Team:Washington-Software/Modeling

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Mathematical Modeling

Alpha

Problem

Given the following construction and point p, or (x,y,z) find the angles θ₁, θ₂, and θ₃.

Note that positive z is the down direction.

Constants

• TR
  • Top radius
• BR
  • Bottom radius
• L
  • Linkage
• CA
  • Control Arm
• φ₁ and φ₂
  • Two angles

Construction

From the top to the bottom:

• A circle centered at the origin with radius TR, named O
• Make three lines A₁, A₂, and A₃ such that:
  • The lines A_x are perpendicular to a tangent of O and a radius of O,
  • The angle between the radii of A₁ and A₂ is φ₁, and
  • The angle between the radii of A₂ and A₃ is φ₂.
• Make a circle centered at point p with radius BR, named P
• Find three points P₁, P₂, and P₃ such that:
  • They are on the circumference of P,
  • The angle between the radius which touches P₁ and the radius that touches P₂ is φ₁,
  • The angle between the radius which touches P₂ and the radius that touches P₃ is φ₂,
  • The ray from the center of P to P_x is parallel to the ray from the center of O to the point which is on A_x and O 's circumference, for all x.
• Construct three line segments CA₁, CA₂, and CA₃ such that:
  • CA_x is in the same plane as A_x, for all x, and
  • The angle between CA_x and A_x is θ_x, for all x.
• The distance between P_x and the end point of CA_x that is not on A_x is L, for all x.

Solution

1. Note that the control arms can only move in a circle, while linkage can move in a sphere.
2. We will calculate θ₁ first, which only involves the points, circles and lines p, CA₁, A₁, and P₁.
3. Find the plane where CA₁ 's circle resides in. Use it to cut the sphere around P₁. For future reference, call CA₁ 's circle C₁ and the circle resulting from the cut C₂
4. Define x_o1, for x offset, for the difference in the x coordinates of the center of C₁ and the point P₁.
   • x_o1 = TR - (BR + x)
   • x_o1 = TR - BR - x
5. Define D₁ to be the distance between the center of C₁ and C₂. It will also be the line connecting the centers.
   • D₁ = sqrt(x₁² + z²)
6. Obviously, D₁, CA₁, and linkage form a triangle. The angle between CA₁ and D₁ is a close approximation to θ₁, but it is not exact. We will call this angle θ₁₁. We use the law of cosines to calculate θ₁₁.
   • L² = CA² + D₁² - CA * D * cos(θ₁₁)
     
     
   • θ₁₁ = cos^-1((D₁² + CA² - L²)/(2*D₁*CA))
     
     
7. One endpoint of D₁ is on A₁. Thus, we can make another triangle, and the angle between D₁ and A₁ is the difference between θ₁ and θ₁₁. We will call this angle θ₁₂. If we have the length of the side on A₁ be z, then the last side will be x_o and the triangle will be a right angle triangle. Since only x_o1 changes sign in the good interval, we should use a trigonometric function that involves x_o in the numerator. Thus,
   • θ₁₂ = sin^-1(x_o1/D₁)
     
     
   • θ₁ = cos^-1((D₁² + CA² - L²)/(2*D₁*CA)) - sin^-1(x_o1/D₁)
8. You just have to rotate the model φ₁ degrees counterclockwise for θ₂, and another φ₂ degrees for θ₃ using the two dimensional rotational matrixes:
   
9. Once you rotate the model, use steps 3-7 to get the next angles.
   • θ₂ = cos^-1((D₂² + CA² - L²)/(2*D₂*CA)) - sin^-1(x_o2/D₂)
   • θ₃ = cos^-1((D₃² + CA² - L²)/(2*D₃*CA)) - sin^-1(x_o1/D₃)

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