equation simplification
semiconductor
I have a question regarding the equations listed below. I want to simplify them as much as possible using trig identities so that I can solve for theta1, theta2, and theta3.
My second question is are there any books that discuss how to solve/simplify such (trig) equations.
Constants: Px,Py,Pz. l1,...,l5. d1,...,d5.
Variables: t1,...,t5 (t=theta)
1) l1*cos(t1)+l2*cos(t1+t2)+(d3+d4+l5*sin(t5))*sin(t1+t2)+cos(t1+t2)*(l3*cos(t3)+l4*cos(t3+t4)+cos(t3+t4)*l5*cos(t5)+d5*sin(t3+t4))=Pz
2) -(d3+d4)*cos(t1+t2)-l5*sin(t5)*cos(t1+t2)+l1*sin(t1)+l2*sin(t1+t2)+sin(t1+t2)*(l3*cos(t3)+l4*cos(t3+t4)+l5*cos(t5)*cos(t3+t4)+d5*sin(t3+t4))=Py
3) d1+d2-d5*cos(t3+t4)+l3*sin(t3)+l4*sin(t3+t4)+l5*cos(t5)*sin(t3+t4)=Pz
Thank you for your help.
Dave Rusin
semiconductor <erh...@hotmail.com> wrote:
>I have a question regarding the equations listed below. I want to simplify them as much as possible using trig identities so that I can solve for theta1, theta2, and theta3.
I hope you don't intend to work with these symbolically, since explicit
expressions for the t_i will be ghastly. But perhaps you intend to do
numerical computations to solve for t1,t2,t3 when all the other
quantities are known numerically? In that case things aren't TOO bad.
You have one equation that involves just one of the three variables:
>3) d1+d2-d5*cos(t3+t4)+l3*sin(t3)+l4*sin(t3+t4)+l5*cos(t5)*sin(t3+t4)=Pz
This can be expanded to A*sin(t3) + B*cos(t3) = C where A,B,C do not
involve t1, t2, or t3. Use
sin(t) = 2*u/(1+u^2), cos(t) = (1-u^2)/(1+u^2) where u = tan(t3/2)
to write your equation as a quadratic in u . Solve with quadratic formula
and use arctan() to express t3 in terms of everything else.
Then you can turn to
>1) l1*cos(t1)+l2*cos(t1+t2)+(d3+d4+l5*sin(t5))*sin(t1+t2)+cos(t1+t2)*(l3*cos(t3)+l4*cos(t3+t4)+cos(t3+t4)*l5*cos(t5)+d5*sin(t3+t4))=Pz
>2) -(d3+d4)*cos(t1+t2)-l5*sin(t5)*cos(t1+t2)+l1*sin(t1)+l2*sin(t1+t2)+sin(t1+t2)*(l3*cos(t3)+l4*cos(t3+t4)+l5*cos(t5)*cos(t3+t4)+d5*sin(t3+t4))=Py
which, after substituting in the value of t3 above, have the form
1) A*cos(t1)+B*cos(t1+t2)+C*sin(t1+t2)=Pz
2) D*sin(t1)+E*cos(t1+t2)+F*sin(t1+t2)=Py
Again I use use the half-angle substitutions (i.e., express in terms of
u1 = tan(t1/2) and u2 = tan( (t1+t2)/2 ) ) and get a pair of biquadratic
equations in u1 and u2. You'll have to solve a quartic polynomial
this time: u2 is a root of
(2*Py*A^2*E+Pz^2*D^2+Py^2*A^2+2*Pz*B*D^2+B^2*D^2+E^2*A^2-A^2*D^2)*u2^4+(-4*F*A
^2*E-4*B*C*D^2-4*Pz*C*D^2-4*F*A^2*Py)*u2^3+(2*Py^2*A^2-2*B^2*D^2+4*A^2*F^2-2*E
^2*A^2+4*C^2*D^2+2*Pz^2*D^2-2*A^2*D^2)*u2^2+(-4*F*A^2*Py-4*Pz*C*D^2+4*F*A^2*E+
4*B*C*D^2)*u2+E^2*A^2+Py^2*A^2+B^2*D^2-2*Pz*B*D^2-2*Py*A^2*E-A^2*D^2+Pz^2*D^2
and then u1 =
-D*(-A-A*u2^2+B*u2^2+Pz*u2^2-B+Pz-2*C*u2)/A/(E*u2^2+Py*u2^2-E+Py-2*F*u2)
(assuming the denominator is nonzero).
So you have 4 {u1,u2} pairs; take arctangents to compute t1 and t1+t2
(and thus t2).
dave
PS -- I'm guessing the "Pz" in equation 1 is supposed to be "Px" ...