I have two objects in a game, which for this purpose can be considered points on a 2d plane, but I use Vector3s because the game itself is 3d.
I have a game camera which I want to align perpendicularly (also on the plane) to the two objects, so that they are both in view of the camera. Due to the nature of the game, the objects could be in any imaginable configuration of positions, so the directional vector between them could have any direction.
Part1: How do I get the perpendicular angle from the two positional vectors?
I have:
Vector3 object1Position; // x and z are relevant
Vector3 object2Position;
I need:
float cameraEulerAngleY;
Part2: Now, because of the way the game's assets are modelled, I want to only allow the camera to view within a 180 degree 'cone'. So if the camera passes a certain point, it should use the exact opposite position the above math might produce.
An image is attached of what I need, the circles are the objects, the box is the camera.
I hope this post is clear and you guys won't burn me alive for being total rubbish at vector math :P
greetings,
Draknir
You'll need to specify a distance from the object line, and an up vector:
Vector3 center = 0.5 * (object2position + object2position)
Vector3 vec12 = object2position - object1position
Vector3 normal = Cross(vec12, up)
normal.Normalize()
Vector3 offset = distance * normal
Vector3 cameraA = center + offset
Vector3 cameraB = center - offset
< choose which camera position you want >
Instead of using Euler angles, you should probably use something like LookAt() to orient your camera.
Assuming Y is always 0 (you mentioned "X and Z" are your relevant components), then you can use some 2-d math for this:
1.Find any perpendicular vector (there are two). You can get this by calculating the difference between the two vectors, swapping the components, and negating one of them.
Vector3 difference = (object1Position - object2Position);
Vector3 perpendicular = new Vector3(difference.z, 0, -difference.x);
2.Using your separating plane's normal, flip the direction of your new vector if it's pointing opposite of intended.
Vector3 separatingPlaneNormal = ...; // down?
if(Vector3.Dot(separatingPlaneNormal, perpendicular ) < 0)
{
perpendicular = -perpendicular ;
}
// done.
Well, for the first bit, if you have points (x1, y1) and (x2, y2) describing the positions of your objects, just think of it in terms of triangles. The angle you're looking for ought to be described by
arctan((y2-y1)/(x2-x1))+90
I don't completely understand what you want to do with the second part, though.
Related
The simple way to get player front position at specific distance
Vectro3 forward = player.transform.forward * distance;
Vector3 newPlayerFronPosition= player.transform.position + new Vector3(forward.x, forward.y, forward.z);
newPlayerFronPosition= new Vector3(newPlayerFronPosition.x, newPlayerFronPosition.y - 4f, newPlayerFronPosition.z);
navigationCanvas.transform.position = newPlayerFronPosition;
navigationCanvas.transform.rotation = camRotationToWatch.transform.rotation;
its actually working fine but the problem is as my player move up or down the navCanvas become appear ver near to my player. How to mainitain spcific distance all the time.?? that no matter player look above or down the navcanvas display at specfic distance.(position)
disatnceUICam = Vector3.Distance(newPlayerFronPosition, player.transform.position);
I also logged the distance and surprisingly it the distance is changing when i am moving up or down. its changing from 6 to 12 as i am looking up to down.
If I've understood you correctly, and you want a point on in front of your player transform on the X Z Plane a set distance from the forward of your player, you should try something like this:
Vector3 horizontalForward = new Vector3(
player.transform.position.x + player.transform.forward.x,
player.transform.position.y,
player.transform.position.z + player.transform.forward.z
).normalized * distance;
I suspect what you're describing is occurring because the transform of your 'player' variable is connected to the direction of your game camera. As the camera looks up, the world position of your forward changes relative to the camera. Using just the X and Z will produce a varying distance as your camera transform rotates around the X Axis. Perhaps this diagram will illustrate what I mean a little better:
Sorry the hypotenuse is a little wonky but you get the idea right?
I'm trying to find a way to handle reflections for a breakout clone.
I would upload an image to the post instead of the following paragraph, however i have not yet gained the privilege of that yet.
If the ball intersects the left hand side i want it to bounce off to the left.
if the ball intersects the right hand side i want it to bounce off to the right. if the ball intersects the middle section i want it to bounce up the way. i want to learn how to make it bounce in a varying direction dependant on what side of the left, right, or middle section was intersected
I would like to not use three separate rectangles for this, i want to learn how to do it with one.
I use a Vector2 for ball velocity, projVel.
It's position is projPos.
A rectangle for the paddle lightRect.
The reason I use proj.collRect for the beginning of the if is because I cannot use the intersect method with Vector2.
This is my makeshift collision handler at present, which does work but the speed changes to an extent which renders the game unplayable. The speed clamp i have only slightly slows it down i think. i have a variable for projSpeed i cannot clamp that or it will never be able to stop.
if (proj.collRect.Intersects(lightSaber.lightRect))
{
proj.projPos.Y = lightSaber.lightRect.Y - proj.projTxr.Height;
proj.projVel.Y *= -1;
proj.projVel.X = 10 * (proj.projPos.X - lightSaber.lightRect.Center.X) / (lightSaber.lightRect.Center.X);
}
proj.projVel.X = Math.Max(-4, Math.Min(proj.projVel.X, 4));
proj.projVel.Y = Math.Max(-4, Math.Min(proj.projVel.Y, 4));
Help me by showing me how I could do this, maybe in the Math. method, or even an alternative to .Intersects so I can use projPos instead of collRect.
I really am not sure where to start, if there is another way I could do it an example would be great.
Instead of manipulating X and Y velocities independently, I recommend that you calculate a reflection angle based on the position and then derive the velocity from the angle and the speed prior to impact.
Example:
// NOTE: this code assumes that positive Y is down
if (proj.collRect.Intersects(lightSaber.lightRect) && projPos.projVel.Y > 0.0f) // only bounce if projectile is moving downward
{
// remember current speed for when we calculate new velocity
var projSpeed = projVel.Length();
// make sure the projectile no longer intersects the bar
proj.projPos = lightRect.Y - proj.projTxr.Height;
// interpolate reflection angle
var t = (proj.projPos.X - lightSaber.lightRect.X) / lightSaber.lightRect.Width;
var reflectDegrees = 150.0f - t * 120f; // straight up +/- 60 degrees
var reflectRadians = reflectDegrees * (float)Math.PI / 180.0f;
// final velocity determined by angle and original projectile speed
proj.projVel = new Vector2((float)Math.Cos(reflectRadians) * projSpeed, -(float)Math.Sin(reflectRadians) * projSpeed);
}
I have two squares in 3D space. I want to find the x,y,z angles between them. I started by finding the normal vectors for both of the squares and I am trying to figure out how to get the angle between them.
I am using XNA (C#) Vector3 objects.
I have calculated the normal vectors as follows:
Vector3 normal1 = (Vector3.Cross(sq1.corners[0] - sq1.corners[1], sq1.corners[0] - sq1.corners[2]));
Vector3 normal2 = (Vector3.Cross(sq2.corners[0] - sq2.corners[1], sq2.corners[0] - sq2.corners[2]));
I want to find the euler rotation that will get normal1 facing the same way as normal2
First, you can calculate the axis and amount of rotation (assuming an arbitrary axis):
Vector3 axis = Vector3.Cross(normal1, normal2);
axis.Normalize();
double angle = Math.Acos(Vector3.Dot(normal1, normal2) / normal1.Length() / normal2.Length());
If the normals are normalized, then the calculation of the angle reduces to
double angle = Math.Acos(Vector3.Dot(normal1, normal2));
Then you can transform this to euler angles with the function from here
i am making 3D Simulation of Solar System
for every planet i have an instance of class Celestial Body. I am using the following code for revolution and rotation of sun, planets & their moons.
world = Matrix.CreateTranslation(0,0,0) * Matrix.CreateRotationY(rotation) * Matrix.CreateTranslation(position) * Matrix.CreateRotationY(revolution);
rotation is my float variable for rotation of planet around its own axis
revolution is my float variable for revolution of planet in orbit
position is to my vector3 variable to put the body in orbit or at its radius from center e.g postion = new Vector3(70,0,0)
Now it works really fine.
But the problem is i need to locate \ get the position of my planet, to where it has been translated after the Matrix multiplication literally in x,y,x co-ordinates.
How To ? get the current X , Y , Z coordinates of my planet
the other option for me would be to use some maths formula that calculates a 2D circle for me.
I think what you're looking for is Matrix.Translation. This gives you the x, y, z co-odinates of the matrix that it's called on in a Vector3.
So, to get the new position, you should use
Vector3 newPosition = world.Translation;
after your calculations.
I've been working on a simple program in C# in which a Ball [X,Y] cordinates are periodical incremented.
I've managed to implement a collision detection method, but I'm trying to determine how to reflect the ball at an angle oposed bouncing it back along the same linear path.
dx = -dx //This bounces the ball back along the same linear path
dy = -dy
Solution
Trigonometry
theta = range between 0<theta<=360 depending on where it bounced
x = cos(theta)*time
y= sin(theta)*time
The whole point of Newtonian physics is that it is not random, it is deterministic. If you throw the same ball against the same wall at the same angle and with the same velocity and the same spin, it goes to the same place every time.
This sort of program is a really great learning opportunity for both programming and physics. What I encourage you to do is to first write a program that simulates very simple bouncing. As you note, when an object is moving straight down and hits a horizontal surface, then you can model the bounce as simply reversing the vertical velocity component. Just get that right; no gravity, no nothing. That's a great start.
Then try adding bouncing off of horizontal walls, the same way.
Then try adding bouncing off of walls that are not aligned with horizontal or vertical directions. That's where you're going to have to learn how vectors and trigonometry work, because you'll have to work out what component of the ball's velocity is changed by striking the wall obliquely.
Then add gravity. Then add friction from the air. Then add the fact that the ball can be spinning. Add elasticity, so that you can model deformation of the ball.
Once you get to that point, if you want to introduce randomness you'll be able to figure out how to do it. For example, you might introduce randomness by saying "well, when the ball strikes the wall and deforms, I'll introduce a random element that changes its deformation by 0-10%". That will then change how the simulation bounces the ball. You can experiment with different kinds of randomness: add random air currents, for instance.
You will have to add in randomness yourself. To rephrase your question: "Deterministically, it bounces off at angle theta. How can I make it bounce back at angle theta + epsilon, where epsilon is some random value?"
To rotate a vector, see this. You will just specify theta.
pseudocode:
RotateVector(vec):
bounce_vec = [-vec.x vec.y]; //deterministic answer is negative x, normal y
bounce_angle = acos(dot(vec,bounce_vec) / (norm(vec)*norm(bounce_vec)));
modified_angle = bounce_angle + random_number();
ca = cos(modified_angle);
sa = sin(modified_angle);
rotation_matrix = [ca -sa; sa ca];
return rotation_matrix * vec;
Line 3 uses the law of cosines to figure out the angle. In line 4, that angle is modified randomly. The rest of the function rotates the original vector by your new angle.
As long as it's a perfect ball with a perfect surface it will not bounce back randomly. Neither vectors nor trigonometry will give you any randomness.
"randomly, though applying to the basic laws of physics" seems like an oxymoron. However...
If you want it to bounce in a random direction, while maintaining its current speed, you might do something like this (pseudocode):
first, bounce back the canonical way (dx = -dx or dy = -dy depending on the collision)
then convert the dx and dy to polar coordinates (theta and r)
jitter theta by a small amount (+ or - a few degrees, according to your taste)
make sure theta isn't heading into a wall that you just bounced off
convert theta and r back to dx and dy
That would be conserving scalar momentum.