英语翻译如翻译不错立即采纳加分.其中很多公式不用翻译也不用管,只需翻英语部分.2.Proposed method2.1.
来源:学生作业帮 编辑:搜搜做题作业网作业帮 分类:英语作业 时间:2024/08/19 01:03:59
英语翻译
如翻译不错立即采纳加分.其中很多公式不用翻译也不用管,只需翻英语部分.
2.Proposed method
2.1.Chord properties
In this section,we introduce the property that relates the center of a circle to its chord and a point on the chord.
Fig.1(a) shows a circle and its chord AB.Let P be an interior point of the circle with its center at O and AB be the chord passing through P in a θ direction.The point A is defined as the endpoint in the quadrant I or II of the Cartesian coordinate system centered at P,and the point B defined as one in the quadrant III or IV.Let θ be a positive angle measured in a counterclockwise direction from PX to AB where θ ∈[0°,180°].Likewise,let φ be an angle measured from PX to OP.Then,
|AP|-|BP|=2|OP|cos(θ-φ).(1)
The proof is straightforward in the following.It is well-known that the perpendicular bisector of any chord of a circle passes through its center O (Rich,1963).Let C be the midpoint of the cord AB.
Then,
|AC|=|BC| and OC⊥AB.
So,
|CP|=|OP|cos(θ-φ).
|AP|-|BP|=(|AC|+|CP|)-(|BC|-|CP|)=2|CP|.
Therefore,
|AP|-|BP|=2|OP|cos(θ-φ)
In Fig.1(a),O(xo,yo) is related with the coordinates of P(xp; yp) as follows:
xo=xp+acosφ,
yo=yp-asinφ (2)
where a=|OP| and the minus sign in the equation of yo reflects that y value increases from top to bottom in the image coordinate system.If we have two chords passing through the point P,two equations are directly acquired from Eq.(1) so that a and φ can be solved with those two equations.As a result,(xo,yo) can be computed from Eq.(2).The details are given in the following section.
2.2.Computation for circle center
Consider two chords A1B1 and A2B2 passing through P(xp,yp) in the θ1 and θ2 directions (θ1≠θ2),respectively,as shown in Fig.1(b).
Let d1=|A1P|-|B1P| and d2=|A2P|-|B2P|.
Then,d1=2acos(θ1-φ) and d2=2acos(θ2-φ) from Eq.(1).
Solving for φ by dividing two sides of the equation of d1 by ones of the equation of d2 and expanding the cosine functions,we get
.一堆公式(略)
Eq.(4) shows us how the center of the circle can be computed from two intersecting chords.Specifically,if we have an interior point P of the circle and two chords passing through it,d1,d2,θ1 and θ2 given by these chords can locate the center of the circle from Eq.(4).One of the simplest cases is for θ1 = 0° and θ2 = 90°.
In that case,it can be easily shown that xo is the midpoint of the horizontal chord of θ1=0° and yo is the midpoint of the vertical chord of θ2=90°.This case is computationally efficient,but may not be robust because it uses only the vertical and horizontal chords.From this reason,the chords with randomly generated directions will be used along with Eq.(4) in this paper.
备注整理:至11-2-22,3 answers,1楼CV党,2楼朋友热心提议,3楼机器.
如翻译不错立即采纳加分.其中很多公式不用翻译也不用管,只需翻英语部分.
2.Proposed method
2.1.Chord properties
In this section,we introduce the property that relates the center of a circle to its chord and a point on the chord.
Fig.1(a) shows a circle and its chord AB.Let P be an interior point of the circle with its center at O and AB be the chord passing through P in a θ direction.The point A is defined as the endpoint in the quadrant I or II of the Cartesian coordinate system centered at P,and the point B defined as one in the quadrant III or IV.Let θ be a positive angle measured in a counterclockwise direction from PX to AB where θ ∈[0°,180°].Likewise,let φ be an angle measured from PX to OP.Then,
|AP|-|BP|=2|OP|cos(θ-φ).(1)
The proof is straightforward in the following.It is well-known that the perpendicular bisector of any chord of a circle passes through its center O (Rich,1963).Let C be the midpoint of the cord AB.
Then,
|AC|=|BC| and OC⊥AB.
So,
|CP|=|OP|cos(θ-φ).
|AP|-|BP|=(|AC|+|CP|)-(|BC|-|CP|)=2|CP|.
Therefore,
|AP|-|BP|=2|OP|cos(θ-φ)
In Fig.1(a),O(xo,yo) is related with the coordinates of P(xp; yp) as follows:
xo=xp+acosφ,
yo=yp-asinφ (2)
where a=|OP| and the minus sign in the equation of yo reflects that y value increases from top to bottom in the image coordinate system.If we have two chords passing through the point P,two equations are directly acquired from Eq.(1) so that a and φ can be solved with those two equations.As a result,(xo,yo) can be computed from Eq.(2).The details are given in the following section.
2.2.Computation for circle center
Consider two chords A1B1 and A2B2 passing through P(xp,yp) in the θ1 and θ2 directions (θ1≠θ2),respectively,as shown in Fig.1(b).
Let d1=|A1P|-|B1P| and d2=|A2P|-|B2P|.
Then,d1=2acos(θ1-φ) and d2=2acos(θ2-φ) from Eq.(1).
Solving for φ by dividing two sides of the equation of d1 by ones of the equation of d2 and expanding the cosine functions,we get
.一堆公式(略)
Eq.(4) shows us how the center of the circle can be computed from two intersecting chords.Specifically,if we have an interior point P of the circle and two chords passing through it,d1,d2,θ1 and θ2 given by these chords can locate the center of the circle from Eq.(4).One of the simplest cases is for θ1 = 0° and θ2 = 90°.
In that case,it can be easily shown that xo is the midpoint of the horizontal chord of θ1=0° and yo is the midpoint of the vertical chord of θ2=90°.This case is computationally efficient,but may not be robust because it uses only the vertical and horizontal chords.From this reason,the chords with randomly generated directions will be used along with Eq.(4) in this paper.
备注整理:至11-2-22,3 answers,1楼CV党,2楼朋友热心提议,3楼机器.
![英语翻译如翻译不错立即采纳加分.其中很多公式不用翻译也不用管,只需翻英语部分.2.Proposed method2.1.](/uploads/image/z/2541374-62-4.jpg?t=%E8%8B%B1%E8%AF%AD%E7%BF%BB%E8%AF%91%E5%A6%82%E7%BF%BB%E8%AF%91%E4%B8%8D%E9%94%99%E7%AB%8B%E5%8D%B3%E9%87%87%E7%BA%B3%E5%8A%A0%E5%88%86.%E5%85%B6%E4%B8%AD%E5%BE%88%E5%A4%9A%E5%85%AC%E5%BC%8F%E4%B8%8D%E7%94%A8%E7%BF%BB%E8%AF%91%E4%B9%9F%E4%B8%8D%E7%94%A8%E7%AE%A1%2C%E5%8F%AA%E9%9C%80%E7%BF%BB%E8%8B%B1%E8%AF%AD%E9%83%A8%E5%88%86.2.Proposed+method2.1.)
(二) 推荐方法
2.1 弦的特性
在本节里,我们介绍圆心与其弦及弦上一点有关的特性.
图1(a)显示一个圆圈与其弦AB.让P为圆圈内的一点,圆心为O,而AB是向θ方向经过P点的弦.以P 为中心的笛卡尔坐标系的四分之一圆周I 或II的端点定义为A,而B则定义为四分之一圆周III 或IV的端点.θ是由PX逆时针测量至AB的正角,θ ∈[0°, 180°].同样地,φ是由PX逆时针测量至OP的正角.那么
|AP|-|BP|=2|OP|cos(θ-φ). (1)
下列的证明很直接.众所周知,任何圆周弦的垂直平分线都会经过圆心O(Rich, 1963).以C作为AB的中心点,那么
|AC|=|BC| and OC⊥AB.
因此,
|CP|=|OP|cos(θ-φ).
|AP|-|BP|=(|AC|+|CP|)-(|BC|-|CP|)=2|CP|.
所以,
|AP|-|BP|=2|OP|cos(θ-φ)
图1(a)显示,O(xo,yo)与P(xp; yp)的坐标有下列的关系:
xo=xp+acosφ,
yo=yp-asinφ
而a=|OP|,yo方程里的减号反映在图像坐标系中,y值从上至下增加.如果我们有两条弦经过P点,可以从Eq. (1)直接得出两个方程,从中可解a 和 φ. 因此,可以从Eq. (2) 计算出(xo,yo).具体细节在下一节展示.
2.2 圆心的计算
考虑犹如图1(b)所示, A1B1 和A2B2两条弦是分别向θ1和θ2 方向 (θ1≠θ2)经过P (xp, yp).
以d1=|A1P|-|B1P| 及d2=|A2P|-|B2P|.
那么从Eq.(1),d1=2acos(θ1-φ) 及d2=2acos(θ2-φ)
将方程d1除以d2以及扩展余弦函数可解φ,我们得到
...一堆公式(略)
Eq. (4) 展示如何从两相交弦计算出圆心.具体地,如果圆周内有个P点被两条弦穿过,利用两弦所提供的d1, d2, θ1 及θ2,可以用Eq. (4) 计算出圆心位置.最简单的实例是θ1 = 0° 及θ2 = 90°.
如果是那样,可以极易地显示θ1=0°横弦的中点是xo,θ2=90°竖弦的中点是yo.这个实例的计算精确,但由于只是使用竖横弦,这种计算方式可能不稳健.因此,本研究将使用随意产生方向的弦来配合Eq. (4)的计算.
【英语牛人团】
2.1 弦的特性
在本节里,我们介绍圆心与其弦及弦上一点有关的特性.
图1(a)显示一个圆圈与其弦AB.让P为圆圈内的一点,圆心为O,而AB是向θ方向经过P点的弦.以P 为中心的笛卡尔坐标系的四分之一圆周I 或II的端点定义为A,而B则定义为四分之一圆周III 或IV的端点.θ是由PX逆时针测量至AB的正角,θ ∈[0°, 180°].同样地,φ是由PX逆时针测量至OP的正角.那么
|AP|-|BP|=2|OP|cos(θ-φ). (1)
下列的证明很直接.众所周知,任何圆周弦的垂直平分线都会经过圆心O(Rich, 1963).以C作为AB的中心点,那么
|AC|=|BC| and OC⊥AB.
因此,
|CP|=|OP|cos(θ-φ).
|AP|-|BP|=(|AC|+|CP|)-(|BC|-|CP|)=2|CP|.
所以,
|AP|-|BP|=2|OP|cos(θ-φ)
图1(a)显示,O(xo,yo)与P(xp; yp)的坐标有下列的关系:
xo=xp+acosφ,
yo=yp-asinφ
而a=|OP|,yo方程里的减号反映在图像坐标系中,y值从上至下增加.如果我们有两条弦经过P点,可以从Eq. (1)直接得出两个方程,从中可解a 和 φ. 因此,可以从Eq. (2) 计算出(xo,yo).具体细节在下一节展示.
2.2 圆心的计算
考虑犹如图1(b)所示, A1B1 和A2B2两条弦是分别向θ1和θ2 方向 (θ1≠θ2)经过P (xp, yp).
以d1=|A1P|-|B1P| 及d2=|A2P|-|B2P|.
那么从Eq.(1),d1=2acos(θ1-φ) 及d2=2acos(θ2-φ)
将方程d1除以d2以及扩展余弦函数可解φ,我们得到
...一堆公式(略)
Eq. (4) 展示如何从两相交弦计算出圆心.具体地,如果圆周内有个P点被两条弦穿过,利用两弦所提供的d1, d2, θ1 及θ2,可以用Eq. (4) 计算出圆心位置.最简单的实例是θ1 = 0° 及θ2 = 90°.
如果是那样,可以极易地显示θ1=0°横弦的中点是xo,θ2=90°竖弦的中点是yo.这个实例的计算精确,但由于只是使用竖横弦,这种计算方式可能不稳健.因此,本研究将使用随意产生方向的弦来配合Eq. (4)的计算.
【英语牛人团】
英语翻译如翻译不错立即采纳加分.其中很多公式不用翻译也不用管,只需翻英语部分.2.Proposed method2.1.
英语翻译如翻译不错立即采纳加分.越快越好!分数不是问题,浮云要多少都可以满足,若嫌分少可以另开若干高分贴让你复制粘贴采纳
英语翻译翻译,不用英语,
英语翻译只要翻译文字部分就好,化学式和公式不用翻译,好的话继续加分,辛苦了!(1) Calculate molecula
初二 翻译 数字不用管
英语翻译如图,不用管那些字,只翻译书上的内容,
英语作文,不用翻译,不要太多生词,必采纳,谢谢
英语翻译部分是韩国人,尽量按标准来翻译.还有如P.也要正确翻出来,除非习惯上不用翻译.Andrew P.M.Samuel
英语翻译不用翻译了,
英语翻译不用逐字翻译
英语高手,请帮忙翻译一下!(乱码不用管)
英语翻译下面是我翻译的部分,是靠谷哥帮我翻译得,很烂,并且里面的专业名称也不准确,不用作为参考.需要一位英语水平不错的人