I have been trying to turn this one conplex equation into code and it appears that I might have done something wrong. Here's the image of the equation:
Here's is the first code I tried using to convert the equation into code.
double answer = 1 - (Math.Pow(f, n) * ((s * l / f) + Math.Pow((20 / f), w) / Math.Pow(20, n)));
Here is the code that I used in my second attempt:
double answer = 1 - Math.Pow(f, n) * ((s * l) / f) + Math.Pow((20 / f), w) / Math.Pow(20, n);
If I assume that every variable of the equation is 2, than I get -.02. But when I ran the code, the first attempt code returned a value of -8, while the second attempt returned -6.75.
Is there anything I'm doing wrong in my code right now? And also sorry if I'm bad at explaining stuffs.
I tested this out and got the result of -0.02. Try splitting up the code to make it more legible. It might help you diagnose the syntax of your complex equation on one line.
double f = 2;
double n = 2;
double s = 2;
double w = 2;
double l = 2;
double A = Math.Pow(f, n);
double B = (s * l) / f;
double C = Math.Pow((20 / f), w);
double bottom = Math.Pow(20, n);
double top = A * (B + C);
double answer = 1 - top / bottom;
In both attempts you just got your brackets in the wrong spot.
Try this:
double answer =
1 - Math.Pow(f, n) * (s * l / f + Math.Pow((20 / f), w)) / Math.Pow(20, n);
Try to use the formula below instead :
double answer = (1 - Math.pow((Math.pow(f,n)*[s*l/f+20/f})),w)/Math.pow(20,f)
I'm trying to find real roots for a cubic equation defined by a set of four coefficients by using Cardano method as described here. The problem is, the roots found by my implementation do not actually work - testing by inserting them in the equation give a significant error (more than the required 10^-6). Is the algorithm implemented wrong, or the error is caused by something else, like rounding accuracy?
static double CubicRoot(double n)
{
return Math.Pow(Math.Abs(n), 1d / 3d) * Math.Sign(n);
}
public static List<double> SolveCubic(double A, double B = 0, double C = 0, double D = 0)
{
List<double> output = new List<double>();
if (A != 0)
{
double A1 = B / A;
double A2 = C / A;
double A3 = D / A;
double P = -((A1 * A1) / 3) + A2;
double Q = ((2.0 * A1 * A1 * A1) / 27.0) - ((A1 * A2) / 3.0) + A3;
double cubeDiscr = Q * Q / 4.0 + P * P * P / 27.0;
if (cubeDiscr > 0)
{
double u = CubicRoot(-Q / 2.0 + Math.Sqrt(cubeDiscr));
double v = CubicRoot(-Q / 2.0 - Math.Sqrt(cubeDiscr));
output.Add(u + v - (A1 / 3.0));
return output;
}
else if (cubeDiscr == 0)
{
double u = CubicRoot(-Q / 2.0);
output.Add(2u - (A1 / 3.0));
output.Add(-u - (A1 / 3.0));
}
else if (cubeDiscr < 0)
{
double r = CubicRoot(Math.Sqrt(-(P * P * P / 27.0)));
double alpha = Math.Atan(Math.Sqrt(-cubeDiscr) / (-Q / 2.0));
output.Add(r * (Math.Cos(alpha / 3.0) + Math.Cos((6 * Math.PI - alpha) / 3.0)) - A1 / 3.0);
output.Add(r * (Math.Cos((2 * Math.PI + alpha) / 3.0) + Math.Cos((4 * Math.PI - alpha) / 3.0)) - A1 / 3.0);
output.Add(r * (Math.Cos((4 * Math.PI + alpha) / 3.0) + Math.Cos((2 * Math.PI - alpha) / 3.0)) - A1 / 3.0);
}
}
return output;
}
A few things
Math.Sign will return zero on zero, which happens to be what you want in this case, but perhaps you are not so lucky with code or algorithm change.
You will have rounding issues and not execute cubeDiscr == 0 branch when you should. You may have rounding issues and execute the wrong > 0 and < 0 branch for the same reason. Test within a delta of zero instead (see below).
But the cubeDiscr == 0 branch is wrong because 1) you didn't calculate v and 2) 2u is an UInt32 with a value of 2, not 2*u.
Calculating alpha is wrong (see below)
(there may be more, but that's all I saw at a quick glance)
On calculating alpha:
double alpha = Math.Atan(Math.Sqrt(-cubeDiscr) / (-Q / 2.0));
is not the same as
double alpha = Math.Atan(Math.Sqrt(-d) / q * 2.0);
if (q > 0) // if q > 0 the angle becomes PI + alpha
alpha = Math.PI + alpha;
What's wrong with using the code included from that page?
public double Xroot(double a, double x)
{
double i = 1;
if (a < 0)
i = -1;
return (i * Math.Exp( Math.Log(a*i)/x));
}
public int Calc_Cardano() // solve cubic equation according to cardano
{
double p, q, u, v;
double r, alpha;
int res;
res = 0;
if (a1 != 0)
{
a = b / a1;
b = c / a1;
c = d / a1;
p = -(a * a / 3.0) + b;
q = (2.0 / 27.0 * a * a * a) - (a * b / 3.0) + c;
d = q * q / 4.0 + p * p * p / 27.0;
if (Math.Abs(d) < Math.Pow(10.0, -11.0))
d = 0;
// 3 cases D > 0, D == 0 and D < 0
if (d > 1e-20)
{
u = Xroot(-q / 2.0 + Math.Sqrt(d), 3.0);
v = Xroot(-q / 2.0 - Math.Sqrt(d), 3.0);
x1.real = u + v - a / 3.0;
x2.real = -(u + v) / 2.0 - a / 3.0;
x2.imag = Math.Sqrt(3.0) / 2.0 * (u - v);
x3.real = x2.real;
x3.imag = -x2.imag;
res = 1;
}
if (Math.Abs(d) <= 1e-20)
{
u = Xroot(-q / 2.0, 3.0);
v = Xroot(-q / 2.0, 3.0);
x1.real = u + v - a / 3.0;
x2.real = -(u + v) / 2.0 - a / 3.0;
res = 2;
}
if (d < -1e-20)
{
r = Math.Sqrt(-p * p * p / 27.0);
alpha = Math.Atan(Math.Sqrt(-d) / q * 2.0);
if (q > 0) // if q > 0 the angle becomes PI + alpha
alpha = Math.PI + alpha;
x1.real = Xroot(r, 3.0) * (Math.Cos((6.0 * Math.PI - alpha) / 3.0) + Math.Cos(alpha / 3.0)) - a / 3.0;
x2.real = Xroot(r, 3.0) * (Math.Cos((2.0 * Math.PI + alpha) / 3.0) + Math.Cos((4.0 * Math.PI - alpha) / 3.0)) - a / 3.0;
x3.real = Xroot(r, 3.0) * (Math.Cos((4.0 * Math.PI + alpha) / 3.0) + Math.Cos((2.0 * Math.PI - alpha) / 3.0)) - a / 3.0;
res = 3;
}
}
else
res = 0;
return res;
}
I had a look at the wave format today and created a little wave generator. I create a sine sound like this:
public static Wave GetSine(double length, double hz)
{
int bitRate = 44100;
int l = (int)(bitRate * length);
double f = 1.0 / bitRate;
Int16[] data = new Int16[l];
for (int i = 0; i < l; i++)
{
data[i] = (Int16)(Math.Sin(hz * i * f * Math.PI * 2) * Int16.MaxValue);
}
return new Wave(false, Wave.MakeInt16WaveData(data));
}
MakeInt16WaveData looks like this:
public static byte[] MakeInt16WaveData(Int16[] ints)
{
int s = sizeof(Int16);
byte[] buf = new byte[s * ints.Length];
for(int i = 0; i < ints.Length; i++)
{
Buffer.BlockCopy(BitConverter.GetBytes(ints[i]), 0, buf, i * s, s);
}
return buf;
}
This works as expected! Now I wanted to swoop from one frequency to another like this:
public static Wave GetSineSwoop(double length, double hzStart, double hzEnd)
{
int bitRate = 44100;
int l = (int)(bitRate * length);
double f = 1.0 / bitRate;
Int16[] data = new Int16[l];
double hz;
double hzDelta = hzEnd - hzStart;
for (int i = 0; i < l; i++)
{
hz = hzStart + ((double)i / l) * hzDelta * 0.5; // why *0.5 ?
data[i] = (Int16)(Math.Sin(hz * i * f * Math.PI * 2) * Int16.MaxValue);
}
return new Wave(false, Wave.MakeInt16WaveData(data));
}
Now, when I swooped from 200 to 100 Hz, the sound played from 200 to 0 hertz. For some reason I had to multiply the delta by 0.5 to get the correct output. What might be the issue here ? Is this an audio thing or is there a bug in my code ?
Thanks
Edit by TaW: I take the liberty to add screenshots of the data in a chart which illustrate the problem, the first is with the 0.5 factor, the 2nd with 1.0 and the 3rd & 4th with 2.0 and 5.0:
Edit: here is an example, a swoop from 200 to 100 hz:
Debug values:
Wave clearly does not end at 100 hz
Digging out my rusty math I think it may be because:
Going in L steps from frequency F1 to F2 you have a frequency of
Fi = F1 + i * ( F2 - F1 ) / L
or with
( F2 - F1 ) / L = S
Fi = F1 + i * S
Now to find out how far we have progressed we need the integral over i, which would be
I(Fi) = i * F1 + 1/2 * i ^ 2 * S
Which give or take resembles the term inside your formula for the sine.
Note that you can gain efficiency by moving the constant part (S/2) out of the loop..
I asked a question about having the Excel's BetaInv function ported to .NET: BetaInv function in SQL Server
now I managed to write that function in pure dependency less C# code and I do get the same results than in MS Excel up to 6 or 7 digits after comma, results are fine for us, the problem is that such code is embedded in a SQL CLR Function and gets called thousands of time from a stored procedure and makes the execution of the whole procedure about 50% slower, from 30 seconds up to a minute if I use that function or not.
here some code of it, I am not asking a deep analysis but is there anybody who sees any major performance issue in the way I am doing this calculations? like for example usage of other data types instead of doubles or whatsoever... ?
private static double betacf(double a, double b, double x)
{
int m, m2;
double aa, c, d, del, h, qab, qam, qap;
qab = a + b;
qap = a + 1.0;
qam = a - 1.0;
c = 1.0; // First step of Lentz’s method.
d = 1.0 - qab * x / qap;
if (System.Math.Abs(d) < FPMIN)
{
d = FPMIN;
}
d = 1.0 / d;
h = d;
for (m = 1; m <= MAXIT; ++m)
{
m2 = 2 * m;
aa = m * (b - m) * x / ((qam + m2) * (a + m2));
d = 1.0 + aa * d; //One step (the even one) of the recurrence.
if (System.Math.Abs(d) < FPMIN)
{
d = FPMIN;
}
c = 1.0 + aa / c;
if (System.Math.Abs(c) < FPMIN)
{
c = FPMIN;
}
d = 1.0 / d;
h *= d * c;
aa = -(a + m) * (qab + m) * x / ((a + m2) * (qap + m2));
d = 1.0 + aa * d; // Next step of the recurrence (the odd one).
if (System.Math.Abs(d) < FPMIN)
{
d = FPMIN;
}
c = 1.0 + aa / c;
if (System.Math.Abs(c) < FPMIN)
{
c = FPMIN;
}
d = 1.0 / d;
del = d * c;
h *= del;
if (System.Math.Abs(del - 1.0) < EPS)
{
// Are we done?
break;
}
}
if (m > MAXIT)
{
return 0;
}
else
{
return h;
}
}
private static double gammln(double xx)
{
double x, y, tmp, ser;
double[] cof = new double[] { 76.180091729471457, -86.505320329416776, 24.014098240830911, -1.231739572450155, 0.001208650973866179, -0.000005395239384953 };
y = xx;
x = xx;
tmp = x + 5.5;
tmp -= (x + 0.5) * System.Math.Log(tmp);
ser = 1.0000000001900149;
for (int j = 0; j <= 5; ++j)
{
y += 1;
ser += cof[j] / y;
}
return -tmp + System.Math.Log(2.5066282746310007 * ser / x);
}
The only thing that stands out for me, and is usually a performance hit, is memory allocation. I don't know how often gammln is called but you might want to move the double[] cof = new double[] {} to a static one time only allocation.
double is usually the best type. Especially since the functions in Math take doubles. Unfortunately I see no obvious improvements to make on your code.
It might be possible to use look up tables to get a better first estimate on which you iterate, but since I don't know the Math behind what you're doing I don't know if that's possible in this specific case.
Obviously larger epsilons will improve performance. So choose it as large as possible while fulfilling your accuracy demands.
If the function gets called repeatedly with the same parameters you might be able to cache results.
One thing that looks odd is the way you force small values for c, d,... to FPMIN. My instinct is that this might lead to suboptimal step sizes.
All I've got is unrolling the j loop in gammln, but it'll make at most a tiny difference.
A more radical thought would be to rewrite in pure T-SQL, since it has everything you use: + - * / abs log are all available.
Trying to find functions that will assist us to draw a 3D line through a series of points.
For each point we know: Date&Time, Latitude, Longitude, Altitude, Speed and Heading.
Data might be recorded every 10 seconds and we would like to be able to guestimate the points in between and increase granularity to 1 second. Thus creating a virtual flight path in 3D space.
I have found a number of curve fitting algorithms that will approximate a line through a series of points but they do not guarantee that the points are intersected. They also do not take into account speed and heading to determine the most likely path taken by the object to reach the next point.
From a physics viewpoint:
You have to assume something about the acceleration in your intermediate points to get the interpolation.
If your physical system is relatively well-behaved (as a car or a plane), as opposed to for example a bouncing ball, you may go ahead supposing an acceleration varying linearly with time between your points.
The vector equation for a constant varying accelerated movement is:
x''[t] = a t + b
where all magnitudes except t are vectors.
For each segment you already know v(t=t0) x(t=t0) tfinal and x(tfinal) v(tfinal)
By solving the differential equation you get:
Eq 1:
x[t_] := (3 b t^2 Tf + a t^3 Tf - 3 b t Tf^2 - a t Tf^3 - 6 t X0 + 6 Tf X0 + 6 t Xf)/(6 Tf)
And imposing the initial and final contraints for position and velocity you get:
Eqs 2:
a -> (6 (Tf^2 V0 - 2 T0 Tf Vf + Tf^2 Vf - 2 T0 X0 + 2 Tf X0 +
2 T0 Xf - 2 Tf Xf))/(Tf^2 (3 T0^2 - 4 T0 Tf + Tf^2))
b -> (2 (-2 Tf^3 V0 + 3 T0^2 Tf Vf - Tf^3 Vf + 3 T0^2 X0 -
3 Tf^2 X0 - 3 T0^2 Xf + 3 Tf^2 Xf))/(Tf^2 (3 T0^2 - 4 T0 Tf + Tf^2))}}
So inserting the values for eqs 2 into eq 1 you get the temporal interpolation for your points, based on the initial and final position and velocities.
HTH!
Edit
A few examples with abrupt velocity change in two dimensions (in 3D is exactly the same). If the initial and final speeds are similar, you'll get "straighter" paths.
Suppose:
X0 = {0, 0}; Xf = {1, 1};
T0 = 0; Tf = 1;
If
V0 = {0, 1}; Vf = {-1, 3};
V0 = {0, 1}; Vf = {-1, 5};
V0 = {0, 1}; Vf = {1, 3};
Here is an animation where you may see the speed changing from V0 = {0, 1} to Vf = {1, 5}:
Here you may see an accelerating body in 3D with positions taken at equal intervals:
Edit
A full problem:
For convenience, I'll work in Cartesian coordinates. If you want to convert from lat/log/alt to Cartesian just do:
x = rho sin(theta) cos(phi)
y = rho sin(theta) sin(phi)
z = rho cos(theta)
Where phi is the longitude, theta is the latitude, and rho is your altitude plus the radius of the Earth.
So suppose we start our segment at:
t=0 with coordinates (0,0,0) and velocity (1,0,0)
and end at
t=10 with coordinates (10,10,10) and velocity (0,0,1)
I clearly made a change in the origin of coordinates to set the origin at my start point. That is just for getting nice round numbers ...
So we replace those numbers in the formulas for a and b and get:
a = {-(3/50), -(3/25), -(3/50)} b = {1/5, 3/5, 2/5}
With those we go to eq 1, and the position of the object is given by:
p[t] = {1/60 (60 t + 6 t^2 - (3 t^3)/5),
1/60 (18 t^2 - (6 t^3)/5),
1/60 (12 t^2 - (3 t^3)/5)}
And that is it. You get the position from 1 to 10 secs replacing t by its valus in the equation above.
The animation runs:
Edit 2
If you don't want to mess with the vertical acceleration (perhaps because your "speedometer" doesn't read it), you could just assign a constant speed to the z axis (there is a very minor error for considering it parallel to the Rho axis), equal to (Zfinal - Zinit)/(Tf-T0), and then solve the problem in the plane forgetting the altitude.
What you're asking is a general interpolation problem. My guess is your actual problem isn't due to the curve-fitting algorithm being used, but rather your application of it to all discrete values recorded by the system instead of the relevant set of values.
Let's decompose your problem. You're currently drawing a point in spherically-mapped 3D space, adjusting for linear and curved paths. If we discretize the operations performed by an object with six degrees of freedom (roll, pitch, and yaw), the only operations you're particularly interested in are linear paths and curved paths accounting for pitch and yaw in any direction. Accounting for acceleration and deceleration also possible given understanding of basic physics.
Dealing with the spherical mapping is easy. Simply unwrap your points relative to their position on a plane, adjusting for latitude, longitude, and altitude. This should allow you to flatten data that would otherwise exist along a curved path, though this may not strictly be necessary for the solutions to your problem (see below).
Linear interpolation is easy. Given an arbitrary number of points backwards in time that fit a line within n error as determined by your system,* construct the line and compute the distance in time between each point. From here, attempt to fit the time points to one of two cases: constant velocity or constant acceleration.
Curve interpolation is a little more difficult, but still plausible. For cases of pitch, yaw, or combined pitch+yaw, construct a plane containing an arbitrary number of points backwards in time, within m error for curved readouts from your system.* From these data, construct a planar curve and once again account for constant velocity or acceleration along the curve.
You can do better than this by attempting to predict the expected operations of a plane in flight as part of a decision tree or neural network relative to the flight path. I'll leave that as an exercise for the reader.
Best of luck designing your system.
--
* Both error readouts are expected to be from GPS data, given the description of the problem. Accounting and adjusting for errors in these data is a separate interesting problem.
What you need (instead of modeling the physics) is to fit a spline through the data. I used a numerical recipies book (http://www.nrbook.com/a has free C and FORTRAN algorithms. Look into F77 section 3.3 to get the math needed). If you want to be simple then just fit lines through the points, but that will not result in a smooth flight path at all. Time will be your x value, and each parameter loged will have it's own cublic spline parameters.
Since we like long postings for this question here is the full code:
//driver program
static void Main(string[] args)
{
double[][] flight_data = new double[][] {
new double[] { 0, 10, 20, 30 }, // time in seconds
new double[] { 14500, 14750, 15000, 15125 }, //altitude in ft
new double[] { 440, 425, 415, 410 }, // speed in knots
};
CubicSpline altitude = new CubicSpline(flight_data[0], flight_data[1]);
CubicSpline speed = new CubicSpline(flight_data[0], flight_data[2]);
double t = 22; //Find values at t
double h = altitude.GetY(t);
double v = speed.GetY(t);
double ascent = altitude.GetYp(t); // ascent rate in ft/s
}
// Cubic spline definition
using System.Linq;
/// <summary>
/// Cubic spline interpolation for tabular data
/// </summary>
/// <remarks>
/// Adapted from numerical recipies for FORTRAN 77
/// (ISBN 0-521-43064-X), page 110, section 3.3.
/// Function spline(x,y,yp1,ypn,y2) converted to
/// C# by jalexiou, 27 November 2007.
/// Spline integration added also Nov 2007.
/// </remarks>
public class CubicSpline
{
double[] xi;
double[] yi;
double[] yp;
double[] ypp;
double[] yppp;
double[] iy;
#region Constructors
public CubicSpline(double x_min, double x_max, double[] y)
: this(Sequence(x_min, x_max, y.Length), y)
{ }
public CubicSpline(double x_min, double x_max, double[] y, double yp1, double ypn)
: this(Sequence(x_min, x_max, y.Length), y, yp1, ypn)
{ }
public CubicSpline(double[] x, double[] y)
: this(x, y, double.NaN, double.NaN)
{ }
public CubicSpline(double[] x, double[] y, double yp1, double ypn)
{
if( x.Length == y.Length )
{
int N = x.Length;
xi = new double[N];
yi = new double[N];
x.CopyTo(xi, 0);
y.CopyTo(yi, 0);
if( N > 0 )
{
double p, qn, sig, un;
ypp = new double[N];
double[] u = new double[N];
if( double.IsNaN(yp1) )
{
ypp[0] = 0;
u[0] = 0;
}
else
{
ypp[0] = -0.5;
u[0] = (3 / (xi[1] - xi[0])) *
((yi[1] - yi[0]) / (x[1] - x[0]) - yp1);
}
for (int i = 1; i < N-1; i++)
{
double hp = x[i] - x[i - 1];
double hn = x[i + 1] - x[i];
sig = hp / hn;
p = sig * ypp[i - 1] + 2.0;
ypp[i] = (sig - 1.0) / p;
u[i] = (6 * ((y[i + 1] - y[i]) / hn) - (y[i] - y[i - 1]) / hp)
/ (hp + hn) - sig * u[i - 1] / p;
}
if( double.IsNaN(ypn) )
{
qn = 0;
un = 0;
}
else
{
qn = 0.5;
un = (3 / (x[N - 1] - x[N - 2])) *
(ypn - (y[N - 1] - y[N - 2]) / (x[N - 1] - x[N - 2]));
}
ypp[N - 1] = (un - qn * u[N - 2]) / (qn * ypp[N - 2] + 1.0);
for (int k = N-2; k > 0; k--)
{
ypp[k] = ypp[k] * ypp[k + 1] + u[k];
}
// Calculate 1st derivatives
yp = new double[N];
double h;
for( int i = 0; i < N - 1; i++ )
{
h = xi[i + 1] - xi[i];
yp[i] = (yi[i + 1] - yi[i]) / h
- h / 6 * (ypp[i + 1] + 2 * ypp[i]);
}
h = xi[N - 1] - xi[N - 2];
yp[N - 1] = (yi[N - 1] - yi[N - 2]) / h
+ h / 6 * (2 * ypp[N - 1] + ypp[N - 2]);
// Calculate 3rd derivatives as average of dYpp/dx
yppp = new double[N];
double[] jerk_ij = new double[N - 1];
for( int i = 0; i < N - 1; i++ )
{
h = xi[i + 1] - xi[i];
jerk_ij[i] = (ypp[i + 1] - ypp[i]) / h;
}
Yppp = new double[N];
yppp[0] = jerk_ij[0];
for( int i = 1; i < N - 1; i++ )
{
yppp[i] = 0.5 * (jerk_ij[i - 1] + jerk_ij[i]);
}
yppp[N - 1] = jerk_ij[N - 2];
// Calculate Integral over areas
iy = new double[N];
yi[0] = 0; //Integration constant
for( int i = 0; i < N - 1; i++ )
{
h = xi[i + 1] - xi[i];
iy[i + 1] = h * (yi[i + 1] + yi[i]) / 2
- h * h * h / 24 * (ypp[i + 1] + ypp[i]);
}
}
else
{
yp = new double[0];
ypp = new double[0];
yppp = new double[0];
iy = new double[0];
}
}
else
throw new IndexOutOfRangeException();
}
#endregion
#region Actions/Functions
public int IndexOf(double x)
{
//Use bisection to find index
int i1 = -1;
int i2 = Xi.Length;
int im;
double x1 = Xi[0];
double xn = Xi[Xi.Length - 1];
bool ascending = (xn >= x1);
while( i2 - i1 > 1 )
{
im = (i1 + i2) / 2;
double xm = Xi[im];
if( ascending & (x >= Xi[im]) ) { i1 = im; } else { i2 = im; }
}
if( (ascending && (x <= x1)) || (!ascending & (x >= x1)) )
{
return 0;
}
else if( (ascending && (x >= xn)) || (!ascending && (x <= xn)) )
{
return Xi.Length - 1;
}
else
{
return i1;
}
}
public double GetIntY(double x)
{
int i = IndexOf(x);
double x1 = xi[i];
double x2 = xi[i + 1];
double y1 = yi[i];
double y2 = yi[i + 1];
double y1pp = ypp[i];
double y2pp = ypp[i + 1];
double h = x2 - x1;
double h2 = h * h;
double a = (x-x1)/h;
double a2 = a*a;
return h / 6 * (3 * a * (2 - a) * y1
+ 3 * a2 * y2 - a2 * h2 * (a2 - 4 * a + 4) / 4 * y1pp
+ a2 * h2 * (a2 - 2) / 4 * y2pp);
}
public double GetY(double x)
{
int i = IndexOf(x);
double x1 = xi[i];
double x2 = xi[i + 1];
double y1 = yi[i];
double y2 = yi[i + 1];
double y1pp = ypp[i];
double y2pp = ypp[i + 1];
double h = x2 - x1;
double h2 = h * h;
double A = 1 - (x - x1) / (x2 - x1);
double B = 1 - A;
return A * y1 + B * y2 + h2 / 6 * (A * (A * A - 1) * y1pp
+ B * (B * B - 1) * y2pp);
}
public double GetYp(double x)
{
int i = IndexOf(x);
double x1 = xi[i];
double x2 = xi[i + 1];
double y1 = yi[i];
double y2 = yi[i + 1];
double y1pp = ypp[i];
double y2pp = ypp[i + 1];
double h = x2 - x1;
double A = 1 - (x - x1) / (x2 - x1);
double B = 1 - A;
return (y2 - y1) / h + h / 6 * (y2pp * (3 * B * B - 1)
- y1pp * (3 * A * A - 1));
}
public double GetYpp(double x)
{
int i = IndexOf(x);
double x1 = xi[i];
double x2 = xi[i + 1];
double y1pp = ypp[i];
double y2pp = ypp[i + 1];
double h = x2 - x1;
double A = 1 - (x - x1) / (x2 - x1);
double B = 1 - A;
return A * y1pp + B * y2pp;
}
public double GetYppp(double x)
{
int i = IndexOf(x);
double x1 = xi[i];
double x2 = xi[i + 1];
double y1pp = ypp[i];
double y2pp = ypp[i + 1];
double h = x2 - x1;
return (y2pp - y1pp) / h;
}
public double Integrate(double from_x, double to_x)
{
if( to_x < from_x ) { return -Integrate(to_x, from_x); }
int i = IndexOf(from_x);
int j = IndexOf(to_x);
double x1 = xi[i];
double xn = xi[j];
double z = GetIntY(to_x) - GetIntY(from_x); // go to nearest nodes (k) (j)
for( int k = i + 1; k <= j; k++ )
{
z += iy[k]; // fill-in areas in-between
}
return z;
}
#endregion
#region Properties
public bool IsEmpty { get { return xi.Length == 0; } }
public double[] Xi { get { return xi; } set { xi = value; } }
public double[] Yi { get { return yi; } set { yi = value; } }
public double[] Yp { get { return yp; } set { yp = value; } }
public double[] Ypp { get { return ypp; } set { ypp = value; } }
public double[] Yppp { get { return yppp; } set { yppp = value; } }
public double[] IntY { get { return yp; } set { iy = value; } }
public int Count { get { return xi.Length; } }
public double X_min { get { return xi.Min(); } }
public double X_max { get { return xi.Max(); } }
public double Y_min { get { return yi.Min(); } }
public double Y_max { get { return yi.Max(); } }
#endregion
#region Helpers
static double[] Sequence(double x_min, double x_max, int double_of_points)
{
double[] res = new double[double_of_points];
for (int i = 0; i < double_of_points; i++)
{
res[i] = x_min + (double)i / (double)(double_of_points - 1) * (x_max - x_min);
}
return res;
}
#endregion
}
You can find an approximation of a line that intersects points in 3d and 2d space using a Hough Transformation algorithm. I am only familiar with it's uses in 2d however but it will still work for 3d spaces given that you know what kind of line you are looking for. There is a basic implementation description linked. You can Google for pre-mades and here is a link to a 2d C implementation CImg.
The algorithm process (roughly)... First you find equation of a line that you think will best approximate the shape of the line (in 2d parabolic, logarithmic, exponential, etc). You take that formula and solve for one of the parameters.
y = ax + b
becomes
b = y - ax
Next, for each point you are attempting to match, you plugin the points to the y and x values. With 3 points, you would have 3 separate functions of b with respect to a.
(2, 3) : b = 3 - 2a
(4, 1) : b = 1 - 4a
(10, -5): b = -5 - 10a
Next, the theory is that you find all possible lines which pass through each of the points, which is infinitely many for each individual point however when combined in an accumulator space only a few possible parameters best fit. In practice this is done by choosing a range space for the parameters (I chose -2 <= a <= 1, 1 <= b <= 6) and begin plugging in values for the variant parameter(s) and solving for the other. You tally up the number of intersections from each function in an accumulator. The points with the highest values give you your parameters.
Accumulator after processing b = 3 - 2a
a: -2 -1 0 1
b: 1
2
3 1
4
5 1
6
Accumulator after processing b = 1 - 4a
a: -2 -1 0 1
b: 1 1
2
3 1
4
4
5 2
6
Accumulator after processing b = -5 - 10a
a: -2 -1 0 1
b: 1 1
2
3 1
4
5 3
6
The parameter set with the highest accumulated value is (b a) = (5 -1) and the function best fit to the points given is y = 5 - x.
Best of luck.
My guess is that a serious application of this would use a http://en.wikipedia.org/wiki/Kalman_filter. By the way, that probably wouldn't guarantee that the reported points were intersected either, unless you fiddled with the parameters a bit. It would expect some degree of error in each data point given to it, so where it thinks the object is at time T would not necessarily be where it was at time T. Of course, you could set the error distribution to say that you were absolutely sure you knew where it was at time T.
Short of using a Kalman filter, I would try and turn it into an optimisation problem. Work at the 1s granularity and write down equations like
x_t' = x_t + (Vx_t + Vx_t')/2 + e,
Vx_t_reported = Vx_t + f,
Vx_t' = Vx_t + g
where e, f, and g represent the noise. Then create a penalty function such as e^2 + f^2 + g^2 +...
or some weighted version such as 1.5e^2 + 3f^2 + 2.6g^2 +... according to your idea of what the errors really are and how smooth you wnat the answer to be, and find the values that make the penalty function as small as possible - with least squares if the equations turn out nicely.