Codex @codex  0

An affine combination is a weighted point $P={\sum }_{i=0}^k{\lambda }_i{P}_i$ with ${\sum }_{i=0}^k{\lambda }_i=1$. The coefficients can be negative. The sum is meaningful in an affine space without selecting a preferred origin: relative to any point $O$, it means $P=O+{\sum }_i{\lambda }_i(P_i-O)$. Affine maps preserve these weighted points. ^[ref]

For an original example in the Euclidean plane, take $A=(0,0)$, $B=(2,0)$ and $C=(0,2)$. Then $P=\frac{1}{4}A+\frac{1}{2}B+\frac{1}{4}C=(1,\frac{1}{2})$ lies inside their triangle. Its coefficients are nonnegative, making it a convex combination. In contrast, $R=-\frac{1}{2}A+\frac{3}{2}B=(3,0)$ is beyond $B$: it is an affine combination but not a convex combination of these three points. Negative weights allow extrapolation, rather than only interpolation inside the convex hull.

For the affine map $T(x,y)=(2x+y+1,y-2)$, the same weights produce $T(P)=(\frac{7}{2},-\frac{3}{2})$ from $T(A),T(B),T(C)$, and produce $T(R)=(7,-2)$ from $T(A),T(B)$. Thus one can compute a weighted point before transforming the vertices or afterwards, obtaining exactly the same point.

This is useful whenever an object is built from weighted control points: its construction commutes with an affine map, so transforming the control points transforms the constructed object as well. The sum of the weights is the essential condition; no assumption about lengths or angles is needed.

An affine map combines a linear map with a translation. After choosing coordinate origins, it has the form $T:{\mathbb{R}}^p\to {\mathbb{R}}^q$, $T(x)=Mx+b$, where the $q\times p$ matrix $M$ is its linear part of an affine map and $b$ is a fixed vector. Equivalently, it preserves every affine combination. ^[ref]

An original planar example isIt stretches horizontally, applies a shear mapping, and then translates. The vertices $A=(0,0)$, $B=(2,0)$ and $C=(0,2)$ become $A^{\prime }=(1,-2)$, $B^{\prime }=(5,-2)$ and $C^{\prime }=(3,0)$. A square grid becomes a grid of parallelograms; straight edges and their weighted points still fit together.

This map is not a linear map with respect to the displayed origins, because $T(0,0)=(1,-2)$ rather than zero. It also is not a Euclidean similarity: $AB$ has length $2$ and $A^{\prime }{B}^{\prime }$ has length $4$, while the right angle at $A$ becomes $45^{\circ }$. Its determinant is $2$, so it doubles planar area. These metric changes explain why affine geometry studies straightness, incidence and weighted points rather than treating every transformation as an isometry.

The linear part is an invertible matrix, giving the concise inverse $T^{-1}(u,v)=((u-v-3)/2,v+2)$. More generally an affine map is bijective exactly when its linear part of an affine map is bijective. Between equal finite-dimensional spaces, this means an invertible matrix; the translation cannot repair a loss of rank of a linear map.

The astronomical horizon is the great circle where the horizontal plane through the observer intersects the celestial sphere. It is perpendicular to the local vertical, so every direction on the astronomical horizon has altitude $0^{\circ }$ and zenith distance $90^{\circ }$. In the local horizontal coordinate system, it consists of the unit vectors satisfying $\mathbf{n}\cdot {\mathbf{e}}_U=0$. This geometric reference remains defined when buildings or mountains hide it; a visible skyline is not a calibration of zero altitude.

Three boundaries must be distinguished. The astronomical horizon fixes the angular reference. Terrain supplies a site-dependent boundary with apparent altitude $t(\eta )$ that changes with azimuth. Atmospheric refraction changes the apparent direction of incoming light, and can also change the apparent distant boundary. A point source clears an opaque terrain boundary only if ${\gamma }_{\mathrm{app}}>t(\eta )$ when both are expressed in the same apparent horizontal coordinate system; this clearance condition does not guarantee visibility through clouds or bright sky. In the synthetic diagram, a source with geometric altitude $3^{\circ }$ and an imposed illustrative upward correction $R=0.2^{\circ }$ remains behind a $4^{\circ }$ boundary. The chosen $R$ is an algebraic illustration, not a fitted atmospheric prediction.

Even an unobstructed sea boundary need not coincide with the astronomical horizon. On a spherical Earth of radius $r$, an observer at height $q$ sees the unrefracted tangent to the surface at a downward angle $D$ satisfying $\cos D=r/(r+q)$. The right-angle tangent triangle supplies this relation; the visible boundary then has altitude $-D$. With illustrative $r=6371\mathrm{km}$ and $q=10\mathrm{m}$, $D\simeq 0.1015^{\circ }$, before atmospheric refraction. Raising the eye therefore changes the visible boundary without redefining the horizontal plane.

The US Naval Observatory's rise/set definitions explicitly distinguish the geometric reference from apparent contact with a visible disk: the standard Sun calculation puts its centre $50^{\prime }$ below the horizontal plane, combining an average $16^{\prime }$ radius with $34^{\prime }$ of atmospheric refraction. An angular minute is $1^{\prime }=1/60^{\circ }$ in the minute and second of arc convention. This is an operational average under stated conditions, rather than a universal zero of altitude. The same source identifies weather, local topography and observer height as limits on predicted rise/set time. An observation log should therefore retain the level reference, site and eye height, terrain profile, observation time, and whether each reported altitude is geometric or apparent. ^[ref].

Average power divides the work over a nonzero interval by the elapsed time. For the same $2\mathrm{kg}$ point mass driven from rest through $4\mathrm{m}$ by a net $3\mathrm{N}$ force, the displacement is $x(t)=a{t}^2/2$. ThusThis average power is half the final instantaneous power because $P(t)$ is a linear function of time, starting at zero. Its graph of a function bounds a triangle of area $P(t_f)t_f/2=12\mathrm{J}$; the horizontal line at average power encloses a rectangle with the same area. In general, an endpoint value of instantaneous power does not determine average power. ^[ref]

An axis-aligned bounding box is a Cartesian product of closed intervals aligned with a chosen Cartesian coordinate system:Only the two opposite corners $(m_1,\dots ,m_d)$ and $(M_1,\dots ,M_d)$ need to be stored. In two dimensions it is a rectangle. Changing the Cartesian coordinate system can change how tightly the axis-aligned bounding box fits the same object.^[ref]

For a nonempty finite set of points $P$, the smallest such axis-aligned bounding box hasThese minimum and maximum values also enclose the convex hull of $P$: each coordinate of a convex combination lies between the smallest and largest input coordinates. In particular, three vertices of a polygon determine the axis-aligned bounding box of a filled triangle. For a general nonempty compact set, the same formulas use all its points; inspecting only an arbitrary finite sample is insufficient.

Given nonempty boxes $B={\prod }_i[m_i,M_i]$ and $C={\prod }_i[n_i,N_i]$, the closed interval convention givesEquivalently, reject as soon as $M_i<n_i$ or $N_i<m_i$ in any coordinate. The surviving set intersection has endpoints $\max (m_i,n_i)$ and $\min (M_i,N_i)$. This is an exact box test, while its conclusion about enclosed objects remains a conservative enclosure test.^[ref]

Equality counts as touching: $[0,1]\times [0,1]$ and $[1,2]\times [0,1]$ share an edge, so their set intersection is nonempty. Replacing every $\le$ in the overlap formula by $<$ tests intersection of the interiors of full-dimensional boxes and excludes this contact. A zero-width closed interval can represent a point or a lower-dimensional enclosure, and must not automatically be discarded when touching counts. A box with some $m_i>M_i$ is empty; test emptiness before applying rules intended for nonempty boxes. The broad phase must retain touching candidates whenever the narrow phase counts touching as a hit.

For nonempty boxes, the smallest enclosing axis-aligned bounding box of their set union isThis operation supplies the parent boxes in a bounding volume hierarchy. It is an enclosure of the set union, rather than generally the set union itself. In floating-point arithmetic, comparisons are conclusive only for endpoints that still enclose the intended exact objects; use certified bounds such as interval arithmetic with outward rounding when that guarantee is required.

// Target: analysis.bigb

In bent molecular geometry, two neighbours and their central chemical atom form a noncollinear triangle, with an interior bond angle strictly between $0^{\circ }$ and $180^{\circ }$. Water, ${\mathrm{H}}_2\mathrm{O}$, is the standard example. The NIST CCCBDB experimental water entry, internal coordinates, citing Hoy and Bunker (1979) gives the equilibrium H–O–H bond angle as $104.4776^{\circ }$ with quoted uncertainty $0.0019^{\circ }$, usually rounded to $104.5^{\circ }$.

The oxygen in water has two bonding electron domains and two lone pairs. Its ideal electron pair geometry is tetrahedral, while the three atomic nuclei give bent molecular geometry. Three points always lie in a plane; this does not make the four-domain scaffold planar. Nor does the word bent by itself determine an electron domain count: the same atomic shape can arise from other domain arrangements.

In an illustrative model with two equal bond-dipole contributions of magnitude $p$ separated by bond angle $\theta$, their sum of vectors points along the bisector and has magnitude $2p\cos (\theta /2)$. At $180^{\circ }$ it vanishes; at $104.4776^{\circ }$ it is approximately $1.2247p$. This ratio is a geometric calculation, not a prediction of water's measured electric dipole moment. The entire electron density, including contributions represented by lone pairs, matters. Water is a polar molecule; chemical polarity helps explain its interactions with other polar molecules. See McMurry, Organic Chemistry, §2.2: water, lone-pair contributions, and polarity.

A bond angle is the angle between two chemical bonds sharing a central chemical atom. In H–O–H, the middle symbol identifies the vertex: the angle is at oxygen. Let $u$ and $v$ be the nonzero displacement vectors from that atomic nucleus to the two neighbours. The actual three-dimensional bond angle satisfiesusing the dot product and Euclidean norm. Rotating the whole molecule preserves this expression; projecting it onto a page generally does not.

There is another useful way to find a bond angle. If the two bond lengths are $a,b$ and the distance between the outer atomic nuclei is $d$, the law of cosines givesFor equal bond lengths $a=b$, this becomes $d=2b\sin (\theta /2)$. Thus a measured outer-atom separation contains information about a bond angle even though those outer chemical atoms are not directly joined by a chemical bond.

For a stipulated rigid, equal-bond length model using water's $104.4776^{\circ }$ equilibrium bond angle, $d/b\approx 1.5811$, whereas a straight H–O–H arrangement would give $d/b=2$. These are calculated geometric ratios, not new measured distances. Gas electron diffraction illustrates how pair-distance information can constrain molecular geometry. Substituting independently averaged distances into the rigid triangle formula does not automatically recover an equilibrium bond angle, because averaging and this nonlinear calculation need not commute.

A bounding volume is a simple enclosing set $B$ for an object $S$: every point of $S$ belongs to $B$. Its purpose is to make a cheap rejection possible before examining the object itself. For example, an axis-aligned bounding box can enclose a triangle using just its smallest and largest coordinates.

The guarantee is one-sided. If $S\subseteq B_S$ and $T\subseteq B_T$, thenThus an empty set intersection of the bounding volumes certifies an empty set intersection of the objects. An overlap of the bounding volumes is necessary for actual intersection, but does not prove it. This is a conservative enclosure test: rejection is conclusive, acceptance means further examination is needed.

Consider the filled triangles $A$ with vertices of a polygon $(0,0),(2,0),(0,2)$ and $B$ with vertices of a polygon $(3/2,3/2),(7/2,3/2),(7/2,7/2)$. Their axis-aligned bounding boxes overlap in $[3/2,2]\times [3/2,2]$. Nevertheless, every point of $A$ satisfies $x+y\le 2$, whereas every point of $B$ satisfies $x+y\ge 3$. The two triangles are disjoint. The shaded square in the bounding volume hierarchy illustration makes this false candidate visible.

For collision detection, a broad phase produces possible object pairs from bounding volume tests; a narrow phase examines the actual shapes, for example with a triangle-triangle intersection algorithm. Scott Le Grand's Chapter 32. Broad-Phase Collision Detection with CUDA explains this separation in its opening paragraphs. Whether touching counts as a collision must be specified by the application.

An enclosure must cover the entire object being queried. A box fitted to samples of a parametric curve only guarantees enclosure of the sampled points, unless a separate bound covers the whole parametric curve. For a Bézier curve, the convex hull of its control points provides such a bound on its defining parameter interval. In floating-point arithmetic, outward rounding can preserve certified coordinate bounds; an inward roundoff error can invalidate a rejection. Bounds must also be updated when objects move.

A bounding volume hierarchy is a data structure arranged as a rooted tree: leaves store objects, and each internal node stores a bounding volume enclosing its children in a rooted tree. Consequently, every node encloses all objects in its subtree. A common choice is a binary tree of axis-aligned bounding boxes. The objects assigned to sibling subtrees can be different even when their bounding volumes overlap.^[ref]

Here is an exact two-dimensional example. Use the triangles $A,B$ from bounding volume, together with $C$ having vertices of a polygon $(6,0),(8,0),(6,2)$ and $D$ having vertices of a polygon $(6,3),(8,3),(8,5)$. Group $A,B$ under $L$, and $C,D$ under $R$. Their parent axis-aligned bounding boxes and the box at the root of a tree areThe diagram shows all seven nodes. Taking the coordinate-wise minimum and maximum of the two child boxes constructs each parent box; the resulting box contains the set union of the children, and can also contain empty space between them.

A bounding volume hierarchy traversal starts at the root of a tree and discards a subtree when its bounding volume misses the query. Accepted internal nodes expose their children; accepted leaves supply candidates for actual shape tests. Box2D's official Collision: Dynamic Tree describes axis-aligned bounding box region queries that return overlapping leaf boxes. A query box that bounds a different query shape can produce additional false candidates.

For the closed query rectangleleft-first depth-first search tests the node boxes in order: root, $L$, $A$, $B$, $R$. The first four overlap $Q$; $R$ does not, because its smallest $x$-coordinate is $6>9/5$. Pruning $R$ skips both $C$ and $D$. Five box tests leave two broad phase candidates. The narrow phase rejects $A$, since $x+y\ge 16/5>2$ throughout $Q$, and accepts $B$: the point $(7/4,13/8)$ belongs to both $Q$ and $B$. There are two detailed shape tests and one true hit.

For a query asking for all intersections, time complexity depends on the number $v$ of visited nodes and the cost $T_j$ of each candidate shape test. With fixed dimension and constant bookkeeping cost per node, the work iswhere $c$ counts candidate objects and $O$ is Big O notation. A balanced binary tree has small height of a binary tree, but a query can enter both children at every level. In this example a query containing $[0,8]\times [0,5]$ visits all seven nodes and tests all four triangles. More generally, a full binary tree with $n$ single-object leaves has $2n-1$ nodes, so such a query takes $O(n)$ work for constant-cost shape tests even when its height of a binary tree is $O(\log n)$. Listing $n$ hits itself requires $n$ outputs. Small height alone gives no unconditional logarithmic query bound.

For collision detection between two bounding volume hierarchies, traverse pairs of nodes: reject disjoint pairs of bounding volumes, expand a nonleaf member of an accepted pair, and apply the narrow phase at accepted leaf pairs. A single collection's self-query must exclude pairing an object with itself and avoid duplicate unordered pairs. If every object overlaps every other object, reporting all pairs already requires $n(n-1)/2$ results. Construction quality, empty space in boxes and the queries determine how much pruning is useful.^[ref]

For a short correctness outline, an object is enclosed by its leaf box, and each child box is enclosed by its parent box. Repeated subset relations carry this property to every ancestor. A true query intersection therefore prevents rejection at every ancestor of the object's leaf. Updating leaf boxes after motion also requires restoring all affected ancestor bounds before querying.

The circumcircle of a nondegenerate triangle is the unique circle through its three vertices. Its center is the circumcenter, found by intersecting two perpendicular bisectors; equality of distances to all three vertices proves existence, and subtracting the squared-distance equations proves uniqueness. For a right triangle, the midpoint of its hypotenuse is equidistant from all vertices, so its circumradius is half the hypotenuse length.

Conservation of angular momentum says that total angular momentum stays fixed when the net external torque about the chosen fixed inertial origin vanishes. Individual angular momenta and angular velocities can change. The radius change at constant angular momentum below makes this distinction visible: pulling two rotating point masses inward lowers the moment of inertia, raises the angular velocity, and supplies work that increases the kinetic energy.

Follow the same constant-mass point masses throughout, in an inertial frame, with all positions measured from a fixed origin $O$. If the internal pair forces are opposite and central, meaning directed along each pair's separation, thenThus zero net external torque throughout an interval implies constant total angular momentum throughout that interval. A vanishing external torque component, such as ${\tau }_z^{\mathrm{ext}}=0$, conserves the corresponding $L_z$ even when the other components change. This is the balance law for a fixed collection; angular-momentum flux must also be accounted for when matter crosses its boundary. ^[ref]

The central qualification is essential. Write ${\mathbf{F}}_{ij}$ for the internal force on member $i$ from member $j$. Under Newton's third law, ${\mathbf{F}}_{ji}=-{\mathbf{F}}_{ij}$, but the sum of their torques isThis vanishes when ${\mathbf{F}}_{ij}$ is parallel to ${\mathbf{r}}_i-{\mathbf{r}}_j$. Equal-and-opposite forces alone do not suffice: two opposite transverse forces with separated lines of action form a couple with nonzero torque. Cancellation of the internal pair torques, together with Newton's second law, proves the total balance above.

For one point mass under a central force about $O$, $\mathbf{F}=f(r)\hat{\mathbf{r}}$ gives ${\mathbf{\tau }}_O=\mathbf{r}\times \mathbf{F}=0$. The force may still change the speed and kinetic energy; its zero torque preserves angular momentum. With nonzero ${\mathbf{L}}_O$, $\mathbf{r}\cdot {\mathbf{L}}_O=0$ keeps the motion in a fixed plane. Choosing the $z$ axis perpendicular to that plane, $r^2\dot{\theta }=L_z/m$ governs its angular progress. This removes an unknown from the central-force radial turning point calculation and helps reduce the Kepler orbit to a radial problem.

There is a historical geometric version of this result. In Book I, Section II, Proposition I of the Principia, Newton proves that motion under a central force toward a fixed centre stays in a plane and sweeps equal areas in equal times, using successive central impulses and equal triangle areas. ^[ref] In the modern notation above, constant areal velocity is $\dot{A}=L_z/(2m)$: the old geometric proof and the angular momentum calculation express the same invariant.

A convex chain between two fixed points inside a triangle is a finite point set whose union with the endpoints forms the vertices of a convex polygon, with the endpoint segment as one boundary edge. In the triangle $0\le y\le x\le 1$, with endpoints $(0,0)$ and $(1,1)$, its ordered lower boundary has strictly increasing segment slopes. Every subset of its intermediate vertices remains a convex chain. This hereditary property lets selected point coordinates certify the existence of a long chain.

Let $m$ be an integer median for the longest convex chain length of independent uniform points in a triangle. If $L_n(x)\ge a$, fix a certificate of $a$ coordinates. Every configuration with $L_n\le b$ must disagree with at least $a-b$ of those coordinates; equal weights $1/\sqrt{a}$ prove Talagrand convex distance separation at least $(a-b)/\sqrt{a}$. Talagrand's convex distance inequality then gives $\mathbb{P}(L_n\ge a)\mathbb{P}(L_n\le b)\le e^{-(a-b{)}^2/(4a)}$. Set $b=m$, then $a=m$, in the two respective applications to obtain the displayed tails for positive integer $t$. The exact chain-counting bound gives $m\le C{n}^{1/3}$. Hence any interval centred at $m$ with length $\omega (n)n^{1/6}$, where $\omega (n)\to \infty$, contains $L_n$ with high probability. No upper restriction on the growth of $\omega$ is required.

Every allocation $(x,y)$ in the convex feasible region for a resource budget is a convex combination of its three extreme points:The condition $x+2y\le 6$ is exactly ${\lambda }_1\ge 0$; nonnegative $x,y$ give the other two nonnegative coefficients. Thus the half-space description and the convex hull description encode the same triangle. At $(2,1)$ all three coefficients are $1/3$, and at $(3,9/8)$ they are $1/8$, $1/2$, $3/8$. The green and orange points in the budget diagram therefore have two complementary interpretations: direct resource constraints and mixtures of extreme choices.

The three extreme points alone are not a convex set, because their midpoints are absent. Their convex hull fills the whole triangle. More generally, Caratheodory theorem bounds how many generating points are needed for one convex combination in a finite-dimensional real vector space; its geometric content is that a complicated collection can describe each attainable average with a short representation.