参数化功能回旋 [英] Parameterized function for clothoid
问题描述
我编码渲染路网，其中基础上， RoadXML 格式。
I'm coding renderer for road network, which based on the RoadXML format.
这是这种格式公路曲线有四种类型：
Road curves from this format has four types:
- 段，
- 圆弧，
- 在多行，
- 克洛索弧。
和我有问题，最后一个。
And I have problem with the last one.
回旋是相同的欧拉和螺旋大角螺旋。在RoadXML克洛索弧是由三个参数给出：
Clothoid is the same with Euler spiral and Cornu spiral. In the RoadXML clotho arc is given by three parameters:
- 在开始弯曲，
- 在结束曲，
- 的长度。
有关弧线三角，我需要类似foo（T），它返回（X，Y）COORDS在t = 0..length功能。我创建了类似的方法圆弧没有问题，但我不能做它克洛索弧。
For arc triangulation I need a function like foo(t), which returns (x, y) coords for t = 0..length. I created similar methods for circle arc without problems, but I can't do it for clotho arc.
问题的部分原因是，我并不完全了解如何应用标准回旋公式开始和结束曲参数。
Part of the problem is that I not totally understand how to apply start and end curvature parameters in standard clothoid formulas.
例如，样本RoadXML道路。
For example, sample RoadXML road.
这是在红色椭圆克洛索曲线项目。它的参数：
This is clotho curve item in the red ellipse. It's parameters:
- 在开始曲率= 0，
- 在年底曲率= -0.0165407，
- 长度= 45.185。
我不知道如何实现这些参数，因为回旋的曲率从0至-0.0165是非常简单。
I don't know how to implement these parameters, because clothoid curvature from 0 to -0.0165 is very straight.
我会高兴，如果你给我这个函数的code（在C ++，C＃，Java和Python或伪code）或只是一个公式，我可以code。
I will happy, if you give me a code of this function (in C++, C#, Java, Python or pseudocode) or just a formula, which I can code.
下面是我的公式：
x(t) ≈ t,
y(t) ≈ (t^3) / 6,
where length = t = s = curvature.
x(-0.0165) = -0.0165,
y(-0.0165) = -7.48688E-07.
Clotho length = 0.0165,
Source length = 45.185.
缩放坐标：
x'(l) = x / clotho_length * source_length = 45.185,
y'(l) = y / clotho_length * source_length = 5.58149E-07 ≈ 0.
x'(0) = 0,
y'(0) = 0.
因此我得到（0，0）（45，0）点，这是非常简单。
Thus I get (0, 0)...(45, 0) points, which is very straight.
在哪里是我的错？我究竟做错了什么？
Where is my mistake? What am I doing wrong?
推荐答案
让我们来看看。您的数据是：
Let's see. Your data is:
start curvature = 0, straight line, R=INF
end curvature = -0.0165407, circular arc, R_c = 1/k_c = 60.4569335
length = 45.185. distance along clothoid, s_c = 45.185
根据维基百科的文章，
R s = const = R_c s_c ( s ~ k = 1/R by definition of clothoid )
d(s) = R d(theta)
d(theta) = k d(s)
d(theta) / d(s) = 1 / R = k = s / R_c s_c
theta = s^2 / 2 R_c s_c = (s/a)^2 = s / 2 R = k s / 2
where ___________________
a = sqrt(2 R_c s_c) (... = 73.915445 )
~~~~~~~~~~~~~~~~~~~
and so theta_c = k_c s_c / 2 (... = 0.37369576475 = 21.411190 degrees )
( not so flat after all !! )
（注：我叫 A 这里的WP文章所称的倒数 A ）。然后，
(note: I call a here a reciprocal of what WP article calls a). Then,
d(x) = d(s) cos(theta)
d(y) = d(s) sin(theta)
x = INT[s=0..s] cos(theta) d(s)
= INT[s=0..s] cos((s/a)^2) a d(s/a)
= a INT[u=0..(s/a)] cos(u^2) d(u) = a C( s/a )
y = a INT[u=0..(s/a)] sin(u^2) d(u) = a S( s/a )
其中， C（T）和 S（T）是菲涅耳积分的。
所以这就是你怎么办缩放。不仅仅是 T = S ，而 T = S / A = SQRT（THETA）。在这里，终点， T_C =开方（k_c s_c / 2）=开方（0.0165407 * 45.185 / 2）= 0.6113066 。
So that's how you do the scaling. Not just t = s, but t = s/a = sqrt(theta). Here, for the end point, t_c = sqrt( k_c s_c / 2) = sqrt( 0.0165407 * 45.185 / 2) = 0.6113066.
现在，<一个href="http://www.wolframalpha.com/input/?i=%7B73.915445+Sqrt%5Bpi%2F2%5D+FresnelC%5B0.6113066%2FSqrt%5Bpi%2F2%5D%5D%2C+73.915445+Sqrt%5Bpi%2F2%5D+FresnelS%5B0.6113066%2FSqrt%5Bpi%2F2%5D%5D%7D"相对=nofollow> WolframAlpha的说， {73.915445的Sqrt的π/ 2] FresnelC [0.6113066 /的Sqrt的π/ 2]，73.915445的Sqrt的π/ 2]菲涅尔[0.6113066 /开方的π/ 2]} = {44.5581，5.57259} 。 （显然Mathematica使用比例与其他的Sqrt [PI一个定义/ 2] 因素。）
你的功能测试它， X 〜= T - ＆GT; A *（S / A） = 45.185 是 〜= T ^ 3/3 - ＆GT; A *（S / A）^ 3/3 = 73.915445 * 0.6113066 ^ 3月3日 = 5.628481 （原文如此！ / 3 不是 / 6 ，你必须有一个错误）。
Testing it with your functions, x ~= t --> a*(s/a) = 45.185, y ~= t^3/3 --> a*(s/a)^3/3 = 73.915445 * 0.6113066^3 / 3 = 5.628481 (sic! /3 not /6, you have an error there).
所以你看，只使用第一项菲涅耳积分的泰勒级数重新presentation是不够的 - 迄今为止。你必须使用更多的，停止只有当达到所希望的precision（即当最后核算的长期低于你的pre-设置precision的幅度值）。
So you see, using just the first term from the Taylor series representation of Fresnel integrals is not enough - by far. You have to use more, and stop only when desired precision is reached (i.e. when the last calculated term is less than your pre-set precision value in magnitude).
请注意，如果你只是实现一般的菲涅耳积分功能为一次性规模回旋计算，你将失去更多的precision时，你会被乘后面的结果（这是10的顺序 2 ... 10 3 通常，对于公路和铁路）。
Note, that if you'll just implement general Fresnel integral functions for one-off scaled clothoid calculation, you'll lose additional precision when you'll multiply the results back by a (which is on the order of 102 ... 103 normally, for roads and railways).
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