versa. The length of the arc l, and the angular displacement γ when we move by a distance x along the arc:
Nearly there! Just one more step - we need to work out how to rotate the point A by angle γ around point C, to get the point we want (call it D):
(Adapted from this Wikipedia page)
Now for some code, in case the above was confusing (it probably was!):
public Vector2 getPointOnArc(Vector2 A, Vector2 B, Vector2 T, double x)
{
// calculate preliminaries
Vector2 BA = B - A;
double d = BA.Length();
double theta = Math.Acos(Vector2.DotProduct(BA, T) / d) - Math.PI * 0.5;
// calculate radius
double r = d / (2.0 * Math.Cos(theta));
// calculate center
Vector2 n = new Vector2(BA.y, -BA.x);
Vector2 C = 0.5 * (A + B + n * Math.Tan(theta));
// calculate displacement angle from point A
double l = (Math.PI - 2.0 * theta) * r;
double gamma = (2.0 * Math.PI * x) / l;
// sign change as discussed
double cross = T.x * BA.y - T.y * BA.x;
if (cross < 0.0) gamma = -gamma;
// finally return the point we want
Vector2 disp = A - C;
double c_g = Math.Cos(gamma), s_g = Math.Sin(gamma);
return new Vector2(disp.X * c_g + disp.Y * s_g + C.X,
disp.Y * c_g - disp.X * s_g + C.Y);
}
I've been trying to figure this relationship out but I can't, maybe I'm just not searching for the right thing. If I project a world-space coordinate to clip space using Vector3.Project, the X and Y coordinates make sense but I can't figure out how it's computing the Z (0..1) coordinate. For instance, if my nearplane is 1 and farplane is 1000, I project a Vector3 of (0,0,500) (camera center, 50% of distance to far plane) to screen space I get (1050, 500, .9994785)
The resulting X and Y coordinates make perfect sense but I don't understand where it's getting the resulting Z-value.
I need this because I'm actually trying to UNPROJECT screen-space coordinates and I need to be able to pick a Z-value to tell it the distance from the camera I want the world-space coordinate to be, but I don't understand the relationship between clip space Z (0-1) and world-space Z (nearplane-farplane).
In case this helps, my transformation matrices are:
World = Matrix.Identity;
//basically centered at 0,0,0 looking into the screen
View = Matrix.LookAtLH(
new Vector3(0,0,0), //camera position
new Vector3(0,0,1), //look target
new Vector3(0,1,0)); //up vector
Projection = Matrix.PerspectiveFovLH(
(float)(Math.PI / 4), //FieldOfViewY
1.6f, // AspectRatio
1, //NearPlane
1000); //FarPlane
Standard perspective projection creates a reciprocal relationship between the scene depth and the depth buffer value, not a linear one. This causes a higher percentage of buffer precision to be applied to objects closer to the near plane than those closer to the far plane, which is typically desired. As for the actual math, here's the breakdown:
The bottom-right 2x2 elements (corresponding to z and w) of the projection matrix are:
[far / (far - near) ] [1]
[-far * near / (far - near)] [0]
This means that after multiplying, z' = z * far / (far - near) - far * near / (far - near) and w' = z. After this step, there is the perspective divide, z'' = z' / w'.
In your specific case, the math works out to the value you got:
z = 500
z' = z * 1000 / (1000 - 999) - 1000 / (1000 - 999) = 499.499499499...
w' = z = 500
z'' = z' / w' = 0.998998998...
To recover the original depth, simply reverse the operations:
z = (far / (far - near)) / ((far / (far - near)) - z'')
Here's 2 methods available;
if(rectangle.Intersects(otherRectangle))
{
//collision stuff
}
Catch: Only works with non-rotating rectangles.
if(Vector2.Distance(player.pos, enemy.pos) < 50)
{
//collision stuff
}
Catch: Only works with circles.
What I want is to calculate x and y in this image:
Facts
The width and length of both rectangles is defined, along with their rotations.
I can calculate D using the Pythagorean theorem.
But the TRUE distance is D - (X + Y).
General approach
Evidently x and y can be calculated using the Cosine rule.
But I only have the width or length and the angle between the two shapes.
Complication
Plus, this needs to work for any rotation.
The rectangle on the left could be rotated in any direction, and x would be different depending on said rotation.
Question
How would I calculate x and y?
I just want an effective collision detection method more complex than bounding boxes and Pythagoras' theorem.
One approach is to rotate the line with the inverse angle and check with the axis-aligned box:
class RotatedBox
{
...
float CalcIntersectionLength(Vector2 lineTo) //assume that the line starts at the box' origin
{
Matrix myTransform = Matrix.CreateRotationZ(-this.RotationAngle);
var lineDirection = Vector2.Transform(lineTo -this.Center, myTransform);
lineDirection.Normalize();
var distanceToHitLeftOrRight = this.Width / 2 / Math.Abs(lineDirection.X);
var distanceToHitTopOrBottom = this.Height / 2 / Math.Abbs(lineDirection.Y);
return Math.Min(distanceToHitLeftOrRight, distanceToHitTopOrBottom);
}
}
Now you can calculate the actual distance with
var distance = (box1.Center - box2.Center).Length
- box1.CalcIntersectionLength(box2.Center)
- box2.CalcIntersectionLength(box1.Center);
Be sure that the rotation direction matches your visualization.
For a screen overlay I am making for a 3-dimensional game, I need to display icons over certain locations in the 3D world. I've managed to get several variables from the game that should make this possible:
Player position (x,y,z)
Camera position (x,y,z)
Point position (x,y,z)
Camera Angle(FOV)
Using these variables I managed to get the right and left edge of the camera view. Using these 2 variables I am able to get the point between 0 and 1 where the icon should be displayed on the x axis. (0 is on the left edge, 1 is right edge) This seems to work quite well, but only if I am aligned to either the X or Z axis, as shown in the following example:
I've been trying to fix this using the following rotation matrix:
[ Math.Cos(angle), -Math.Sin(angle) ]
[ Math.Sin(angle), Math.Cos(angle) ]
What I do is, I put the player position, camera position and the camera edge positions in the matrix with as rotation point the world point. The problem is, as soon as I put the angle amount at 90 degrees, the X and Y are being flipped. I've been trying to find a solution for this for a few days now, but I can't find it, so I hope anyone can push me in the right direction here. Below are a few parts of my code that might help in finding the solution:
float PCDistXZ = (float)Math.Sqrt(deltaPCx * deltaPCx + deltaPCz * deltaPCz); // X/Z distance from the world point to the camera
Point fakeAvatarPos = RotateAround(new Point((int)point.x, (int)point.z), new Point((int)avatar.x, (int)avatar.z), (int)90);
Point fakeCameraPos = RotateAround(new Point((int)point.x, (int)point.z), new Point((int)camera.x, (int)camera.z), (int)90);
double edgeRight = fakeC.X + (Math.Sin(45) * PCDistXZ);
double edgeLeft = fakeC.X - (Math.Sin(45) * PCDistXZ);
float edgeTest_ScreenPositionX = (1 - (float)((edgeRight - P.x) / (edgeRight - edgeLeft))) * screenWidth;
public static Point RotateAround(Point pCenter,Point pPoint, float pAngle)
{
double angle = (pAngle * Math.PI) / 180;
double[,] matrix = new Double[2, 2] {
{ Math.Cos(angle), Math.Sin(angle) },
{ Math.Sin(angle), Math.Cos(angle) }
};
double xOffset = pPoint.X - pCenter.X;
double yOffset = pPoint.Y - pCenter.Y;
Point newPoint = new Point(
(int)(((pPoint.X - xOffset) * matrix[0, 0]) - ((pPoint.Y - xOffset) * matrix[0, 1])),
(int)(((pPoint.X - yOffset) * matrix[1, 0]) + ((pPoint.Y - yOffset) * matrix[1, 1]))
);
newPoint.X += (int)xOffset;
newPoint.Y += (int)yOffset;
return new Point(newPoint.X,newPoint.Y);
}
Note: I've changed the names of some of the variables to more understandable one, so it could be possible that there are inconsistencies in the names.
EDIT: I found out about view- and projection matrices. I might be able to use those to convert the 3D position to screen. I'm not sure if it's possible to make this matrices with the limited information I have though.