Convert linear distance to steering angle

Question

I need to calculate the angle of the front steering wheel using a collapsible piston(linear sensor). 'x' is used to represent the length in inches of the movable part of the sensor and is the independent variable.

θ represents the steering angle. The angle is 0 when the wheels are perfectly straight and increases as the wheels turn to the left and decreases as the wheels turn to the right.

This is the image where the θ is zero:

diagram 1

This is when it's steering all the way to the left:

diagram 2

and this is when it's steering all the way to the right:

diagram 3

I'm trying to solve it using the parametric equations of the circle; Relating the piston with the equations using the euclidean distance. As it turns out, I'm having a hard time solving for theta.

In the following two images you can see my derivations:

image 1 image 2

I'm stuck here:

$$b\cdot cos(\theta) - a\cdot sin(\theta) = \frac{a^2 + b^2 - d^2 + r^2}{2\cdot r}$$

Any help would me appreciate it!

Answer 1

What you could do is use the cosine rule in the triangle formed by the steering wheel mount point, the piston-wheel pin joint and the piston mount point. In this triangle, assuming $d$ is distance between both the fixed mounting points, and that for an extension $x$ the angle opposite to the piston is $\alpha$, we have

$$(x+5)^2 = 5^2 + d^2 -10d\cos\alpha$$

Also, you can relate $\theta$ to $\alpha$ as $\theta = \alpha - \alpha_0$, the resting position angle that can be calculated by simple geometry.

Does this help in any way?

Answer 2

Just taking your calculation further:

$$b\cdot cos(\theta) - a\cdot sin(\theta) = \frac{a^2 + b^2 - d^2 + r^2}{2\cdot r} =Q $$

$$\dfrac {b\cdot cos(\theta) - a\cdot sin(\theta)}{\sqrt{a^2+b^2}} = \dfrac{Q}{\sqrt{a^2+b^2}}= R$$

$$ \sin \beta \ cos\theta - \cos \beta \sin \beta = R$$

$$ \sin (\beta-\theta)=R$$

$$ \theta =\beta-\sin^{-1}R$$ where

$$ \beta = \tan^{-1}\frac{b}{a}$$