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From: Nicolaj Baramsky on 10 Aug 2010 05:52
Thank you very much!!

Nicolaj
From: Nicolaj Baramsky on 10 Aug 2010 10:58
The program works very fine for one single calculation.
Do I have to set a couple of vars back before i start the routine again?

I have a Matrix which contains four rows of pairs of data:

xy = [
x11 y11 x21 y21 x31 y31 x41 y41
x12 y12 x22 y22 x32 y32 x42 y42
x13 y13 x23 y23 x33 y33 x43 y43
x13 y14 x24 y24 x34 y34 x44 y44
]

soemthing like that.
I want to finde the middle points for each row with the following code:
(durchmesser = diameter; mittelpunkt = middle point)

mittelpunkte = zeros(4,2);
ssqs = zeros(4,1);

err = 0.002;
cp = [0.8,-4.9];
r2 = 5;
for i =1:4
j=2*i-1;
res = @(p) (xy(:,j)-p(1)).^2 + (xy(:,j+1)-p(2)).^2 - r2;
[p,ssq,cnt] =LMFnlsq(res,[-1;-5],'Display',-1,'xTol',1e-8,'FunTol',1e-8);
mittelpunkte(i)=p(1);
mittelpunkte(i+4)=p(2);
ssqs(i)=ssq;
end



But the result is, that only the calculation of the first row the iteration works fine:

The middle points are (x and y cooridnates)
0.368374615862839 -4.977713121786886
0.867764471106047 -4.914736713757876
0.866466160216724 -4.913231160578434
0.864868475942583 -4.911810869157897

SSQ for earch iterations are:
0.000481104234919 <--- see the huge difference to the other ones!
0.023268759661560
0.020660058447488
0.021206672812456



What did i forget?
Making more steps for the iterations changes nothing. So it must be a kind of initial problem.


Thanks a lot!
Nicolaj









*************************************************
itr nfJ sum(r^2) x dx
*************************************************
0 1 3.3770e+003 -1.0000e+000 0.0000e+000
-5.0000e+000 0.0000e+000
1 2 6.8409e+002 3.5038e-001 -1.3504e+000
-5.1610e+000 1.6102e-001
2 3 2.1454e-001 3.6805e-001 -1.7673e-002
-4.9810e+000 -1.8001e-001
3 4 4.8113e-004 3.6837e-001 -3.2410e-004
-4.9777e+000 -3.3010e-003
4 5 4.8110e-004 3.6837e-001 -1.0901e-007
-4.9777e+000 -1.1111e-006

5 6 4.8110e-004 3.6837e-001 -2.9573e-011
-4.9777e+000 2.4418e-012

*************************************************
itr nfJ sum(r^2) x dx
*************************************************
0 1 5.6133e+003 -1.0000e+000 0.0000e+000
-5.0000e+000 0.0000e+000
1 2 2.5388e+003 8.6903e-001 -1.8690e+000
-5.2638e+000 2.6379e-001
2 3 2.6187e+000 8.6781e-001 1.2231e-003
-4.9263e+000 -3.3751e-001
3 4 2.3272e-002 8.6776e-001 4.1860e-005
-4.9148e+000 -1.1529e-002
4 5 2.3269e-002 8.6776e-001 4.6656e-008
-4.9147e+000 -1.3483e-005

5 6 2.3269e-002 8.6776e-001 -1.8488e-010
-4.9147e+000 -2.0236e-011

*************************************************
itr nfJ sum(r^2) x dx
*************************************************
0 1 5.6365e+003 -1.0000e+000 0.0000e+000
-5.0000e+000 0.0000e+000
1 2 2.5298e+003 8.6764e-001 -1.8676e+000
-5.2617e+000 2.6169e-001
2 3 2.5989e+000 8.6650e-001 1.1326e-003
-4.9247e+000 -3.3695e-001
3 4 2.0664e-002 8.6647e-001 3.8629e-005
-4.9132e+000 -1.1491e-002
4 5 2.0660e-002 8.6647e-001 4.4456e-008
-4.9132e+000 -1.3394e-005

5 6 2.0660e-002 8.6647e-001 1.7052e-010
-4.9132e+000 -1.4988e-011

*************************************************
itr nfJ sum(r^2) x dx
*************************************************
0 1 5.6576e+003 -1.0000e+000 0.0000e+000
-5.0000e+000 0.0000e+000
1 2 2.5192e+003 8.6592e-001 -1.8659e+000
-5.2596e+000 2.5956e-001
2 3 2.5791e+000 8.6490e-001 1.0213e-003
-4.9233e+000 -3.3629e-001
3 4 2.1210e-002 8.6487e-001 3.4814e-005
-4.9118e+000 -1.1445e-002
4 5 2.1207e-002 8.6487e-001 3.9691e-008
-4.9118e+000 -1.3289e-005

5 6 2.1207e-002 8.6487e-001 8.9575e-011
-4.9118e+000 -1.1246e-011
From: Nicolaj Baramsky on 10 Aug 2010 11:03
I should add:

The iteration works fine for all data! When i change the second row with the first row, all works fine for the new first row.

So it is no problem of the data also.
From: Miroslav Balda on 10 Aug 2010 16:34
"Nicolaj Baramsky" <nicolaj.baramskyREMOVETHIS(a)tu-harburg.de> wrote in message <i3rpdt$at3$1(a)fred.mathworks.com>...
SNIP

> mittelpunkte = zeros(4,2);
> ssqs = zeros(4,1);
>
> err = 0.002;
> cp = [0.8,-4.9];
> r2 = 5;
> for i =1:4
> j=2*i-1;
> res = @(p) (xy(:,j)-p(1)).^2 + (xy(:,j+1)-p(2)).^2 - r2;
> [p,ssq,cnt] =LMFnlsq(res,[-1;-5],'Display',-1,'xTol',1e-8,'FunTol',1e-8);
> mittelpunkte(i)=p(1);
> mittelpunkte(i+4)=p(2);
> ssqs(i)=ssq;
> end

SNIP

The code looks strange. You defined mittelpunkte as a matrix 4x2, however you
address it as if it were a vector. I thing, that the proper code of the cycle could be as it follows (it has not been tested):

j=1:2:size(mittelpunkte,2);
for i =1:4
res = @(i,p) (xy(i,j)-p(1))'.^2 + (xy(i,j+1)-p(2))'.^2 - r2;
[p,ssqs(i),cnt] =LMFnlsq(res,[-1;-5],'Display',-1,'xTol',1e-8,'FunTol',1e-8);
mittelpunkte(i,:) = p';
end

Since all 4 rows are giving virtually the same result, it could be solved as a single problem for getting an average center point:

xy = [......];
r = ......;
r2 = r*r;
XY = xy';
XY = XY(:);
i = 1:2:length(XY);
res = @(p) (XY(i)-p(1)).^2 + (XY(i+1)-p(2)).^2-r2;
[mittelpunkte,ssqs,cnt] =LMFnlsq(res,[-1;-5],'Display',-1,'xTol',1e-8,'FunTol',1e-8);

If the data are well spread, the initial guess fo the solution can be set before calling the function LMFnlsq by a command:

p0 = [mean(XY(i));mean(XY(i+1))];

I hope that there is no error.
Best regards.

Mira
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