Kinematics of a linkage system of 4 bars

[iroZn]

Homework Statement

Tabulate and plot the angular position, velocity and acceleration of θ4 for t=0 to t=10 in increments of 0.1

r1= 30 mm
r2 = 12 mm
r3 = 39 mm
r4 = 36 mm

θ2 = 0.1t (radians)

The Attempt at a Solution

Well first I wrote this down:


r1 = r2cos(θ2) + r3cos(θ3) - r4cos(θ4)

It didn't really get me anywhere so after some research I found Freudenstein's Eqn:



K 1 cos θ 2 + K2 cos θ 5 + K 3 = cos ( θ 2 - θ 5 )

K1 = l1 / l4 K2 = l 1 / l 2 K3 = ( l 32 - l 12 - l 22 - l 2 4 ) / 2 l 2 l 4



Inputting the values I ended up with this:

(30/36)cosθ2 + (30/12)cosθ5 - (91/96) = cos(θ2 - θ5)

I have no idea how to solve this equation in terms of θ5 though!

(once I find θ5 I'll just use θ5 = 360 - θ4 to get θ4)





Am I doing this right at all? Any help would be soooooooo much appreciated!

Let me know if you need any more info or if I posted wrong or anything!
Thanks again!

 

[pongo38]

r2 and r4 have fixed points of rotation, but r3 is floating. Can you find a relationship concerning that the rotating ends of r2 and r4 have a fixed distance r3 between them?
Another approach is to let t=0.1 Are you then able to calculate the new geometry? Obviously, you could take days to slog through this 0.1 seconds at a time, but the process may indicate to you what you have to do to make it more general, and easier.

 

[iroZn]

Yeah, well what I ended up doing is just throwing Freudenstein's eqn into Mathematica, have it solve it for θ5 and then just use that... it wasn't a pretty equation, but it seemed to have gotten the job done:

y = -cos^(-1)((160 cos^2(x)-582 cos(x)-2 sqrt(9216 sin^4(x)+49319 sin^2(x)+2816 sin^2(x) cos^2(x)-31520 sin^2(x) cos(x))+455)/(48 (4 sin^2(x)+4 cos^2(x)-20 cos(x)+25)))

(y is θ5, x is θ2)

Still this solution is obviously not how they wanted me to do it...

If anyone knows a more elegant way of solving this problem, I've already handed it in but I'd love to know how to do it!