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semiconductor  
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 More options Oct 28 2005, 7:12 am
Newsgroups: sci.math.symbolic
From: semiconductor <erh...@hotmail.com>
Date: Fri, 28 Oct 2005 07:12:35 EDT
Local: Fri, Oct 28 2005 7:12 am
Subject: equation simplification
Print | Individual message | Show original | Report this message | Find messages by this author
Hi,

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(t 3)+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+t 2)*(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.


 
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Dave Rusin  
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 More options Oct 28 2005, 1:08 pm
Newsgroups: sci.math.symbolic
From: ru...@vesuvius.math.niu.edu (Dave Rusin)
Date: 28 Oct 2005 17:08:50 GMT
Local: Fri, Oct 28 2005 1:08 pm
Subject: Re: equation simplification
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In article <28131408.1130497985592.JavaMail.jaka...@nitrogen.mathforum.org>,

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(t 3)+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+t 2)*(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" ...


 
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