Question

I’m coding renderer for road network, which based on the RoadXML format.

Road curves from this format has four types:

• segment,
• circle arc,
• poly line,
• clotho arc.

And I have problem with the last one.

Clothoid is the same with Euler spiral and Cornu spiral. In the RoadXML clotho arc is given by three parameters:

• start curvature,
• end curvature,
• length.

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.

Part of the problem is that I not totally understand how to apply start and end curvature parameters in standard clothoid formulas.

For example, sample RoadXML road.
RoadXML sample http://img560.imageshack.us/img560/8172/bigroandabout.png

This is clotho curve item in the red ellipse. It’s parameters:

• start curvature = 0,
• end curvature = -0.0165407,
• length = 45.185.

I don’t know how to implement these parameters, because clothoid curvature from 0 to -0.0165 is very straight.

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.

Here is my equations:

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.

Scaled coords:

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.

Thus I get (0, 0)…(45, 0) points, which is very straight.

Where is my mistake? What am I doing wrong?

Answer 1

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

according to Wikipedia article,

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 !! )

(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 )

where C(t) and S(t) are Fresnel integrals.

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.

Now, WolframAlpha says, {73.915445 Sqrt[pi/2] FresnelC[0.6113066/Sqrt[pi/2]], 73.915445 Sqrt[pi/2] FresnelS[0.6113066/Sqrt[pi/2]]} = {44.5581, 5.57259}. ^{(Apparently Mathematica uses a definition scaled with the additional Sqrt[pi/2] factor.)}

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).

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).

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).