I have a rectangle on the scene and I want to rotate it with mouse.
The rectangle has his own origin point. Clicking on the scene represents start of rotation and mouse moving represent angle of rotation.
where:
O - origin of rotation point
A - anchor point (saved in OnMousePress event)
C - current point (from OnMouseMove event)
so I calculate the angle in next steps:
Fistly, I get lengths of triangle sides:
AO = sqrt( (O.x - A.x)^2 + (O.y - A.y)^2 )
CO = sqrt( (O.x - C.x)^2 + (O.y - C.y)^2 )
AC = sqrt( (C.x - A.x)^2 + (C.y - A.y)^2 )
and then I calculate the angle (a):
a = arccos ( (AO^2 + CO^2 - AC^2) / (2 * AO * CO) )
it works, but this calculation look too complicated taking into account that I need to repeat on it all OnMouseMove call.
So my question - is there another way to calculate the angle? I write it in c++ so some code snippet will be apprecated.
You can find angle between vectors OA and OC through their scalar product and cross product:
OA = (OA.X, OA.Y) = (A.X-O.X, A.Y-O.Y)
OC = (OC.X, OC.Y) = (C.X-O.X, C.Y-O.Y)
SP = OA * OC = OA.X*OC.X+OA.Y*OC.Y
CP = OA x OC = OA.X*OC.Y-OA.Y*OC.X
Angle = atan2(CP, SP)
Example: O = (0,0), A = (-1, 0), C = (-2, 1)
SP = 2, CP = -1, Angle = -0.463
This method allows to avoid sqrt calculations, and determines rotation direction (unlike arccos)
You use a dot product of vectors OA and OC divided by their magnitude to calculate cosine of the angle and then use acos() function to find the angle.
float cosAngle = (x1 * x2 + y1 * y2) / sqrt(x1*x1 + y1*y1) * sqrt(x2*x2 + y2*y2);
float angle = acos(cosAngle);
Related
There is a camera represented by a quaternion [w, x, y, z] associated with a position [x, y, z].
The camera is looking at a mirror (plane) defined as a normal and a point residing on the mirror. This mirror may also be represented as the plane equation AX + BY + CZ + D = 0.
Using the information above, how would the reflected/virtual camera be found?
The quaternion roll should be preserved, ie if you are looking directly into a mirror, the quaternion looking back at you should have the same roll and not be upside down. Ray vs plane can be used to find the virtual camera position, the reflected vector origin and 3D direction, but I'm not sure that helps in finding the quaternion. Finding the virtual camera position was easy enough as shown below (I prefer this over the ray vs. plane tracing method due to parallel or looking away case):
//plane equation ax + by + cz + d = 0
float plane_equation[4];//a, b, c, d
plane_equation[0] = mirror_normal[0];
plane_equation[1] = mirror_normal[1];
plane_equation[2] = mirror_normal[2];
float mults[3];
mults[0] = plane_equation[0] * mirror_position[0];
mults[1] = plane_equation[1] * mirror_position[1];
mults[2] = plane_equation[2] * mirror_position[2];
plane_equation[3] = -1 * (mults[0] + mults[1] + mults[2]);
//find the virtual camera position for reflection using the plane equation
//a(p0 + aT) + b(p1 + bT) + c(p2 + cT) + d = 0
//ap0 + bp1 + cp2 + T(a^2 + b^2 + c^2) + d = 0
//U + TV + d = 0
//T = (-d - U) / V
float t_parts[2];
//U
t_parts[0] = plane_equation[0] * cam_pos[0] + plane_equation[1] * cam_pos[1] + plane_equation[2] * cam_pos[2];
//V
t_parts[1] = plane_equation[0] * plane_equation[0] + plane_equation[1] * plane_equation[1] + plane_equation[2] * plane_equation[2];
float t = (-plane_equation[3] - t_parts[0]) / t_parts[1];
//t gets us to the intersection point. 2t gets us to the virtual point
virtual_cam_pos[0] = 2 * t * plane_equation[0] + cam_pos[0];
virtual_cam_pos[1] = 2 * t * plane_equation[1] + cam_pos[1];
virtual_cam_pos[2] = 2 * t * plane_equation[2] + cam_pos[2];
One solution, it could probably be simplified but this works:
Do ray vs plane. There are 3 cases that need to be taken into account. First is looking towards the mirror so that there is an intersection. Second is looking parallel to the mirror. Third is looking away from the mirror (the mirror may still be visible by the camera, do flipped ray vs plane)
Find the reflected vector
Use the distance from the camera position to the mirror intersection point, intersection point, and reflected vector to find the virtual camera position
Use the vector from the virtual camera to the mirror intersection point to create a quaternion
Apply a local roll to the quaternion. An exact equation to find the appropriate amount of roll for all possible cases is a little unclear to me, but this gets most of the way there
I'm trying to write an algorithm to determine if point is located inside a triangle or on it's edge in 3D coordinate space.
For example, I try to reach such results for different cases
I've figured out how to check if point P inside the triangle, I calculated normal vectors for triangles ABP, BCP, CAP and checked if they are similar.
Can someone explain how to check if a point is on the edge of a triangle (but not outside of a triangle)? You can provide formulas or code as you wish.
Make vectors:
r = p - A (r.x = p.x - A.x, r.y = p.y - A.y, r.z = p.z - A.z)
s = B - A
q = C - A
Calculate normal to ABC plane:
n = s x q (vector product)
Check if p lies in ABC plane using dot product:
dp = n.dot.r
If dp is zero (or has very small value like 1.0e-10 due to the floating point errors, then p is in the plane, and we can continue
Decompose vector p by base vectors s and q. At first check if z-component of normal (n.z) is non-zero. If so, use the next pair of equations (otherwise choose equations for x/z or y/z components):
px = a * sx + b * qx
py = a * sy + b * qy
Solve this system
a = (sy * qx - sx * qy) / (py * qx - px * qy)
b = (px - a * sx) / qx
If resulting coefficients a and b fulfill limits:
a >= 0
b >= 0
a + b <= 1.0
then point p lies in triangle plane inside it.
I am making my first raycasting engine, and would like to rotate a line over an angle θ
How does one do this? Would it be possible to show me some basic C++ code or some pseudocode?
This image describes my problem:
Optional question
I decided to make all of this in graphics.h, because it is the simplest graphics header for C/C++.
You want:
B = P + M * (A - P)
Where M is a 2D rotation matrix:
M = | cos(ϴ) -sin(ϴ) |
| sin(ϴ) cos(ϴ) |
In C++ it could be written as:
float c = cos(theta), s = sin(theta);
float dx = ax - px, dy = ay - py;
float bx = px + c * dx - s * dy;
float by = py + s * dx + c * dy;
One simple algorithm:
Move the circle -P, so that P is at (0, 0).
Rotate A by the angle by multiplying it by the rotation matrix.
Move the circle P to restore its original position.
All these three steps can be done using one 3x3 matrix multiplication.
The scalar product of two vectors have the following property:
vec(PA) . vec(PB) = rho cos theta
Taking the definition of our two vectors:
vec(PA) = (x_a-x_p, y_a-y_p)
vec(PB) = (x_b-x_p, y_b-y_p)
We can get:
(x_a-x_p)(x_b-x_p) + (y_a-y_p)(y_b-y_p) = rho cos theta (1)
Since PA=PB, we also have:
(x_a-x_p)^2 + (y_a-y_p)^2 = (x_b-x_p)^2 + (y_b-y_p)^2 (2)
From (1) and (2) you can derive x_band y_b with some arithmetic autopilot.
my question is regarding working out the direction of the smallest angle between two vectors in 2D. I am making a game in C++ where one of the obstacles is a heat seeking missile launcher. I have it working by calculating the vector between the target and bullet, normalising the vector and then multiplying it by its speed. However, I am now coming back to this class to make it better. Instead of instantly locking onto the player I want it to only do so only when the bullets vector is within a certain angle (the angle between the bullets vector and the vector bulletloc->target). Otherwise I want it to slowly pan towards the target by a degrees thus giving the player enough space to avoid it. I have done all this (in a vb.net project so i could simplify the problem, work it out then re write in in C++). However the bullet always rotates clockwise towards the target even if the quickest route would be counter clockwise. So the problem is working out the direction to apply the rotation in so the smallest angle is covered. Here is my code so you can try and see what I am describing:
Function Rotate(ByVal a As Double, ByVal tp As Point, ByVal cp As Point, ByVal cv As Point)
'params a = angle, tp = target point, cp = current point, cv = current vector of bullet'
Dim dir As RotDir 'direction to turn in'
Dim tv As Point 'target vector cp->tp'
Dim d As Point 'destination point (d) = cp + vector'
Dim normal As Point
Dim x1 As Double
Dim y1 As Double
Dim VeritcleResolution As Integer = 600
tp.Y = VeritcleResolution - tp.Y 'modify y parts to exist in plane with origin (0,0) in bottom left'
cp.Y = VeritcleResolution - cp.Y
cv.Y = cv.Y * -1
tv.X = tp.X - cp.X 'work out cp -> tp'
tv.Y = tp.Y - cp.Y
'calculate angle between vertor to target and vecrot currntly engaed on'
Dim tempx As Double
Dim tempy As Double
tempx = cv.X * tv.X
tempy = cv.Y * tv.Y
Dim DotProduct As Double
DotProduct = tempx + tempy 'dot product of cp-> d and cp -> tp'
Dim magCV As Double 'magnitude of current vector'
Dim magTV As Double 'magnitude of target vector'
magCV = Math.Sqrt(Math.Pow(cv.X, 2) + Math.Pow(cv.Y, 2))
magTV = Math.Sqrt(Math.Pow(tv.X, 2) + Math.Pow(tv.Y, 2))
Dim VectorAngle As Double
VectorAngle = Acos(DotProduct / (magCV * magTV))
VectorAngle = VectorAngle * 180 / PI 'angle between cp->d and cp->tp'
If VectorAngle < a Then 'if the angle is small enough translate directly towards target'
cv = New Point(tp.X - cp.X, tp.Y - cp.Y)
magCV = Math.Sqrt((cv.X ^ 2) + (cv.Y ^ 2))
If magCV = 0 Then
x1 = 0
y1 = 0
Else
x1 = cv.X / magCV
y1 = cv.Y / magCV
End If
normal = New Point(x1 * 35, y1 * 35)
normal.Y = normal.Y * -1
cv = normal
ElseIf VectorAngle > a Then 'otherwise smootly translate towards the target'
Dim x As Single
d = New Point(cp.X + cv.X, cp.Y + cv.Y)
a = (a * -1) * PI / 180 'THIS LINE CONTROL DIRECTION a = (a*-1) * PI / 180 would make the rotation counter clockwise'
'rotate the point'
d.X -= cp.X
d.Y -= cp.Y
d.X = (d.X * Cos(a)) - (d.Y * Sin(a))
d.Y = (d.X * Sin(a)) + (d.Y * Cos(a))
d.X += cp.X
d.Y += cp.Y
cv.X = d.X - cp.X
cv.Y = d.Y - cp.Y
cv.Y = cv.Y * -1
End If
Return cv
End Function
One idea I had was to work out the bearing of the two vectors and if the difference is greater than 180 degrees, rotate clockwise otherwise rotate counter clockwise, any ideas would be helpful. Thanks.
EDIT: I would like to add that this site is very helpful. I often use questions posed by others to solve my own problems and I want to take the chance to say thanks.
As you've written in your code, the angle between two (normalized) vectors is the inverse cosine of their dot product.
To get a signed angle, you can use a third vector representing the normal of the plane that the other two vectors lie on -- in your 2D case, this would be a 3D vector pointing straight "up", say (0, 0, 1).
Then, take the cross-product of the first vector (the one you want the angle to be relative to) with the second vector (note cross-product is not commutative). The sign of the angle should be the same as the sign of the dot product between the resulting vector and the plane normal.
In code (C#, sorry) -- note all vectors are assumed to be normalized:
public static double AngleTo(this Vector3 source, Vector3 dest)
{
if (source == dest) {
return 0;
}
double dot; Vector3.Dot(ref source, ref dest, out dot);
return Math.Acos(dot);
}
public static double SignedAngleTo(this Vector3 source, Vector3 dest, Vector3 planeNormal)
{
var angle = source.AngleTo(dest);
Vector3 cross; Vector3.Cross(ref source, ref dest, out cross);
double dot; Vector3.Dot(ref cross, ref planeNormal, out dot);
return dot < 0 ? -angle : angle;
}
This works by taking advantage of the fact that the cross product between two vectors yields a third vector which is perpendicular (normal) to the plane defined by the first two (so it's inherently a 3D operation). a x b = -(b x a), so the vector will always be perpendicular to the plane, but on a different side depending on the (signed) angle between a and b (there's something called the right-hand rule).
So the cross product gives us a signed vector perpendicular to the plane which changes direction when the angle between the vectors passes 180°. If we know in advance a vector perpendicular to the plane which is pointing straight up, then we can tell whether the cross product is in the same direction as that plane normal or not by checking the sign of their dot product.
Based on #Cameron's answer, here is the python translation i've used:
As bonus, i've added the signed_angle_between_headings function to directly return the 'quickest' turn angle between two north-referenced headings.
import math
import numpy as np
def angle_between_vectors(source, dest):
if np.array_equal(source, dest):
return 0
dot = np.dot(source, dest)
return np.arccos(dot)
def signed_angle_from_to_vectors(source, dest, plane_normal):
angle = angle_between_vectors(source, dest)
cross = np.cross(source, dest)
dot = np.dot(cross, plane_normal)
return -angle if dot < 0 else angle
def signed_angle_between_headings(source_heading, destination_heading):
if source_heading == destination_heading:
return 0
RAD2DEGFACTOR = 180 / math.pi
source_heading_rad = source_heading / RAD2DEGFACTOR
dest_heading_rad = destination_heading / RAD2DEGFACTOR
source_vector = np.array([np.cos(source_heading_rad), np.sin(source_heading_rad), 0])
dest_vector = np.array([np.cos(dest_heading_rad), np.sin(dest_heading_rad), 0])
signed_angle_rad = signed_angle_from_to_vectors(source_vector, dest_vector, np.array([0,0,1]))
return signed_angle_rad * RAD2DEGFACTOR
I have this algorithm here:
pc = # the point you are coloring now
p0 = # start point
p1 = # end point
v = p1 - p0
d = Length(v)
v = Normalize(v) # or Scale(v, 1/d)
v0 = pc - p0
t = Dot(v0, v)
t = Clamp(t/d, 0, 1)
color = (start_color * t) + (end_color * (1 - t))
to generate point to point linear gradients. It works very well for me. I was wondering if there was a similar algorithm to generate radial gradients. By similar, I mean one that solves for color at point P rather than solve for P at a certain color (where P is the coordinate you are painting).
Thanks
//loop through vector
//x and y px position
int x = i%w;
int y = i/w;
float d = distance(center,int2(x,y));
//if within the grad circle
if(d < radius)
{
//somehow set v[i] alpha to this:
float a = d/r;
}
Linerise over atan2(dy,dx) where dx is x-center, and dy is y-center.
cx # center x
cy # center y
r1 # ring is defined by two radius
r2 # r1 < r2
c1 # start color
c2 # stop color
ang # start angle
px # currect point x,y
py
if( px^2 + py^2 <= r2^2 AND px^2 + py^2 >= r1^2 ) # lies in ring?
t= atan2(py-cy,px-cx)+ang
t= t+ pi # atan2 is from -pi to pi
if (t > 2* pi) # it might over 2pi becuse of +ang
t=t-2*pi
t=t/(2*pi) # normalise t from 0 to 1
color = (c1 * t) + (c2 * (1 - t))
Problem whit this algorhitm is that ang is actualy wrong and should be rotated by pi and normalized between 0 and 2pi.
Based on the comment, what you want can still be viewed as a linear gradient -- i.e. you have a line from the center to the outside of the circle, and you have a linear gradient along that line. As such, the calculation is virtually identical to what you already had.
Edit: Okay, apparently I misunderstood what you want. To figure a gradient running around a radius, you still basically linearize it -- figure out the circumference at that radius (2*Pi*R), and then do a linear interpolation along a line of that length.