Let's say I had my player at position 1, and my enemy at position 2. What would I need to do with the two points's positions to get the rotation needed for the player to look at the enemy, assuming that the player starts at zero degrees?
x1 = player's X coordinate
y1 = player's Y coordinate
x2 = enemy's X coordinate
y2 = enemy's Y coordinate
angle = math.atan2(y2 - y1, x2 - x1) * 180 / math.pi()
from there, you can get the angle you needed to turn : )
Related
I have a simple maths/physics problem here: In a Cartesian coordinate system, I have a point that moves in time with a known velocity. The point is inside a box, and bounces orthognally on its walls.
Here is a quick example I did on paint:
What we know: The red point position, and its velocity which is defined by an angle θ and a speed. Of course we know the dimensions of the green box.
On the example, I've drawn in yellow its approximate trajectory, and let's say that after a determined period of time which is known, the red point is on the blue point. What would be the most efficient way to compute the blue point position?
I've tought about computing every "bounce point" with trigonometry and vector projection, but I feel like it's a waste of resources because trigonometric functions are usually very processor hungry. I'll have more than a thousand points to compute like that so I really need to find a more efficient way to do it.
If anyone has any idea, I'd be very grateful.
Apart from programming considerations, it has an interesting solution from geometric point of view. You can find the position of the point at a specific time T without considering its temporal trajectory during 0<t<T
For one minute, forget the size and the boundaries of the box; and assume that the point can move on a straight line for ever. Then the point has constant velocity components vx = v*cos(θ), vy = v*sin(θ) and at time T its virtual porition will be x' = x0 + vx * T, y' = y0 + vy * T
Now you need to map the virtual position (x',y') into the actual position (x,y). See image below
You can recursively reflect the virtual point w.r.t the borders until the point comes back into the reference (initial) box. And this is the actual point. Now the question is how to do these mathematics? and how to find (x,y) knowing (x',y')?
Denote by a and b the size of the box along x and y respectively. Then nx = floor(x'/a) and ny = floor(y'/b) indicates how far is the point from the reference box in terms of the number of boxes. Also dx = x'-nx*a and dy = y'-ny*b introduces the relative position of the virtual point inside its virtual box.
Now you can find the true position (x,y): if nx is even, then x = dx else x = a-dx; similarly if ny is even, then y = dy else y = b-dy. In other words, even number of reflections in each axis x and y, puts the true point and the virtual point in the same relative positions, while odd number of reflections make them different and complementary.
You don't need to use trigonometric function all the time. Instead get normalized direction vector as (dx, dy) = (cos(θ), sin(θ))
After bouncing from vertical wall x-component changes it's sign dx = -dx, after bouncing from horizontal wall y-component changes it's sign dy = -dy. You can see that calculations are blazingly simple.
If you (by strange reason) prefer to use angles, use angle transformations from here (for ball with non-zero radius)
if ((ball.x + ball.radius) >= window.width || (ball.x - ball.radius) <= 0)
ball.theta = M_PI - ball.theta;
else
if ((ball.y + ball.radius) >= window.height || (ball.y - ball.radius) <= 0)
ball.theta = - ball.theta;
To get point of bouncing:
Starting point (X0, Y0)
Ray angle Theta, c = Cos(Theta), s = Sin(Theta);
Rectangle coordinates: bottom left (X1,Y1), top right (X2,Y2)
if c >= 0 then //up
XX = X2
else
XX = X1
if s >= 0 then //right
YY = Y2
else
YY = Y1
if c = 0 then //vertical ray
return Intersection = (X0, YY)
if s = 0 then //horizontal ray
return Intersection = (XX, Y0)
tx = (XX - X0) / c //parameter when vertical edge is met
ty = (YY - Y0) / s //parameter when horizontal edge is met
if tx <= ty then //vertical first
return Intersection = (XX, Y0 + tx * s)
else //horizontal first
return Intersection = (X0 + ty * c, YY)
#Problem
Hey I wondered how we can find the intersection between a point with a vector direction, assuming that the point is in the circle.
In Other words, how can I find where the particle will hit the circle circumference when I know the following: Circle position, radius, Particle Position and direction (velocity).
#Implementation
I am currently creating a flight radar, all flights are stored in a Queue and I need to sort the queue based on the time untill the flight leaves the cirlce (Radar Radius)
So I need to return a Point, thus I can just get the distance and calculate the time based on distance and velocity.
At first subtract center cordinates to simplify calculations (now circle center is coordinate origin)
x11 = x1 - x0
y11 = y1 - y0
Point position against time is
x = x11 + vx * t
y = y11 + vy * t
where vx, vy are components of velocity
Point is at circumference when
x^2 + y^2 = R^2
so
(x11 + vx * t)^2 + (y11 + vy * t)^2 = R^2
(vx^2+vy^2)*t^2 + (2*vx*x11+2*vy*y11)*t + (x11^2+y11^2-R^2) = 0
This is quadratic equation for unknown t. Solve it, find roots - none for outside moving, one for single touch case, two roots for regular intersection - larger one is needed time of intersection t_int
Substitute t into point position equations and get coordinates:
x_int = x1 + t_int * vx
y_int = y1 + t_int * vy
background
I am working on a 3d game, where I need to transfer my mouse-input to a certain height in the 3d world, so I can move a unit across a surface.
This is done from the perspective of a tilted camera (45 degress tilted). the playcanvas API supplies me with a function on my camera, which can translate my mouse input on the 2d screen, into the 3d world using a set depth.
when I set my depth to 0, I get the exact coordinates of my camera (which in this case is (0,80,80), and if I set the depth to 100, I get ~(0.09, 11.52, 7.06)...
I need a way to figure out either the coordinates where the line between the two points hits the height (y-axis of 2), OR I could find a way to calculate the depth I need to supply the camera in order to get back the coordinates I need
which leads me to:
I have 2 coordinates in 3d space: let's say p1(0, 80, 80) and p2(0.09, 11.52, 7.06).
I want to figure out how to calculate a 3rd point, where y = 2. This point should be aligned with the 2 other points.
Any suggestions to how I can achieve this? I am looking for a solution with as few calculations as possible.
One way to do it is to write a family of equations that describes all of the points in space that are collinear with p1 and p2.
x = x1*t + x2*(1-t)
y = y1*t + y2*(1-t)
z = z1*t + z2*(1-t)
... Where x1,y1,z1 are the coordinates of p1, ditto for p2, and t is any real number.
We can find the coordinates of our particular desired point by solving for t. We know that y = 2, so we'll rearrange that equation.
y = y1*t +y2 - y2*t
y - y2 = y1*t - y2*t
y - y2 = (y1 - y2)*t
(y - y2) / (y1 - y2) = t
t = (y - y2) / (y1 - y2)
Now that you know t, you can plug it into the two remaining equations to get your x and z values.
I am breaking my head trying to find an appropriate formula to calculate a what sounds to be an easy task but in practice is a big mathematical headache.
I want to find out the offset it needs to turn my vector's angle (X, Y, Angle) to face a coord ( X, Y )
My vector won't always be facing 360 degrees, so i need that as a variable as well..
Hoping an answer before i'm breaking my pc screen.
Thank you.
input
p1 = (x1,y1) point1 (vector origin)
p2 = (x2,y2) point2
a1 = 360 deg direction of vector
assuming your coodinate system is: X+ is right Y+ is up ang+ is CCW
your image suggest that you have X,Y mixed up (angle usually start from X axis not Y)
da=? change of a1 to match direction of p2-p1
solution 1:
da=a1-a2=a1-atanxy(x2-x1,y1-y1)
atanxy(dx,dy) is also called atan2 on some libs just make sure the order of operands is the right one
you can also use mine atanxy in C++
it is 4 quadrant arctangens
solution 2:
v1=(cos(a1),sin(a1))
v2=(x2-x1,y2-y1)
da=acos(dot(v1,v2)/(|v1|*|v2|))
or the same slightly different
v1=(cos(a1),sin(a1))
v2=(x2-x1,y2-y1)
v2/=|v2| // makes v2 unit vector, v1 is already unit
da=acos(dot(v1,v2))
so:
da=acos((cos(a1)*(x2-x1)+sin(a1)*(y2-y1)/sqrt((x2-x1)*(x2-x1)+(y2-y1)*(y2-y1)));
[notes]
just change it to match your coordinate system (which you did not specify)
use radians or degrees according to your sin,cos,atan dependencies ...
The difference between the vectors is also a vector.
Then calculate the tangens (y part / x part) and invert it to an angle.
Of course use the sign of y if x = 0.
if the coord to face is (x2 ,y2)
deltaY = y2 - y1
deltaX = x2 - x1
You have the angle in degrees between the two points using this formula...
angleInDegrees = arctan(deltaY / deltaX) * 180 / PI
subtract the original angle of your vector and you will get the correct offset!
I have a Vector2 in my 2D Game and what I would like to do now is set my vector2 x and y by calculating them using rotation in degrees
Do I need to use PI to calculate new X and Y coordinates then add move distance per second in order to get the correct coordinates?
Example : Lets say degree is 90, which means my gameobject would move forward,at 5 floating units per second, then Y would be 5,10,15 and if degree would be 180 then X would increase by 5 every second, this is simple, but how to do it for other degrees such as 38,268 etc?
The usual convention is that 0 degrees points in the positive X direction and as the angle increases you rotate the direction anti-clockwise. Your convention seems to be that 0 degrees points in the negative X direction and the angle increases clockwise, so first of all you must translate your angle, say alpha, into one with the usual convention, say beta
beta = 180.0 - alpha
Next, trigonometric functions assume radians which run from 0 to 2π rather than from 0 to 360, so you must translate beta into an angle in radians, say theta
theta = 2.0*PI*beta/360.0
Finally, cos(theta) gives the change in X for a move of 1 unit in the direction given by theta and sin(theta) gives the change in Y. So you need
X = X + D * cos(theta)
Y = Y + D * sintheta)
for a distance D. Using your convention this translates to
X = X + D * cos(2.0*PI*(180.0-alpha)/360.0)
Y = Y + D * sin(2.0*PI*(180.0-alpha)/360.0)