Team:Washington-Software/Modeling

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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 are basing 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, 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

Note that the locus of the control arms are circles, while the locus of the linkage is a sphere.
We will calculate θ₁ first, which only involves the points, circles and lines p, CA₁, A₁, and P₁.
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
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
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)
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 θ_1₁. We use the law of cosines to calculate θ_1₁.
- L² = CA² + D₁² - CA * D * cos(θ_1₁)
- θ_1₁ = cos^-1((D₁² + CA² - L²)/(2*D₁*CA))
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 θ_1₁. We will call this angle θ_1₂. 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,
- θ_1₂ = sin^-1(x_o₁/D₁)
- θ₁ = cos^-1((D₁² + CA² - L²)/(2*D₁*CA)) - sin^-1(x_o₁/D₁)
You just have to rotate the model φ₁ degrees counterclockwise for θ₂, and another φ₂ degrees for θ₃ using the two dimensional rotational matrixes:
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)

@@ Line 53: / Line 53: @@
 ====Construction====
+We are basing 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''