Exploring the Bezier curves.docx

1、Exploring the Bezier curvesExploring the Bezier curvesA Bezier curve is a parametric curve which usually used in computer graphics to model smooth curves that can be scaled indefinitely and also used in animation as a tool to control motion. Bezier curves are also used in the time domain, particular

2、ly in animation and interface design, e.g., Bezier curves can be used to specify the velocity over time of an object such as an icon moving from A to B, rather than simply moving at a fixed number of pixels per step. When animators or interface designers talk about the physics or feel of an operatio

3、n, they may be referring to the particular Bezier curves used to control the velocity over time of the move in question. ( http:/en.wikipedia.org/wiki/B%C3%A9zier_curve )This investigation is aimed to use the knowledge of vectors and parametric equations to explore the mathematics of computer aided

4、design curves. In this investigation, coordinate axes, graphs, including axes intercepts; vectors, including vector addition, scalar multiplication and ratio division; quadratic and cubic polynomials will be used. First, a simple example will be used to show how the parametric equation works. Then,

5、our knowledge and calculator will be used to explore more about Bezier curves.To begin to give a simple explanation about how the parametric equation works. There is a line segment start from point A (-3, 4) to point B (2, 5). On this line segment, there is a point P moving from A to B. So P divided

6、 AB in the ratio t to 1-t, and. When t = 0, P is at the initial point A. When t=1, it is at the end point B.tAPBO According to the diagram, and use the division formula, we can find that. Because it is a arbitrary point on AB, so its parametric equation of AB is,. Show it in graphic calculator isABT

7、hen, use the same method and follow the same step to find another line segment BC, where C is (4, 6).Now, Q is the moving point on BC. Use the same method, we can find that. So the parametric equation of BC is. Show it in graphic calculator isCBAIn the graph, both points P and Q move from the initia

8、l point to the end point, and there is an invisible line segment between P and Q. The line segment created by PQ is moving too. Assume that there is another point S lies on PQ, and S moves from P to Q just like P moves from A to B and Q moves from B to C with t changing from 0 to 1. It means that wh

9、en P starts at A (-3, 4), S also starts at A (-3, 4), and when Q arrives at C (4, -6), S also arrives at C (4, -6). In order to get the path of S, previous results will be used. Since S also divides PQ in the ratio of t to 1-t,So the parametric equation for the arc AC isThen enter the equation on th

10、e graphic calculation, and graph the curve. We can getCBAFrom the graph, we can find that AB and BC are the tangent lines of arc AC.Use the parametric equation, the coordinates of the curve where it cuts the axes can be found.When x=0:So it cuts the Y-axes at (0, 10/3)When y=0:So it cuts the X-axes

11、at (-14/3, 0)Now we are going to use the knowledge we get previously to draw a picture of Nemo. The picture is a combination of six arcs. The coordinate of points are:H (-5,1) ; I (-1,5) ; J (4,2) ; K (1,-6) ; L (-1,-2) ; M (-2,-4) ; N (1,-3) ; R (2,1.5) ; T (1,-2) ; V (5,6) ; W (7,-1) ; X (4,-2)RJW

12、VITXLHMNKTake the dorsal fin into consideration first. It has an initial point E (-2, 3) and end point G (2, 3). Then, use our ruler, we can find the sharp point F (3, 8). We can image that P is always the point moving from initial point to sharp point and Q is always moves from sharp point to end p

13、oint. Then S is always moves from P to Q.Using the same method used above, the parametric equation of EF and GF can be found. And therefore, the parametric equation for arc EG can be found. So EF= So FG= So the parametric equation of arc EG isEnter the equation into the calculator will getGFERepeat

14、the process for the remaining and graph all six arcs with the calculator. So the parametric equation of arc HJ is So the parametric equation of arc HJ2 is So the parametric equation of arc LN is So the parametric equation of arc TN is So the parametric equation of arc VW is Enter these six arcs into

15、 calculator will get a picture of Nemo.Because all curves above are parts of parabolas, they are called parabolic arcs.Next step is going to do more about the parabolic curves.Given that initial point is L (-2, 4), sharp point M (0, -4) and end point N (2, 4).Use the same method:Enter it into calcul

16、ator and add the line segment LM and MN can get:MNLFrom the graph, LM，MN both are the tangent line of curve LN.So arc LN is a part of parabolic curve Another parabolic arc has initial point U (0, 0), sharp point V (0, 1) and end point W (4, 2). Enter it into calculator and add the line segment UV an

17、d VW can get:WUVLine segments UV and VW are tangent lines of curve UW.From the parametric equation, Cartesian equation can be found.So the Cartesian equation is Because parabolic arcs are all have three points, so the parametric equation can be found in a regular way. Assume that there is a paraboli

18、c arc with initial point A, sharp point B and end point C. The parametric equation can be found.So the parametric equation is This equation can be used to check the work we have done.Arc AC=Arc EG=Arc LN=Arc UW=This equation also can be used to find a new arc with initial point (-1, 5), sharp point

19、(1, 5) and end point (3, 1). So it is a part of the parabola: All of the arcs found above are called parabolic arcs. They are all controlled by three points. In order to create a more flexible curve, another point can be added to the graph. And they can create a non-parabolic arc called Bezier curve

20、. First arc can be used again to find a Bezier curve. Then another point D (6, 0) will be added.Use the equation to find arc BD: Arc AC is Then enter them into calculate.BDCAAssume that there is a point T moves from B to D on arc BD and S from A to C on arc AC. As t increases from 0 to 1, line segme

21、nt ST starts as AB when t=0 and finishes as CD when t=1. Now assume again that there is a point Z on line segment ST and dividing ST in the ratio t to 1-t. Then use the same method, the parametric equation of arc AD can be found.So the parametric equation for arc AD is: CDBAThe arc AD is called the

22、Bezier curve with initial point A, sharp point B and C and end point D.With the work done before, the general equation also can be determined.Let D be S is T is So the parametric equation is:In conclusion, if an arc is controlled by an initial point, a sharp point and an end point, we call this arc parabolic arc. Its equation is.If there is a fourth point also control one arc, the arc can be more flexible and non-parabolic. It is called Bezier curve. And its equation isBibliography: http:/en.wikipedia.org/wiki/B%C3%A9zier_curve