Eccentric circles problem

[Kataeb]

yes, we have some competitors that have done this with a plc. the control is not that fast (the diameter is adjusted by inflating or deflating the bubble cage).
i think that it is done with a simplified formula (that we do not have) with acceptable errors.

 

[Peter Nachtwey]

Why don't you provide some more information?

What is fast?
How accurate must this be?
Obviously the other PLC solutions are not perfect solutions but maybe you don't need that.

What is the radius of ring that the sensors are mounted on?
What is the range of values for r0?
What is the range of values for x0?
What is the range of values for y0?

Consider an iterative technique called a grid search.
Let's say:
r0 varies from 100 to 200 mm
x0 varies from -100 mm to +100 mm
y0 varies from -100 mm to +100 mm

Now try 3 loops
BestSSE=10000000; // big number

 For r0 =100 to 200 step 10
   For x0=-100 to +100 step 10
      for y0=-100 to +100 step 10
          SSE = The equation above.
          if BestSSE > SSE THEN
             BestSSE=SSE
             BestR0=R0
             BestX0=X0
             BestY0=y0
          Endif
       EndFor
   EndFor  
 EndFor

I call this brute force and ignorance but it will get the job done. It is called a grid search. Notice that you will have to go through 4000+ iterations to find the best solution. At 10 ms per scan that is a long wait. Once you find the Best values you can narrow the search grid to around the best solution to get the answer within a mm.

The are other better ways. I suggest you plug my formulas for computing the errors in a spread sheet and see what happens when you change r0,x0 and y0. Do this manually at first until you get a feel for how the SSE change with the changes in r0,x0 and y0.

Another search that may work well is the simplex search. It doesn't require a whole lot of fancy math and will probably converge MUCH faster than the grid search.
http://www.boomer.org/c/p3/c11/c1102.html
Try this in excel. Make a plot showing the errors as a function of x0, y0 and r0 as shown in the link. This will provide a feel for what you are doing.

If you are using a S7 then buy SCL for this. Don't skimp on the hardware or software or you will waste many times the cost saved in man hours in developing and fixing bugs. Do you really want to wait 40 seconds for a solution if a faster PLC can do it in half the time?

 

[Peter Nachtwey]

There is a solution for the radius on.

sci.math.symbolic
http://groups.google.com/group/sci.math.symbolic/browse_frm/thread/41731828042e31a4#

Danial Lichtblau works for Wolfram which makes what is probably the best math software available, Mathematica.

Yesterday I had worked out y0 in terms of r0. The equation wasn't too bad. Then one substitutes again to find x0. Easy

I still don't think you want to plug that into a PLC.

 

[Electrically_Bound]

Just use a tape measure.

 

[Peter Nachtwey]

Aw you are no fun.

BTW, I just checked my e-mail and someone from the sci.math.symbolic sent me the whole solution, I think. It was done using Mathematica and the person who e-mailed me said it took a few seconds to generate. It is too big to be useful. So much for the simple formula idea but now we know.

 

[Peter Nachtwey]

Now one person has a 3D solution

A professor that writes his own math software thought the problem was too easy so the added a another dimension. In his problem the 4 ultrasonic sensors are arranged in a tetrahedron and instead of finding the location and radius of a circle he can now find the radius and location of a sphere. He said that the solution is extreme big. The text for the solution takes 10 megs of ram.

http://groups.google.com/group/sci.math.symbolic/browse_frm/thread/41731828042e31a4#

These are the guys I take my hat off to. The know of algorithms and techniques I haven't heard of.
Now I must research [FONT=Courier, Monospaced]Dixon-EDF.[/FONT]

I wonder how Kataeb is doing with his iterative calculations.

 

[Kataeb]

Dear Sirs, I am working on this issue only in my spare time (tons of other projects and administrative work,…). But I made some serious simplifications and reductions to reach a simple formula (equilateral position of the sensors, origin of the x and y axis is the center of the sensors equilateral triangle,…) . I think I did reach some promising results. But need some time to verify the calculation and double check it in autocad software…
Thanks all…

 

[Peter Nachtwey]

WOW, I was doing some math and came across this

http://en.wikipedia.org/wiki/Heron%27s_formula
http://en.wikipedia.org/wiki/Incenter

I needed to compute where the center of a circle is knowing the radius that is tangent to two lines. This is for making CNC machine types of turns

It just goes to show that knowing the algorithms is crucial because brute force and computer software will not always be the right solution. Know on the news group knew this either.

I am still digesting this. Note, Heron lived in 60 AD. Wow!!!

I can solve my problem but it is messy so was looking for simpler/faster solutions.

Never mind, this is not the solution to Kataeb's problem. The outer circles only touch the tangent lines, not the inner circle. Heron's formula is still cool.

 

[rootboy]

... it turns into a pretty pattern; I am sure I can see a face in there

Funny, all I see now is blonde, brunette, redhead...

 

[JOLTRON]

Funny, all I see now is blonde, brunette, redhead...

Do not try and bend the spoon. That's impossible. Instead... only try to realize the truth.

There is no spoon.