[TOC]
A quadratic equation has the form $$ Ax^2 + Bx + C = 0 $$
To solve the quadratic equation analytically, we first divide by A: $$ x^2 + \frac{B}{A}x + \frac{C}{A} = 0 $$ Then we “complete the square” to group terms: $$ {(x + \frac{B}{2A})}^2 - \frac{B^2}{4A^2} + \frac{C}{A} = 0 $$ Moving the constant portion to the right-hand side and taking the square root gives $$ x + \frac{B}{2A} = \pm \sqrt{\frac{B^2}{4A^2} - \frac{C}{A}} $$ Subtracting B/(2A) from both sides and grouping terms with the denominator 2A gives the familiar form: $$ x = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A} \ D \equiv B^2 - 4AC $$ which is called the discriminant of the quadratic equation. If D > 0, there are two real solutions (also called roots). If D = 0, there is one real solution (a “double” root). If D < 0, there are no real solutions.
An angle is defined by the length of the arc segment it cuts out on the unit circle.A common convention is that the smaller arc length is used, and the sign of the angle is determined by the order in which the two half-lines are specified.
Each of these angles is the length of the arc of the unit circle that is “cut” by the two directions.
The conversion between degrees and radians is $$ degrees = \frac{180}{\pi}radians \ radians = \frac{\pi}{180}degrees \ $$
Shifting identities
$$
sin( − A) = − sin A \
cos( − A) = cos A \
tan( − A) = − tan A \
sin(π/2 − A) = cos A \
cos(π/2 − A) = sin A \
tan(π/2 − A) = cot A
$$
Pythagorean identities
$$
{sin}^2A + {cos}^2A = 1 \
{sec}^2A - {tan}^2A = 1 \
{csc}^2A - {cot}^2A = 1 \
$$
Addition and subtraction identities
$$
sin(A + B) = sin A cos B + sin B cos A \
sin(A − B) = sin A cos B − sin B cos A \
sin(2A) = 2 sin A cos A \
cos(A + B) = cos A cos B − sin A sin B \
cos(A − B) = cos A cos B + sin A sin B \
cos(2A) = {cos}^2A - {sin}^2A \
tan(A + B) = \frac{tanA + tanB}{1 - tanAtanB} \
tan(A - B) = \frac{tanA - tanB}{1 + tanAtanB} \
tan(2A) = \frac{2tanA}{1 - {tan}^2A}
$$
Half-angle identities
$$
{sin}^2(A/2) = (1 - cosA) /2 \
{cos}^2(A/2) = (1 + cosA) / 2\
$$
Product identities
$$
sin A sin B = − (cos(A + B) − cos(A − B))/2 \
sin A cos B = (sin(A + B) + sin(A − B))/2 \
cos A cos B = (cos(A + B) + cos(A − B))/2 \
$$
Geometry for triangle laws
The following identities are for arbitrary triangles with side lengths a, b, and c, each with an angle opposite it given by A, B, C, respectively $$ \frac{sinA}{a} = \frac{sinB}{b} = \frac{sinC}{c} $$
The area of a triangle can also be computed in terms of these side lengths:
triangle area =
A vector describes a length and a direction.
Two vectors are equal if they have the same length and direction even if we think of them as being located in different places
Vectors will be represented as bold characters, e.g.,
A vector’s length is denoted$\left | a \right | $.
A unit vector is any vector whose length is one.
The zero vector is the vector of zero length. The direction of the zero vector is undefined.
Vectors can be used to store an offset, also called a displacement. Vectors can also be used to store a location, another word for position or point.
Two vectors are equal if and only if they have the same length and direction. Two vectors are added according to the parallelogram rule(平行四边形定则).
vector addition is commutative:
subtraction:
also:
a vector
by the Pythagorean theorem, the length of a is:
$$
\left | \mathbf{a} \right | = \sqrt{{x_a}^2 + {y_a}^2}
$$
By convention we write the coordinates of a either as an ordered pair
The dot product of a and b is denoted a · b and is often called the scalar product because it returns a scalar.
$$
\mathbf{a} . \mathbf{b} = \left | \mathbf{a} \right | \left | \mathbf{b} \right | cos\phi
$$
The dot product is related to length and angle and is one of the most important formulas in graphics.
The most common use of the dot product in graphics programs is to compute the cosine of the angle between two vectors.
The dot product can also be used to find the projection of one vector onto another. This is the length a → b of a vector a that is projected at right angles onto a vector b
$$
\mathbf{a}→\mathbf{b} = \left | \mathbf{a} \right |cos\phi = \frac{\mathbf{a}.\mathbf{b}}{ \left | \mathbf{b} \right |}
$$
The projection of a onto b is a length found by Equation
The dot product obeys the familiar associative and distributive properties（结合律和分配律) we have in real arithmetic: $$ \mathbf{a}.\mathbf{b} = \mathbf{b}.\mathbf{a} \ \mathbf{a} . (\mathbf{b} + \mathbf{c}) = \mathbf{a}.\mathbf{b} + \mathbf{a}.\mathbf{c} \ (k\mathbf{a}).\mathbf{b} = \mathbf{a}.(k\mathbf{b}) = k\mathbf{a}.\mathbf{b} \ $$ If 2D vectors a and b are expressed in Cartesian coordinates, we can take ad- vantage of x · x = y · y = 1 and x · y = 0 to derive that their dot product is $$ \mathbf{a}.\mathbf{b} = (x_a\mathbf{x} + y_a\mathbf{y}) . (x_b\mathbf{x} + y_b\mathbf{y}) \ = x_ax_b(\mathbf{x}.\mathbf{x}) + x_ay_b(\mathbf{x}.\mathbf{y}) + x_by_a(\mathbf{y}.\mathbf{x}) + y_ay_b(\mathbf{y}.\mathbf{y}) \ = x_ax_b + y_ay_b $$ in 3D we can find: $$ \mathbf{a}.\mathbf{b} = x_ax_b + y_ay_b + z_az_b $$
The cross product a × b is usually used only for three-dimensional vectors;
The cross product returns a 3D vector that is perpendicular(垂直的) to the two arguments of the cross product. The length of the resulting vector is related to
it is standard to assume that $$ \mathbf{z} = \mathbf{x} \times \mathbf{y} $$ The cross product has the nice property that: $$ \mathbf{a}\times\mathbf{b} =-( \mathbf{b}\times\mathbf{a} )\ \mathbf{a} \times (\mathbf{b} + \mathbf{c}) = \mathbf{a}\times\mathbf{b} + \mathbf{a}\times\mathbf{c} \ \mathbf{a}\times(k\mathbf{b}) = k(\mathbf{a}\times\mathbf{b}) \ $$ In Cartesian coordinates, we can use an explicit expansion to compute the cross product: $$ \mathbf{a} \times \mathbf{b} = \ (x_a\mathbf{x} + y_a\mathbf{y} + z_a\mathbf{z}) \times (x_b\mathbf{x} + y_b\mathbf{y} + z_b\mathbf{z}) = \ x_ax_b\mathbf{x} \times \mathbf{x} + x_ay_b\mathbf{x} \times \mathbf{y} + x_az_b\mathbf{x} \times \mathbf{z} + \ y_ax_b\mathbf{y} \times \mathbf{x} + y_ay_b\mathbf{y} \times \mathbf{y} + y_az_b\mathbf{y} \times \mathbf{z} + \ z_ax_b\mathbf{z} \times \mathbf{x} + z_ay_b\mathbf{z} \times \mathbf{y} + z_az_b\mathbf{z} \times \mathbf{z} = \ (y_az_b - z_ay_b)\mathbf{x} + (z_ax_b - x_az_b)\mathbf{y} + (x_ay_b - y_ax_b)\mathbf{z} $$ so in coordinate form, $$ \mathbf{a} \times \mathbf{b} = (y_az_b - z_ay_b, z_ax_b - x_az_b, x_ay_b - y_ax_b) $$
Intuitively, a curve(曲线) is a set of points that can be drawn on a piece of paper without lifting the pen. A common way to describe a curve is using an implicit equation. An implicit equation in two dimensions has the form $$ f(x,y) = 0So you evaluate f to decide whether a point is “inside” a curve. $$ So you evaluate f to decide whether a point is “inside” a curve.
The gradient（梯度） vector ∇ f(x, y) is given by： $$ ∇f(x,y) = (\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}) $$ The gradient vector evaluated at a point on the implicit curve f(x, y) = 0 is perpendicular(垂直) to the tangent (切线) vector of the curve at that point. This perpendicular vector is usually called the normal vector to the curve.
a more general form is often useful:
$$ Ax + By + C = 0, $$ for real numbers A, B, C.
the distance from a point to the line is the length of the vector
If f is instead a quadratic function of x and y, with the general form $$ Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0 $$ Two-dimensional quadric curves include ellipses(椭圆) and hyperbolas(双曲线), as well as the special cases of parabolas(抛物线), circles(圆), and lines.
Examples of quadric curves include the circle with center
implicit equations implicitly define a set of points that are on the surface: $$ f(x,y,z) = 0 $$ Any point (x, y, z) that is on the surface results in zero when given as an argument to f. Any point not on the surface results in some number other than zero.
You can check whether a point is on the surface by evaluating f, or you can check which side of the surface the point lies on by looking at the sign of f, but you cannot always explicitly construct points on the surface.
Using vector notation, we will write such functions of
A surface normal is a vector perpendicular to the surface. Each point on the surface may have a different normal vector.
the surface normal at a point p on an implicit surface is given by the gradient of the implicit function: $$ \mathbf{n} = ∇f(\mathbf{p}) = (\frac{\partial f(\mathbf{p})}{\partial x},\frac{\partial f(\mathbf{p})}{\partial y}, \frac{\partial f(\mathbf{p})}{\partial z} ) $$ The reasoning is the same as for the 2D case: the gradient points in the direction of fastest increase in f, which is perpendicular to all directions tangent to the surface,in which f remains constant.
The gradient vector points toward the side of the surface where f(p) > 0,
As an example, consider the infinite plane through point a with surface normal n. The implicit equation to describe this plane is given by
$$
(\mathbf{p} - \mathbf{a}).\mathbf{n} = 0
$$
Any of the points p shown are in the plane with normal vector n that includes point a if Equation (2.2) is satisfied.
Sometimes we want the implicit equation for a plane through points a, b, and c. The normal to this plane can be found by taking the cross product of any two vectors in the plane.
One such cross product is $$ \mathbf{n} = (\mathbf{b} - \mathbf{a}) \times (\mathbf{c} - \mathbf{a}) $$ This allows us to write the implicit plane equation: $$ (\mathbf{p} - \mathbf{a}).((\mathbf{b} - \mathbf{a}) \times (\mathbf{c} - \mathbf{a})) = 0 $$ A geometric way to read this equation is that the volume of the parallelepiped defined by p − a, b − a, and c − a is zero, i.e., they are coplanar(共平面的).
The full-blown Cartesian representation for this is given by the determinant: $$ \begin{vmatrix} x-x_a & y-y_a & z-z_a\ x_b-x_a & y_b-y_a & z_b-z_a\ x_c-x_a & y_c-y_a & z_c-z_a\ \end{vmatrix} = 0 $$
quadratic polynomials in x, y, and z define quadric surfaces in 3D. For instance, a sphere can be written as $$ f(\mathbf{p}) = (\mathbf{p} - \mathbf{c})^2 - r^2 = 0 $$ and an axis-aligned ellipsoid may be written as $$ f(\mathbf{p}) = \frac{(x - x_c)^2}{a^2} + \frac{(y - y_c)^2}{b^2} + \frac{(z - z_c)^2}{c^2} - 1 = 0 $$
A 3D curve can be constructed from the intersection of two simultaneous implicit equations: $$ f(\mathbf{p}) = 0 \ g(\mathbf{p}) = 0 $$
A parametric curve is controlled by a single parameter that can be considered a sort of index that moves continuously along the curve. Such curves have the form、 $$ \begin{bmatrix} x \ y \end{bmatrix} = \begin{bmatrix} g(t) \ h(t) \end{bmatrix} $$ Often we can write a parametric curve in vector form, $$ \mathbf{p} = f(t) $$ where f is a vector-valued function.
A parametric line in 2D that passes through points
A circle with center$(x_c, y_c)$ and radius r has a parametric form: $$ \begin{bmatrix} x \ y \end{bmatrix} = \begin{bmatrix} x_c + rcos\phi \ y_c + rsin\phi \end{bmatrix} $$ An axis-aligned ellipse can be constructed by scaling the x and y parametric equations separately: $$ \begin{bmatrix} x \ y \end{bmatrix} = \begin{bmatrix} x_c + acos\phi \ y_c + bsin\phi \end{bmatrix} $$
A 3D parametric curve operates much like a 2D parametric curve: $$ x = f(t),\ y = g(t),\ z = h(t),\ $$ In vector form we can write $$ \begin{bmatrix} x \ y \ z \end{bmatrix} = \mathbf{p}(t) $$
write it in vector form: $$ \mathbf{p} = \mathbf{o} + t\mathbf{d} $$ The way to visualize this is to imagine that the line passes through o and is parallel to d.
These surfaces have the form $$ x = f(u,v), \ x = g(u,v), \ z = h(u,v), \ $$ or, in vector form, $$ \begin{bmatrix} x \ y \ z \end{bmatrix} = \mathbf{p}(u,v) $$ spherical coordinates:
Consider the function $ \mathbf{q}(t) = \mathbf{p}(t, v_0 )$.This function defines a parametric curve obtained by varying u while holding v fixed at the value
The derivative of q gives a vector tangent to the curve, and since the curve lies in the surface the vector
since both are tangent to the surface, their cross product, which is perpendicular to both tangents, is normal to the surface. $$ \mathbf{n} = \mathbf{p}_u \times \mathbf{p}_v $$
an example of linear interpolation of position to form line segments in 2D and 3D, where two points a and b are associated with a parameter t to form the line
Another common linear interpolation: we have an associated height, y i . We want to create a continuous function y = f(x) , The parameter t is just the fractional distance between$ x_i$ and
Triangles in both 2D and 3D are the fundamental modeling primitive in many graphics programs.
Often information such as color is tagged onto triangle vertices, and this information is interpolated across the triangle.
The coordinate system that makes such interpolation straightforward is called barycentric coordinates(重心坐标系)
If we have a 2D triangle defined by 2D points a, b, and c, we can first find its area: $$ area = \frac{1}{2} \begin{vmatrix} x_b - x_a & x_c - x_a \ y_b - y_a & y_c - y_a \ \end{vmatrix} \ = \frac{1}{2}(x_ay_b + x_by_c + x_cy_a - x_ay_c - x_by_a - x_cy_b) $$ This area will have a positive sign if the points a, b, and c are in counterclockwise order and a negative sign, otherwise.
There are a variety of ways to assign a property, such as color, at each triangle vertex and smoothly interpolate the value of that property across the triangle, the simplest is to use barycentric coordinates(重心坐标系).
the coordinate origin is a and the vectors from a to b and c are the basis vectors. With that origin and those basis vectors, any point p can be written as $$ \mathbf{p} = \mathbf{a} + \beta(\mathbf{b} - \mathbf{a}) + \gamma(\mathbf{c} - \mathbf{a}) $$ to get: $$ \mathbf{p} = (1 - \beta - \gamma)\mathbf{a} + \beta\mathbf{b} + \gamma\mathbf{c} $$ which yields the equation $$ \mathbf{p}(\alpha, \beta, \gamma) = \alpha\mathbf{a} + \beta\mathbf{b} + \gamma\mathbf{c} $$ with the constraint that $$ \alpha + \beta + \gamma = 1 $$
- A particularly nice feature of barycentric coordinates is that a point p is inside the triangle formed by a, b, and c if and only if
- If one of the coordinates is zero and the other two are between zero and one, then you are on an edge.
- If two of the coordinates are zero, then the other is one, and you are at a vertex.
Another nice property of barycentric coordinates is that
So if the line$ f_{ac} (x, y) = 0$ goes through both a and c, then we can compute β for a point (x, y) as follows: $$ \beta = \frac{f_{ac}(x,y)}{f_{ac}(x_b, y_b)} $$ and we can compute γ and α in a similar fashion. $$ f_{ab}(x,y) \equiv (y_a - y_b)x + (x_b - x_a)y + x_ay_b - x_by_a = 0
$$ 代入即可得到$\alpha, \beta, \gamma$的值
Another way to compute barycentric coordinates is to compute the areas
Barycentric coordinates obey the rule
$$
\alpha = A_a / A, \
\beta = A_b / A, \
\gamma = A_c / A, \
$$
where A is the area of the triangle,
One wonderful thing about barycentric coordinates is that they extend almost transparently to 3D. If we assume the points a, b, and c are 3D, then we can still use the representation $$ \mathbf{p} = (1 - \beta - \gamma)\mathbf{a} + \beta\mathbf{b} + \gamma\mathbf{c} $$ The normal vector to a triangle can be found by taking the cross product of any two vectors in the plane of the triangle. It is easiest to use two of the three edges as these vectors, for example, $$ \mathbf{n} = (\mathbf{b} - \mathbf{a}) \times (\mathbf{c} - \mathbf{a}) $$ The area of the triangle can be found by taking the length of the cross product: $$ area = \frac{1}{2} \left | (\mathbf{b} - \mathbf{a}) \times (\mathbf{c} - \mathbf{a}) \right | $$ suggest the formulas: $$ \alpha = \frac{\mathbf{n}.\mathbf{n}_a}{\left | \mathbf{n}^2 \right |} \ \beta = \frac{\mathbf{n}.\mathbf{n}_b}{\left | \mathbf{n}^2 \right |} \ \gamma = \frac{\mathbf{n}.\mathbf{n}_c}{\left | \mathbf{n}^2 \right |} \ $$ where: $$ \mathbf{n}_a = (\mathbf{c} - \mathbf{b}) \times (\mathbf{p} - \mathbf{b}) \ \mathbf{n}_b = (\mathbf{a} - \mathbf{c}) \times (\mathbf{p} - \mathbf{c}) \ \mathbf{n}_c = (\mathbf{b} - \mathbf{a}) \times (\mathbf{p} - \mathbf{a}) \ $$