About the Nonlinear System Solver

Hi, I am trying to solve a nonlinear system that is a little complicated to get the solution manually:
B*Sin(P)/(2*sin(B)*b)=0
B*Sin(W)/(2*sin(A)*b)=0
A+B+P+W-180=0
I tried to use the Nonlinear System Solver entering the formulas and
all I get is an error code -23001 (Error -23001 occurred at an
unidentified location Possible reason(s):
Analysis: Syntax error of parser. "
I don't know what I am doing wrong, I am doing it similarly to the
attached VI, with my own equations and controls. I even tried to change
the equations so I would delete the "sin" (sine), just to make a test
(about syntax and the solver not understanding that command?), and I
got the same error.
Help is much appreciated,
Thanks

Also remember that the argument to trigonometric functions must be in radians here.
Are you sure that A and Beta are not [0<A<π/2], [0<Beta<π/2].
How about the "180" in the last formula. Shouldn't it be π instead?
Is b 83 degrees or just a constant of 82?
LabVIEW Champion . Do more with less code and in less time .

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Distance along an ellipse arc (nD Nonlinear System Solver (VI).vi)

I'm hoping to get some help solving an issue I've come up against. I appreciate that what I'm asking requires some of your time, so I've tried hard to do my due diligence before posting here. I'll really appreciate any help I can get. I'm using LabVIEW 2014 Professional on Windows 7.
I've stripped away all the details of the specific application I'm working on and tried hard to isolate just the issue that is causing me difficulties.
I've posed the problem below and I attached some investigation code (sorry, there is quite a bit) that represents my current solution (with the issue).
Problem: Given an ellipse (with a major and a minor axis and a frame of reference), a distance to travel and any point on the ellipse, calculate the end point on the ellipse where the elliptical arc length between the two points equals the given distance to travel.
I'll restate the problem, but with physical values. Given an ellipse, centered on the origin, with axes of 10mm and 8mm, and a starting point (5mm, 0mm); what is the end point on the ellipse after travelling 1 mm along the ellipse?
This initially seemed like a simple calculation (and it is for a circle), but for an ellipse, the math turns out to be somewhat more complicated and requires numerical computation.
Issue: I was given some python code that solves this problem and embarked on a LabVIEW solution. The python code uses the function optimize.fsolve from the scipy library to solve one aspect of the problem. I'm not too familiar with the LabVIEW VIs available to do the equivalent, but settled on "LabVIEW 2014\vi.lib\gmath\zero.llb\nD Nonlinear System Solver (VI).vi" (my code still has other options disabled in the code). When I use this solver VI, I do not always get the same answer from this VI. Sometimes it finds one root and sometimes two. For points separated by less than half the circumference of the ellipse, there should always be 2 roots (one in the clock-wise and one in the counter clock-wise direction). Any ideas why the solver does not always return 2 roots?
Current Solution: My investigation VI takes two points on an ellipse (defined by a start angle and end angle) and calculates the distance between them along the ellipse. Then, I take the start point and the distance and try to calculate the end angle (which I can use to find the end point). Run once, it usually finds the right answer, but run 100 times it often returns the answer 19. I should mention that I'm not 100% positive there are no issues with the code used to support this calculation and I'm open to any suggested changes or improvements, but the issue currently causing problems is the inconsistent operation of the solver VI.
Again, I appreciate this is long and thank you for reading this far; I hope you find this as fun a problem as I do!
Chris
Solved!
Go to Solution.
Attachments:
NI Forums.zip ‏265 KB

Hello Mark,
Thanks for your reply, I was concerned that I had sent up an air ball on this one! And thank you for taking the time to review the code.
You are correct, I am having an issue with the results of the nD Nonlinear System Solver (VI).vi. And that is a good poiint you raise, regarding the choice of the starting points. I'm not intending to use random values, I think I'm using a number very close to the start angle (in this case 20.001 degrees) as a start and 2pi (or 360 degrees) as the end. When I read the description of the start and end from the documentation:
Start is an array describing the left corner of the n-dimension interval. The randomly chosen start points of the zero-finding algorithm can be found in the n-dimensional rectangle spanned by Start and End.
End is an array describing the right corner of the n-dimension interval. The randomly chosen start points of the zero-finding algorithm can be found in the n-dimensional rectangle spanned by Start and End.
I have to admit I don't *understand* what this means and I am currently guessing for the choices of start and end. I will think on this some more and review your links.
As far as my final goal, I'm going to use this code I shared for generating an ellipitical motion profile on the PC to send to a cRIO and have executed on a FPGA.
I'll report back after I work on this some more, but if you have some more insights, please share.
Chris
(BTW, I have an open support item for this #7444649 and it was suggested I also post on the forums)

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To whom it may concern,
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About the Nonlinear System Solver

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