Increasing accuracy of solution of transcendental

Question

I have a specific kinematics as a part of a more complex machine and need to compute some physical parameters that are very hard (more like impossible) to measure with proper accuracy with instruments I have at my disposal

[kinematics]

At first look it is a simple 1 degree of freedom arm (black) which can rotate around x axis. It has a weight to force it to go always up until it hit the mechanic endpoint (angle a0) or some tube (blue) with radius r0. Arm rotation center is at y0. The tube can be moved to any y(t) height.

[usage]

This is used to measure the radius of a tube for further processing. The radius can be computed (by basic goniometry) which leads to equation in the bottom of image. The constants a0,y0,z0 are very hard to measure (it is inside complex machinery) so the measurement accuracy for distances is min 0.1 mm and angle 0.1 deg and even that is questionable.

[calibration]

So I decided to try compute these parameters from set of measurements done by the machine itself (auto-calibration). So I have calibration tube with known radius r0. All green parameters can be handled as constants. Now I position the tube along y axis to cover as much angles of arm as I could. Sadly the range is only about 20 degrees (for current machine setup) remembering measured a(t) for preset y(t) ... as n point dataset. This gives me system of n transcendent equations. From this I try/guess "all" possibilities of a0,y0,z0 remembering the best solution (closest to r0)

[approximation of a0,y0,z0]

approximation is based on this class of mine:

//---------------------------------------------------------------------------
class approx
    {
public:
    double a,aa,a0,a1,da,*e,e0;
    int i,n;
    bool done,stop;

    approx()            { a=0.0; aa=0.0; a0=0.0; a1=1.0; da=0.1; e=NULL; e0=NULL; i=0; n=5; done=true; }
    approx(approx& a)   { *this=a; }
    ~approx()           {}
    approx* operator = (const approx *a) { *this=*a; return this; }
    //approx* operator = (const approx &a) { ...copy... return this; }

    void init(double _a0,double _a1,double _da,int _n,double *_e)
        {
        if (_a0<=_a1) { a0=_a0; a1=_a1; }
        else          { a0=_a1; a1=_a0; }
        da=fabs(_da);
        n =_n ;
        e =_e ;
        e0=-1.0;
        i=0; a=a0; aa=a0;
        done=false; stop=false;
        }
    void step()
        {
        if ((e0<0.0)||(e0>*e)) { e0=*e; aa=a; }         // better solution
        if (stop)                                       // increase accuracy
            {
            i++; if (i>=n) { done=true; a=aa; return; } // final solution
            a0=aa-fabs(da);
            a1=aa+fabs(da);
            a=a0; da*=0.1;
            a0+=da; a1-=da;
            stop=false;
            }
        else{
            a+=da; if (a>a1) { a=a1; stop=true; }       // next point
            }
        }
    };
//---------------------------------------------------------------------------

It search the full range of single variable by some initial step then find the min deviation point. After that change the range and step to close area of this point and recursively increase accuracy.

The solution itself looks like this:

// (global) input data
#define _irc_calib_n 100
#define _irc_approx_n 5
int    irc_calib_ix; // number of measured points
double irc_calib_y[_irc_calib_n]; // y(t)
double irc_calib_a[_irc_calib_n]; // a(t)
double irc_calib_r; // calibration tube radius + arm radius

// approximation
int ix=0;
double e,a,deg=M_PI/180.0;
approx aa,ay,az;
//           min       max       step     recursions    ErrorOfSolutionVariable
for (aa.init(-90.0*deg,+90.0*deg,10.0*deg,_irc_approx_n,&e);!aa.done;aa.step())
for (ay.init(  0.0    ,200.0    ,10.0    ,_irc_approx_n,&e);!ay.done;ay.step())
for (az.init( 50.0    ,400.0    ,10.0    ,_irc_approx_n,&e);!az.done;az.step())
    {
    for (e=0.0,ix=0;ix<_irc_calib_n;ix++) // test all measured points (e is cumulative error)
        {
        a=irc_calib_a[ix]+aa.a;
        if (a> pi) a-=pi2;
        if (a<-pi) a+=pi2;
        if (fabs(a)>0.5*pi) { e=100.0; break; } // ignore too far angles
        e+=fabs(+(cos(a)*(irc_calib_y[ix]-ay.a))
                -(sin(a)*(az.a))
                -(irc_calib_r));
        }
    }
// here aa.a,ay.a,az.a holds the result

This leads to solution close to the measured values but inside simulation the result is still not accurate enough. It is from 0.1 mm to 0.5 mm depending on number of points and angle range. If I measure properly z0 and ignore its approximation then the precision is boosted significantly leaving y0 without error (in simulation) and a0 with error around 0.3 degree

Q1 how can I further improve accuracy of the solution?

I cannot increase the angular range. The number of points is best around 100 the more the better accuracy but above 150 the result is unstable (for some radiuses it is completely off). Have absolutely no clue why. The recursions number above 6 has not much effect

Could help weighting the deviations according to angular distance from 0 degree ? But sadly a(t) range does not necessarily include 0 degrees

desired accuracy is 0.01 mm for y0,z0 and 0.01 degree for a0

Q2 is there something I have missed?

Like wrongly nested approximations or some math simplification or different approach

[notes]

The angle must be in form of a(t)+a0 because it is measured by IRC with SW reset (16000 steps/round). It is reseted when in a0 position I do not count vibrations and calibration tube eccentricity they are taken care of already and my first goal is to make this work in simulation without them. Tube y(t) can be positioned at free will and the a(t) measurement can be done at will.

Right now the calibration process scan points along y axis (movement from a0 down). Computation with 6 recursions take around 35 seconds (so be patient). 5 recursions take around 22 seconds

[edit1] here how the simulation is done

approx aa; double e;
for (aa.init(-90.0*deg,+90.0*deg,10.0*deg,6,&e);!aa.done;aa.step())
 e=fabs(+(cos(aa.a)*(y(t)-y0))
        -(sin(aa.a)*(z0))
        -(irc_calib_r));
if (aa.a<a0) aa.a=a0;

[edit2] some values

Just realized that I had only 4 recursions in simulation code to match the input IRC accuracy then there must be 6 recursions. After changing it (also in previous edit) here are some results

                | a0[deg]| y0[mm] | z0[mm] | 
    simulated   | -7.4510|191.2590|225.9000|
    z0 known    | -7.4441|191.1433|225.9000|
    z0 unknown  | -7.6340|191.8074|225.4971|

So the accuracy with z0 measured is almost in desired range but with z0 unknown the error is still ~10 times bigger then needed. Increasing simulation accuracy has no effect above 6 recursions and also no sense because real input data will not be more accurate either.

Here the simulated/measured points for testing with above simulated settings:

 ix   a [deg]    y [mm]
  0   -0.2475 +105.7231 
  1   -0.4500 +104.9231 
  2   -0.6525 +104.1231 
  3   -0.8550 +103.3231 
  4   -1.0575 +102.5231 
  5   -1.2600 +101.7231 
  6   -1.4625 +100.9231 
  7   -1.6650 +100.1231 
  8   -1.8675  +99.3231 
  9   -2.0700  +98.5231 
 10   -2.2725  +97.7231 
 11   -2.4750  +96.9231 
 12   -2.6775  +96.1231 
 13   -2.8575  +95.3077 
 14   -3.0600  +94.5154 
 15   -3.2625  +93.7231 
 16   -3.4650  +92.9308 
 17   -3.6675  +92.1385 
 18   -3.8700  +91.3462 
 19   -4.0725  +90.5538 
 20   -4.2750  +89.7615 
 21   -4.4877  +88.9692 
 22   -4.6575  +88.1769 
 23   -4.8825  +87.3615 
 24   -5.0850  +86.5154 
 25   -5.2650  +85.7000 
 26   -5.4675  +84.9077 
 27   -5.6700  +84.1154 
 28   -5.8725  +83.3231 
 29   -6.0750  +82.5308 
 30   -6.2775  +81.7000 
 31   -6.5025  +80.8462 
 32   -6.6825  +80.0462 
 33   -6.8850  +79.2538 
 34   -7.0875  +78.4615 
 35   -7.2900  +77.6538 
 36   -7.5159  +76.7692 
 37   -7.6725  +75.9769 
 38   -7.8750  +75.1846 
 39   -8.1049  +74.3692 
 40   -8.2800  +73.5000 
 41   -8.4825  +72.7077 
 42   -8.6850  +71.9154 
 43   -8.9100  +71.0308 
 44   -9.0900  +70.2231 
 45   -9.2925  +69.4308 
 46   -9.5175  +68.5462 
 47   -9.6975  +67.7462 
 48   -9.9000  +66.9462 
 49  -10.1025  +66.0615 
 50  -10.3148  +65.2692 
 51  -10.4850  +64.3769 
 52  -10.6875  +63.5846 
 53  -10.9125  +62.7462 
 54  -11.0925  +61.9077 
 55  -11.2950  +61.0846 
 56  -11.4975  +60.2231 
 57  -11.7000  +59.3923 
 58  -11.9025  +58.5308 
 59  -12.1288  +57.6692 
 60  -12.3075  +56.8385 
 61  -12.5100  +55.9462 
 62  -12.7125  +55.1538 
 63  -12.9150  +54.2615 
 64  -13.1175  +53.4000 
 65  -13.2975  +52.5769 
 66  -13.5000  +51.6846 
 67  -13.7025  +50.7923 
 68  -13.9050  +50.0000 
 69  -14.1075  +49.1077 
 70  -14.3100  +48.2154 
 71  -14.5350  +47.3615 
 72  -14.7150  +46.5308 
 73  -14.9175  +45.6385 
 74  -15.1200  +44.7462 
 75  -15.3225  +43.8538 
 76  -15.5250  +42.9615 
 77  -15.7490  +42.0692 
 78  -15.9075  +41.2769 
 79  -16.1100  +40.3846 
 80  -16.3125  +39.4923 
 81  -16.5150  +38.6000 
 82  -16.7175  +37.7077 
 83  -16.9200  +36.8154 
 84  -17.1225  +35.9231 
 85  -17.3250  +34.9308 
 86  -17.5275  +34.0385 
 87  -17.7300  +33.1462 
 88  -17.9325  +32.2538 
 89  -18.1350  +31.3615 
 90  -18.3405  +30.4692 
 91  -18.5175  +29.4769 
 92  -18.7200  +28.5846 
 93  -18.9225  +27.6923 
 94  -19.1250  +26.8000 
 95  -19.3275  +25.8077 
 96  -19.5300  +24.9154 
 97  -19.7325  +23.9231 
 98  -19.9350  +23.0308 
 99  -20.1375  +22.1385 

[edit3] progress update

some clarification for @Ben

how it works

the colored equation under the first image gives you the radius r0 it is made from 2 joined 90 degree triangles (basic trigonometry)

red stuff:

• y(t) is motor position and it is known
• a(t) is IRC state also known

green stuff:

• a0,y0,z0 are mechanical dimensions and are known but not precise so I measure many a(t) for different positions of y(t) with known calibration tube r0 and compute the a0,y0,z0 with higher precision from it

further accuracy improvement

I actually managed to get it more precise by measuring y1=y0+z0*cos(a0) from special calibration movement with precision around 0.03 mm and better. It is the height of intersection between arm in a0 position and tube y movement axis. It is measured and interpolated from situation when arm get first time contact when tube coming from up to down but the real position must be recomputed by used radius and a0... because contact point is not on this axis ... (unless r0=0.0). This also eliminates one approximation loop from calibration because y1,a0,z0 are dependent and can be computed from each other. Also removing double aliasing from measurement of IRC due to discontinuous manner of measurement and a(t),y(t) positions helped a lot to increase precision and computation stability (on real machine). I can not reliably asses accuracy right now because by analysis of many measured cycles I found some mechanical problems on the machine so I wait until it is repaired. Anyway the calibration vs. simulation accuracy for r0=80.03 mm with accounting both approaches and _irc_calib_n=30 is now:

    ;      computed     simulated  |delta|
    a0=  -6.915840 ;  -6.916710   +0.000870 deg
    y0=+186.009765 ;+186.012822   +0.003057 mm
    y1=+158.342452 ;+158.342187   +0.000264 mm
    z0=+228.102470 ;+228.100000   +0.002470 mm

The bigger the calibration r0 the less accuracy (due to more limited a(t) range) this is by computing all a0,y0,(y1),z1 nothing is measured directly or known. This is already acceptable but as I wrote before need to check on machine when it is ready. Just to be complete here is how simulated measurements looks like now:

[edit4] see How approximation search works

Answer 1

If I understand this correctly, you're trying to infer (but not measure) the radius r0 of the tube from measurements for y and a.

Applying the usual error propagation to your formula for r0, one obtains (an estimate for) the error of the resulting r0. In the limit of small angles (applicable here, since a(t) is limited to 20 degrees), this gives roughly (using the small-angle approximation for the trigonometic functions)

dr0^2 ~= dy^2 + z0^2 (pi*da/180)^2

Thus, in the case of r0 much smaller than z0, the relative error on r0 is always much larger than the relative errors of y and z0*sin(a). This is already clear from your graph: the measured quantities depend only weakly on r0.

In other words, this is not a clever way to determine the radius r0. There is not much you can do about this fundamental limitation (except you can increase the range of angle a). Making many measurements (the usual method to beat down noise/errors) presumably won't help, because these measurements aren't independent of each other due to the internal workings of your machine. So, the only help would be more accurate measurements.

To analyse the situation, I recommend to make plots/figures of, say, the inferred r0 as function of y or of y as function of a for fixed r0.