I am working on a ray tracer and I got around to adding cylinders to the scene. The point I am stuck at is finding the surface normal vector in the point the ray hits. I need this to be able to do the diffuse lighting. What I have at this point is the 3d point where the camera ray hits the cylinder and the actual cylinder which is defined with a point on the central axis, the vector representing the direction of the axis and the radius. So to sum up my question, how do I find the normal vector in a point having the cylinder hit point, the radius, a point on its axis and the direction vector of the axis?
The cylinder normal vector starts at the centerline of the cylinder at the same z-height of the point where the ray intersects the cylinder, ends at the radial point of intersection. Normalize it and you have your unit normal vector.
If the cylinder centerline is not along the global z-direction of the scene you'll have to transform to cylinder coordinates, calculate the normal vector, and transform that back to global coordinates.
There are three possible situations:
the hit_pt is on the TOP CAP of the cylinder:
if (length(hit_pt - cy.top_center) < cy.radius)
surface_normal = cy.ori;
the hit_pt is on the BOTTOM CAP of the cylinder:
if (length(hit_pt - cy.bottom_center) < cy.radius)
surface_normal = -1 * cy.ori;
the hit_pt is on the SIDE of the cylinder. We can use dot product to find the point 'pt' on the center line of the cylinder, so that the vector (hit_pt - pt) is orthogonal to the cylinder's orientation.
t = dot((hit_pt - cy.bottom_center), cy.ori); // cy.ori should be normalized and so has the length of 1.
pt = cy.bottom_center + t * cy.ori;
surface_normal = normalize(hit_pt - pt)));
Given a plane(in my case a triangle) normal N_T and a reference Normal N_R, both have the length 1.
I calculated the rotation_normal
N = N_T x N_R
and now i need to calculate the angle around this rotation_normal, which i get with the following calculation:
angle = acos(<N_T, N_R>), with <x,y> is the dotproduct of x and y
This angle is in the interval of [0°, 180°] and is the smallest angle between both normals.
So my problem is that if i want to rotate my triangle in a manner that its normal is equal to the reference normal, i need to know in which direction (positive or negative) the calculated angle is.
Does anybody know how to get this direction or how to solve this problem in general?
you need to use atan2 (4-quadrant arc tangens)
create reference plane basis vectors u,v
must be perpendicular to each other and lie inside plane
preferably unit vectors (or else you need to account for its size)
so let N=N_T x N_R; ... reference plane normal where the rotation will take place
U=N_T;
V= N x U; ... x means cross product
make them unit U/=|U|; V/=|V|; if they are not already
compute plane coordinates of N_R
u=(N_R.U); ... . means dot product
v=(N_R.V);
compute angle
ang=atan2(v,u);
if you do not have atan2 then use ang=atanxy(u,v);
this will give you angle in range ang=<0,2*M_PI>
if you want signed angle instead then add
if (ang>M_PI) ang-=2.0*M_PI; ... M_PI is well known constant Pi=3.1415...
now if you want the opposite sign direction then just use -ang
Given circle centre: vectorA and another Vector on the circle's perimeter:vectorB, how can you determine the shorter route for vectorB to translate to another point on the circle's perimeter that is variable:vectorC? Will the shorter route be clockwise or counter clockwise rotation?
If it helps think of a clock. If the times is a random point on the clock's perimeter eg. 6, and the minute hand position is known, eg. 4. Does the hand need to rotate around the clock's centre point clockwise or counter clockwise to reach the random point (6)?
See also:
Vec1 = Circle centre, Vec2 = mousepos, find the point on the circle between Vec1, Vec2
Just compute winding direction of triangle ABC
so if you compute normal n=(B-A)x(C-B) where x is cross product then n.z sign determine the direction.
n.z = ((B.x-A.x)*(C.y-B.y)) - ((B.y-A.y)*(C.x-B.x))
if (n.z<0.0) dir=CW else dir=CCW;
that is all you need (CW means clockwise and CCW counter clockwise) of coarse if your coordinate system is different then the rotation can be negated
[Notes]
if (n.z==0) then the points B,C are either opposite or identical so direction does not matter because both ways the angular distance is the same
I have a Point on the surface of the earth which I am converting to a Vector from Earth Center.
I have a True North Heading in degrees describing the path the point will travel on the surface of the earth.
I need to calculate a Vector which is perpendicular to the plane created by the path of this point along the earths surface.
I have tried calculating an arbitrary point along the path using the method described here
and then taking the cross product of the two vectors however it does not seem to be quite accurate enough and seems like more overhead than is necessary.
This is related to my other post ray-polygon-intersection-point-on-the-surface-of-a-sphere.
I'm assuming you're trying to compute a vector lying in the plane of the path, not perpendicular to it (since you've already got one - namely the vector from the origin to your point).
You first need to compute vectors lying in that plane that point due north and due east. To do this, let's call P your point, O the origin, and N = (0, 0, R) is the point at the top of your sphere. Then
e = cross(N - P, P - O)
is a vector that points due east, and is tangent to the sphere because it's perpendicular to P - O, a radius of the sphere.
For similar reasons
n = cross(e, P - O)
will point due north, and will be tangent to the sphere.
Now normalize n and e, and you've got an orthonormal basis for the tangent space at your point. To find a vector in a direction theta (say, counterclockwise from the positive east axis, to simplify the math), just take a little of e and a little of n:
v = cos(theta) * e + sin(theta) * n
Here's my understanding of your problem:
You have a point on the Earth's surface, specified as latitude/longitude coordinates
The direction "true north" is the direction that a person at that point would travel to reach the (geographic) North Pole by the most direct possible route. That is, the "true north vector" is tangent to the Earth's surface at your chosen point and points directly north, parallel to a line of longitude.
The direction of the point's motion will be (initially) tangent to the Earth's surface at your chosen point.
You have an angle in degrees from true north which specifies the heading at which this point is going to move.
This angle is the angle between the "true north vector" and the direction of motion of the point.
You want to calculate a vector that is tangent to the Earth's surface at that point but perpendicular to the direction of motion of the point.
If I've understood all that correctly, you can do it as follows:
The "true north vector" at latitude lat, longitude lng is given by [-sin(lat) * cos(lng), -sin(lat) * sin(lng), cos(lat)]
A vector perpendicular to the "true north vector" which points along a line of latitude (to the east) is given by [-sin(lng), cos(lng), 0]
Since these two vectors identify the plane tangent to the Earth's surface, and the vector specifying the direction of motion of your point is also in that plane, your motion vector is a linear combination of the previous two: [
-(sin(lat) * cos(lng) * cos(th) + sin(lng) * sin(th))
-(sin(lat) * sin(lng) * cos(th) - cos(lng) * sin(th))
cos(lat) * cos(th)
] where th is your heading angle.
To find a vector perpendicular to that motion vector, you can just take the cross product of the radius vector (that is, the vector pointing from the center of the Earth to your point,[cos(lat) * cos(lng), cos(lat) * sin(lng), sin(lat)] with the motion vector. (That math would be messy, best to let the computer handle it)
You already have 2 vectors:
N = (0,0,1) points straight up from the origin.
P = (a,b,c) points from the origin to your point.
Calculate the unit vector to your point
U = P/|P|
Calculate a unit vector perpendicular to U and N
E = U X N
Calculate a unit vector perpendicular to U and E (this will be tangent to the sphere)
T = U X E
T could be pointing either North or South, so
if T.z < 0, multiply T by -1.
T now points due north, and is parallel to the plane tangent to the sphere at P.
You now have enough information to construct a rotation matrix (R), so you can rotate T around U. You can find how to make a matrix for rotation around any axis on wikipedia:
Using R, you can calculate a vector pointing in the direction of travel.
A = RT
A is the answer you are looking for.
I'm trying to make a triangle (isosceles triangle) to move around the screen and at the same time slightly rotate it when a user presses a directional key (like right or left).
I would like the nose (top point) of the triangle to lead the triangle at all times. (Like that old asteroids game).
My problem is with the maths behind this. At every X time interval, I want the triangle to move in "some direction", I need help finding this direction (x and y increments/decrements).
I can find the center point (Centroid) of the triangle, and I have the top most x an y points, so I have a line vector to work with, but not a clue as to "how" to work with it.
I think it has something to do with the old Sin and Cos methods and the amount (angle) that the triangle has been rotated, but I'm a bit rusty on that stuff.
Any help is greatly appreciated.
The arctangent (inverse tangent) of vy/vx, where vx and vy are the components of your (centroid->tip) vector, gives you the angle the vector is facing.
The classical arctangent gives you an angle normalized to -90° < r < +90° degrees, however, so you have to add or subtract 90 degrees from the result depending on the sign of the result and the sign of vx.
Luckily, your standard library should proive an atan2() function that takes vx and vy seperately as parameters, and returns you an angle between 0° and 360°, or -180° and +180° degrees. It will also deal with the special case where vx=0, which would result in a division by zero if you were not careful.
See http://www.arctangent.net/atan.html or just search for "arctangent".
Edit: I've used degrees in my post for clarity, but Java and many other languages/libraries work in radians where 180° = π.
You can also just add vx and vy to the triangle's points to make it move in the "forward" direction, but make sure that the vector is normalized (vx² + vy² = 1), else the speed will depend on your triangle's size.
#Mark:
I've tried writing a primer on vectors, coordinates, points and angles in this answer box twice, but changed my mind on both occasions because it would take too long and I'm sure there are many tutorials out there explaining stuff better than I ever can.
Your centroid and "tip" coordinates are not vectors; that is to say, there is nothing to be gained from thinking of them as vectors.
The vector you want, vForward = pTip - pCentroid, can be calculated by subtracting the coordinates of the "tip" corner from the centroid point. The atan2() of this vector, i.e. atan2(tipY-centY, tipX-centX), gives you the angle your triangle is "facing".
As for what it's relative to, it doesn't matter. Your library will probably use the convention that the increasing X axis (---> the right/east direction on presumably all the 2D graphs you've seen) is 0° or 0π. The increasing Y (top, north) direction will correspond to 90° or (1/2)π.
It seems to me that you need to store the rotation angle of the triangle and possibly it's current speed.
x' = x + speed * cos(angle)
y' = y + speed * sin(angle)
Note that angle is in radians, not degrees!
Radians = Degrees * RadiansInACircle / DegreesInACircle
RadiansInACircle = 2 * Pi
DegressInACircle = 360
For the locations of the vertices, each is located at a certain distance and angle from the center. Add the current rotation angle before doing this calculation. It's the same math as for figuring the movement.
Here's some more:
Vectors represent displacement. Displacement, translation, movement or whatever you want to call it, is meaningless without a starting point, that's why I referred to the "forward" vector above as "from the centroid," and that's why the "centroid vector," the vector with the x/y components of the centroid point doesn't make sense. Those components give you the displacement of the centroid point from the origin. In other words, pOrigin + vCentroid = pCentroid. If you start from the 0 point, then add a vector representing the centroid point's displacement, you get the centroid point.
Note that:
vector + vector = vector
(addition of two displacements gives you a third, different displacement)
point + vector = point
(moving/displacing a point gives you another point)
point + point = ???
(adding two points doesn't make sense; however:)
point - point = vector
(the difference of two points is the displacement between them)
Now, these displacements can be thought of in (at least) two different ways. The one you're already familiar with is the rectangular (x, y) system, where the two components of a vector represent the displacement in the x and y directions, respectively. However, you can also use polar coordinates, (r, Θ). Here, Θ represents the direction of the displacement (in angles relative to an arbitary zero angle) and r, the distance.
Take the (1, 1) vector, for example. It represents a movement one unit to the right and one unit upwards in the coordinate system we're all used to seeing. The polar equivalent of this vector would be (1.414, 45°); the same movement, but represented as a "displacement of 1.414 units in the 45°-angle direction. (Again, using a convenient polar coordinate system where the East direction is 0° and angles increase counter-clockwise.)
The relationship between polar and rectangular coordinates are:
Θ = atan2(y, x)
r = sqrt(x²+y²) (now do you see where the right triangle comes in?)
and conversely,
x = r * cos(Θ)
y = r * sin(Θ)
Now, since a line segment drawn from your triangle's centroid to the "tip" corner would represent the direction your triangle is "facing," if we were to obtain a vector parallel to that line (e.g. vForward = pTip - pCentroid), that vector's Θ-coordinate would correspond to the angle that your triangle is facing.
Take the (1, 1) vector again. If this was vForward, then that would have meant that your "tip" point's x and y coordinates were both 1 more than those of your centroid. Let's say the centroid is on (10, 10). That puts the "tip" corner over at (11, 11). (Remember, pTip = pCentroid + vForward by adding "+ pCentroid" to both sides of the previous equation.) Now in which direction is this triangle facing? 45°, right? That's the Θ-coordinate of our (1, 1) vector!
keep the centroid at the origin. use the vector from the centroid to the nose as the direction vector. http://en.wikipedia.org/wiki/Coordinate_rotation#Two_dimensions will rotate this vector. construct the other two points from this vector. translate the three points to where they are on the screen and draw.
double v; // velocity
double theta; // direction of travel (angle)
double dt; // time elapsed
// To compute increments
double dx = v*dt*cos(theta);
double dy = v*dt*sin(theta);
// To compute position of the top of the triangle
double size; // distance between centroid and top
double top_x = x + size*cos(theta);
double top_y = y + size*sin(theta);
I can see that I need to apply the common 2d rotation formulas to my triangle to get my result, Im just having a little bit of trouble with the relationships between the different components here.
aib, stated that:
The arctangent (inverse tangent) of
vy/vx, where vx and vy are the
components of your (centroid->tip)
vector, gives you the angle the vector
is facing.
Is vx and vy the x and y coords of the centriod or the tip? I think Im getting confused as to the terminology of a "vector" here. I was under the impression that a Vector was just a point in 2d (in this case) space that represented direction.
So in this case, how is the vector of the centroid->tip calculated? Is it just the centriod?
meyahoocomlorenpechtel stated:
It seems to me that you need to store
the rotation angle of the triangle and
possibly it's current speed.
What is the rotation angle relative to? The origin of the triangle, or the game window itself? Also, for future rotations, is the angle the angle from the last rotation or the original position of the triangle?
Thanks all for the help so far, I really appreciate it!
you will want the topmost vertex to be the centroid in order to achieve the desired effect.
First, I would start with the centroid rather than calculate it. You know the position of the centroid and the angle of rotation of the triangle, I would use this to calculate the locations of the verticies. (I apologize in advance for any syntax errors, I have just started to dabble in Java.)
//starting point
double tip_x = 10;
double tip_y = 10;
should be
double center_x = 10;
double center_y = 10;
//triangle details
int width = 6; //base
int height = 9;
should be an array of 3 angle, distance pairs.
angle = rotation_angle + vertex[1].angle;
dist = vertex[1].distance;
p1_x = center_x + math.cos(angle) * dist;
p1_y = center_y - math.sin(angle) * dist;
// and the same for the other two points
Note that I am subtracting the Y distance. You're being tripped up by the fact that screen space is inverted. In our minds Y increases as you go up--but screen coordinates don't work that way.
The math is a lot simpler if you track things as position and rotation angle rather than deriving the rotation angle.
Also, in your final piece of code you're modifying the location by the rotation angle. The result will be that your ship turns by the rotation angle every update cycle. I think the objective is something like Asteroids, not a cat chasing it's tail!