機械外文翻譯文獻翻譯制訂YD-65油鋸右箱工序卡及銑鏜結(jié)合面夾具設(shè)計相關(guān)外文翻譯

1、 畢業(yè)設(shè)計(論文)外文資料翻譯學(xué)院 (系)： 機械工程院 專 業(yè)： 機械工程及自動化 姓 名： 周雷 學(xué) 號： 0601510172 外文出處： The Parametric and Nonparametric Splines 附 件： 1.外文資料翻譯譯文；2.外文原文。 指導(dǎo)教師評語：此翻譯文章翻譯用詞比擬準確，文筆也較為通順，為在以后工作中接觸英文資料打下了根底 簽名： 年 月 日注：請將該封面與附件裝訂成冊。附件1：外文資料翻譯譯文與的比照 利用三次樣條函數(shù)的好處如下是： 1. 他們簡化計算的必要條件和數(shù)字的不穩(wěn)定性由高階的曲線引起的。 2. 他們允許有轉(zhuǎn)折點的最低階的三維曲線。 3.

2、 他們在空間中有能力扭曲。在這章中我們將提出兩種類型的樣條參量性的和非參量性的樣條，我們在這里負責(zé)解釋根本的數(shù)學(xué)推導(dǎo)和舉例論證他們的工具的任務(wù)。 4.7 拋物線的三次樣條函數(shù)考慮在平面內(nèi)由隨著變化描繪所得的一組數(shù)據(jù)點。我們的結(jié)果是要在所有的這些點之間通過一參量性的三次樣條函數(shù)。參量性的三次樣條函數(shù)是表示為一或多個參量的函數(shù)的曲線。在任何兩點之間參量性的三次樣條函數(shù)等式是根據(jù)參數(shù)得到的，如下： (4.56) 和 根據(jù)邊界條件和曲線的連續(xù)性和穩(wěn)定性而決定的常數(shù)。注意在任何兩點之間如何定義精確的距離。如果距離是標準的，因此它的涵義是從0到1。在時，樣條與系數(shù)相等。從而， for (4.57)我們在這

3、個時候目標是要求在每一時間間隔之間常數(shù)的值。參數(shù)的弦長定義為 當 (4.58)求其它常數(shù)的值的方法如下。考慮這三點，, 和 。讓在 和 之間的弦長為和在 和 之間的弦長為。讓為在 和 之間參量性的三次樣條函數(shù)和為在 和 之間參量性的三次樣條函數(shù)。因為在開始和在結(jié)束，的涵義是應(yīng)該在從0開始和在以結(jié)束。實際上當它們是被定義點所需要的時，在等式4.56中定義常數(shù)有 和 成分。按照x軸向和y軸向分量兩者所表示的參量性樣條函數(shù)的一般關(guān)系式如下被表達： (4.59)式中 和 再次注意到當我們在如何求的值以及它的導(dǎo)數(shù)的時候，我們得到 (4.60) (4.61)因此 (: 控制頂點) ( 未知) (4.62)

4、同樣地，我們以在點 和 寫入導(dǎo)數(shù) (4.63) (4.64) (4.65)我們由等式4.56定義三次樣條函數(shù),當我們代替常數(shù) 和 同從等式4.60和4.61獲得的 和 的時候，采取以下的形式： (4.66)在控制頂點 的連續(xù)性使我們得出 (4.67)從那我們求出 和 。因為的 和 和 是的函數(shù)，它是更多符合需要的表示它們 (4.68)現(xiàn)在我們可以求出適合 和 的表達式當作和的函數(shù)。利用等式4.67和4.68，我們得到 (4.69) (4.70)因此，那樣條函數(shù)在 和 之間可以簡單的表示為 (4.71)在計算機圖形處理的環(huán)境中和通用算法的開展中，我們需要問以下問題： 1.我們怎樣才能形成為所有的

5、三次函數(shù)解決 和 的方法？ 2.我們怎樣選擇, 和 而得到數(shù)據(jù)集點？ 3.我們怎樣確定在節(jié)點中樣條函數(shù)之間的連續(xù)性？總之，等式4.71能對于任何兩個相鄰的立方局部進行歸納而得到解答，例如當時的 和 ，為數(shù)據(jù)點的數(shù)目。為一般的數(shù)據(jù)集改寫等式4.71，我們得 (4.72)答復(fù)前面的問題，我們首先指出那個確定在立方局部之間的連續(xù)性，我們必須計算和的第二階導(dǎo)數(shù)與在他們的相應(yīng)的相連點方面把他們等同起來。從等式4.56，我們得到 (4.73) (4.74) (4.75)我們也能分辨那邊界條件 (4.76)利用等式4.73，4.74，和4.75與等式4.40，4.41，和4.42一起我們得到 4.77在矩陣

6、形式中，等式4.77是被明確地寫成的，顯示了那等式的重要特征。簡而言之， (4.78)等式4.78得到含未知數(shù)的個等式是顯而易見的。本質(zhì)上，為了求出未知數(shù)，我們必須按照的兩個附加等式。另一方面，如果端點 和 ，通常就是這樣在射束偏轉(zhuǎn)中分析,然前方程組結(jié)果形成一致的聯(lián)立方程式為我們求出所有的未知數(shù)?，F(xiàn)在我們能檢驗邊界條件使上述問題徹底地解決。邊界條件自然樣條函數(shù)。 亦稱衰減條件，自然樣條函數(shù)由根據(jù)時間函數(shù)從開始階段和到0結(jié)束所設(shè)定第二階導(dǎo)數(shù)來確定。因此， (4.79) (4.80)按照寫出這些條件，我們得到兩個等式 (4.81和 (4.82)增加等式4.81和4.82給等式4.49個方程，我們因

7、此能求出所有的定位樣條函數(shù)。 為這樣條函數(shù)提出的邊界條件是以致于在 和 的第一階導(dǎo)數(shù)斜率被確定。它們必須在等式4.78里構(gòu)成附加的其他兩個等式。 總結(jié) 在任何兩點之間參量性的三次樣條函數(shù)構(gòu)造如下： 1. 求出最大弦長和計算。 2. 利用等式4.78和相應(yīng)的邊界條件求出。 3. 利用等式4.62，4.69，和4.70求出組成參量性三次樣條函數(shù)的系數(shù)。范例4.5為以下數(shù)據(jù)集(1,1), (1.5,2), (2.5,1.75), 和(3.0,3.25)，求出參數(shù)的三次樣條函數(shù)在基點的兩個末端假設(shè)一種衰減形式。解答圖表4.7 i xiyi ti+1 0 1 1 1.118 1 1.5 2.0 1.03

8、1 2 2.5 1.75 0.707 3 3.0 2.25 我們首先計算弦的跨度 計算所必須的精確等式從等式4.74獲得 i=0) i=1) i=2) i=3) (4.83)利用等式4.81和4.82得到邊界條件，與上述等式4.54或者簡單地利用等式4.78一起，我們得 (4.84)式中 (4.85)= (4.86)為了給作解答，我們經(jīng)過倍增等式4.84，使其自動地得到常數(shù)，得出 式中 (4.87)現(xiàn)在我們使用等式4.69得到 for i=1,2,3 (4.88) (4.89)相似地,等式4.68給出常數(shù) (4.90) (4.91)總之，我們已經(jīng)得到聯(lián)接所有的4個數(shù)據(jù)點的所有的三條樣條和他們在

9、他們的明確形式中被表達.。 (4.92)在圖4.10中表示。圖4.10 參數(shù)的三次曲線是等式4.92得出的。范例4.5結(jié)束4.8 非參量性的三次樣條函數(shù)非參量性的三次樣條函數(shù)被定義為是有唯一的單參數(shù)的函數(shù)的曲線。非參量性的三次樣條函數(shù)允許在參數(shù)值和那三次樣條函數(shù)的數(shù)值之間的直線變量的關(guān)系式來決定。這從它的數(shù)學(xué)表達式中可看出： (4.93)從等式4.93，我們注意到那三次樣條函數(shù)是單獨的函數(shù)。如此，我們可以說在范圍內(nèi)的時間間隔中適合于的數(shù)據(jù)集點的判定，我們必須建立經(jīng)過所有這些點的樣條。讓每一子區(qū)間由來表示；因此，我們的任務(wù)將求出這些間隔中的每個三次樣條函數(shù)。再次，我們必須得到一個為常數(shù), 和作解

10、答的算法。三次樣條函數(shù)是由三次局部樣條組成。每個點有 和 數(shù)值；因此，那的函數(shù)是為所有的點定義。對那間隔，我們可以寫 (4.94) (4.95)考慮那三次樣條函數(shù)的平滑性和連續(xù)性，從以下的情況得到： (4.96) (4.97)那非參量性的三次樣條函數(shù)適合于任何間隔 可以表示為 (4.98)它的第一和第二導(dǎo)數(shù)是 (4.99) (4.100)由等式4.94到4.100得到樣條的利用標準，我們推斷出以下的： (4.101) (4.102)和 (4.103) (4.104) (4.105)式中 因為所有的的值是的，我們可以利用等式4.102和4.105求出： (4.106) 本質(zhì)上，上述等式適合于利用

11、 和 求出的結(jié)果。類似這樣的函數(shù)，如果我們使用 和 ，我們將得到另一個表達式，如下： (4.107)等式4.106和4.107定義同樣的。一旦我們使他們相等，他們就會變成一個依據(jù)未知的等式 (4.108)當系數(shù)被確定時，我們再一次在一矩陣形式中寫出上述等式 (4.109)等式4.109由等式同未知數(shù)組成；因此,它不能被求解。但是，樣條的端點 和 通常被認為必須滿足完全的邊界條件。通過 和 ，等式4.109用于通過值求出其余的。上述等式可以用緊湊結(jié)構(gòu)來表示，如下 for i=1, (4.110) 和 可以當作它們隨精確地系數(shù)來分別地計算依次，樣條的等式確定通過由等式4.76得出的隨著由等式4.1

12、06得出的來計算。 (4.111) for i=0, (4.112)4.9 邊界條件自然樣條函數(shù)在自然樣條函數(shù)的邊界條件中由在曲線的開始和末端的點設(shè)定第二導(dǎo)數(shù)為0時而被求出。因此， (4.113)當代入等式(4.66)時使我們得出 (4.114) 定位樣條函數(shù)定位的邊界條件由在at 和 確定第一導(dǎo)數(shù)斜率來求出。簡而言之， (4.115)和 (4.116)當是一個指定函數(shù)。以下范例說明這種方法計算非參量性的三次樣條函數(shù)和使它的有用的局部最顯著。注意在唯一的一個簡單化的方法中我們已經(jīng)采用樣條的根本概念；它被留給讀者去研究在這個最主要的曲線擬合法之后數(shù)學(xué)的推進。范例4.6為在下面的表格中顯示出的點求

13、出非參量性的三次樣條函數(shù)自然樣條函數(shù)。 i 0 1 1 0.5 1 1.5 2 1 n=2 2.5 1.75 解答：第1步：控制點,時間間隔和第2步：求出：自然樣條(=0) 第3步：求出 和 for i = 0, for i = 1, for i = 0, 結(jié)果如下 i 0 1 0.5 1 2.375 0 -1.5 1 1.5 1.0 2 1.25 -2.25 0.75 2.5 1.75 0 圖4.11 非參量性的三次樣條函數(shù)。4.10 貝塞爾曲線貝塞爾曲線的形狀是由那位置上的交點來定義的，并且那曲線不能與除邊界點之外的所有的點相交。在確實的情況中，存在著不適當?shù)慕稽c或不適合地定位點，那三次樣

14、條函數(shù)的方法在不判定更多點時不能形成平滑曲線。貝塞爾曲線允許非限制曲線的彎曲度適當?shù)刎灤┧械狞c。這樣可設(shè)想曲線的形狀適合由一系列的點所定義的多邊形。貝塞爾曲線的數(shù)學(xué)根底影響彎曲的形狀的重量因素與伯恩斯坦根底相關(guān)，通過以下 (4.117)式中 和定義為 (4.118)在有序集合中是多項式的階和是特殊的極點在0到之間。那曲線的交點被定義為 (4.119)從 到 ，和等相當于不同的點的矢量分量。 為了建立貝塞爾曲線，我們必須求的值，它有參數(shù)的函數(shù)。它是假設(shè)在時存在最大值的函數(shù)和是由其給予的 (4.120)以下范例說明貝塞爾曲線的曲線擬合法。范例4.7判定貝塞爾曲線經(jīng)過以下各點： 求出貝塞爾曲線經(jīng)過

15、這些點的間距。解答 我們注意到4個點構(gòu)成貝塞爾曲線的多邊形。因為我們有4個判定極點，那么。利用等式(4.117)我們可求出函數(shù)的值，式中 (4.121)因此， (4.122)因為不同的的值，在圖表4.8中可以求出貝塞爾曲線適合的系數(shù)。結(jié)果是的函數(shù)是求出方式 那答案被繪制在圖4.12。 圖4.12 貝塞爾曲線表示為圖表4.8 圖表4.8 貝塞爾曲線函數(shù)的賦值 按照參數(shù) t 0 1 0 0 0 0.15 0.614 0.325 0.0574 0.0034 0.35 0.275 0.444 0.239 0.043 0.5 0.125 0.375 0.375 0.125 0.65 0.043 0.23

16、9 0.444 0.275 0.85 0.0034 0.0574 0.325 0.614 1 0 0 0 1范例4.7結(jié)束4.11 貝塞爾曲線等式的區(qū)別貝塞爾曲線利用一階乘積的表達式和需要為了簡單化的計算而需要限制緊湊結(jié)構(gòu)的的特征函數(shù)。我們知道任何曲線的函數(shù)必須是在交點那個時候的區(qū)別，是定位最小值或最大值，斜率，和邊界點的必要條件。讓我們從那貝塞爾曲線正如等式(4.119) 所定義的開始，式中 (4.123)利用由等式(4.123) 局部地區(qū)別，我們獲得 (4.124)讓我們尋找一個由 和 兩者都為了完成我們的區(qū)別的表達式。注解變?yōu)榱闶且驗樗跁r表示出的特征值，和 (4.125)代替等式(4.

17、125)變成等式(4.124)使我們得出貝塞爾曲線的區(qū)別 (4.126)上述等式注意到左邊和右邊的關(guān)系和標志在 和 通過 隨著 在左邊關(guān)系簡單地轉(zhuǎn)換使我們可以引導(dǎo)的數(shù)值。簡而言之 (4.127)式中 和 然后等式(4.101)取以下的式子 (4.128)4.12 B樣條曲線在貝塞爾曲線中B樣條曲線被引入克服一些薄弱局部?？雌饋砜刂祈旤c的數(shù)目影響曲線的真實度。此外，在貝塞爾曲線中的混合函數(shù)的性質(zhì)無法考慮到一個比擬容易的方法修正當前曲線的形狀。B樣條曲線被定義為如下： (4.129)式中 (4.130)和 其余的全部 (4.131)此外， 被認為是節(jié)點值并需要求值是為可獲得所有的的函數(shù)。知道是一常

18、數(shù)；所以，根據(jù)等式(4.130)是比例為1的一個函數(shù)。類似這樣式子我們可看見有比例為，當是更大的時就要求按1的比例。的數(shù)值明確表示B樣條曲線的種類。這有兩種節(jié)點的類型：a) 周期性的節(jié)點： (4.132)b) 非周期性的節(jié)點： (4.133)當明確表示周期性的節(jié)點的曲線無法經(jīng)過第一個和最后一個點時，就由貝塞爾曲線來確定，反之非周期性的節(jié)點確定第一個和最后一個點經(jīng)過曲線。這個是由于k函數(shù)的雙重節(jié)點在第一和最后的節(jié)點如等式(4.133)描述。范例4.8用例3定義B樣條曲線適合非周期性統(tǒng)一的節(jié)點。因為控制點的彎曲由,和 給予的。解答：當那相鄰的節(jié)點之間的間隔總是1時，這個定義那統(tǒng)一的節(jié)點情況。從等式

19、(4.133) 我們獲得那節(jié)點值如下： 和 (注解 and )從等式(4.130)我們得到的函數(shù)。我們需要求出的函數(shù)相當于按要求排序點1, 2, 和 3。從 到 我們定義的混合函數(shù)。種類1. 讓我們求解全部可能的函數(shù)。 ( 4.134) 從以上可知在 和 時我們需要選擇非零的函數(shù)。因為我們選擇是在范圍(0,1)內(nèi)有特征值為1的僅有的非零的函數(shù)。種類2. 我們獲得種類2的函數(shù)如下： (4.135) 附件2：外文原文復(fù)印件FIGURE4.9 Graph for versus . The benefits of using cubic splines are as follows: 1. They

20、reduce computational requirements and numerical instabilities that arise from higher-order curves. 2. They have the lowest-degree space curve that allows inflection points. 3. They have the ability to twist in space.In this chapter we will introduce two types of splines (the parametric and the nonpa

21、rametric splines), where we undertake the task of explaining the basic mathematical derivation and provide examples to demonstrate their implementation.4.7 Parabolic Cubic Splines Consider a set of data points described in the -plane by with . Our objective is to pass a parametric cubic spline betwe

22、en all these points. A parametric cubic spline is a curve that is represented as a function of one or more parameters.The parametric cubic spline equation between any two points is given in terms of a parameter as follows: (4.56) and are constants that are determined from the boundary conditions and

23、 the continuity and smoothness of the curve. Observe how defines a precise length between any two points. So its value goes from 0 to 1 if the length is normalized. At , the spline is equal to the coefficient . Hence, for (4.57)Our objective at this point is to evaluate the constants between each in

24、terval.Parameter cord length is defined by for (4.58)The procedure for evaluating the other constants is as follows.Consider three points, , and . Let the chord length between and be and the one between and be . Let be the parametric cubic spline between and and the one between and . Because starts

25、at and ends at , the value of should start from 0 at and end with at . In reality the constants defined in Equation (4.56) have and components as they are needed to define points. A general form of the parametric spline expressed in terms of both and components can be expressed as follows: (4.59)whe

26、re and Again observe how when we evaluate at as well as its derivative we obtain (4.60) (4.61)Therefore (Knowns: control points) ( Unknowns) (4.62)Similarly, we write the derivatives at points and as (4.63) (4.64) (4.65)Our cubic spline defined by equation (4.56), when we substitute the constants an

27、d with and obtained from Equations (4.60) and (4.61), takes the following form: (4.66)Continuity at the control points yields (4.67)from which we solve for and . Because is known and and are functions of it is more desirable to express them as (4.68)Now we can find the expression for and as function

28、s of and . Using Equations (4.67) and (4.68), we get (4.69) (4.70)Therefore, the spline function between and could simple be expressed as (4.71) In the context of computer graphics and general-purpose algorithm development, we need to ask the following questions: 1. How can we generate a solution fo

29、r and for all cubic functions ? 2. How do we select , and for a given set of data points? 3. How do we ensure continuity between the splines at knots ? In any case, the solution given by Equation (4.71) can be generalized for any two adjacent cubic segments such as and for , where is the number of d

30、ata points. Rewriting Equation (4.71) for a general data set we get (4.72)To answer the foregoing questions, we first note that to ensure continuity between the cubic segments, we need to compute the second derivative of and and equate them at their corresponding connecting points. From Equation (4.

31、56), we obtain (4.73) (4.74) (4.75)We also know from the boundary conditions that (4.76)Using Equations (4.73), (4.74), and (4.75) together with Equations (4.40), (4.41), and (4.42) we obtain 4.77In matrix form, Equation (4.77) can be written explicitly to show the important feature of the equation.

32、 That is, (4.78)It is obvious that Equation (4.78) yields equations with unknowns. Essentially we need two additional equations in terms of in order to solve for the unknowns. On the other hand, if end points and are know, as is the case in beam-deflection analysis, then the system of equations resu

33、lts in a consistent set of equations for which we can solve for all the unknowns. Now we can examine the boundary conditions to complete the solution to the above problem.Boundary Conditions Natural Spline. Also known as relaxed conditions, natural splines are determined by setting the second deriva

34、tives of with respect to time at the beginning and end to 0. Thus, (4.79) (4.80)Writing these conditions in term of , we obtain two equations (4.81)and (4.82)Adding Equations (4.81) and (4.82) to the equations given by Equation (4.49), we can then solve for all the . Clamped Spline. The boundary con

35、ditions for this spline are such that the first derivatives (slope) at and are specified. Hence, they form the additional two other equations needed in Equation (4.78). SummaryThe parametric cubic spline between any two points is constructed as follows: 1. Find the maximum cord length and determine

36、. 2. Use Equation (4.78) together with the corresponding boundary conditions to solve for the . 3. Solve for the coefficients that make up the parametric cubic splines using Equations (4.62), (4.69), and (4.70).Example 4.5For following data set (1,1), (1.5,2), (2.5,1.75), and (3.0,3.25), find the pa

37、rametric cubic spline assuming a relaxed condition at both ends of the data.Solution:TABLE4.7 i xiyi ti+1 0 1 1 1.118 1 1.5 2.0 1.031 2 2.5 1.75 0.707 3 3.0 2.25 we first compute the cord length The explicit equations needed to evaluate the are obtained from Equation (4.74) i=0) i=1) i=2) i=3) (4.83

38、)Using the boundary conditions given by Equations (4.81) and (4.82), together with either the above Equation (4.54) or simply making use of Equation (4.78), we get (4.84)where (4.85)= (4.86)To solve for we multiply Equation (4.84) by which automatically yields the constants given by where s (4.87)We

39、 now use Equation (4.69) to find for i=1,2,3 (4.88) (4.89)In a similar fashion, Equation (4.68) gives the coefficients (4.90) (4.91)In conclusion, we have derived all three splines joining all four data points and they are expressed in their explicit forms. (4.92)The display of this is given in Figu

40、re 4.10. FIGURE 4.10 Parametric cubic curve given by Equation (4.92)End of Example 4.54.8 Nonparametric Cubic SplineA nonparametric cubic spline is defined as a curve having a function of only one parameter. Nonparametric cubic splines allow a direct variable relationship between the parameter value

41、 and the value of the cubic spline function to be determined. This is seen from its mathematical representation: (4.93) From Equation (4.93), we see that the cubic spline is a function of alone. Thus, we could say that for a given set of data points defined in the interval in the domain , we need to

42、 construct the spline that passes through all these points. Let each subinterval be denoted by ; hence, our task is to find the cubic spline function for each of these intervals.Once more, we must find an algorithm to solve for the constants , and . Cubic spline is composed of cubic segment splines.

43、 Each point has an and value; hence, the function is defined for all points. For the interval , we can write (4.94) (4.95) By considering the smoothness and continuity of the cubic splines, the following conditions are derived: (4.96) (4.97)The nonparametric cubic spline function for any interval co

44、uld be expressed as (4.98)Its first and second derivatives are (4.99) (4.100)Making use of the criteria of the spline given by Equations (4.94) to (4.100), we deduce the following: (4.101) (4.102)and (4.103) (4.104) (4.105)where Because all the values are known, we can solve for using Equations (4.1

45、02) and (4.105): (4.106)In essence, the foregoing equation for was the result of using and . In a similar fashion, if we use and , we will get another expression as follows: (4.107)Equations (4.106) and (4.107) define the same . Once we equate them they result into an equation in terms of the unknow

46、n (4.108)Once again we write the above equation in a matrix form, where the coefficients are to be determined (4.10)Equation (4.109) consists of equations with unknowns; therefore, it cannot be solved. However, end points and of the spline are usually known through the boundary conditions that must

47、be supplied. By knowing and , Equation (4.109) is then used to solve for the remaining through values. The above equation can be expressed in a compact form as for i=1, (4.110) and can be computed separately as they depend strictly on known coefficients.In turn, the equation of the splines can be de

48、termined by computing the from Equation (4.76) followed by the from Equation (4.106). (4.111) for i=0, (4.112)4.9 Boundary Conditions Natural SplinesThe boundary conditions in natural splines are found by setting the second derivatives at both the beginning and end points of the curve to 0. Therefor

49、e, (4.113)which when substituted into Equation (4.66) yields (4.114) Clamped SplinesThe clamped end conditions are determined by specifying the first derivatives (slope) at and . That is, (4.115)and (4.116)where is a specified function. The following example illustrates the method used to evaluate t

50、he nonparametric cubic splines and highlights its usefulness. Note that we have introduced the concepts of splines in only a simplistic way; it is left for the reader to explore further the mathematics behind this most important curve-fitting method.Example 4.6Find the nonparametric cubic spline (na

51、tural spline) for the points shown in the table below. i 0 1 1 0.5 1 1.5 2 1 n=2 2.5 1.75 Solution:Step 1: Control points, intervals, and Step 2: Solve for : natural spline (=0) Step 3: Solve for and for i = 0, for i = 1, for i = 0, The results are compiled in the following table and shown in Figure

52、 4.11. i 0 1 0.5 1 2.375 0 -1.5 1 1.5 1.0 2 1.25 -2.25 0.75 2.5 1.75 0 FIGURE 4.11 Nonparametric cubic spline function.4.10 Bezier CurvesThe shapes of Bezier curves are defined by the position of the points, and the curves may not intersect all the given points except for the endpoints. In certain c

53、ircumstances, where there are insufficient points or awkwardly located points, the cubic spline method may not provide a smooth curve without defining more points. Bezier curves allow the flexibility of not constraining the curve to fit through all the points. One can imagine the shape of the curve

54、to fit in a polygon defined by a series of points. The mathematical bases (the weighing factor that affects the shape of the curve) of the Bezier curve is related to the Bernstein basis given by (4.117)where and is defined as (4.118)where is the degree of the polynomial and is the particular vertex

55、in the ordered set (between 0 and ). The curve points are defined by (4.119)where to , and the contain the vector components of the various points. In order to construct the Bezier curve, we need to evaluate the , which are functions of parameter . It is seen that the maximum value of the function o

56、ccurs at and is given by (4.120)The following example illustrates the Bezier curve method of curve fitting.Example 4.7Define the Bezier curve that passes through the following points: Find the Bezier curve space that passes through these points.Solution We note that the four points form the Bezier p

57、olygon. Because we have four defined vertices, then . Using Equation (4.117) we evaluate the function, where (4.121)Therefore, (4.122)For various values of , the coefficients for the Bezier curve are found in Table 4.8. The resulting function is then found as The results are plotted in Figure 4.12.F

58、IGURE 4.12 Bezier curve representation for Table 4.8 TABLE 4.8 Evaluation of the Bezier function in terms of the parameter t 0 1 0 0 0 0.15 0.614 0.325 0.0574 0.0034 0.35 0.275 0.444 0.239 0.043 0.5 0.125 0.375 0.375 0.125 0.65 0.043 0.239 0.444 0.275 0.85 0.0034 0.0574 0.325 0.614 1 0 0 0 1End of E

59、xample 4.74.11 Differentiation of Bezier Curve EquationThe Bezier curve makes use of an expression that uses a factorial and a function that needs to be kept in a compact form in order to simplify the calculation. We know that any curve function needs to be differentiated when intersection points, m

60、inimum or maximum, slopes, and endpoint conditions are sought. Let us begin with the Bezier curve as defined by Equation (4.119), where (4.123)using differentiation by parts for Equation (4.123) we obtain (4.124)Let us seek an expression for both and in order to complete our differentiation. Note th