Partner: James Cameron
Reality Check 1: The Stewart Platform in 2 Dimensions
A Stewart Platform consists of six variable length struts supporting a payload. As a six-degree-of-freedom robot, the Stewart platform can be placed at any point and inclination in three dimensional space that's within its reach. The Stewart Platform we are dealing with is a two-dimensional version one. There will be three struts making up a triangle platform with lengths L1, L2, and L3 and 3 variable struts with lengths p1, p2, and p3 that move the move the triangle. Given the three strut lengths, the problem of finding the position of the platform is called the forward, or direct, kinetic problem, which is the problem of computing (x,y) and θ for each p1,p2,p3.
The best way to solve this problem is to use the geometry that you can see in the above picture.
p12=x2+y2
p22=(x+A2)2+(y+B2)2
p32=(x+A3)2+(y+B3)2
where
A2=L3cosθ-x1
B2=L3sinθ
A3=L2cos(θ+γ-x2=L2(cosθcosγ-sinθsinγ)-x2
B3=L2sin(θ+γ-y2=L2(cosθcosγ+sinθsinγ)-y2
After doing some algebra, it can be shown that:
x=N1/D=[B3(p22-p12-A22-B22)-B2(p32-p12-A32-B32)]/2(A2B3-B2A3)
y=N2/D=[-A3(p22-p12-A22-B22)+A2(p32-p12-A32-B32)]/2(A2B3-B2A3)
As long as D=2(A2B3-B2A3)≠0
Using more algebra, it can be shown that
f=N12+N22-p12D2=0
with only one unknown: θ. Our goal for this problem was to write a function file for f(θ) and then solve various problems to find θ.
We wrote a code (f.m) to plot various poses using different strut lengths. You can see code by clicking the image for Step 2.
Step 2: Plot f(θ) on the interval [-π, π] with the following parameters: p1 = p2 = p3 = √5 and L1 = 2, L2 = L3 = √2 and γ = π/2 where the p1, p2, p3 are lengths of the struts and L1, L2, L3 are the lengths of the sides of the triangle. There are zeros at: θ = -π/4 and θ = π/4
Step 3: We plotted the Stewart Platforms with the paramenters listed in Step 2 using the zeros (θ):
θ = -π/4 and θ = π/4
Step 4: We changed the paramenters and solved for the zeros of f(θ) when p1 = 5, p2 = p1, p3 = 3 and L1 = 3, L2 = L1, L3 = 3√2 and γ = π/4. We found the zeros to be at: (with the corresponding Steward Platform poses)
θ = 2.115909014086806 θ = 5.562336102719566
θ = 12.235365430075944 θ = 18.128706717077911
Step 5: We found a different value for p2 such that there are 6 poses. We picked p2 = 7.02 and p1 = 5, p3 = 3 and L1 = 3, L2 = L1, L3 = 3√2 and γ = π/4. We found the zeros to be at: (with the corresponding Stewart Platform poses)
θ = -0.585659711118751 θ = -0.486464280623295
θ = 0.074749032623913 θ = 0.464513052646907
θ = 0.977150550043916 θ = 2.515984235616088
Step 6: We found a strut length for p2 such that there are only 2 poses: We used p2 = 8 with the zeros at: (with the corresponding Stewart Platform poses)
θ = 1.070918998332672 θ = 2.544259100061936
