I have an irregular tetrahedron using 4 vertices.
I need to find out the altitude given that one specific vertex is the top and the others are the base.
Basically the height would be the shortest distance from the top to its base creating a 90 degrees angle. It should be a simple math question but I cannot find anything on Google.
I am looking for an optimized function that looks like this :
float GetPyramidAltitude (Vector3 top, Vector3 baseA, Vector3 baseB, Vector3 baseC) {
...
}
Thank you for your help.
This is equivalent to finding the distance between a point and a plane. The plane is defined by the three points comprising the base. There is a detailed explanation on determining a plane given three points and finding the minimal distance between a point and a plane.
Disclaimer: I don't know Unity3D, so I'm kind of making up the syntax below. If something's not clear, let me know; otherwise you're going to have to translate into something that will compile using the Unity3D API.
The first step is to determine the equation of the plane given three points. The plane normal is given by:
n = cross(baseB-baseA, baseC-baseA);
n = n / norm(n);
Where cross returns the cross product of the two arguments, and norm returns the l2 norm (vector magnitude). The offset term in the plane equation is given by:
d = -n.x*baseA.x - n.y*baseA.y - n.z*baseA.z;
This will result in the plane equation:
n.x*X + n.y*Y + n.z*Z + d = 0
To find the distance between the top and the plane is then given by
D = dot(n, top) + d;
where dot is the dot product of the unit vector normal of the plane n and top and d is defined earlier. When D > 0 top is "above" the plane defined by the three base points where the normal points "up". When D < 0 the top is below the plane. So, in your case, you may want to take the absolute value of D to get the distance.
Thanks, however I found a solution from a method in Unity.
Basically we only need 3 parameters : the top vertex, one vertex from the base and the base's face normal which I already had.
Here's my solution :
float GetPyramidAltitude (Vector3 top, Vector3 baseA, Vector3 baseNormal) {
Vector3 topToBase = Vector3.Project(baseA - top, baseNormal);
return topToBase.magnitude;
}
Related
I have some point on a 2D grid (x, y) and I need to find all points that are n distance away from that point. The way I'm measuring distance is by using the distance formula between the two points. Anyone know how to do this?
Edit: Just for reference, what I'm trying to do is to write some AI path finding that will maintain some distance away from a target in a system that uses grid based locations. Currently I'm using A* path finding, but I'm not sure if that matters or makes a difference since I'm kind of new to this stuff.
Here's what I would do:
First filter out all points that are further than D on either x or y. These are certainly outside the circle of radius D. This is a much simpler computation, and it can quickly eliminate a lot of work. This is a outer bounding-box optimization.
You can also use an inner bounding-box optimization. If the points are closer than D * sqrt(2)/2 on either x or y, then they're certainly within the circle of radius D. This is also cheaper than calculating the distance formula.
Then you have a smaller number of candidate points that may be within the circle of radius D. For these, use the distance formula. Remember that if D = sqrt(Δx2+Δy2), then D2 = Δx2+Δy2.
So you can skip the cost of calculating square root.
So in pseudocode, you could do the following:
for each point
begin
if test 1 indicates the point is outside the outer bounding box,
then skip this point
if test 2 indicates the point is inside the inner bounding box,
then keep this point
if test 3 indicates the point is inside the radius of the circle,
then keep this point
end
This problem is known as range query. The brute force solution is just as you described: computed the distance of all points from the reference point and return those whose distance is less than the desired range value.
The brute force algorithm is O(N^2). There are, however, more efficient algorithms that employ spatial indexes to reduce algorithm complexity and the number of distance calculations. For example, you can use a R-Tree to index your points.
Its called nearest neighbor search. More at http://en.wikipedia.org/wiki/Nearest_neighbor_search
There are open libraries for that. I have used one written for C and recommend it: http://www.cs.umd.edu/~mount/ANN/. ANN stands for Approximate Nearest Neighbor, however, you can turn the approximation off and find the exact nearest neighbors.
This wouldn't use the distance formula, but if you're looking for points exactly n distance away, perhaps you could use sin/cos?
In pseudocode:
for degrees in range(360):
x = cos(degrees) * n
y = sin(degrees) * n
print x, y
That would print every point n away in 360 degree increments.
Java implementation:
public static Set<Point> findNearbyPoints(Set<Point> pts, Point centerPt, double radius) {
Set<Point> nearbyPtsSet = new HashSet<Point>();
double innerBound = radius * (Math.sqrt(2.0) / 2.0);
double radiusSq = radius * radius;
for (Point pt : pts) {
double xDist = Math.abs(centerPt.x - pt.x);
double yDist = Math.abs(centerPt.y - pt.y);
if (xDist > radius || yDist > radius)
continue;
if (xDist > innerBound || yDist > innerBound)
continue;
if (distSq(centerPt, pt) < radiusSq)
nearbyPtsSet.add(pt);
}
return nearbyPtsSet;
}
I am trying to find circumcenter of Given Three point of Triangle……..
NOTE: all these three points are with X,Y and Z Co-Ordinate Means points are in 3D
I know that the circumcenter is the point where the right bisectors intersect….
But for that I have to find middle point of each side then the right bisectors and then intersection point of that …..this is long and error some process……
Is there not any formula which just takes as input these three points of triangle and giving us the Circumcenter of Triangle ……?
Thanks………
The wiki page on Circumscribed circle has it in terms of dot and cross products of the three vertex vectors. It also has a formula for the radius of the circle, if you are so interested.
First of all, you need to make sure that points are not collinear. i.e. do not lie in the same line. For that you need to find the direction cosines of the lines made by three points, and if they have same direction cosines, halt, you can't get circle out of it.
For direction cosine please check this article on wikipedia.
(A way of finding coordinate-geometry and geometry -- based on the theorem that, a perpendicular line from center of circle bisects a chord)
Find the equation of the plane.
This equation must reduce to the form
and the direction cosines (of the line perpendicular to plane determines the plane), so direction cosines of the line perpendicular to this line is
given by this link equations -- 8,9,10 (except replace it for l, m, n).
Find the equation of the lines (all three) in 3-d
(x-x1)/l=(y-y1)/m=(z-z1)/n (in terms of direction cosines) or
(x-x1)/(x2-x1)=(y-y1)/(y2-y1)=(z-z1)/(z2-z1)
Now we need to find the equation of line
a) this perpendicular to the line, from 2 (let l1, m1, n1 be direction cosines of this line)
b) must be contained in place from 1 (let l2, m2, n2 be direction cosines of this line perpendicular to plane)
Find and solve (at least two lines) from 3, sure you will be able to find the center of the circle.
How to find out equation ??? as we are finding the circum-center, we will get our points (i.e. it is the midpoint of the two points) and for a) we have
l1*l+m1*m+n1*n = 0 and l2*l+m2*m+n2*n = 0 where l, m, n are direction cosines of our, line, now solving this two equation, we can get l, m interms of n. And we use this found out (x1,y1,z1) and the value of l, m, 1 and we will have out equation.
The other process is to solve the equation given in this equation
https://stackoverflow.com/questions/5725871/solving-the-multiple-math-equations
Which is the deadliest way.
The other method is using the advantage of computer(by iteration) - as I call it (but for this you need to know the range of the coordinates and it consumes lot of memory)
it's like this (You can make it more precise by incrementing at 1/10) but certainly bad way.
for(i=minXrange, i>=maxXrange; i++){
for(j=minYrange, j>=maxYrange; j++){
for(i=minZrange, i>=maxZrange; k++){
if(((x1-i)^2 + (y1-j)^2 + (z1-k)^2) == (x2-i)^2 + (y2-j)^2 + (z2-k)^2) == for z)){
return [i, j, k];
}
}
}
}
I have two points in 2D space, centred on origin (0,0). The first point represents the starting location and the second represents the end location. I need to calculate the angle of rotation between the two points, my problem being that the hypoteneuse from each point to (0,0) is not equal.
Could someone tell me how to work out the angle between the two points, bearing in mind that they could be anywhere relative to (0,0).
Many thanks,
Matt.
Let's say point 1 is (x1,y1) and point 2 is (x2,y2)
The tangent of the Angle from X axis to point 1, relative to (0,0) is y1/x1
The tangent of the Angle from X axis to point 2, relative to (0,0) is y2/x2
Take the arc tangent (is that the right term? Tan-1 on a calculator) to get the actual angle for each, then subtract to get the answer you're looking for
This is easily accomplished taking the arccosine of the normalized inner product of the two vectors. That is, given u = (ux, uy) and v = (vx, vy), the angle between the two is given by θ = acos(u·v/|u||v|), where u · v = uxvx + uyvy is the dot product of the two and the | | operator is the l2 normal given by |u| = sqrt(ux2 + uy2). This will result in the smallest rotation that can be applied to one of the vectors that will make them linear multiples of each other. Therefore, you may need to fiddle with the sign of θ to make sure you're going in the right direction if you have one you want to start from.
I'm using CML to manage the 3D math in an OpenGL-based interface project I'm making for work. I need to know the width of the viewing frustum at a given distance from the eye point, which is kept as a part of a 4x4 matrix that represents the camera. My goal is to position gui objects along the apparent edge of the viewport, but at some distance into the screen from the near clipping plane.
CML has a function to extract the planes of the frustum, giving them back in Ax + By + Cz + D = 0 form. This frustum is perpendicular to the camera, which isn't necessarily aligned with the z axis of the perspective projection.
I'd like to extract x and z coordinates so as to pin graphical elements to the sides of the screen at different distances from the camera. What is the best way to go about doing it?
Thanks!
This seems to be a duplicate of Finding side length of a cross-section of a pyramid frustum/truncated pyramid, if you already have a cross-section of known width a known distance from the apex. If you don't have that and you want to derive the answer yourself you can follow these steps.
Take two adjacent planes and find
their line of intersection L1. You
can use the steps here. Really
what you need is the direction
vector of the line.
Take two more planes, one the same
as in the previous step, and find
their line of intersection L2.
Note that all planes of the form Ax + By + Cz + D = 0 go through the origin, so you know that L1 and L2
intersect.
Draw yourself a picture of the
direction vectors for L1 and L2,
tails at the origin. These form an
angle; call it theta. Find theta
using the formula for the angle
between two vectors, e.g. here.
Draw a bisector of that angle. Draw
a perpendicular to the bisector at
the distance d you want from the
origin (this creates an isosceles
triangle, bisected into two
congruent right triangles). The
length of the perpendicular is your
desired frustum width w. Note that w is
twice the length of one of the bases
of the right triangles.
Let r be the length of the
hypotenuses of the right triangles.
Then rcos(theta/2)=d and
rsin(theta/2)=w/2, so
tan(theta/2)=(w/2)/d which implies
w=2d*tan(theta/2). Since you know d
and theta, you are done.
Note that we have found the length of one side of a cross-section of a frustrum. This will work with any perpendicular cross-section of any frustum. This can be extended to adapt it to a non-perpendicular cross-section.
I'm looking for an algorithm to find the common intersection points between 3 spheres.
Baring a complete algorithm, a thorough/detailed description of the math would be greatly helpful.
This is the only helpful resource I have found so far:
http://mathforum.org/library/drmath/view/63138.html
But neither method described there is detailed enough for me to write an algorithm on.
I would prefer the purely algebraic method described in the second post, but what ever works.
Here is an answer in Python I just ported from the Wikipedia article. There is no need for an algorithm; there is a closed form solution.
import numpy
from numpy import sqrt, dot, cross
from numpy.linalg import norm
# Find the intersection of three spheres
# P1,P2,P3 are the centers, r1,r2,r3 are the radii
# Implementaton based on Wikipedia Trilateration article.
def trilaterate(P1,P2,P3,r1,r2,r3):
temp1 = P2-P1
e_x = temp1/norm(temp1)
temp2 = P3-P1
i = dot(e_x,temp2)
temp3 = temp2 - i*e_x
e_y = temp3/norm(temp3)
e_z = cross(e_x,e_y)
d = norm(P2-P1)
j = dot(e_y,temp2)
x = (r1*r1 - r2*r2 + d*d) / (2*d)
y = (r1*r1 - r3*r3 -2*i*x + i*i + j*j) / (2*j)
temp4 = r1*r1 - x*x - y*y
if temp4<0:
raise Exception("The three spheres do not intersect!");
z = sqrt(temp4)
p_12_a = P1 + x*e_x + y*e_y + z*e_z
p_12_b = P1 + x*e_x + y*e_y - z*e_z
return p_12_a,p_12_b
Probably easier than constructing 3D circles, because working mainly on lines and planes:
For each pair of spheres, get the equation of the plane containing their intersection circle, by subtracting the spheres equations (each of the form X^2+Y^2+Z^2+aX+bY+c*Z+d=0). Then you will have three planes P12 P23 P31.
These planes have a common line L, perpendicular to the plane Q by the three centers of the spheres. The two points you are looking for are on this line. The middle of the points is the intersection H between L and Q.
To implement this:
compute the equations of P12 P23 P32 (difference of sphere equations)
compute the equation of Q (solve a linear system, or compute a cross product)
compute the coordinates of point H intersection of these four planes. (solve a linear system)
get the normal vector U to Q from its equation (normalize a vector)
compute the distance t between H and a solution X: t^2=R1^2-HC1^2, (C1,R1) are center and radius of the first sphere.
solutions are H+tU and H-tU
A Cabri 3D construction showing the various planes and line L
UPDATE
An implementation of this answer in python complete with an example of usage can be found at this github repo.
It turns out the analytic solution is actually quite nice using this method and can tell you when a solution exists and when it doesn't (it is also possible to have exactly one solution.) There is no reason to use Newton's method.
IMHO, this is far easier to understand and simpler than trilateration given below. However, both techniques give correct answers in my testing.
ORIGINAL ANSWER
Consider the intersection of two spheres. To visualize it, consider the 3D line segment N connecting the two centers of the spheres. Consider this cross section
(source: googlepages.com)
where the red-line is the cross section of the plane with normal N. By symmetry, you can rotate this cross-section from any angle, and the red line segments length can not change. This means that the resulting curve of the intersection of two spheres is a circle, and must lie in a plane with normal N.
That being said, lets get onto finding the intersection. First, we want to describe the resulting circle of the intersection of two spheres. You can not do this with 1 equation, a circle in 3D is essentially a curve in 3D and you cannot describe curves in 3D by 1 eq.
Consider the picture
(source: googlepages.com)
let P be the point of intersection of the blue and red line. Let h be the length of the line segment along the red line from point P upwards. Let the distance between the two centers be denoted by d. Let x be the distance from the small circle center to P. Then we must have
x^2 +h^2 = r1^2
(d-x)^2 +h^2 = r2^2
==> h = sqrt(r1^2 - 1/d^2*(r1^2-r2^2+d^2)^2)
i.e. you can solve for h, which is the radius of the circle of intersection. You can find the center point C of the circle from x, along the line N that joins the 2 circle centers.
Then you can fully describe the circle as (X,C,U,V are all vector)
X = C + (h * cos t) U + (h * sin t) V for t in [0,2*PI)
where U and V are perpendicular vectors that lie in a plane with normal N.
The last part is the easiest. It remains only to find the intersection of this circle with the final sphere. This is simply a plug and chug of the equations (plug in for x,y,z in the last equation the parametric forms of x,y,z for the circle in terms of t and solve for t.)
edit ---
The equation that you will get is actually quite ugly, you will have a whole bunch of sine's and cosine's equal to something. To solve this you can do it 2 ways:
write the cosine's and sine's in terms of exponentials using the equality
e^(it) = cos t + i sin t
then group all the e^(it) terms and you should get a quadratic equations of e^(it)'s
that you can solve for using the quadratic formula, then solve for t. This will give you the exact solution. This method will actually tell you exactly if a solution exists, two exist or one exist depending on how many of the points from the quadratic method are real.
use newton's method to solve for t, this method is not exact but its computationally much easier to understand, and it will work very well for this case.
Basically you need to do this in 3 steps. Let's say you've got three spheres, S1, S2, and S3.
C12 is the circle created by the intersection of S1 and S2.
C23 is the circle created by the intersection of S2 and S3.
P1, P2, are the intersection points of C12 and C13.
The only really hard part in here is the sphere intersection, and thankfully Mathworld has that solved pretty well. In fact, Mathworld also has the solution to the circle intersections.
From this information you should be able to create an algorithm.
after searching the web this is one of the first hits, so i am posting the most clean and easy solution i found after some hours of research here: Trilateration
This wiki site contains a full description of a fast and easy to understand vector approach, so one can code it with little effort.
Here is another interpretation of the picture which Eric posted above:
Let H be the plane spanned by the centers of the three spheres. Let C1,C2,C3 be the intersections of the spheres with H, then C1,C2,C3 are circles. Let Lij be the line connecting the two intersection points of Ci and Cj, then the three lines L12,L23,L13 intersect at one point P. Let M be the line orthogonal to H through P, then your two points of intersection lie on the line M; hence you just need to intersect M with either of the spheres.