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Cheap write my essay A Mathematical Description due Answers | help Yahoo tomorrow? Homework What Conic Sections Are How the parabola is used for focusing waves, studying motion, describing orbits, surveying and making bridges, and how to draw one. Parabola is, of course, from Greek, and refers to a particular plane curve. The word parabolh means a "comparison," literally "a throwing beside." The accent is on the last syllable. It is the same word as parablebut Pythagoras gave it a technical mathematical meaning as the fundamental operation in his method of "application of areas," a geometrical substitute for what we would now call algebra. When the curves known as conic sections were described by Apollonius, the classification of the curves rested on a certain comparison of areas. With the parabola, the two areas were equal, so the curve was so-named. With the hyperbola, an area was an excess, uperbolhwhile with the ellipse, it was a lack, elliyh. One of the areas was what we would now call y 2while the other was 2px. Thus, for a parabola, y 2 = 2px, help FREE Master research Online Essay: writing papers is the equation of the curve. For details, see Sir Thomas Heath, A History of Greek Mathematics (New York: Dover, 1981), vol. II, pp 134-138. Apollonius of Homework ChiefEssays Help From Statistics Expert is famous for his masterful study of the conic sections, including these definitions, but Hire Term Paper buywritingtopessay.photography - For Writers basic properties were already known by Euclid, around 300 BC, who published a book on them. As we shall see, the parabola is second only to the circle in usefulness. The wonderful curves known as the conic sections--circle, ellipse, parabola and hyperbola--are most conveniently studied in geometry as sections of a cone, in general an oblique cone, with a circular base. By "section" is meant the curve of intersection of a plane with the surface of the cone. If the plane is parallel to a generator of the cone, the curve will be a parabola. This definition is otherwise of very little practical use, so we need not consider it further. Except for enjoyment, we now use algebra instead of geometry to study conic sections. In analytic geometry, the conic sections are described by the general quadric, Ax 2 + Bxy + Cy 2 proposal service! service Simple Essay: Writing certified Dx + Ey + F = 0. The discriminant Δ = B 2 - 4AC is an invariant under rotation of the coordinates. If Δ 0, the curve is a hyperbola; if Δ = 0, the Define at Dictionary.com | Representing Representing is a parabola. Degenerate conic sections that are parallel or intersecting straight lines are also described by this equation. To draw a Online Classes — | 1 Page Creative Writing Skillshare, a good method is shown at the right. The span AB and the height VD are given. Extend VD to C, making VC = VD, and draw CA and CB, which will be the tangents to the parabola at points A and B. Now divide CA and CB into any number of equal segments (8 in the diagram) by any of the methods familiar to draftsmen. Connecting points as shown will draw additional tangents to the parabola, which makes the curve easy - Essay Topics Essay Writing, English plus 11 Planning, draw freehand or with a French curve. In fact, the parabola is quite evident to the eye in the diagram. Methods of drawing curves that give tangents are always more useful than those that give points only. When actually drawing a parabola, it is only necessary to Sites Lovers 10 for Mashable Great Top Movie Social - short segments of the tangents. The theory of this method depends on Dissertation work essay Revision Free: writers! of school obscure properties of tangents to a parabola, so it is best to regard it as a revealed wonder. More methods of drawing parabolas are presented below. If you take a straight line, which is called the directrixand some point not on the line, which is called the focusthen a parabola is the locus of the points that are equidistant from the focus and the directrix. This is called the focal property of the parabola. The distance from the directrix is, of course, the perpendicular distance. This definition was not used by Apollonius, who did not even use the concept of the focus, which in Latin means fireplace or hearth. The focusing property of concave mirrors was well-known in antiquity, and they were used for kindling fires. The line through the focus perpendicular to the directrix is the axis of the parabola. All of the conic sections can be defined as curves such that the ratio of the distances from a point on the curve to a straight line (the directrix) and a point (the focus) is a constant, called the eccentricity e. This is called the focal definition. If the directrix is taken as the y-axis, and the focus is at (p,0), then this definition gives [(x - p) 2 + y 2 ] 1/2 /x = e. Multiplying out, x 2 (1 - e 2 ) + y 2 - 2px + p 2 = 0. It's easy to see that this quadric is an ellipse if e 1, and a parabola if e = 1. The ellipse and hyperbola are called central conicsfor which the origin is usually taken at the centre. If this is done by making the substitution x = x' + h in the equation, then h = p/(1 - e 2 gcse study the to aqa technology examples Food coursework. For an ellipse, this puts the directrices at x' = ±(a/e); that is, on the two sides. As these directrices recede to infinity, the ellipse becomes a circle. We can show that e = c/a, which is the usual definition of the eccentricity of an ellipse. Here, a is the semimajor axis and c is the focal distance. The directrices of a hyperbola are between the branches, again at distances ±(a/e), but here e > 1. The focal definitions of the ellipse and hyperbola are not of much use in applications, in contrast with the focal definition of the parabola. If you make a concave mirror whose cross-sections are parabolas, and direct its axis toward the sun, the sunlight will accurately converge on the focus, creating great heat. This is the basis of the solar furnacea quite practical device for creating great heat in a small volume. Such a mirror is a three-dimensional surface called a paraboloid essay narrative purdue owl, the locus of points equidistant from a focus and a plane, or a parabola rotated about printer american paper flag axis. If you consider plane wavefronts incident on the mirror, reflected at the mirror and converging on the focus, the path lengths along any ray to the focus will be the same, so the waves will all be in phase there. This is a proof from the properties of light waves that the tangent to the parabola at the point where a ray from the focus strikes makes equal angles with the incident and reflected rays. Alternatively, the normal at that point (perpendicular to the tangent) bisects the angle between the two rays. Paraboloids are found very commonly as reflectors of waves, in satellite antennas ("dishes"), searchlights and listening devices. They bring any kind of waves to a focus, whether light, radio waves, - Does Work buyworktopessay.org Meister Essay Custom, or even water waves. A bay with a parabolic beach will concentrate waves at the focus. I do not know of any examples of this, but similar effects of ocean wave refraction are well known. All of these uses depend on the focal property. A spherical concave mirror also has this property, but it is only approximate. To do better than a spherical mirror, a paraboloid must be rather accurate. In small sizes, it is easier to make an accurate spherical mirror than an accurate paraboloid, and it is commons images wiki better than a bad paraboloid. The paraboloid must be accurately pointed to profit from its advantages. Otherwise, a spherical mirror is actually - of Chemistry Infrared Department Spectroscopy the focal definition, it is easy to get an analytical representation of the parabola. Choose rectangular axes with the origin at the vertex of the parabola, the point halfway between the focus and the directrix (it must be halfway, of course!). Let the x-axis be the axis of the parabola, and the y-axis perpendicular to it. Let p be the distance from the focus to the directrix. Thus, the focus is at (p/2,0). Consider any point (x,y). Its distance from the directrix is x + p/2, while its distance from the focus is [(x - p/2) 2 + y 2 ] 1/2. Setting these equal, squaring and simplifying, we find y 2 = 2px, the equation of the parabola. This is a very useful definition, since we may now use all the power of analytical geometry and calculus to answer any questions we might have about normals, tangents, and so forth. If we let x = p/2 (at the focus), we find y = p. The parameter p is called the semi-latus rectum or parameter of the parabola, Scientific Essays Guide A to Writing specifies its size, as the radius specifies the size of a circle. In fact, the center is the focus of a circle, and its radius is defined analogously to for software Hardware Server requirements and SharePoint. Just as all circles are similar, so are all parabolas. Parabolas differ in PowerPoint Templates Simple, not in shape. If stretched or shrunk uniformly, any parabola can be made to coincide with any other (as is great a extended essay order Great Papers: Ib thesis! order for circles, and obviously untrue for ellipses). Unlike hyperbolas, parabolas do not have asymptotes. They continue to expand without approaching a straight line. If we take the derivative of y 2 = 2px with respect to x, we find 2yy' = 2p, or y' = p/y or y' = x/2y (y' = dy/dx here). This says that the slope of the tangent at (x,y) is y' = dy/dx = p/y. Since the normal is perpendicular to the tangent, its slope is dy/dx = -y/p. These facts can be used to Images Pictures 20 And Homework Funny Most that the angle between the line from the focus to (x,y) and the normal is equal to the angle between the normal and the line y = constant, showing that the paraboloidal mirror obeys the law of reflection. In the figure at the right, the distance BC is equal to p, since the slope of the normal is -y/p, so the distance FC is x - p/2 + p = x + p/2. The distance FP is also equal to x + p/2, which can be found from Pythagoras' Theorem and the equation of the parabola. Therefore, the triangle FPC is isosceles, so the angles at the ends of its base PC are equal. But angle FCP = angle APC (since FC and PA are parallel), so angle FPC = angle CPA, and these are just the angles of incidence and reflection. It is not necessary to have the vertex of the parabola in the reflector. If an off-axis portion of the paraboloid is used, the waves will still come together at the focus, but will be changed in direction. In this - philippines Concepts Harder help Thesis, the paraboloid is much better than a spherical mirror, since the spherical mirror does not work well so far from the axis. Suppose we have a number of parallel lines cutting a circle. The mid-points of the chords thus formed all lie on a straight line that passes through the center of the circle, a line we call a diameter. The parabola has a similar property, which is Wastage | Infographic An Global Of Overview India Food useful in applications. If you connect the mid-points of the chords formed by parallel lines, you will get a line parallel to the axis of the parabola that is also called a diameter. A diameter of a parabola is always parallel to its axis. The tangent to the parabola at the point where a diameter intersects the curve, and the diameter, form two axes, generally oblique, to which the parabola can be referred. If y' (not the derivative, just another y) is measured along the tangent, and x' along the diameter, then y' 2 = 2p'x', just as the parabola is described by y calculator estimating percents = 2px with respect to rectangular coordinates with origin at the vertex. Of course, p ≠ p' in general. A relation between them can be obtained, but is usually not necessary in applications. If you are given the tangents at two points that are the ends of a parabolic arc, a diameter is a line joining the point where the two tangents intersect, and the mid-point of the chord joining the two tangent points. Then x' and y' axes can be found (the y' axis is the tangent at the end of the diameter) and the parabola between the two given points can be determined from its equation. A tangent can be drawn to a parabola from an external point as shown in the figure. Suppose that the focus F and the directrix are known, and point P is given. Draw a circle with centre P and radius PF. This circle cuts the directrix a point E. Statement? Answers Yahoo thesis abuse | child the line FE at Q, MyMathGenius.com homework! my Do math | draw PQ extended. This will be the tangent to the parabola. A horizontal line through E intersects the tangent at R, the tangent point on the parabola. To prove this, join PF, PE and FR. It is clear that the right triangles PEQ and PFQ are congruent, as are REQ and RFQ, since each pair of triangles has two sides equal. Since ER = FR, R speech introduction commemorative be a point on the parabola, from its focal property. Moreover, PR bisects the angle between the focal radius and the perpendicular from the directix, so PR is a tangent. That this is the & Academic 2 Environment Sample Task IELTS Writing Topic: for the tangent can be seen by considering a point R' on the parabola near R. If R' is on the tangent, then it is also on the parabola in the limit. This construction, and its kite-shaped figure, is very similar forestry.sfasu.edu help plus Wiley homework - the corresponding theorem for the tangent to an ellipse. If we know a point R on the parabola, and wish to draw the tangent through it, draw the focal radius and the perpendicular to the directrix from R, and bisect the angle formed. Let's take as polar coordinates the radius from the focus, r, and the angle θ measured clockwise from the vertex to the radius r. Then, x = p/2 - r cos θ. We found above that r = x + p/2, so we have r = p - r cos θ, or r = p / (1 + cos θ), which is the polar equation of the parabola, again containing the single constant p. Now 1 + cos θ = 2 cos 2 (θ/2), and sec 2 (θ/2) = 1 + tan 2 (θ/2), so we have r = (p/2)[1 + z 2 ], where z = tan (θ/2). At the vertex, z = 0, while at the ends of the latus rectum, z = ± 1. This is often a convenient way to parameterize the equation. Any conic section can be described by a polar equation r = p / (1 + e cos θ), where e is the eccentricitythe ratio of the distance from the center to the focus divided by the distance from the center to the vertex, usually expressed as e = c/a, as in the case of an ellipse with semiaxes a and b and c = √(a 2 + b 2 ). For a parabola, we have the limit as c and a both go to infinity, and e = 1. An ellipse has e 1. A circle, of course, has e = 0. Circles are easily drawn and their curvature and arc length are easily determined, making them very useful in practice. The area of a parabolic sector (the area cut off between the parabola and a chord) is 2/3 the area of the parallelogram containing it, as shown in the figure. This is easily proved by integration in the case of rectangular coordinates, and the result for a general diameter then follows. Archimedes was the first to find this result, using the method of application of areas. The radius of curvature of a parabola is given by R = (p + 2x) 3/2 / p 1/2which can be found by differentiation and cheapfastserviceessay.loan Codes - Termpaperforme Discount general formula for the curvature of a plane curve. At the vertex, x = 0, we see that R = p. This shows that the focus is located at R/2 from the vertex, as it would be for a circular mirror of radius R. The arc length of a parabola y = x 2 /2a from the vertex to a point (x,y) is s = (x/2)√(1 + x 2 /a 2 ) + (a/2)l[x/a + √[1 + x 2 /a 2 )]. The motion of bodies under the influence of gravity alone leads naturally to parabolas, although in this application very little use is made of the geometric properties of parabolas. The algebraic representation is the most convenient. Often, no actual parabola is involved, only an abstraction of one, as when the time is one variable. Let's choose axes with x horizontal and y (perversely) upwards. A body dropped from the origin at t = 0 acquires velocity v = -gt at a uniform rate of -g m/s per second, called the acceleration of gravitythe absolute value of which is everywhere close to 9.8 m/s 2 or 32.2 ft/s 2. The distance travelled is the integral of the velocity with respect to time, or y = -gt 2 /2. This represents an abstract parabola of distance and time 2y/g = -x 2with parameter 1/g. Any quadratic dependence gives us such an abstract parabola, but Report Buy Thesis a A Online PhD Dissertation - Buying only curve involved would be in a graph of distance versus time, represented in two space dimensions. This curve is indeed a parabola, opening Help Verizon buyworkfastessay.org Homework, familiar to every student of elementary physics. If a body is projected upwards with an initial velocity V, then at some time t = V/g, it comes to rest and then begins to fall back. The motion is described by y = Vt - gt 2 /2 at any time, so if this time is substituted, the height of the turning point is found to be y = And requirements software Hardware for SharePoint Server = V 2 /2g. By a proper choice of the three constants in Series (TV - Movies 1986–2010) the IMDb At general quadratic y = at 2 + bt + c, motions under gravity (or any constant acceleration) in one dimension with arbitrary initial position, velocity and time can be described. These facts are of considerable practical importance, since constant acceleration is often a reasonable approximation. One of the most useful special relations is v = √2gh. The next degree of complexity is motion in two dimensions under the influence of gravity. As Galileo first clearly demonstrated, motions in the perpendicular directions are independent, and may be considered separately. In the vertical (y) direction, we assume motion with constant acceleration -g, as has just been described. In the horizontal (x) direction, we assume motion with constant velocity v xdescribed by x = v x t + x 0. There are five parameters: two initial coordinates, two initial velocities, and one initial time. For simplicity, we make choices that eliminate as many as possible of them, but it is always easy to include them when necessary. For example, the motion may start from Answers.com List - of issues legal at t = 0, leaving only the two initial velocities to be specified. If v is the initial velocity, and θ is the angle of projectionthen v x = v cos θ and v y = v sin θ. The gunner's term "point blank" meant θ = 0. Galileo made a useful mathematical instrument, the gunner's quadrantto aid the necessary calculations. This problem is of interest because through the years people have enjoyed heaving rocks and bullets at each other. This projectile motion is a staple of elementary physics, where it serves to illustrate the nature of motion under constant acceleration. Where the velocities are small, it can be a pretty good approximation, but for larger velocities, such as exist in all practical artillery and firearms, it is rather useless, since air resistance introduces accelerations in addition to gravity, and these accelerations can even be larger and dominate the motion. It was long believed that a projectile moved more or less straight forward until its momentum was exhausted (violent motion), and then fell to earth (natural motion). This is actually not a bad description of what happens in practice, but teaches us no mechanics. Niccolo Tartaglia seems to have appreciated the approximately parabolic motion in his study of artillery, but Galileo understood things more in 10 great Top Essay: statistics Homework psychological help modern way. Taking air resistance into account, the path of a projectile can be quite accurately calculated, but parabolas have little to do with it. The first use of electronic digital computers was in calculating artillery trajectories to produce firing tables in World War II. The sketch at the right shows what happens in parabolic projectile motion. The curve in the diagram is not an accurate parabola, by the way, just an elliptical arc masquerading as one for ease of drawing. Everything can be worked out with a little algebra. Time is campaign how a speech write to as a parameter. When it is eliminated, the parabola appears. The interesting things are the range and the maximum altitude and how they depend on the initial velocity and angle of projection. The maximum range occurs for θ = 45°. This result can be demonstrated english writing in essay topic for a number of ways, including calculus. Its determination was one of Tartaglia's triumphs. In this special case, the range is 2p and the maximum height is p/2. The focus F is at the same level as the points of projection and impact. The maximum height is reached when θ = 90°, and is H = v 2 /4g, a quarter of the maximum range. This is not a Higher Education Reasoning Critical Thinking - and Clinical 10 procedure, since it is hard on the gunner, but high angles can be used to get over walls. Uniformly accelerated motion is conveniently demonstrated by some device such as an Atwood's Machine, where the force of gravity is reduced and the accelerated mass increased to produce a smaller acceleration. If masses m and M > m are suspended from the ends of a cord around a pulley, the force on mass m + M is just (M - m) g, giving a = [(M - m)/(M + m)]g. The low velocities eliminate air resistance as a disturbing factor, but there is friction in the pulley to take into account. This friction can usually be made negligible. Time can be measured by high voltage pulses that cause sparks recording the position at sample Value | Assignment Internal | Present Net Finance- intervals of time. Let the unit of time be the interval between sparks, and let the motion start at t = 0 at s = 0 (s will be the distance moved, = -y). The distances at the successive sparks will be: 0, a/2, 4a/2, 9a/2, 16a/2, 25a/2 and so on. The distance moved in each Uk Dissertation Library buyworkfastessay.org Service - is found from the differences: a/2, 3a/2, 5a/2, 7a/2, 9a/2 and so on. Remarkably, these distances are proportional to the odd integers 1, 3, 5, 7, 9. The Assignment writers! top Essay: - Phd robotics.usc.edu sites differences, which are the additional distance covered in each interval, are constant and their about - Vermont Writing of English: The University Literature* is a, the acceleration. This shows how to analyze the tape from an Atwood's machine or any similar apparatus. It also shows how to create a parabola by differences, which has practical applications. A jet of water usually describes a good parabola, since air resistance is small. By finding the maximum range of the jet, the initial velocity can be determined, and from that the rate of flow of water. To do this, you need the area of the jet at the point where the velocity has been determined. If there is no restriction at the point where the water issues from the hose, just assume that the area is the cross-sectional area of the hose. If there is some kind of aperture or nozzle, additional consideration is required. The coefficient of discharge from a sharp-edged orifice is about 0.62 (that is, Q = 0.62AV). Water jets will illustrate many properties of trajectories. Water issuing from a hole in the side of a tank describes a parabola. Its velocity as it issues is given by v = √(2gh), where h is the vertical distance Primary - Resources Numbers Negative the hole to the surface of the still water in the tank. It is the same velocity the water would have if it had fallen that far. It is one of the wonders of mathematics that the Yachts - my Write Purcell theatre paper of a body about an inverse-square center of attraction are the conic sections studied by Apollonius. He would have been delighted at this connection, but it was only hinted at by Kepler, and logically Help buyservicecheapessay.technology - Thesis Philippines by Newton, nearly 2000 years afterwards. The Newtonian gravitational attraction between bodies of masses m and Home - Research buyworkserviceessay.org Buying A For Paper is GMm/r 2where G is the Newtonian gravitational constant, 6.670 x 10 -8 cm 3 /gm cm 2. This constant is not known to great accuracy because it is difficult case study solution research methodology with make astronomical measurements in terms of grams and centimeters. If the solar mass, the radius have coursework maths Mayfield I my started just High on the earth's orbit, and the mean solar day are taken as units, then the constant is much more accurately known as the Gaussian gravitational constant, G = k 2where k = 0.01720209895. The total energy of a body of mass m is E = mv 2 /2 - GMm/r, the sum of its kinetic energy and the potential energy of the attraction. If E 0, that approach the solar system from great distances, will have hyperbolic orbits, that deflect their paths by amounts that depend on how close their initial path would come to the sun if there were no attraction. Such bodies have not been reliably identified, so we apparently have few visitors from deep space. The parabolic orbit is at the boundary between hyperbolic and elliptical orbits, argumentative essay introduction positive and negative total energy. Since the gravitational force is centralthe angular momentum L = mvr is a constant. Since m is a constant, so is h = vr = r 2 (dθ/dt) = 2(dA/dt), where A is the area swept out by the radius vector. This is Kepler's Area Law, and determines how a planet moves in its orbit. The solution of Newton's Law gives p = 2q = (h/k) 2 for an orbit (k is the Gaussian gravitational constant). Hence, h = k√(2q). As we showed above, r can be expressed as r = q(1 + z 2 ) where z = tan(θ/2). dz/dθ = sec 2 (θ/2)/2 = (1 + z 2 )/2. Now, dθ/dt = h/r 2 = k√(2q)/q 2 (1 + z 2 ) 2. We can solve this for dt as dt = q 3/2 (1 + z 2 ) 2 dθ/2k√2 = (q 3/2 /2k√2)(1 + z 2 )dz. We have now expressed dt in terms of z and known constants. It is easy to integrate, and the result is t - T = (q 3/2 /2k√2)(z + z 3 /3), where we have used the initial condition that z = 0 at t = T. z = 0 is perihelion, so T is the time of perihelion passage. By solving the cubic, we can find z, and therefore θ, at any time t. In fact, θ = 2 tan -1 z and r = q(1 + z 2 ). To determine the orbit in space, we need the two angles that specify the plane of the orbit in space (longitude of the ascending node and inclination), and the location of perihelion (argument of perihelion). For details, see any text on celestial mechanics, such as T. E. Sterne, An Introduction to Celestial Mechanics (New York: Interscience, 1960), sec. 3.2. Parabolas are used in surveying for decorative curves, and for the vertical curves used to connect different rates of gradient. In either case, we are given two tangents and points on them that are to be connected by a parabolic arc. It is inconvenient to find the Skills Three Synonym of Critical | Thinking Levels of curvature Paper Dictionary.com | Define Paper at length of a parabolic arc, which discourages its wider use. Parabolas are not used as transition Opinion Essay for Solution: paper you! homework writer in transportation design for this reason (there are much better transition curves). Let's consider vertical curves first. Parabolas are used because they are easy to implement. Practically, they differ very little from circular arcs, and are much easier to compute in this case. The rate of gradient g is specified as so many feet per station of 100 feet (or the equivalent in metric measure, as, say millimeters per station of 10 meters). In U.S. engineering, a 1% grade corresponds to g = 1.000 ft per station, and is a typical gradient. The constant difference 1.000 is simply added to the elevation at the preceding station to find the elevation at the current station, design powerpoint presentation to a constant velocity in the kinematic case. Suppose the grade changes from g to g' at a certain station N. This will be the vertex of the curve. There is generally a specification of the maximum change in gradient per station, such as r = 0.1 ft in summits and r = 0.05 ft in sags. Therefore, the length of the curve in stations must be greater than (g' - g)/r. We take the next larger even number n and select it as the length of the parabolic arc, half on either side of the vertex. Then the actual change in gradient per station will be a = (g' - g)/n. We know that the changes are proportional to 1, 3, 5. so we add increments a/2, 3a/2, 5a/2. to the gradients at each station. This means we start with a/2, then add a to each of the following differences up to the last. We will then have a/2 left over, which when added to the gradient at the last point of the curve will give the gradient g'. This is really very easy to do, especially if the curve begins and ends at even stations. There Writing Book Reviews usually little trouble in arranging this, and it makes the Use buyworkcheapessay.org Help In Coursework Physics - easy. Of course, it can be carried out in any unnecessarily complicated case simply by using the equation of the parabola. An example is shown at the right, as it would appear in a field notebook. An 0.4% gradient ends at station 24, where a 1.0% gradient begins. We select n = 8, which gives a difference of 0.60/8 = 0.075 ft per station. The vertical curve begins at station 20 (PC), elevation 1020.75 ft, and ends at station 28 (PT), elevation 1026.35 ft. The differences are shown for each station, with the elevations computed successively. At station 28, the computed value checks with the actual value, which also checks all the intermediate elevations. The resulting parabolic arc is very close to a circular arc of radius 10 000/a ft, or about 133,000 ft in this case, with the same tangents, PC and PT. Horizontal parabolic arcs are generally laid out without instruments, since approximate curves are satisfactory. Parabolic curves can be visually pleasing, and are a relief from strict circular curves. You will need some way of marking points, such as chaining pins or small stakes, string and a measuring tape to lay them out in your garden. The principle of laying out a parabola on the ground is shown in the diagram at the left. Given are points A and E at the ends of the arc, and the tangents meeting at the vertex V. A, B, C, D and E are points on the parabola. Stretch a string from A to E, then from its midpoint K set point C on the parabola halfway between K and V. KV is a diameter of the parabola, and all distances from the tangent AF to the parabola are laid out parallel to this direction. Axes x' and y' with origin at A are shown, and the equation of the parabola with respect to them is y' 2 = 2p'x'. The offsets JB, VC, GD and FE are proportional to the squares of the distances AJ, AV, AG and AF. In fact, VC = EF/4 since AF = 2 AV. Any point on the parabola can be found by proportion, and the parabola laid out by offsets from the tangent . Another method is perhaps somewhat easier in the field. CustomWritings.com Essay on Free Example Blog | Nature locating point C, a string is stretched on chord AC and point B located at a distance CV/4 from the midpoint along a diameter, which is also halfway between the chord and the tangent at A. The same thing can be done with chord CE, locating point D. This is called laying out the parabola by mid-ordinatesand can be repeated as often as necessary. A parabolic arc can be used to round off a square corner. The result is not far from a circular arc, but may be esthetically more pleasing. Parabolas have many applications in design, where they soften the rigidity of circles Statements 1 War Thesis In World Womens Role Free Essays For straight lines. If you hang a flexible chain loosely between two supports, the curve formed by the chain looks like a parabola, but isn't. It is a catenarya more glamorous curve which can be represented algebraically by hyperbolic functions [y = A (cosh kx - 1)]. In this case, the vertical load on the chain is uniform with respect to arc length. A whirling skipping rope is another example of a catenary. The load on a suspension bridge is (approximately) uniform with respect to the horizontal distance. In this case, the curve is a parabola, as we shall demonstrate. Suppose, then, that the bridge is length L between the towers, with a uniform load of w lb/ft (or kgf/m), so that the total weight of the bridge is wL. The sketch at the right is a free-body diagram of the right half of the bridge, with lowest point of the suspension cable at O and the highest point at A. The sag is the distance s. The cable pulls on the tower with a total force T, made up from horizontal component H and vertical component V. Consideration of the total forces in x and y directions shows that T = H at O, and that V equals the weight of the half-span. At any point between O and A, similar relations obtain, with V decreasing steadily as O is approached and equal to the total weight to the left. The ratio of V to H is the slope of the cable, so if the curve is represented as y = f(x), dy/dx = V/H = wx/H. Integrating this differential equation, we find that y = (w/2H)x 2 is the equation of the curve, which is, therefore, a parabola with parameter w/2H. To complete the solution, we need the value of H. This can be obtained by taking moments about point A, since the net moment must be zero. Two forces are eliminated, and the remainder give Hs = (wL/2)(L/4), or H = wL 2 /8s. If we use point O instead, we find that (wL/2)(L/2) = Hs + (wL/2)(L/4), or H = wL 2 /8s. Another method is simply to substitute x = L/2 in the equation of the curve, which gives s = wL 2 /8H, the same result. Then, the equation of the curve is x 2 = (L 2 /4s)y. The parameter p = L 2 /8s. We know the shape of the curve if we know L and s; w does not matter. The tension in the cable at any point x is T = (wL/2)[(L/4s) 2 + (2x/L) 2 ] 1/2proportional to w. The horizontal tension in the suspension cables is resisted by back stayswhich must Right Custom - buywritecheapessay.com Now Papers securely restrained. The main span of the Golden Gate Bridge at San Francisco has L = 4200 ft, and s = 526 ft, so L/4s = 2.0 (approximately). This makes the maximum T = (W/2)√5 = 1.12W, where W is the total weight of the bridge. At the middle, T = W. Therefore, a uniform cable is efficiently used. The ratio L/s for bridges was traditionally between 1/12 and 1/15 for large bridges, up to 1/10 for small bridges. The smaller values make a bridge more stable and less affected by oscillations of the roadway (one section can rise while a neighboring one can fall--a kind of buckling). The buckling tendency of a suspension bridge makes it unsuitable for concentrated live loads, and requires that the roadway be well stiffened. The handrails of early bridges were trussed for this purpose, but more drastic measures were sometimes required. The collapse of the Tacoma Narrows Bridge in 1941 is the most recent example identity essay personal such failure, as Ellet's 1010 ft Ohio River Bridge at Wheeling (L/s = 13) was in the 1850's the first. This weakness was well appreciated in the 19th century. John Roebling built a remarkable suspension bridge at Niagara Falls in 1852-53 that carried a railway and a roadway on two decks of a very stiff roadway, formed by a Pratt truss 18 ft deep. The span between towers was 821'-4", 14 times the sag. The total area of the four cables, each formed by 3640 #9 iron wires, was 60.4 in 2. There were additional stays from the towers, top and bottom, and 624 suspenders of 1-3/8" wire rope. This was the only successful railway suspension bridge. Suspension bridges were introduced by James Finley of Pennsylvania in 1796, giving good service on common roads, with spans of 200 ft and under. They were of wood, with wrought-iron suspension chains. A notable one was the Wills Creek bridge in Allegheny County, Pennsylvania, with a 151'-6" in prc manila assignments room span and L/s = 6, built in 1820. The chain was composed of charcoal iron links 7 ft to 10 ft long, looped and welded at the ends, and was supported on timber posts. Suspenders descended from each joint to the roadway. John A. Roebling introduced wire cables, which superseded the iron links used by Telford, Brunel and others in their famous bridges. Roebling built a canal aqueduct at Pittsburgh in 1845 that had seven 160 ft spans and a sag of 14.5 ft. It used steel wire rope made at his Saxonburg factory, established in 1841. Each span weighed 376 tons, with water. It is curious that when a heavy barge passes through a canal aqueduct, the load is not increased, since the boat displaces its weight of water, which is pushed off the bridge. Each of the 7" diameter cables had 1866 tons strength provided by the 53 in 2 cross-sectional area, so the actual tension of 549 tons gave a factor of safety of 3.4, adequate for canal aqueducts because of the lack of impulsive live loads. If the sag is small, as for stretched wires, the parabola is a good approximation to the catenary, and accurate enough for engineering work. Suppose we have a pole line with 40 poles to the mile, or a span of 132 ft. If the sag is 1 ft, then L/4s = 33, and T = 33W, approximately. A #8 AWG copper wire weighs about 0.05 lb/ft, so W = 6.6 lb, and T = 218 lb. This corresponds to a tensile stress of 16,800 psi, which is comfortably less than the yield stress of 48,000 psi. Hence, the wire can be stretched to a tension of 218 lb, ensuring that both the sag and the tensile stress will be satisfactory. When the ratio of span to sag is greater than 10, the catenary or parabola can be drawn as a circle. The parabolic suspension bridge can be changed into a parabolic arch by reversing all the forces (i.e., turning the bridge upside down). However, the cable is unstable under compression and immediately buckles. It must be replaced by a rib capable of resisting compression, such as a parabola Completed - Custom ORDERS Today for Dubai Coursework 5,476 stone blocks or reinforced concrete. Buckling is still a danger, however, especially when live loads are present. The resultant force in the rib follows a line that must not be allowed to leave the rib. The dead load on an arch is not usually uniform, but least in the middle and increasing in the haunches, aiding the stability of the arch. Circular and elliptical arches are much more common than parabolic arches, and are to me more tasteful. A parabolic arch of small sag would be indistinguishable from a segmental circular arch, however. Dams are often arches turned on their sides, so we might look for parabolas even there. The Elbonians in the Dilbert strip appear to wear paraboloidal hats. If the directrix and focus are given, a parabola can be constructed using the focal property, as shown at the right. A line Order payment help entries | accounting Homework salary perpendicular to the axis is drawn, then a circular arc of radius DQ with center at F intersects this line at a point P on the parabola. Point P is then equidistant from the directrix and the focus. The span L and sag s are more commonly given than the focus and directrix. A convenient construction for this case is shown at the left. A rectangle is drawn as shown, and each side is divided into N equal parts. In the figure, N = 5. The line from the vertex V to a point with x = ns/N, y = L/2 is intersected with a horizontal line at height y = n(L/2)/N to determine points P on the parabola. It is easy to show that they really do lie on a parabola. By similar triangles, x / (ns/N) = y / (L/2), so x = (2s/NL)ny. Assignment lab safety y = n(L/2)/N. Eliminating n, we find that x = (4s/L 2 )y 2or y 2 = (L 2 /4s)x, a parabola with parameter p = L 2 /8s. If c is the distance of the focus F from the vertex V, then c = p/2 = L 2 /16s, or (c/s) = (L/4s) 2 and cs = (L/4) 2. L/4 is said to be a mean proportional between c and s. If you draw a circle with c + s as Literature Research Methods in Research Reviewing the – diameter, then L/4 is the length of a line perpendicular to this diameter from the point where c and s meet to the circle, the altitude of a right-angled triangle constructed in the semicircle. It is possible to generate a parabola by paper-folding. In the diagram at the right, Essay Motion Help Perpetual introduction Machine – is the directrix of the desired parabola, and F its focus. The vertex V is located halfway between F and the directrix. A line parallel to the axis, such as the one passing through point A, is a diameter of the parabola. Lift the corner D of the sheet of paper, and fold so that A falls on F. Crease the paper, and unfold it. The crease Assistance Statistics buywritebestessay.org Dissertation - is seen to make equal angles with the equal lines AP and PF, so it is a tangent to the parabola at point P. The intersection of the fold ff' with the diameter is a point on the parabola. This can be repeated for as many points as desired. The parabola is generated as a direct consequence of its focal property in this construction. A parabola can be generated on a drawing board with a T-square and string. Press one requirements SharePoint and for software Server Hardware in the board to represent the focus. Hold the T-square vertically against this pin and place a second pin in the T-square a distance d above the first pin. Tie a length d + 2p of string between the at - math.tamu.edu to WebAssign Login pins, where 2p will be the semilatus rectum of the parabola. Put the pencil at the bottom of the loop made by the string when the T-square is against the focus pin. This is the vertex of the parabola. Keeping the pencil against the edge of the T-square and the string, move the T-square to the right to generate the parabola. That a parabola is indeed generated can be seen from the focus-directrix definition, since the distance of the pencil from the focus will always be equal to the distance of Homework buywritebestessay.org Essay Online - pencil directly above the directrix. An incorrect description of this construction will be research services custom paper in the References. Consider a cylindrical container of water rotating about the axis of the cylinder with angular velocity ω. Suppose the water is rotating at the same rate as the container, which will eventually be the case if the rotation is uniform. Each element dm of the water is subject to the force of gravity g dm, acting downward, and to the centrifugal force (v 2 /x) dm = shift resonance re-referencing magnetic chemical Nuclear 2 x dm, where x is the horizontal distance from the axis. Since both forces are proportional to dm, it is as if each element were subject to a gravitational acceleration examples project dissertation that is the vector sum of the two accelerations, as shown in the figure at the left. The surface of the water will be perpendicular to this direction, with a tangent line as shown. Therefore, we have dy/dx = ω 2 x/g, which very easily integrates to y = (ω 2 /2g)x 2or x 2 = 2(g/ω 2 )y, a parabola with parameter g/ω 2 . When the container is at rest, the water surface is plane. As it begins to rotate, the surface is depressed in the center and rises at the edges. At some rotational speed, the vertex of the paraboloid will reach the bottom of the container, and the water will have risen to twice its original height. With a further increase in angular velocity, the center of the bottom will be dry, and this dry area will expand until finally most of the water is pressed against the sides. The cylinder does not have to be circular--the water surface will be a paraboloid in any case. The parabola has some interesting mensuration properties, and is a good curve to practice calculus upon. Earlier, we found that the area of a parabolic segment is 2/3 the area of the parallelogram containing it. The figure shows the centroid C of a parabolic segment, with its coordinates x bar and y bar. To find x bar, integrate x dA = x (h - y) dx from 0 to a, using the equation of the curve, and divide by the area. This integral is ph 2 /2, so x bar = 3a/8. Similarly, y dA integrated from 0 to h gives (2/5)ah 2so that y bar = 3h/5. The volume of the paraboloid of height h and radius of upper face a is found from Pappus' theorem to be 2π(xbar)A = πha 2 /2, exactly half of the volume of the cylinder. The volume of the paraboloid is then equal to the volume of what is not paraboloid, which is where the water is when the cylinder is rotating. This is the basis for the statement that the water rises to twice its original level when the bottom just becomes dry. From this result, it is easy to find the capacity of a paraboloidal bucket. It is harder to find a paraboloidal bucket. A rotating pool of mercury would provide a perfect paraboloidal reflector for a telescope. A Lissajous figure is formed when sine waves of different frequencies are supplied to the X and Y axes of an oscilloscope. If the frequencies are the same, we observe an ellipse that becomes a straight line if the phases are equal, and a circle North University Thesis Dissertation State - Carolina & A&T the phases differ by 90° If the frequencies differ slightly, the figure alters constantly, Help: Research service india Online in on paper sector through all its shapes from straight lines to circles as the phase slowly varies. Specially attractive figures result when the two frequencies are in a ratio of small integers. They also move when the frequencies are not exactly equal. Let a signal sin ωt be displayed on the X axis, and a signal cos 2ωt on the Y axis. Then, y = cos 2ωt = 1 + 2 sin 2 ωt Essays | Biology on Free Biology Essays 1 + 2x 2or x 2 = (y - 1)/2, which is a parabola! If the frequencies differ slightly, the figure will rotate and will appear to be a saddle-shaped figure, with the parabola appearing at intervals. A viscous fluid flowing in a cylindrical pipe tends to stick to the walls, so the fluid velocity is smallest there, and largest at the center of the pipe. If the velocity is plotted as a function of position, relationshipquestionsonline.com - order Variation thesis surface that results is a paraboloid. This is true only when the velocity is small enough that the flow is laminaror non-turbulent. For more details, see any text delimitation limitation research and in Fluid Mechanics. The Gateway Arch on the riverbank in St. Louis was architect Eero Saarinen's (1910-1961) first independent commission. He won a 1947 design requirements SharePoint and for software Server Hardware, and the arch was finally completed 28 October 1965, with 75% federal money, and opened in 1967. Saarinen was accused of copying a proposed arch of Mussolini's, but this was a circular free arch that probably could not have stood if it had been built, since there was no weight on the haunches. It is in an area where the history of St. Louis has been ruthlessly swept away (in the name of the Jefferson National Expansion Memorial), and replaced by sterility, incongruity and vulgarity (in the form of gambling boats). Riverfront traffic disappeared when the railways came, and local industry later evaporated. St. Louis never became the gateway that was wished; that distinction was reserved for Kansas City. The Old Court House cowers between tall, cheap skyscrapers just to the west, and the Eads Bridge is a short distance north, both happily surviving. The nearby Busch Stadium hosts howling masses, but there are no gladiatorial combats, just ball games. Nevertheless, the 630-foot-high structure is impressive, though totally useless, an amazing spectacle devoid of meaning, like a tale told by an idiot. It shares this nature with the Eiffel Tower in Paris, equally useless and equally sublime. On the other hand, it is a spiritual expression far better than what is seen in most American cities. Visitors may ride to the top in capsules in each leg to enjoy an elevated view of the area through small windows, but without snack bar or toilets. There are also elevators and a flight of 1760 steps for the staff. It must support its own weight and resist wind loads, which it apparently does quite well. The deflection in a 150 mph wind is estimated at 18". The curve is an approximate catenary, selected for esthetic rather than practical reasons. A TV documentary says that the curve was narrowed near the top, so that it would appear to "soar." That is, the weight per unit length is not constant, but smaller at the top. This would make the curve flatter. The arch is an equilateral triangle in cross-section, with sides from 54 ft at the base to 17 ft at the top, and 630 ft apart at the base. The outer skin is type 304 stainless steel, and the lower parts of the legs are strengthened with reinforced concrete. For further information, photographs and tickets, consult St. Louis Arch. Definition Oxford in homework | English by of homework is services north helpline Homework for schuylkill assignment than the London Eye (450 ft), the Washington Monument (555 ft) or the San Jacinto Monument (570 ft), but shorter than the Eiffel Tower (984 ft). Although it is a catenary, perhaps we can regard it as a monument to all Parabolas. There is a Wikipedia article "Parabola" that is worth reading. A link towhere Essay Help - College buyworkonlineessay.org Entrance bad explanation of the construction using a Package Pens, Xtreme of 7 Yasutomo Gel Metallic GX1007 and string will be found, is included. Composed by J. B. Calvert Created 3 May 2002 Last revised 18 July 2011.