Wednesday, October 13, 2010

Wedding Decor Branches

The conical

When a right circular cone is cut by a plane that passes through the vertex can be generated the following curves:
Ellipse (e): the cutting plane is not parallel to any generating and cuts all the generators of the cone. Is a closed curve without points at infinity.
Parable (p): the cutting plane is parallel to a generator and has a point at infinity.
Hyperbola (h): the cutting plane is parallel to 2 generative and has two points at infinity.
Circumference: the plane is orthogonal to the axis of revolution and is a particular case of the ellipse in which the two axes are equal.





















Generation
projective conics.

As homology across all sets of points correspond to other series and straight beams correspond to other beam lines, a conic becomes another conical homology.
In the figure we see several properties of homologies on conical points aligned counterparts SS'están the center of projection O, the homologous lines aa 'cut into the shaft, the axis points are double or counterparts of themselves , which means that if a figure cut to stop homology in the points, his counterpart goes through these two points, resulting in these unchanged.
The tangents to a conic are also to its counterpart. In two peer reviews on a pole and polar have their counterparts in other pole and polar of the other.
Any conic can be transformed into a circle, so we get points on the circumference counterparts thus obtaining new points of the conic.


Given two sets of points on a conic SAB EDC, if we join each of them with the other set, except that faces it, we at the intersection of all three lines MNO points are aligned.
As we have seen this procedure used to obtain new points of the conic, defined by five points to get the sixth, seventh, and so on.
As in geometry points and lines are correlative terms, can be exchanged for each other, so a conic defined by five points can be transformed into another set of five straight. Any of these five elements can be transformed with the result that we can build any cone with five elements between points and lines.


abcd Given four lines and a point M incident in one of them d determine the conic through the point is tangent to given lines. Make a circle with a diameter either (in yellow) tangent to the incident point on the line d M is going to be the focus. Taking one of the lines d as the axis of homology, the counterpart of the first line c is the tangent to the circle from the intersection of the line d axis c. The intersection of each line c and its counterpart c 'with the other line by its counterpart b' adjacent respectively, determine a new line that is part of a radiation (green lines) in which vertex is the center of homology.
If we get new points of the ellipse, we take any two homologous points, eg RR '. Using any straight line passing through R 'we see that intersects the circle at P', we need to align the center of homology with P 'until it intersects the axis of the point V, and V with R we align the intersection with OP "a point of the ellipse P.


Another example similar to above but with the outer edge of the cone. We have several points on the conic and a tangent ABCE m on a point of it A.
Making a yellow circle tangent to the given line m and passing through point A, and aligning the center of homology to the points of the ellipse ECB, we at the intersection with the circumference of their counterparts B'C'E '. Using any straight line passing through 2 points of the circle C 'D' intersects the axis at a point L, which together with the known homologous point of the ellipse C have at the intersection of this line CL with the double line AD 'a new point on the ellipse D.



Correlate the previous year, as data have four points and one incident on a tangent to the conic. Trying to determine the conic that passes through the points BCDE and tangent to the line to point B. If we take the point B as the center of homology and make a circle (in yellow) to the line tangent to this point B, we build double lines (lines through the center of homology B and given points) passing through points CDE. These lines intersect the circle at points C'D'E ', respectively. EC counterparts
C'E'se Lines intersect at a point of Y axis, another pair of lines CD C'D 'intersect at another point on the axis S, two points determine YS shaft position.
To construct the tangents to the conic from any point T, we f tangent from that point on the circumference. Align the point of tangency S 'with the center of homology B until it intersects the axis at a point. We calculate the counterpart of this item and get the point of tangency S with the tangent cone and tangent homologous fu f 'to the circle.


In previous cases, the elements, fixed points or are tangent lines were always incidents (where there was a tangent, there was a point of the cone that was about it) and therefore there was a single cone passing for points and was tangent to the lines. In the event that the items are no incidents, the solution is not just a single cone but can be, for example, if we have three points and two tangents to the conic, as is the statement of this drawing, exercise has four solutions. In the exercise free
But one solution appears resolved, one in which the yellow circle is tangent to two given lines and passing through two points of the conic.
If you align the center point of homology E intersects the circle O at point E '. At this point you align with a point T 'of the circle and intersects the axis at a point determined together with E T at the intersection of the line OT. " T is a conical point and if we join his counterpart T 'to the point of tangency G' we have an intersection point on the axis have joined the T point in prolonging the G-spot at the intersection with the tangent b.

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The parable

The parabola is the locus points which are equidistant from the other, called the focus F and a line called the directrix d. Similarly is the locus of the centers of tangent circles incidents adequate focus.
Throughout the vertex V parabola equidistant from focus and guideline: dV = VF, remains the guideline a line perpendicular to the axis of the curve, which is its axis of symmetry.
To trace perpendicular to the axis are different (eg straight), take the distance perpendicular to the guidance given, and that radio is a circle with center at F and da radio. The 2 points of intersection are given and symmetrical points of the curve.











We can see an example with numbers: for any point, 5 for example, make a perpendicular to the axis. 5H with distance (from point 5 to the guideline D) center in F do and where to cut perpendicular through the point 5 are the 2 points of the parabola.












2 lines that are cut are listed in order of increasing and decreasing in the same number of points. They join the following points: 1 with 1, 2 to 2, etc., The resulting curve is the envelope of the parabola tangent to the 2 original lines.













2 are divided horizontal and vertical lines in the same number of points as in Fig.
points on the horizontal and vertical are plotted on the horizontal points are straight to the top of the parabola. The intersection of the 2 radiation are the points of the parabola.












2 intersecting straight fall into the same number of points in the order of the figure, for example from 0 to 9. Point 0 of a line joins with the other line, and point 9 of the other with all points on the other line. The intersection of the 2 radiation are the points of a parabola.











equidistant lines and circles in the intersection of each line in each circle are the points of parabolic curves.























If two axes ab P whose intersection is divided into equal segments (1, 2, 3 , 4 ,...) from P and a random point M a (to the right of a, like in the picture) we circumferences passing through 1, 2, 3, 4, and M get circles that cut the line b in D, V, etc. At the intersection of these circles and b we cut horizontal to vertical in 1 2 3 4 points of the parabola.













All parabolas have the same shape in the drawing parabolas look the same but at different distances. If we take one of them and always comes out just climbed to the previous form and they are all proportional. Are therefore the parables in this curve equal to the circumference, They are also always the same way although they may have different sizes.















Given the focus of the parabola and two points of the AB, to determine the axis, the vertex and directrix of the parabola. We can
arc of 90 ° for AF, that is, make a circle whose diameter is the focus F and a point A. We
another circle passing through the focus F and the other point B.
construct a tangent line, in red, the two circles and we focus perpendicular to the tangent obtained in this way at their intersection the vertex V of the parabola. The line passing through the vertex and the focus is the core of it, to get the guidance we center at the vertex and taking the distance as the radius of the vertex V to the focus F, we make an arc intersecting the axis at this distance VF . At that intersection make a perpendicular to the axis-green-and that is the directrix of the parabola. The resolution of the exercise is based on the whole tangent to the parabola intersects the red line that passes through the vertex, a point which, together with the focus, we have that is perpendicular to the tangent.


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The hyperbola

The hyperbola is the locus of points whose difference distances to other 2 fixed (called foci F1, F2) is constant and equal to the long axis V1, V2.
In the picture you can see that the distance of a point F2 (blue) minus the distance of F1 at that point (in green) is equal to V1-V2.
points are taken to build (from F1 to infinity and infinity to F2), for example from A to V1 and made her a bow with center F1. Then you take the distance from A to V2 and made an arc centered at F2. The intersection of the 2 arches is a point on the hyperbola.
















can draw a branch by hyperbolic cone section of a dihedral system: the vertical plane cutting the cone with vertex V determines the branch of hyperbola. To calculate m we a vertical plane through the center of the cone, it cuts the circumference of the cone base in P1, point P2 project to the sum obtained on the ground line. The plane cut at T m the plane to at this point until we make a vertical cut in G. P2-V The other points are determined the same way. To determine the vertex E is a circle with center O is tangent to the plane, it intersects the ground line in Z. By Z is a vertical until it intersects the generatrix of the cone shape of H and by this point a short horizontal VO in E.














The intersection of a set of equidistant concentric circles are hyperbolic curves.












Given any 2 vertices of a hyperbola AB, construct a square with vertex at A and divide the sides into equal parts as provided in the figure. From B draw a radiation passing through points 1, 2, 3. For points across the square to make other radiation A.
The intersection of two radiations are the points of the curve.















The points to take to build the hyperbola should be, as is the Ellipse, between the two foci. The difference is that in the ellipse are points within the segment between two points, while in the hyperbola the chosen points ranging from a focus to infinity and from infinity to the other focus. Keeping
that every line has a point at infinity, it follows that the hyperbola, like the parable has a single point at infinity, as it pertains to the same curve and is located on the same line, albeit in homology we have that when the boundary line cuts the circle in two points, its counterpart must have given these two points at infinity.












To calculate the asymptotes or tangents to the hyperbola at infinity do a circle c with center O and O-F1 radio. At the intersection of the vertical by V1-V2 and circumference c we have 4 points which together with O define the asymptotes of the hyperbola.











A numerical example we have with drawing the hyperbola: From V1 to a point measuring 28.69, centered at F2 is an arc with that measure. V2
From that point is 13.6, with center in F1 is another arch with that as at the intersection with the anterior arch have a point on the hyperbola. The difference between the 2 variables is the distance between vertices: 15.09, according to the metric concept hyperbola.

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The ellipse

The ellipse is the locus of points whose sum of distances of each point to 2 fixed points, called foci, is equal to the major axis AB. Given
axes AB and CD, we center C with the measurement of semi higher OA cuts an arc AB in two points, these are the foci F and F '. We
points between spots at different distances, for example, point 2, with the distance 2A and F center we arc and the distance from B to F and center on F 'we take another bow. The intersection of the two arches are two points of the ellipse symmetrical to AB. Two others can be obtained by symmetry about the axis CD.

















In dihedral system can calculate the ellipse as a section of a cone left
A pink plane to the cone sections determining and the major axis, we take a generating m cutting the base at point E. This generating touches the cutting plane as we see from the elevation in J2, went down this plant and get the J1. Vertically until it intersects J1 m1 determines Y1 is a point on the ellipse. In the folded down flat ellipse to see their true form. Right
get the axes, the shear plane AB in the standard determines the major axis. By the midpoint of A2-B2 and spend the vertex V2 straight m. Line m intersects the base of the cone in L2, we obtain the projection on the ground of this point (L1) and united with V1, where this line intersects the vertical obtain H. T1 The minor axis of the ellipse is TH.
In the folded down this axis to standard see the true shape of the ellipse.















According to Pascal's theorem, if we take six points on a conic, and the join of pairs of the form shown in the drawing of the upper left -Yellow-we have at the intersection of these three new radiation XYZ points are always aligned.
This theorem helps us to build new parts of the cone and shows that the same five elements of the set. These elements can be points and / or straight, either.
Given the five points of the conic ABDMN, establishing new points of it.
1-On the point B draw a line any hour, which cuts the line through the points in a point P. DN
2 - That the join point P to point L, which is the intersection of the lines through the points AN and BM. We have in this way the line through the points LP.
3 - The line through the points intercepts PL that passes through the points in the point S. MD AS the line intersects the line h at the point Q, which is the solution. Other points are calculated similarly.















Throughout we have a circle ellipse is tangent to the vertices of the axis of it is called head circumference-red color in the illustration. If we in the top center of any point of the minor axis of the ellipse with the ratio of the circumference, we get the foci F1 F2 at the intersection with the axis of the ellipse to the circle we call on the CP picture .
focal circumference is one that is one of the focal center of the ellipse, for example the focus F2-radio and the distance between the vertices of the axis of the ellipse.
The ellipse is the locus of the centers of the circles-a, b, c, d - which are tangent to the circle focal CF and at the same time going through one of the foci F1. Thus we have the circumferences b (yellow) (green) c have their centers at points on the ellipse, tangent to the focal circle and one of its passes through the center of the ellipse F1. It follows as a special case that the circumference d that determines the foci of the ellipse is centered at a vertex of the minor axis it is tangent to the focal circle.















Head circumference of an ellipse is the locus of intersection points that determine perpendiculars from the foci to the tangent of the ellipse. This means that you can draw rectangles infinite tangent to the ellipse at two opposite sides and whose other side is always inserted in the headlights so that the vertices of these rectangles will always pass by the head circumference, which is one whose diameter matches the diameter of the ellipse.














In the drawing we see the tangents to the ellipse rectangles whose sides pass through the foci and vertices whose impact on the circumference.














An ellipse we can build by a method of affinity, different diameters are drawn through the center of the ellipse and the intersection of the two smaller diameters and circumferences of the ellipse we draw horizontal and vertical lines, respectively. The intersection of these lines are points of the ellipse.














Here we see a numerical example for the construction of the ellipse: the diameter of the ellipse has been separated into two dimensions: 20.84 and 43 65. Making center in the radio spots and these two dimensions have a point on the ellipse.















If we make the tangents to an ellipse we can see that these are the axis of symmetry of a curve through a focus and a curve is an arc of focal circle, the form of pink and yellow are symmetrical and its axis of symmetry is the line of contact between them that is tangent to the ellipse. This is another way of constructing the ellipse, if we are given the circumference c and take its center focal F1 as one of the foci of the ellipse, just fold the paper so that we match a point on the circumference c to the other focus F2 . The line along which we have doubled the paper is a tangent to the ellipse as it is the axis of symmetry of the curve by bending the paper is overlapped with the focus F2.













Construction of projective bundles
Draw a quadrilateral parallelogram sides are divided into equidistant parts as in Fig. The intersection of beams emerging from the midpoint of the upper and lower sides to equidistant divisions of the figure are the points of the curve.
As the procedure is based on a method of projective geometry is valid for any projection.















As a projective method is preserved by projection, however be tempted to use the cylindrical projection and must be divided equally segments and this would be laborious in a conic.
















If a quadrilateral in which to inscribe the ellipse vertical sides split in equal parts, for example six, and join a point with C, and do the same with the axis of the ellipse, divided into six equal parts and join with those point D points as shown in Fig. The intersection of the line through point D and point one of the major axis of the ellipse with the line through point c and point a vertical segment tangent to the ellipse at point B, is a point M the ellipse.















Since the diameter of an ellipse determine the other conjugate diameter. Is a line parallel to the given segment b and the midpoint M is attached to the center of O given diameter, which is the center of the ellipse. OM segment determines the conjugate diameter of the ellipse.
If we project the two diameters of an ac -circle at the top to 'C'-we can see that the two diameters of the above are processed perpendicular.















To construct an ellipse by another method of affinity, is sufficient to affect similar triangles with vertices at the points of the minor axis of the ellipse or the diameter the circumference, as they are consistent.
Given the circumference to be transformed into the ellipse and diameter, since any point of the ellipse Z counterpart center of the circle O, determine the affine ellipse of this circle.
We join the point P to point Z and we have the red triangle OPZ. For a given point or the diameter of the circle make another similar triangle-green color, this means that he make a straight point U m 'parallel to line m. UT do another line parallel to the line OP until it intersects the circle at point T. For the T make a parallel point d 'to d until it cuts straight to the line m', the intersection of the two lines m 'd' we determine the point Z ', which is a point on the ellipse.
To find new points of the ellipse is sufficient to the previous similar triangles, ie triangles with sides parallel and whose vertices pass through the line to and from points on the circle.

















Given the conjugate axis of the ellipse (the projection corresponding to the orthogonal axes of the circumference), determine the major axis and minor ellipse. Mannheim

Method: Given
conjugate axes of the ellipse (light blue w, q), we center in O (intersection of the conjugate axis) with distance OB (B is the end of the semi-conjugated) and draw the arc a.
create a pink t perpendicular to OB and where this short arc to obtain Z, join with N (end of the other conjugate diameter). V We center, midpoint of ZN and make a circle of radius VZ. We join with O V
obtained as the intersection with the circle with center point P V: PN (m) is the direction of the axis of the ellipse and perpendicular m 'the direction of another axis.
The length of the two axes is: OP for the minor axis and OF (F is the intersection of m 'with OP) to the major axis. The axis, determined graphically as the intersection of C (green circle) with k (axis parallel am ') and C' (red circle) with N (axis parallel am).

















Dandelin Theorem 1 or 2 areas exist Tangent to a cone and a section plane. This map identifies the focus of the conic in the points of contact with the spheres.
The planes defining the intersection of the areas with the cone (2 circles) and the plane of the conic determined by the same guidelines.















observe 2 In this profile areas (in pink and yellow) with a plane p tangent to both. This map p Sectioning the cone as the blue ellipse major axis m of TY (ends of the cone section.) The intersection of p and the planes where the spheres are tangent to the cone, j, k, are 2 lines parallel to the axis.
we project the p plane and folded down in true form for the ellipse. TY the major axis of the ellipse is transformed to project it into T'Y ', the center of the ellipse is the midpoint of O' to be a symmetrical curve on the two main axes. The foci are the points of contact DF areas with the plane p projected onto the ellipse: D'F '. For the semi-minor axis of the ellipse is a plane LO and through O perpendicular to the axis of the cone in which we observe folded section in real scale. Then place it on the blue ellipse orthogonal to T'Y 'and from the center of the ellipse: L'O'.















The case of the parable.

note in the drawing a sphere tangent to the parabolic section of the cone at the focus of the parabola. The axis of the parabola, and passes through the point of tangency of the sphere with the plane of section and cut the plane where the cone and sphere are tangent, ie the plane through the line M.
The two planes, which divides the cone and the contact between the sphere and the cone- the two lines pass through em, respectively, and cut in a straight dy is the guideline of the parable. Dandelin
theorem we easily determine the focus of a section of the cone. The solution to the exercises for the determination of the foci of the conic is tangent to a sphere and the cone section and point of contact with the tapered area is a focus of the conic.















In this picture we see in a profile cone separated into two pieces by a court, with the sphere tangent to the plane of section in a.
also note that the m-blue plane is the plane through which the cone and sphere are tangent, cut the section plane of the cone on the line d, which is the guideline that the parable.













Tangents to conics. To
t1 t2 tangents to a conic from any point P, two secants are drawn s1 s2 to the same and the four points of intersection with the diagonal are cone n, m and drying d, a. The intersection of nm and gives two points through which the polar. The polar intersects the conic in the points of contact where we can t1 t2 draw tangents from P.
















Practical applications of conic sections and their existence in nature.

The parable is given in the headlights of cars in the spotlight, on radio and television antennas in radio telescopes, all of which are served in the parable whose rays leaving or entering the projection parallel to the axis of the parabola from direction or focus.
There is also the arches of the bridges, the design of warheads, in cables of suspension bridges with uniform load distribution, in electric stoves, reflecting telescopes, in a rotating fluid in a container by centrifugal force effect, in exceptional planetary orbits hinges between ellipses and hyperbolas-between slow and fast movement, respectively, on body surfaces and fuselages for its character streamlined in the path of the curve that describes an object being thrown with homogeneous gravitational field and neglecting air resistance, such as a missile.
Parable no mistake with the catenary curve, similarly, that is the curve takes a rope hanging from two points for his weight.
When several projectiles are fired from a cannon in different directions, all the paths that follow these shells define an enclosure that is never exceeded by any of the paths. The envelope is a parabola of revolution called security (ie the surface generated by the revolution of a semiparábola).
The envelope of all the parables has its tangent at the vertex a line that is guideline of all the parables, that is, a line perpendicular to all axes of the parables.



The ellipse is given in the orbits of celestial bodies that revolve around another intercontinental missile with a background in which there is strength and where g is not constant.
The ellipse is given in sections of vaults and arches of different forms, also in view of our perception we can see the circles into ellipses,
The vaulted ceiling of the subway platform has an elliptic curve, which makes a good sound.
The rocks eroded by water from the river are approximate ellipsoidal forms whose sections are close in shape to elliptical sections. The oblique view from a parallel projection onto a plane is an ellipse, if the projection is tapered is a conic anyone.
spend if we want to play the ball inside a building without ever having to pick up the ball, we have a site elliptical shape. We stand at the foci of the ellipse and when the ball bounce with the ellipse-shaped walls will always go to the other position where the other person at the other focus of the ellipse.

The hyperbola is given in rounding parts, in projecting the shadows of almost all the lights and screens on the house, at the intersection of equidistant circles causes two stones thrown into the water, thermal power plants cooling towers.
The hyperbola is the curve of a border or surround audio zone: a plane that moves with constant speed and linear motion spreads the sound of the engine in growing areas. According to spheres move that leaves behind are increasing and the envelope curve of these areas is a paraboloid. The orthogonal projection of these areas and the paraboloid is respectively a set of circles and hyperbole surround them.
When a stick is stuck orthogonally on the floor and is illuminated by the sun, the shade produced by the end of the stick is a hyperbola.


conic curves are given in general in almost all designs that are to be smooth surfaces, ie not notice the transition between the binding surfaces such as hull surfaces ships, aircraft fuselages, auto body, and so on.
also occur generically in the movements of celestial bodies, which the force of attraction can generate any of the conic, if the object follows a straight path through space, neglecting the curvature of space and is attracted by the orbit another describes a hyperbolic shape, exceptional dish. If the object that attracts him is strong enough to incorporate it into their own orbit, it describes an elliptic curve whose focus is the object. The vaults are
section tapers and arcs tend to have generally parabolic, elliptical or circular.
As in the laws of mechanics, that cause the flight of the ball out of a focus of the ellipse and crashed into a wall of the elliptical field and go to another center, so with the laws of reflection and refraction in optics with all conics.
A ray of light that bounces out of a focus on the wall of an elliptical mirror and returns to the other focus. In the case of the parabola if the point of light is the focus of it, the lightning, touching the parabolic surface, leaves reflected in a direction parallel to the axis of revolution of the paraboloid while if the surface is hyperbolic, the beam of light that bounces out of focus with the surface and follows a direction defined by that point and the other focus.

Sunday, October 10, 2010

How Do I Open This /poptropica