Team:Washington-Software/Modeling

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Latest revision as of 22:18, 18 October 2009

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Mathematical Modeling

Solving Reverse Triangulation in 3-D Space with Polar Coordinate Constraints

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

We based the mathematical construction on the physical structure of ALPHA.

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,
  • The length of CA_x is CA, 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 locus of the control arms are circles of radius CA, while the locus of the linkages are spheres of radius L.
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
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 find the error in the next step. 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 following three 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_o3/D₃)

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