What is the best way to solve the system of following trigonometric equations with respect to $w_l, w_r, b_l, b_r$?

$$ F_x=w_l^2sin(b_l)+w_r^2sin(b_r) $$ $$ F_z=w_l^2cos(b_l)+w_r^2cos(b_r) $$ $$ M_x=w_r^2cos(b_r)-w_l^2cos(b_l) $$ $$ M_y=w_l^2sin(b_l)+w_r^2sin(b_r) $$ $$ M_z=w_r^2sin(b_r)-w_l^2sin(b_l) $$ I need to find relation between $w_l, w_r, b_l, b_r$ and symbolic coefficients $F_x,F_z,M_x,M_y,M_z$.

  • $\begingroup$ May I clarify if you are looking to express $w_l, w_r, b_l, b_r$ in terms of $F_x, F_z, M_x, M_y, M_z$? $\endgroup$ – Yiyuan Lee Sep 12 '15 at 12:59
  • $\begingroup$ Do you have $F_x=M_y$ or is there a typo? $\endgroup$ – David Quinn Sep 12 '15 at 13:00
  • $\begingroup$ @LeeYiyuan yes It shouldn't be all 5 coefficients, of course. As there are only 4 unknowns and 5 equations. $\endgroup$ – Andrew Sep 12 '15 at 13:03
  • $\begingroup$ @DavidQuinn It is not a typo. The full version of equations looks like $M_y/h*k_f$, but i replaced this with $M_y$ to simplify. $\endgroup$ – Andrew Sep 12 '15 at 13:07

Note that $$F_z + M_x = 2w_r^2\cos b_r$$ $$F_x + M_z = 2w_r^2\sin b_r$$

thus for $b_r$ and $w_r$ we have (dividing one equation by another, using $\sin ^2 \theta + \cos^2 \theta = 1$):

$$\tan b_r = \frac{F_x + M_z}{F_z + M_x}$$

$$4w_r^4 = (F_x + M_z)^2 + (F_z + M_x)^2$$

Note also that

$$F_z - M_x = 2w_l^2\cos b_l$$ $$F_x - M_z = 2w_1^2\sin b_l$$

thus for $b_l$ and $w_l$ we have

$$\tan b_l = \frac{F_x - M_z}{F_z - M_x}$$ $$ 4w_l^4 = (F_x - M_z)^2 + (F_z - M_x)^2$$

(As mentioned in the comments by OP, $M_y$ turns out to be unused indeed.)

  • $\begingroup$ Thank you. Actually this is simplified version of full model with first and second differential terms, and few other forces and moments. But since they are only 1/10 of the full force/torque vectors so i omitted them in simplified model shown above. This model will be used in microcontroller's closed loop, so it must be as simple as possible to guarantee fast response. $\endgroup$ – Andrew Sep 12 '15 at 13:22
  • $\begingroup$ Perhaps $w_l, w_r, b_l, b_r$ can indeed be expressed even more easily in terms of the components of the other 9/10 vectors. However, with just these components available, I think the representation given is rather simple indeed. $\endgroup$ – Yiyuan Lee Sep 12 '15 at 13:28

If you ignore the fourth equation, since $F_x$ and $ M_y$ are identical, then number the remaining equations 1-4:

Take successively the sum of equations 1 and 4,and the difference of 1 and 4.

Call these equations A and B. Do the same operations on equations 2 and 3, call these C and D.

Now dividing B and D gives $$\tan \beta_l=\frac{F_x-M_z}{F_z-M_x}$$ Squaring and adding the same gives $$4\omega_l^4=(F_x-M_z)^2+(F_z-M_x)^2$$

Now do the same set of operations on equations A and C and you're done.


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