Solving three equations with three unknowns with matlab 2011

Hi!

I'm having trouble with solving these equations:

L1=1,66
L2=7,01
L3=3,58

t1, t2 and t3 are the unknown angles.

1: cos(t1)*(L3*cos(t2 + t3) + L2*cos(t2))=4.600
2: sin(t1)*(L3*cos(t2 + t3) + L2*cos(t2))=0.200
3: L1 + L3*sin(t2 + t3) + L2*sin(t2)=1.400

Answer 1

[Walter Roberson]

There are at least 4 solutions:

t1 = 0.4345089537e-1, t2 = -.5282401133, t3 = 2.516094426

t1 = -3.098141759, t2 = -2.613352541, t3 = -2.516094426

t1 = 0.4345089539e-1, t2 = .4154231405, t3 = -2.516094425

t1 = -3.098141759, t2 = 2.726169514, t3 = 2.516094425

Answer 2

[Roger Stafford]

I can sketch a method of solving them which doesn't involve the 'solve' function. Divide the second equation by the first one to get:

 tan(t1) = .2/4.6
 t1 = atan(.2/4.6)

which means that t1 has two solution between -pi and +pi, namely

 t1 = 0.043450895 and t1 - -0.043450895 and
 sin(t1) = 0.043437224 or -0.043437224

Then the second equation can be written either as

 L3*cos(t2 + t3) + L2*cos(t2) = 0.2/0.043437224 = 4.604345773 or
 L3*cos(t2 + t3) + L2*cos(t2) = -4.604345773

Combining this with the third equation we have two equations of the form

 L3*cos(t2 + t3) + L2*cos(t2) = K1
 L3*sin(t2 + t3) + L2*sin(t2) = K2

where K1 is either 4.604345773 or -4.604345773 and K2 = 1.4-1.66 = -.26

These equations can be solved this way.

 L3^2 = L3^2*sin(t2+t3)^2+L3^2*cos(t2+t3)^2 =
 (K1-L2*sin(t2))^2+(K2-L2*cos(t2))^2 = 
 K1^2+K2^2+L2^2-2*K1*L2*sin(t2)-2*K2*L2*cos(t2)

 K1*sin(t2)+K2*cos(t2) = (K1^2+K2^2+L2^2-L3^2)/(2*L2) = K3

Let x be such that

 K1/sqrt(K1^2+K2^2) = sin(x)
 K2/sqrt(K1^2+K2^2) = cos(x)

which we can solve using the 'atan' or 'atan2' functions and we then have

 cos(t2-x) = K3*sqrt(K1^2+K2^2)

which will have two solutions between -pi and +pi using the acos function.

With t1 and t2 determined to these four possible combinations, two of the three equations can be used to determine sin(t2+t3) and cos(t2+t3) uniquely which will then determine t3 to one of four possibilities within -pi to +pi.

I leave the remaining details to you. You can undoubtedly use the 'solve' function to shortcut at least some of the steps above.

Note: I see that Walter has already obtained four solutions, but you may find the above types of manipulations useful in dealing with problems of this kind in the future.

[Walter Roberson]

The methods I used would not necessarily inform me if additional solutions existed; the methods you used are better for making those kinds of decisions.