THE REALITY, NOW AND UNDERSTANDING

HIGHACC.TXT: HIGH ACCURACY

1.0 INTRODUCTION

In this chapter three concepts are explained:

The use of accumulators in order to improve accuracy
How to calculate a particular position based on three measurements
How to calculate the maximum position based on three measurements

The second concept is used in order to calculate a position in the past. The third concept is used to find the maximum position of Mercury.

2.0 DESCRIPTION

In order to simulate Newtons Law three parameters are calculated:

        acceleration : a, ax and ay
        speed        : v, vx and vy
        position     : s, x and y

At each iteration or simulation cycle those three parameters are more or less calculated as follows:

        an = function of m1, m2 and distance
        vn = vn + an * t
        sn = sn + vn * t

The problem with this method is accuracy. When the initial values are zero this is not a problem. On the other hand when the initial speed is 100 and acceleration small then after each iteration the speed stays more or less constant but the position not. The position increases linear with at each iteration becoming less accurate because the influence of a is decimated.

Two improve accuracy in x direction 3 accumulators are defined for each of the objects used: two for acceleration (Sa and SSa) and one for speed (Sv). The same for y direction

As a result for each iteration only the values of the accumulators have to be calculated and not speed and position. The position is only necessary if you want to display the place an object. The accuracy of the distance between two objects (necessary for acceleration) is drastically improved.

2.1 ACCUMULATORS

Initial conditions

        m0 = mass of object 0
        m1 = mass of object 1
        s0 = distance between object 0 and object 1 at t = 0 or iteration 0
        v0 = speed of object 0 at t = 0 or iteration 0
        t = delta t = time period between each iteration (calculation cycle)
At iteration 1 the following values are calculated:

        a0 = m1 * G / r                                r = s0 * s0
        a0x = a0 * dx / s0
        v1 = v0 + a0 * t
        s1 = s0 + v1 * t  = s0 + v0 * t + a0 * t       t = t * t

At iteration 2 the following values are calculated:

        a1 = m1 * G / r                                r = s1 * s1 
        a1x = a1 * dx / s1
        v2 = v1 + a1 * t  = v0 + (a0 + a1) * t 
        s2 = s1 + v2 * t  = s0 + v0 * t + a0 * t + v0 * t + (a0 + a1) * t
           = s0 + 2 * v0 * t + (2 * a0 + a1) * t

At iteration 3 the following values are calculated:

        a2 = m1 * G / r                                r = s2 * s2
        a2x = a2 * G / s2
        v3 = v2 + a2 * t = v0 + (a0 + a1 + a2) * t
        s3 = s2 + v3 * t =
           = s0 + 3 * v0 * t + (3 * a0 + 2 * a1 + a0) * t

At iteration n the following values are calculated:

        an   = m1 * G / r                              r = sn-1 * sn-1 
        anx  = an * dx / sn-1
        any  = an * dy / sn-1
        San  = a0  + a1  + a2  ....  an =  San-1 + an
        SSan = Sa0 + Sa1 + Sa2 .... San = SSan-1 + San
        Svn  = n * v0                   =  Svn-1 + v0
        vnt = v0 + San * t
        snt = s0 + Svn * t + SSan * t

for x and y the following applies

        vxnt = vx0 + Sanx * t                                (2.1)
        vynt = vy0 + Sany * t                                (2.2)
        sxnt = x0  + Svnx * t + SSanx * t                   (2.3)
        synt = y0  + Svny * t + SSany * t                   (2.4)

        Sanx  =  a0x +  a1x +  a2x ....  anx =  San-1x + anx    (2.5)
        SSanx = Sa0x + Sa1x + Sa2x .... Sanx = SSan-1x + Sanx   (2.6)
        Svnx  = n * v0x                      =  Svn-1x + v0x    (2.7)
        Sany  =  a0y +  a1y +  a2y ....  any =  San-1y + any    (2.8)
        SSany = Sa0y + Sa1y + Sa2y .... Sany = SSan-1y + Sany   (2.9)
        Svny  = n * v0y                      =  Svn-1y + v0y    (2.10)

For each iteration only the equations 2.5 until 2.10 have to be calculated.

2.1.1 DISTANCE

Using equation 2.3 the distance between two objects 1 and 2 is calculated as follows:

        s = (x01 - x02) + (Svnx1 - Svnx2) * t + (SSanx1 - SSanx2) * t  (2.11)  

The brackets are important. First all the values in between the brackets are calculated and then the sum is calculated thereby minimising the influence of x0

2.2 PREDICTION

2.2.1 PREDICTION BASED ON 2 VALUES

Two values are measured:

        x0  at iteration = 0    
        x1  at iteration = 1
What is the value xn for 0
        xn = a + bn
        x0 = a       for  n = 0
        x1 = a + b   for  n = 1
        b = x1 - x0
        xn = x0 + (x1 - x0) * n

for yn the following applies:

        yn = y0 + (y1 - y0) * n

2.2.2 PREDICTION BASED ON 3 VALUES

Three values are measured:

        x0 at n = 0
        x1 at n = 1
        x2 at n = 2
What is the value xn for 0
        xn = a + bn + cn
        x0 = a             for n = 0
        x1 = a + b +  c    for n = 1
        x2 = a + 2b + 4c   for n = 2

        x0 = a
        x1 = x0 + b + c         * -2
        x2 = x0 + 2b + 4c       * 1
        x2 - 2x1 = -x0 + 2c
        c = x0/2 - x1 + x2/2

        b =  x1 - x0 - c   = x1 - x0 - x0/2 + x1 - x2/2
          =  -3x0/2 + 2x1 -x2/2
        xn = x0 + (-3x0 + 4x1 - x2) * n / 2  + (x0 - 2x1 + x2) * n / 2  (2.12)

For yn the following applies:

        yn = y0 + (-3y0 + 4y1 - y2) * n / 2  + (y0 - 2y1 + y2) * n / 2  (2.13)

Example:

        x0 = x0
        x1 = x0 + v1.t
        x2 = x1 + v2.t = x0 + v2.t + v1.t

        xnt = x0 + (-3x0 + 4x0 + 4v1.t  - x0 - v2.t - v1.t) * n / 2
                 + (  x0 - 2x0 - 2v1.t  + x0 + v2.t + v1.t) * n / 2 

        xnt = x0 + (3v1.t - v2.t ) * n / 2 + (- v1.t + v2.t ) * n / 2 

2.2.3 PREDICTION BASED ON 3 VALUES WITH ACCUMULATORS

Three values are measured:

        xn  at t = n
        xn1 at t = n-1
        xn2 at t = n-2
What is the value xnt for n
        xnt = xn + (-3xn + 4xn1 - xn2) * n/2  + (xn - 2xn1 + xn2) * n/2

        xn   = x0 + Svx.t   + SSax.t
        xn-1 = x0 + Svx-1.t + SSax-1.t
        xn-2 = x0 + Svx-2.t + SSax-2.t

       xnt = x0 + Svx.t + SSax.t
             +{ -3x0 -3Svx.t -3SSax.t +4x0 +4Svx-1.t +4SSax-1.t
                                       - x0 - Svx-2.t - SSax-2.t } * n/2
             +{   x0 + Svx.t + SSax.t -2x0 -2Svx-1.t -2SSax-1.t
                                       + x0 + Svx-2.t + SSax-2.t } * n/2
        xnt = x0 + Svx.t + SSax.t
             +{ -3Svx.t -3SSax.t +4Svx-1.t +4SSax-1.t 
                                  - Svx-2.t - SSax-2.t } * n/2
             +{   Svx.t + SSax.t -2Svx-1.t -2SSax-1.t 
                                  + Svx-2.t + SSax-2.t } * n/2

     Svx   = n     * vx0 
     Svx-1 = (n-1) * vx0
     Svx-2 = (n-2) * vx0

     -3Svx + 4Svx-1 - Svx-2 = -3n*vx0 +4n*vx0 -4vx0 -n*vx0 +2vx0 = -2vx0
       Svx - 2Svx-1 + Svx-2 =   n*vx0 -2n*vx0 +2vx0 +n*vx0 -2vx0 =  0                      

        xnt = x0 + Svx.t + SSax.t
             +{ -2vx0.t -3SSax.t +4SSax-1.t - SSax-2.t } * n/2
             +{           SSax.t -2SSax-1.t + SSax-2.t } * n/2

     SSax   = SSax-1 + Sax
     SSax-1 = SSax-2 + Sax-1
     SSax   = SSax-2 + Sax-1 + Sax

        xnt = x0 + Svx.t + SSax.t
             +{ -2vx0.t - 3SSax-2.t -3Sax-1.t -3Sax.t
                         +4SSax-2.t +4Sax-1.t - SSax-2.t } * n/2
             +{         +  SSax-2.t + Sax-1.t + Sax.t 
                         -2SSax-2.t -2Sax-1.t + SSax-2.t } * n/2
Remove SSax-2.t

        xnt = x0 + Svx.t + SSax.t
              +{ -2vx0.t -3Sax-1.t -3Sax.t +4Sax-1.t} * n/2
              +{         + Sax-1.t + Sax.t -2Sax-1.t} * n/2

        xnt = x0 + Svx.t + SSax.t
              +{ -2vx0.t + Sax-1.t -3Sax.t} * n/2
              +{         - Sax-1.t + Sax.t} * n/2

Equation in y direction is identical.

EXAMPLE
        xnt      (n=0)  = x0 + Svx.t + SSax.t
        xn-1t    (n=1)  = x0 + Svx.t + SSax.t  -vx0.t   -Sax.t
        xn-2t    (n=2)  = x0 + Svx.t + SSax.t  -2vx0.t  -Sax.t -Sax-1.t

2.4 MAXIMUM CALCULATION

For a function :

                f(t) = a + b * t + c * t  

the maximum is at :

                df/dt = 0
                b + 2c * t = 0
                t = - b / 2c

in equation 2.12 the values of of a, b and c are:

                s0 = sqrt (x0+ y0)   
                etc  for s1 and s2
                a = s0
                b = (-3s0 + 4s1 - s2)/2 
                c = (s0 - 2s1 + s2) / 2
Using those values the maximum is at :

                    3s0 - 4s1 + s2
                t = ---------------
                    2s0 - 4s1 + 2s2   

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