# Team:Washington-Software/Modeling

### From 2009.igem.org

## Contents |

## Mathematical Modeling

### Alpha

#### Problem

Given the following construction and point *p*, or (*x*,*y*,*z*) find the angles *θ _{1}*,

*θ*, and

_{2}*θ*.

_{3}Note that positive *z* is the down direction.

#### Constants

*TR*- Top radius

*BR*- Bottom radius

*L*- Linkage

*CA*- Control Arm

*φ*and_{1}*φ*_{2}- Two angles

#### Construction

From the top to the bottom:

- A circle centered at the origin with radius
*TR*, named*O* - Make three lines
*A*,_{1}*A*, and_{2}*A*such that:_{3}- The lines
*A*are perpendicular to a tangent of_{x}*O*and a radius of*O*, - The angle between the radii of
*A*and_{1}*A*is_{2}*φ*, and_{1} - The angle between the radii of
*A*and_{2}*A*is_{3}*φ*._{2}

- The lines
- Make a circle centered at point
*p*with radius*BR*, named*P* - Find three points
*P*,_{1}*P*, and_{2}*P*such that:_{3}- They are on the circumference of
*P*, - The angle between the radius which touches
*P*and the radius that touches_{1}*P*is_{2}*φ*,_{1} - The angle between the radius which touches
*P*and the radius that touches_{2}*P*is_{3}*φ*,_{2} - The ray from the center of
*P*to*P*is parallel to the ray from the center of_{x}*O*to the point which is on*A*and_{x}*O*'s circumference, for all*x*.

- They are on the circumference of
- Construct three line segments
*CA*,_{1}*CA*, and_{2}*CA*such that:_{3}*CA*is in the same plane as_{x}*A*, for all_{x}*x*, and- The angle between
*CA*and_{x}*A*is_{x}*θ*, for all_{x}*x*.

- The distance between
*P*and the end point of_{x}*CA*that is not on_{x}*A*is_{x}*L*, for all*x*.

#### Solution

- Note that the control arms can only move in a circle, while linkage can move in a sphere.
- We will calculate
*θ*first, which only involves the points, circles and lines_{1}*p*,*CA*,_{1}*A*, and_{1}*P*._{1} - Find the plane where
*CA*'s circle resides in. Use it to cut the sphere around_{1}*P*. For future reference, call_{1}*CA*'s circle_{1}*C*and the circle resulting from the cut_{1}*C*_{2} - Define
*x*, for x offset, for the difference in the_{o1}*x*coordinates of the center of*C*and the point_{1}*P*._{1}*x*_{o1}= TR - (BR + x)*x*_{o1}= TR - BR - x

- Define
*D*to be the distance between the center of_{1}*C*and_{1}*C*. It will also be the line connecting the centers._{2}*D*_{1}= sqrt(x_{1}^{2}+ z^{2})

- Obviously,
*D*,_{1}*CA*, and linkage form a triangle. The angle between_{1}*CA*and_{1}*D*is a close approximation to_{1}*θ*, but it is not exact. We will call this angle_{1}*θ*. We use the law of cosines to calculate_{11}*θ*._{11}*L*^{2}= CA^{2}+ D^{2}- CA * D * cos(θ_{11})

*θ*_{11}= cos^{-1}((D^{2}+ CA^{2}- L^{2})/(2*D*CA))

- One endpoint of
*D*is on_{1}*A*. Thus, we can make another triangle, and the angle between_{1}*D*and_{1}*A*is the difference between_{1}*θ*and_{1}*θ*. We will call this angle_{11}*θ*. If we have the length of the side on_{12}*A*be_{1}*z*, then the last side will be*x*and the triangle will be a right angle triangle. Since only_{o}*x*changes sign in the good interval, we should use a trigonometric function that involves_{o1}*x*in the numerator. Thus,_{o}*θ*_{12}= sin^{-1}(x_{o1}/D_{1})

*θ*_{1}= cos^{-1}((D^{2}+ CA^{2}- L^{2})/(2*D*CA)) - sin^{-1}(x_{o1}/D_{1})

- You just have to rotate the model
*φ*degrees counterclockwise for_{1}*θ*, and another_{2}*φ*degrees for_{2}*θ*using the two dimensional rotational matrixes_{3}

{cos(φ}_{2}),-sin(φ_{2})R={ } {_{1}sin(φ}_{2}) cos(φ_{2})

{cos(φ}_{3}) -sin(φ_{3})R={ } {_{2}sin(φ}_{3}) cos(φ_{3})