Basic Vector Calculus and Geometry

Line Segment to Line Segment

If we are only concerned with two line segments intersecting then the division can be eliminated.

Let $p_{1} - q_{1} = k$

$0 \leq s \leq 1$
$0 \leq \frac{- k b^{- 1}}{a b^{- 1}} \leq 1$

$0 \leq t \leq 1$
$0 \leq \frac{k a^{- 1}}{b a^{- 1}} \leq 1$

Case $a b^{- 1} < 0$
Let $a b^{- 1} = - α$
$α > 0$

Consider when $0 \leq s \leq 1$
$0 \leq \frac{- k b^{- 1}}{- α} \leq 1$
$0 \leq k b^{- 1} \leq α$

Consider when $0 \leq t \leq 1$
$0 \leq \frac{k a^{- 1}}{α} \leq 1$
$0 \leq k a^{- 1} \leq α$

So if these conditions are satisfied then the two line segments are intersecting. I will repeat the results without the derivation for easy reference.

For the case $a b^{- 1} < 0$
$α = - a b^{- 1}$ and
$0 \leq k b^{- 1} \leq α$
$0 \leq k a^{- 1} \leq α$
must be true.

For the case $a b^{- 1} > 0$
$α = a b^{- 1}$ and
$0 \leq - k b^{- 1} \leq α$
$0 \leq - k a^{- 1} \leq α$
must be true.

Orthogonal Projection

Given a set of orthogonal vectors $v_{i}$ , express $c$ as $\sum a_{i} v_{i}$ .

Let
$c = \sum a_{i} v_{i}$
$v_{k} v_{i} = [i == k] v_{k}^{2}$
$v_{k} c = v_{k} \sum a_{i} v_{i}$
$v_{k} c = v_{k}^{2} a_{k}$
$a_{k} = \frac{v_{k} \cdot c}{v_{k} \cdot v_{k}}$

Reflection Theory

A beam of light reflects of a surface. Let $n$ be the normal to the surface and $c$ the vector reflecting off the surface. By resolving $c$ in terms of orthogonal vectors where $n$ is one of these vectors the problem is trivially solved.

Let
$c = a_{0} n + \sum_{1}^{n - 1} a_{i} v_{i}$
Let $c'$ be the reflected ray.
$c' = - a_{0} n + \sum_{1}^{n - 1} a_{i} v_{i}$

Area of Triangle

$\frac{1}{2} base . height$

$\frac{1}{2} b c \sin A = \frac{1}{2} b c (b^{2} c^{2} - (b \cdot c)^{2})^{\frac{1}{2}}$
$= \frac{1}{2} | b \times c |$

With matrices and the determinant.

$\frac{1}{2} | x_{1} - x_{0}, x_{2} - x_{0} |$

The area between a line segment and the x-axis.

$(x_{1} - x_{0}) . y_{0} + \frac{1}{2} (y_{1} - y_{0}) (x_{1} - x_{0}) = (x_{1} - x_{0}) \frac{1}{2} (y_{1} - y_{0})$

Linear combination of trapezoids. Really the area enclosed by the lines and the axis added and subtracted.

$\frac{1}{2} | x_{1} (y_{3} - y_{2}) + x_{2} (y_{1} - y_{3}) + x_{3} (y_{2} - y_{1}) |$

Volume of Tetrahedron

$\frac{1}{6} | x_{1} - x_{0}, x_{2} - x_{0}, x_{3} - x_{0} |$

Point to Line Distance Minimized

I picked up this technique in linear algebra where the teacher said it was possible to use matrix algebra to solve these sort of problems. There was no need for use to use it, maths is egalitarian.

Now in the comsci literature you do not see this or much in maths because no one can argue with the most primitive variables because it is correct (eg Newton published Principa with geometry, not his calculus). So instead of promoting a better way of doing things we avoid any arguments by doing the job in a sub-optimal way. Conservatism is a killer.

In this algebra do not divide. For the moment consider the multiplication operator defined as a dot product. For a single variable and vector multiplication just multiply the variable in each of the vectors components.

Let $P_{1}$ and $P_{2}$ be two points defining a line with direction from 1 to 2.
Let
$(- P_{1} + P_{2}) t + P_{1} = A t + C$
$X$ be the point to the line to be minimized.

$d = (A t + C - X)^{2}$

Consider the distance between the two points. Take the derivative in t and setting to zero for minimum t.

$d' = 2 (A t + C - X) A$
$0 = A^{2} t + (C - X) A$
$t A^{2} = (X - C) A$
$t = \frac{(X - C) A}{A^{2}}$

Bisecting an Angle with Ratios

$p_{1} \frac{b}{a + b} + p_{2} \frac{a}{a + b}$

For some reason I hate angles. Some people say hate is a very strong word. Despite the ancients propensity for geometry very few of them actually used algebra to do it.

I made the error of mistaking the midpoint for this point. By using this representation you can turn a problem involving bisection to one with algebra. For example calculating the inscribed circle/sphere, see The Inner Circle.

Point to Plane Distance Minimized

Let multiplication of vectors be a dot product. Use the result that the minimum distance is perpendicular to the plane.

Let $A$ and $B$ be arms of the plane at the origin.
$P$ be a point on the plane.
Let $W$ be the point being minimized for.

$N = A \times B$
$N X = c$ defines a plane
where $c = N P$

Let $N t + W$ be equation of line perpendicular to the plane and passing through $W$ .

Find the line and planes intersection point.

$N (N t + W) = c$
$t = \frac{d - N W}{N^{2}} = \frac{N (P - W)}{N^{2}}$

Now to get a distance function from the point to the point on the plane let d be the general distance function. Assume some nice properties of d such as subtracting a constant between the two components, and taking out a common factor.

$d (W, N \frac{N (P - W)}{N^{2}} + W)$
$d (0, N \frac{N (P - W)}{N^{2}})$
$N^{2} d = d (0, N (N (P - W)))$

From here we can get the distance from a point without the square root by realizing a component and squaring it, then sum these.

Another thing we can get is signed distance. Because if a dot product is used for multiplication then each components $N (P - W)$ tests positive when the point is on the other side of the half space. As $W$ goes nearer to the plane $NW$ approaches $c$ and $NP - NW = c - NW$ becomes smaller in magnitude.

For me this was surprising. The point is that you can use signed distance and save a dot product. Further it will indicate which side of the plane the point is on.

Anticlockwise Point Ordering Test

Point Ordering Test with Determinants

$[\begin{matrix} x_{0} & y_{0} & 1 \\ x_{1} & y_{1} & 1 \\ x_{2} & y_{2} & 1 \end{matrix}] > 0$

This may generalize in N-dimensions. I suspect the equality operator alternates.

Anticlockwise Point Ordering Test with Half Spaces

A simpler test is with a half space test. This generalizes for simplexes in any dimensions.

Only one halfspace needs to be tested. For if the simplex winding breaks the opposite point will be visible.

If the simplex is degenerate then test for a zero dot product.

Zero Test

Numerical instability is something everyone knows but is addressed differently by different people. I worked on tessellation algorithms for quite some time before I hit a real problem with doubles that the inbuilt rounding algorithms missed.

Let zero be a small positive number.

$(x == 0) =$
$(0 < x + zero < zero + zero)$
or
$(0 < x + zero) (x < zero)$

Many algorithms are made stable with the explicit realization of zero as a very small quantity. Whole theories of calculus are built on this that in your wildest dreams you could not imagine. Even though this test looks harmless enough at its heart is an operation which can not be reversed.

Barycentric Coordinates

Barycentric coordinates can describe a point in a simplex. They are a linear average of the simplexes points.

Here is a point inside a triangle. The variables u_i range from 0 to 1.

$u_{0} + u_{1} + u_{2} = 1$
$p (u_{0}, u_{1}, u_{2}) = u_{0} p_{0} + u_{1} p_{1} + u_{2} p_{2}$

For a tetrahedron there is one more dimension so there is one more variable u₃.

$0 \leq u_{i} \leq 1$
$\sum u_{i} = 1$
$p (u_{0}, u_{1}, ...) = \sum u_{i} p_{i}$

Line and Simplex Intersection

Since Barycentric Coordinates describes a point and a vector equation describes a point equating the two together and solving finds their intersection point. A line and a triangle in 2D does not have a unique solution because there is one extra variable. This corresponds to the fact that the intersection is not a point but a line through the triangle. For a triangle in 3D space the intersection point is generally unique.

$0 \leq u_{i} \leq 1$
$\sum u_{i} = 1$
$a + b t = \sum u_{i} p_{i}$

Real Time Rendering[1] describes an algorithm for computing the Ray/Triangle Intersection in 3D. The matrix inverse is calculated with Cramer's rule and the determinants are replaced by expressions with dot and cross products. The beauty of the algorithm is that only one division needs to be done.

If you do not need to compute the barycentric coordinates u and v and instead just want to know if the intersection occurred then the one division can be avoided altogether.

Consider $a$ to be the determinant,
$u = \frac{1}{a} s \cdot p$
$0 \leq u \leq 1$ becomes
$0 \leq s \cdot p \leq a$

Similarly substituting in the solution for v and multiplying out the denominator in the conditional equality removes the division and gets an expression that uses dot products only.

Consider $a$ to be the determinant,
$v = \frac{1}{a} d \cdot q$
$0 \leq v \leq 1$ becomes
$0 \leq p \cdot q \leq a$

[1] solved the equation

$o + t d = (1 - u - v) v_{0} + u v_{1} + v v_{2}$

As I am quoting I will keep their symbols but the use of o as a point is ridiculous. However their bold faced font on the vectors is very readable and if I could present half as well as the authors I would be happy. Expressing this equation as a matrix please forgive my use of MathML as the current MathML rendering is not adequate.

$(- d, v_{1} - v_{0}, v_{1} - v_{0}) (t, u, v)^{T} = o - v_{0}$
Let $e_{1} = v_{1} - v_{0}$
Let $e_{2} = v_{2} - v_{0}$
Let $s = o - v_{0}$
then
$(- d, e_{1}, e_{2}) (t, u, v)^{T} = s$

$(t, u, v)^{T} = \frac{1}{\det (- d, e_{1}, e_{2})} (\det (s, e_{1}, e_{2}), \det (- d, s, e_{2}), \det (- d, e_{1}, s))^{T}$

$\det (a, b, c) = - (a \times c) \cdot b = - (c \times b) \cdot a$

Let
$p = d \times e_{2}$
$q = s \times e_{1}$

$\det (- d, e_{1}, e_{2}) = (d \times e_{2}) \cdot e_{1} = p \cdot e_{1}$ $\det (s, e_{1}, e_{2}) = (s \times e_{1}) \cdot e_{2} = q \cdot e_{2}$
$\det (- d, s, e_{2}) = (d \times e_{2}) \cdot s = p \cdot s$
$\det (- d, e_{1}, s) = (s \times e_{1}) \cdot d = q \cdot d$

$(t, u, v)^{T} = \frac{1}{p \cdot e_{1}} (q \cdot e_{2}, p \cdot s, q \cdot d)^{T}$

This is easy to implement. Simply reject if the determinant is zero with a Zero Test, calculate u and v and test their bounds to ensure that the intersection point is in the triangle.

Ray and Box

A ray can be used to describe a line with thickness or a line in the computer science literature. For example [1] and [2] are both from the same book but the latter describes a ray as the area between two parallel lines in 2D and the former a line in 3D space.

Polygon Area

Where the points have the same winding, the area of the convex polygon can be calculated.

$area = \sum_{0}^{n - 1} p_{i, x} p_{i + 1, y} - p_{i + 1, x} p_{i, y}$

This can be derived directly by summing the triangles about a point as a sum of cross products ie the individual triangles. The cross product is expanded and when simplified the above result is given.

What is not so obvious is how this generalizes. By considering halfspaces the last two successive points generate a halfspace. Note that all successive halfspaces see the next point.

In 3D can the volume be calculated by having the last 3 successive points generate a halfspace which sees the next point? Perhaps <TODO>

Here is the sum to expand and simplify. Let 0 be the pivot point which is being rotated about. Let p_i be ordered such that the next point is visible on the surface. The area calculated by [1,2,3], [3,2,4], [3,4,5], [5,4,6], [5,6,7], ... This ordering is to have the triangles on the surface have consistent winding.

Heres some thoughts. In 2D there is only one direction of movement. In 3D a surface move can be left or right. Structure the surface transversal so a series of 1's and 0's to represent a right and a left move. Let the last move use the last two points. The path is described. While its length in unknown all the variation is expressed in the vector or 1's and 0's.

Rotating a Plane

Often we work in the xy plane and rotate the result. The best way to describe the planes orientation is with its normal. Working in the xy plane the normal is the z-axis vector. Given that we want to rotate the plane to the new position with normal N how is this done?

Assuming we have a function that rotates around a line through the origin, find the angle between z and N and rotate about their cross product.

$θ = \cos^{- 1} (\frac{N · z}{| N |})$
$rotate (θ, z \times N)$

References

T.Akenine-Moller E.Haines(2002). Real-Time Rendering (Second Edition). ISBN 1-56881-182-9. Pages 578-581.