I am writing a custom animation for wpf and as a non math guy I have a couple questions...
If I am given two Point3D's, the From and To, and assuming the origin is at 0,0,0 how do I calculate a curve between the two points?
And once I have the curve 'plotted' and I know its length (how to do that too?) how can I calculate the x,y,z coords at some given distance along the line?
Thanks!
To get a straight line vector from point A to point B:
B - A
which would translate to:
vector.x = b.x - a.x;
vector.y = b.y - a.y;
vector.z = b.z - a.z;
The length is:
length = Math.Sqrt(vector.x * vector.x +
vector.y * vector.y +
vector.z * vector.z);
To get a point a certain distance along the vector you need to make the vector a unit vector (length 1):
vector.x = vector.x / length;
...
and then multiply by your distance:
vector.x = distance * vector.x;
...
This is all from memory so might not compile straight away.
There's A Vector Type for C# on CodeProject which will do a lot of this for you.
If you want a curve, then you'll need:
a) to define what type of curve you want (arc, spline, etc.)
b) more points (centres, control points etc.)
You'll probably want to express your curve as a set of parametric functions of some other variable:
x = f(t)
y = g(t)
z = h(t)
where 0 <= t <= 1, and
f(0) = from.x, f(1) = to.x
g(0) = from.y, g(1) = to.y
h(0) = from.z, h(1) = to.z
There are an infinite number of curves connecting any two points, so you'll need more
information to decide what form f(t), g(t), and h(t) should take. To move a point
along the curve, you just let t vary between 0 and 1 and calculate the x, y, and z
coordinates. One approach is to define a set of control points that you'd like your
curve to pass through (or near), then express your parametric equations in terms of
spline functions. You won't need to know the arc length of the curve in order to do this.
So I just wanted to follow up with my solution- while it is true there are an infinite number of curves- my (poorly worded) question was how to plot between two points on a curve- the shortest distance, assuming an origin of 0,0,0 and two 3d points. What I did was to convert my points from cartesian to polar, calculate the spherical point at a given time and then convert that point back to cartesians. If anyone wants me to post the actual C# code let me know.
Related
I'm currently building a basic raytracing algorithm and need to figure out which system of handling the intersections would be best performance-wise.
In the method I'm checking for a intersection of the ray and the object I'm returning a struct with the distance of the ray traveled to the hit, the position vector of the hit and the normal vector or -1 for the distance if there is no intersection.
For the next step I have to find the shortest distance of all intersections and exclude the ones with a negative distance.
I even thought about having 2 structs, one with only negative distances and one full struct to reduce the amount of space needed, but thought this wouldn't really make a difference.
My options so far:
first go over the array of the intersections and exclude the ones with negative distances, then find the shortest distance from the remainings via a sorting algorithm (probably insertion sort due to quick implementation).
Or put them together in one algorithm and test in each sort step if the distance is negative.
typedef Point3f float[3];
typedef struct {
float distance;
Point3f point;
Point3f normal;
} Intersection;
Intersection intersectObject (Ray-params, object) {
Intersection intersection;
//...
if (hit) {
intersection.distance = distance;
intersection.point = point;
intersection.normal = normal;
} else {
intersection.distance = -1.0f;
}
return intersection;
}
//loop over screen pixel
Intersection* intersections;
int amountIntersections;
//loop over all objects
//here I would handle the intersections
if (amountIntersections) {
//cast additional rays
}
I can't really figure out what would be the best way to handle this, since this would be called a lot of times. The intersection array will probably be a dynamic array with the amountIntersections as the length variable or an array with the most expected amount of intersections which then have intersections in it with negative distances.
Here is the approach I've succesfully used for a huge number of objects. (Especially for ball-and-stick atomic models; see my Wikipedia user page for the equations I used for those.)
First, transform the objects to a coordinate system where the eye is at origin, and the projected plane is parallel to the xy plane, with center on the positive z axis. This simplifies the equations needed a lot, as you can see from the above linked page.
As an example, if you have a unit ray n (so n·n = 1) and a sphere of radius r centered at c, the ray intersects the sphere if and only if h ≥ 0,
h = (n·c)2 + r2 - (c·c)
and if so, at distance d,
d = n·c ± sqrt(h)
If you work out the necessary code, and use sensible temprary variables, you'll see that you can reject non-intersecting spheres using eight multiplications and six additions or subtractions, and that this vectorizes across objects easily using SSE2/AVX intrinsics (#include <x86intrin.h>). (That is, do not try to use an XMM/YMM vector register for n or c, and instead use each register component for a different object, calculating h for 2/4/8 objects at a time.)
For each ray, sort/choose the objects to be tested according to their known minimum z coordinate (say, cz - r for spheres). This way, when you find an intersection at distance d, you can ignore all objects with minimum z coordinate larger than d, because the intersection point would necessarily be further out, behind the already known intersection.
Similarly, you should ignore all intersections where the distance is smaller than the distance to the projection plane (which is zd / nz, if the plane is at z = zd, and only needs to be computed once per ray), because those intersections are between the eye and the projection plane. (Technically, you've "crashed into" something then, if you think of the projection plane as a camera.)
i have two Objects in a 3D World and want to make the one object facing the other object. I already calculated all the angles and stuff (pitch angle and yaw angle).
The problem is i have no functions to set the yaw or pitch individually which means that i have to do it by a quaternion. As the only function i have is: SetEnetyQuaternion(float x, float y, float z, float w). This is my pseudocode i have yet:
float px, py, pz;
float tx, ty, tz;
float distance;
GetEnetyCoordinates(ObjectMe, &px, &py, &pz);
GetEnetyCoordinates(TargetObject, &tx, &ty, &tz);
float yaw, pitch;
float deltaX, deltaY, deltaZ;
deltaX = tx - px;
deltaY = ty - py;
deltaZ = tz - pz;
float hyp = SQRT((deltaX*deltaX) + (deltaY*deltaY) + (deltaZ*deltaZ));
yaw = (ATAN2(deltaY, deltaX));
if(yaw < 0) { yaw += 360; }
pitch = ATAN2(-deltaZ, hyp);
if (pitch < 0) { pitch += 360; }
//here is the part where i need to do a calculation to convert the angles
SetEnetyQuaternion(ObjectMe, pitch, 0, yaw, 0);
What i tried yet was calculating the sinus from those angles devided with 2 but this didnt work - i think this is for euler angles or something like that but didnt help me. The roll(y axis) and the w argument can be left out i think as i dont want my object to have a roll. Thats why i put 0 in.
If anyone has any idea i would really appreciate help.
Thank you in advance :)
Let's suppose that the quaternion you want describes the attitude of the player relative to some reference attitude. It is then essential to know what the reference attitude is.
Moreover, you need to understand that an object's attitude comprises more than just its facing -- it also comprises the object's orientation around that facing. For example, imagine the player facing directly in the positive x direction of the position coordinate system. This affords many different attitudes, from the one where the player is standing straight up to ones where he is horizontal on either his left or right side, to one where he is standing on his head, and all those in between.
Let's suppose that the appropriate reference attitude is the one facing parallel to the positive x direction, and with "up" parallel to the positive z direction (we'll call this "vertical"). Let's also suppose that among the attitudes in which the player is facing the target, you want the one having "up" most nearly vertical. We can imagine the wanted attitude change being performed in two steps: a rotation about the coordinate y axis followed by a rotation about the coordinate z axis. We can write a unit quaternion for each of these, and the desired quaternion for the overall rotation is the Hamilton product of these quaternions.
The quaternion for a rotation of angle θ around the unit vector described by coordinates (x, y, z) is (cos θ/2, x sin θ/2, y sin θ/2, z sin θ/2). Consider then, the first quaternion you want, corresponding to the pitch. You have
double semiRadius = sqrt(deltaX * deltaX + deltaY * deltaY);
double cosPitch = semiRadius / hyp;
double sinPitch = deltaZ / hyp; // but note that we don't actually need this
. But you need the sine and cosine of half that angle. The half-angle formulae come in handy here:
double sinHalfPitch = sqrt((1 - cosPitch) / 2) * ((deltaZ < 0) ? -1 : 1);
double cosHalfPitch = sqrt((1 + cosPitch) / 2);
The cosine will always be nonnegative because the pitch angle must be in the first or fourth quadrant; the sine will be positive if the object is above the player, or negative if it is below. With all that being done, the first quaternion is
(cosHalfPitch, 0, sinHalfPitch, 0)
Similar analysis applies to the second quaternion. The cosine and sine of the full rotation angle are
double cosYaw = deltaX / semiRadius;
double sinYaw = deltaY / semiRadius; // again, we don't actually need this
We can again apply the half-angle formulae, but now we need to account for the full angle to be in any quadrant. The half angle, however, can be only in quadrant 1 or 2, so its sine is necessarily non-negative:
double sinHalfYaw = sqrt((1 - cosYaw) / 2);
double cosHalfYaw = sqrt((1 + cosYaw) / 2) * ((deltaY < 0) ? -1 : 1);
That gives us an overall second quaternion of
(cosHalfYaw, 0, 0, sinHalfYaw)
The quaternion you want is the Hamilton product of these two, and you must take care to compute it with the correct operand order (qYaw * qPitch), because the Hamilton product is not commutative. All the zeroes in the two factors make the overall expression much simpler than it otherwise would be, however:
(cosHalfYaw * cosHalfPitch,
-sinHalfYaw * sinHalfPitch,
cosHalfYaw * sinHalfPitch,
sinHalfYaw * cosHalfPitch)
At this point I remind you that we started with an assumption about the reference attitude for the quaternion system, and the this result depends on that choice. I also remind you that I made an assumption about the wanted attitude, and that also affects this result.
Finally, I observe that this approach breaks down where the target object is very nearly directly above or directly below the player (corresponding to semiRadius taking a value very near zero) and where the player is very nearly on top of the target (corresponding to hyp taking a value very near zero). There is a non-zero chance of causing a division by zero if you use these formulae exactly as given, so you'll want to think about how to deal with that.)
I want to animate a sine wave as if it's generated from the center and moves outwards (to the left and right).
I started with this library: SISinusWaveView and made some adjustments. Currently I have this naive code for calculating the Y position of the curve depending on the X, phase and frequency:
float width = view.width; float mid = width / 2;
float adjustedX = x < mid ? x : width - x;
float y = maxAmplitude * sinf(2 * M_PI *(adjustedX / mid) * frequency + phase);
// phase increases every frame
Quite obviously, this causes a sharp angle at the middle of the sine wave, as seen below:
I would like to make it so the horizontal center of the animation is a smooth curve rather than a sharp angle, while keeping the animation horizontally symmetric. How would I approach this? Any mathematical insight for achieving this is appreciated.
EDIT
I tried to implement #TheBigH suggestion, but the parabola section is not seamlessly continuing the sine curve. Here's what I tried (implemented on Mathematica for quick visualization):
amp = 10;
freq = 1.5;
phase = 0.5;
Z = 1;
plotSine = Plot[amp*Sin[freq*x + phase], {x, Z, 2 Pi}];
aPara = amp*freq*Cos[phase]/(2 Z);
bPara = 0;
cPara = amp*Sin[c] - aPara*Z^2;
plotPara =
Plot[aPara*x^2 + bPara*x + cPara, {x, -Z, Z },
PlotRange -> {{-Z, Z}, {-20, 20}}];
Show[plotPara, plotSine, PlotRange -> {{-2 Pi, 2 Pi}, {-20, 20}}
Which results in this:
Changing the sign of the parabola didn't quite work either:
EDIT 2
I see now that the problem was assuming s(0) = p(Z) and s'(0) = p'(Z); instead of s(z) = p(Z) and s'(Z) = p'(Z). Moving the sine wave so it would start exactly at the end of the parabole would fix the problem, but it's more convenient solving the parabola such as s(z) = p(Z) and s'(Z) = p'(Z) as that would simplify the implementation. How to do this?
EDIT 3
See this math.stackexchange.com answer for the final solution.
Since OP has asked me to elaborate, here's my take as an answer.
Most generally, you're plotting the function s(x) = a sin(bx + c), where a, b and c come from the original problem. Later we will shift the sine curve by some offset Z but I'll leave it out of the sine curve for now as it will complicate the mathematics.
The new parabolic section will have equation p(x) = Ax^2 + Bx + C (A, B and C are different variables than a,b and c).
You want the two equations to join up cleanly, which means s(0) = p(Z). You also want the slopes to join up nicely so that there are no corners. That means you also need s'(0) = p'(Z). Also, since the parabola is centered about the origin, B = 0.
Thus you have two simultaneous equations for A, C given that you already know a, b, c and Z
a sin( c ) = A Z^2 + C
ab cos( c ) = 2AZ
or
A = ab cos( c ) / (2Z)
C = a sin (c) - A Z^2
This gives you the equation of the parabola, which you plot between -Z and Z. Then all you have to do is plot the sine curves, now adding in that offset. Let me know if any of this is unclear.
EDIT: I see there is a vertical shift of the sine wave as well. This does not pose any problems; just leave it out to begin with and add it to the parabola and sine waves at the end.
We're trying to create a curve editor program for calibrating hardware. As such, Bezier curves seem to be the easiest UI for a person to work with. However, the problem with bezier curves is it's possible to create a curve where there is more than one Y value for a particular X, or vice versa.
Now I know I can constrain the control points (P1, P2) to the region defined by the anchors (P0, P3) which would block an S from being created, but it also limits other otherwise-valid curves.
What I'm wondering is if there is some test you can run, short of walking the curve manually, that you can tell if your curve has an S-bend in it and if so, to reject that curve.
BTW, this is for a cross-platform app, hence WPF and NSBezierCurve being keywords.
The good news is that we can find out all sorts of things about bezier curves. There are several shape interrogation techinques (works fairly well for curves of lower degree, and cubic is on the lower side :)).
S shaped cubic bezier curve
First, an S shaped cubic bezier curve, has an inflection point. Inflection points are points on the curve where the concavity changes to convexity or the other way around.
This also means that at Inflection points, curvature of the curve is zero (Well, if any curve has to change from concave to convex, there has to be some place in between where the curve is straight —just look at "S").
https://en.wikipedia.org/wiki/Inflection_point
The Equation for curvature of a parametric curve is
We can ask Mathematica to simplify this, so that we can do some quick tests.
Here is some Javascript code (using paper.js library) which does the same thing:
// Method1 - Find inflection points
function hasInflection(p0x, p0y, p1x, p1y, p2x, p2y, p3x, p3y){
var CURV_EPSILON = 1e-3;
var e2 = 1 + CURV_EPSILON;
var roots = [], a, b, c;
// Solve for Numerator[k] == 0
// Where, Curvature k = (dx d^2y - d^2x dy) / (dx^2 - dy^2)^(3/2)
// Coefficients from Mathematica
a = -p1x*p0y + 2*p2x*p0y - p3x*p0y + p0x*p1y - 3*p2x*p1y + 2*p3x*p1y - 2*p0x*p2y + 3*p1x*p2y - p3x*p2y + p0x*p3y - 2*p1x*p3y + p2x*p3y;
b = (2*p1x*p0y - 3*p2x*p0y + p3x*p0y - 2*p0x*p1y + 3*p2x*p1y - p3x*p1y + 3*p0x*p2y - 3*p1x*p2y - p0x*p3y + p1x*p3y);
c = (-p1x*p0y + p2x*p0y + p0x*p1y - p2x*p1y - p0x*p2y + p1x*p2y);
// Use any method to solve the quadratic equation (a x^2 + b x + c = 0)
Numerical.solveQuadratic(a, b, c, roots, 0, 1);
// Find the root where, the curve changes from concave to convex
for (i = 0, l = roots.length; i < l; ++i) {
r = roots[i];
if( r > -CURV_EPSILON && r < e2 ){
// Here we basically evaluate the signed normals and look if they are of different signs
if( Curve.evaluate(v,r-CURV_EPSILON,3) * Curve.evaluate(v,r+CURV_EPSILON,3) < 0 )
return true;
}
}
return false;
}
Or see the complete example here
Curves where, more than a single Y value for any X
From your question, I think there is room for a better definition of what exactly is that you want.
The condition of more than a Y value for any X (or vice versa) can occur even if the curve is not S shaped —the curvature never changes sign.
The same is true even if the curve does not self-intersect. So that is probably not a good good test for what you are looking for.
If you need curves that are monotone in X and Y direction —that is, the X and Y value are either decreasing or increasing monotonically along the length of the curve; you can solve the derivative of the cubic equation instead (in both X and Y).
I have the coordinates (x,y) of 2 points. I want to build the third point so that these 3 points make an equilateral triangle.
How can I calculate the third point?
Thank you
After reading the posts (specially vkit's) I produced this simple piece of code which will do the trick for one direction (remember that there are two points). The modification for the other case shold be trivial.
#include<stdio.h>
#include<math.h>
typedef struct{
double x;
double y;
} Point;
Point vertex(Point p1, Point p2){
double s60 = sin(60 * M_PI / 180.0);
double c60 = cos(60 * M_PI / 180.0);
Point v = {
c60 * (p1.x - p2.x) - s60 * (p1.y - p2.y) + p2.x,
s60 * (p1.x - p2.x) + c60 * (p1.y - p2.y) + p2.y
};
return v;
}
You could rotate the second point 60° around first to find the location of the third point.
Something like this:
//find offset from point 1 to 2
dX = x2 - x1;
dY = y2 - y1;
//rotate and add to point 1 to find point 3
x3 = (cos(60°) * dX - sin(60°) * dY) + x1;
y3 = (sin(60°) * dX + cos(60°) * dY) + y1;
Let's call your two points A and B. Bisect AB, call this point C. Find the slope of AB (YA-YB / XA-XB), call it m. Find the perpendicular to that (-1/m) and call it m2. Then compute a segment CD whose length is sin(60) * length(AB), at the slope m2 (there will be two such points, one to each side of AB). ABD is then your equilateral triangle.
That, obviously, is a "constructive" method. You should also be able to do it by solving a set of linear equations. I haven't tried to figure out the right system of equations for this case, but this approach tends to be somewhat more stable numerically, and has fewer special cases (e.g., with the constructive version, a slope of 0 has to be dealt with specially).
For BlueRaja's challenge go to end of post:
Answer using translation and rotation:
Says points are P(x1,y1) and Q(x2,y2).
Since it is graphics, you can use tranforms to get the point.
First translate axes so P is the origin.
Next rotate Q around P by 60 degrees (or -60 to get the other possible point).
This gives you the coordinates of the third point say R, when P is the origin.
Translate back and there you have it.
You can use standard graphics API which take care of precision etc issues for you. No headaches.
Of course you could do the math and actually come up with a formula and use that and that might be faster, but then the question might get closed as off-topic ;-)
To take up BlueRaja's challenge: Here is a method which does not use trigonometry.
Given points P(x1,y1) and Q(x2,y2)
Say the point we need (R) to find is (x3,y3).
Let T be midpoint of PQ.
We know the area of triangle PQR (as it is equilateral and we know the side)
and we know the area of triangle PRT (1/2 the earlier area).
Now area of a triangle can be written as a determinant having the co-ordinates as entries:
2*Area = |D|
where
| 1 x1 y1|
D = | 1 x2 y2|
| 1 x3 y3|
We have two such equations (which are linear), solve for x3 and y3.
pc <- c((x1+x2)/2,(y1+y2)/2) #center point
ov <- c(y2-y1,x1-x2) #orthogonal vector
p3 <- pc+sqrt(3/4)*ov #The 3dr point in equilateral triangle (center point + height of triangle*orthogonal vector)