I am currently doing:
Eigen::Vector2d polar(2.5, 3 * M_PI / 4);
Eigen::Vector2d cartesian = polar.x() * Vector2d(cos(polar.y()), sin(polar.y()));
but I'm not sure if this is the correct way to use Eigen or if there is some better built in way.
Thanks!
That looks right to me if you're wanting to stick with using Eigen.
Generally though since the polar representation has angles, it might be good to avoid using the Eigen::Vector2d just for the sake of reducing mistakes that may be made in the future (like adding multiple angles together and not dealing with the fact that 0 == 2*PI). Maybe you could do it with structs instead:
struct Polar { double range; double angle; };
struct Cartesian { double x; double y; };
Cartesian to_cartesian(const Polar& p) {
double c = cos(p.angle);
double s = sin(p.angle);
return {p.range * c, p.range * s};
}
Polar to_polar(const Cartesian& c) {
return {std::sqrt(c.x * c.x + c.y * c.y), std::atan2(c.y, c.x)};
}
I'm looking to make a simple function that rotates a vector's point b around point a for a given number of degrees.
What's odd is that my code seems to work somewhat - the vector is rotating, but it's changing length pretty drastically.
If I stop erasing the screen every frame to see every frame at once, I see the lines producing a sort of octagon around my origin.
Even weirder is that the origin isn't even in the center of the octagon - it's in the bottom left.
Here's my code:
struct Point { int x, y; };
struct Line {
Point a, b;
void rotate(double);
};
void Line::rotate(double t)
{
t *= 3.141592 / 180;
double cs = cos(t);
double sn = sin(t);
double trans_x = (double)b.x - a.x;
double trans_y = (double)b.y - a.y;
double newx = trans_x * cs - trans_y * sn;
double newy = trans_x * sn + trans_y * cs;
newx += a.x;
newy += a.y;
b.x = (int)newx;
b.y = (int)newy;
}
Using the olc::PixelGameEngine to render, which is why I'm using ints to store coordinates.
Problem
I am writing a ray tracer as a use case for a specific machine learning approach in Computer Graphics.
My problem is that, when I try to find the intersection between a ray and a surface, the result is not exact.
Basically, if I am scattering a ray from point O towards a surface located at (x,y,z), where z = 81, I would expect the solution to be something like S = (x,y,81). The problem is: I get a solution like (x,y,81.000000005).
This is of course a problem, because following operations depend on that solution, and it needs to be the exact one.
Question
My question is: how do people in Computer Graphics deal with this problem? I tried to change my variables from float to double and it does not solve the problem.
Alternative solutions
I tried to use the function std::round(). This can only help in specific situations, but not when the exact solution contains one or more significant digits.
Same for std::ceil() and std::floor().
EDIT
This is how I calculate the intersection with a surface (rectangle) parallel to the xz axes.
First of all, I calculate the distance t between the origin of my Ray and the surface. In case my Ray, in that specific direction, does not hit the surface, t is returned as 0.
class Rectangle_xy: public Hitable {
public:
float x1, x2, y1, y2, z;
...
float intersect(const Ray &r) const { // returns distance, 0 if no hit
float t = (y - r.o.y) / r.d.y; // ray.y = t* dir.y
const float& x = r.o.x + r.d.x * t;
const float& z = r.o.z + r.d.z * t;
if (x < x1 || x > x2 || z < z1 || z > z2 || t < 0) {
t = 0;
return 0;
} else {
return t;
}
....
}
Specifically, given a Ray and the id of an object in the list (that I want to hit):
inline Vec hittingPoint(const Ray &r, int &id) {
float t; // distance to intersection
if (!intersect(r, t, id))
return Vec();
const Vec& x = r.o + r.d * t;// ray intersection point (t calculated in intersect())
return x ;
}
The function intersect() in the previous snippet of code checks for every Rectangle in the List rect if I intersect some object:
inline bool intersect(const Ray &r, float &t, int &id) {
const float& n = NUMBER_OBJ; //Divide allocation of byte of the whole scene, by allocation in byte of one single element
float d;
float inf = t = 1e20;
for (int i = 0; i < n; i++) {
if ((d = rect[i]->intersect(r)) && d < t) { // Distance of hit point
t = d;
id = i;
}
}
// Return the closest intersection, as a bool
return t < inf;
}
The coordinate is then obtained using the geometric interpolation between a line and a surface in the 3D space:
Vec& x = r.o + r.d * t;
where:
r.o: it represents the ray origin. It's defined as a r.o : Vec(float a, float b, float c)
r.d : this is the direction of the ray. As before: r.d: Vec(float d, float e, float f).
t: float representing the distance between the object and the origin.
You could look into using std::numeric_limits<T>::epsilon for your float/double comparison. And see if your result is in the region +-epsilon.
An alternative would be to not ray trace towards a point. Maybe just place relatively small box or sphere there.
I'm new in CUDA, and cannot understand what I'm doing wrong.
I'm trying to calculate the distance of object it has id in array, axis x in array and axis y in array to find neighbors for each object
__global__
void dist(int *id_d, int *x_d, int *y_d,
int *dist_dev, int dimBlock, int i)
{
int idx = threadIdx.x + blockIdx.x*blockDim.x;
while(idx < dimBlock){
int i;
for(i= 0; i< dimBlock; i++){
if (idx == i)continue;
dist_dev[idx] = pow(x_d[idx] - x_d[i], 2) + pow(y_d[idx] - y_d[i], 2); // error here
}
}
}
Is pow not defined in kernel code?
Your problem is that while pow is defined in the CUDA math API (see here), it is not template specialised for integer arguments, ie. there is no version like this:
__device__ int pow ( int x, int y )
This is why you are getting an error. You will need to explicitly cast the base argument to a floating point type like this:
dist_dev[idx] = pow((double)(x_d[idx] - x_d[i]), 2.0) +
pow((double)(y_d[idx] - y_d[i]), 2.0);
Having said that, using double precision floating point exponential in your example for a integer square will be poor from an efficiency point of view. It would be preferable to perform the calculation using integer multiplication instead:
int dx = x_d[idx] - x_d[i];
int dy = y_d[idx] - y_d[i];
dist_dev[idx] = (dx * dx) + (dy * dy);
I'm making a application for school in which I have to click a particular object.
EDIT: This is being made in 2D
I have a rectangle, I rotate this rectangle by X.
The rotation of the rectangle has made my rectangles (x,y,width,height) become a new rectangle around the rotated rectangle.
http://i.stack.imgur.com/MejMA.png
(excuse me for my terrible paint skills)
The Black lines describe the rotated rectangle, the red lines are my new rectangle.
I need to find out if my mouse is within the black rectangle or not. Whatever rotation I do I already have a function for getting the (X,Y) for each corner of the black rectangle.
Now I'm trying to implement this Check if point is within triangle (The same side technique).
So I can either check if my mouse is within each triangle or if theres a way to check if my mouse is in the rotated rectangle that would be even better.
I practically understand everything written in the triangle document, but I simply don't have the math skills to calculate the cross product and the dot product of the 2 cross products.
This is supposed to be the cross product:
a × b = |a| |b| sin(θ) n
|a| is the magnitude (length) of vector a
|b| is the magnitude (length) of vector b
θ is the angle between a and b
n is the unit vector at right angles to both a and b
But how do I calculate the unit vector to both a and b?
And how do I get the magnitude of a vector?
EDIT:
I forgot to ask for the calculation of the dotproduct between 2 cross products.
function SameSide(p1,p2, a,b)
cp1 = CrossProduct(b-a, p1-a)
cp2 = CrossProduct(b-a, p2-a)
if DotProduct(cp1, cp2) >= 0 then return true
else return false
Thank you everyone for your help I think I got the hang of it now, I wish I could accept multiple answers.
If you are having to carry out loads of check, I would shy away from using square root functions: they are computationally expensive. for comparison purposes, just multiply everything by itself and you can bypass the square rooting:
magnitude of vector = length of vector
If vector is defined as float[3] length can be calculated as follows:
double magnitude = sqrt( a[0]*a[0] + a[1]*a[1] + a[2]*a[2] );
However that is expensive computationally so I would use
double magnitudeSquared = a[0]*a[0] + a[1]*a[1] + a[2]*a[2];
Then modify any comparative calculations to use the squared version of the distance or magnitude and it will be more performant.
For the cross product, please forgive me if this maths is shaky, it has been a couple of years since I wrote functions for this (code re-use is great but terrible for remembering things):
double c[3];
c[0] = ( a[1]*b[2] - a[2]*b[1] );
c[1] = ( a[2]*b[0] - a[0]*b[2] );
c[2] = ( a[0]*b[1] - a[1]*b[0] );
To simplify it all I would put a vec3d in a class of its own, with a very simple representation being:
class vec3d
{
public:
float x, y, z;
vec3d crossProduct(vec3d secondVector)
{
vec3d retval;
retval.x = (this.y * secondVector.z)-(secondVector.y * this.z);
retval.y = -(this.x * secondVector.z)+(secondVector.x * this.z);
retval.z = (this.x * secondVector.y)-(this.y * secondVector.x);
return retval;
}
// to get the unit vector divide by a vectors length...
void normalise() // this will make the vector into a 1 unit long variant of itself, or a unit vector
{
if(fabs(x) > 0.0001){
x= x / this.magnitude();
}
if(fabs(y) > 0.0001){
y= y / this.magnitude();
}
if(fabs(z) > 0.0001){
z = / this.magnitude();
}
}
double magnitude()
{
return sqrt((x*x) + (y*y) + (z*z));
}
double magnitudeSquared()
{
return ((x*x) + (y*y) + (z*z));
}
};
A fuller implementation of a vec3d class can be had from one of my old 2nd year coding excercises: .h file and .cpp file.
And here is a minimalist 2d implementation (doing this off the top of my head so forgive the terse code please, and let me know if there are errors):
vec2d.h
#ifndef VEC2D_H
#define VEC2D_H
#include <iostream>
using namespace std;
class Vec2D {
private:
double x, y;
public:
Vec2D(); // default, takes no args
Vec2D(double, double); // user can specify init values
void setX(double);
void setY(double);
double getX() const;
double getY() const;
double getMagnitude() const;
double getMagnitudeSquared() const;
double getMagnitude2() const;
Vec2D normalize() const;
double crossProduct(Vec2D secondVector);
Vec2D crossProduct(Vec2D secondVector);
friend Vec2D operator+(const Vec2D&, const Vec2D&);
friend ostream &operator<<(ostream&, const Vec2D&);
};
double dotProduct(const Vec2D, const Vec2D);
#endif
vec2d.cpp
#include <iostream>
#include <cmath>
using namespace std;
#include "Vec2D.h"
// Constructors
Vec2D::Vec2D() { x = y = 0.0; }
Vec2D::Vec2D(double a, double b) { x = a; y = b; }
// Mutators
void Vec2D::setX(double a) { x = a; }
void Vec2D::setY(double a) { y = a; }
// Accessors
double Vec2D::getX() const { return x; }
double Vec2D::getY() const { return y; }
double Vec2D::getMagnitude() const { return sqrt((x*x) + (y*y)); }
double Vec2D::getMagnitudeSquared() const { return ((x*x) + (y*y)); }
double Vec2D::getMagnitude2 const { return getMagnitudeSquared(); }
double Vec2d::crossProduct(Vec2D secondVector) { return ((this.x * secondVector.getY())-(this.y * secondVector.getX()));}
Vec2D crossProduct(Vec2D secondVector) {return new Vec2D(this.y,-(this.x));}
Vec2D Vec2D::normalize() const { return Vec2D(x/getMagnitude(), y/getMagnitude());}
Vec2D operator+(const Vec2D& a, const Vec2D& b) { return Vec2D(a.x + b.x, a.y + b.y);}
ostream& operator<<(ostream& output, const Vec2D& a) { output << "(" << a.x << ", " << a.y << ")" << endl; return output;}
double dotProduct(const Vec2D a, const Vec2D b) { return a.getX() * b.getX() + a.getY() * b.getY();}
Check if a point is inside a triangle described by three vectors:
float calculateSign(Vec2D v1, Vec2D v2, Vec2D v3)
{
return (v1.getX() - v3.getX()) * (v2.getY() - v3.getY()) - (v2.getX() - v3.getX()) * (v1.getY() - v3.getY());
}
bool isPointInsideTriangle(Vec2D point2d, Vec2D v1, Vec2D v2, Vec2D v3)
{
bool b1, b2, b3;
// the < 0.0f is arbitrary, could have just as easily been > (would have flipped the results but would compare the same)
b1 = calculateSign(point2d, v1, v2) < 0.0f;
b2 = calculateSign(point2d, v2, v3) < 0.0f;
b3 = calculateSign(point2d, v3, v1) < 0.0f;
return ((b1 == b2) && (b2 == b3));
}
In the code above if calculateSign is in the triangle you will get a true returned :)
Hope this helps, let me know if you need more info or a fuller vec3d or 2d class and I can post:)
Addendum
I have added in a small 2d-vector class, to show the differences in the 2d and 3d ones.
The magnitude of a vector is its length. In C++, if you have a vector represented as a double[3], you would calculate the length via
#include <math.h>
double a_length = sqrt( a[0]*a[0] + a[1]*a[1] + a[2]*a[2] );
However, I understand what you actually want is the cross product? In that case, you may want to calculate it directly. The result is a vector, i.e. c = a x b.
You code it like this for example:
double c[3];
c[0] = ( a[2]*b[3] - a[3]*b[2] );
c[1] = ( a[3]*b[1] - a[1]*b[3] );
c[2] = ( a[1]*b[2] - a[2]*b[1] );
You can calculate the magnitude of vector by sqrt(x*x + y*y). Also you can calculate the crossproduct simpler: a x b = a.x * b.y - a.y * b.x. Checking that a point is inside triangle can be done by counting the areas for all 4 triangles. For example a is the area of the source triangle, b,c,d are areas of other ones. If b + c + d = a then the point is inside. Counting the area of triangle is simple: we have vectors a, b that are vertexes of triangle. The area of triangle then is (a x b) / 2
One simple way without getting into vectors is to check for area.
For example ,lets say you have a rectangle with corners A,B,C,D. and point P.
first calculate the area of rectangle, simply find height and width of the rectangle and multiply.
B D
| /
| /
|/____ C
A
For calculating the height,width take one point lets say A, find its distance from all other three points i.e AB,AC,AD 1st and 2nd minimum will be width,and height, max will be diagonal length.
Now store the points from which you get the height, width, lets says those points are B,C.
So now you know how rectangle looks, i.e
B _____ D
| |
|_____|
A C
Then calculate the sum of area of triangles ACP,ABP,BDP,CDP (use heros formula to compute area of rectangle), if it equals to the area of rectangle, point P is inside else outside the rectangle.