Mechanical Advantage through series of links

[KLoux]

Hello,

I'm trying to come up with a method for calculating the mechanical advantage of a vehicle's suspension (i.e. how much does the spring compress if I move the wheel center by some infinitesimally small amount?). I have a method that I thought was correct, but I recently compared my calculations to results from a trusted software package and found that my numbers are off a bit. Can anyone spot an error in my theory or suggest what my pitfall might be?

I'm assuming that I'm working with a double A-arm suspension, which is essentially a 4-bar linkage. Attached to the lower A-arm is a pushrod (attached via a spherical bearing). This pushrod goes to a bellcrank (lever) that rotates around an axis. Also attached to this bellcrank are the spring and damper. So to compress the spring, the wheel must move up, which makes the lower control arm push on the pushrod, the pushrod makes the bellcrank rock on it's axis of rotation, and the other end of the bellcrank compresses the spring. I know all of the locations of the joints and the axis of rotation in 3-space. My method is this:

1. Find the equations of the planes that contain the upper and lower control arms.

2. The wheel's instantaneous axis of rotation must be the axis created by the intersection of those two planes. Find that axis.

3. Find the points on that axis that is closest to the wheel center and the location of the attachment between the pushrod and lower control arm.

4. Define a "virtual force" vector that points upwards to be applied to the center of the wheel.

5. The sum of the moments around the wheel's instantaneous axis of rotation must be zero. We can write this equation:
(Vector_from_wheel_center_to_axis) cross (virtual_force_vector) + (vector_from_pushrod_end_to_axis) cross (reaction_force_at_pushrod) = 0
Everything except the reaction at the pushrod is known.

6. We know the direction of the reaction force because the spherical bearing can't transmit torques (safe assumtion) - the force must be in the direction of the pushrod. We can re-write the equation:
(Vector_from_wheel_center_to_axis) cross (virtual_force_vector) + (vector_from_pushrod_end_to_axis) cross (magnitude_of_reaction_force * direction_of_reaction_force_at_pushrod) = 0
For more clarity, we re-write again:
Moment_due_to_virtual_force + (direction_of_reaction_moment * moment_arm_length) * magnitude_of_reaction_force = 0

7. Now the only unknown is the magnitude of the reaction force. I think it is safe to divide any of the three components (so long as they are non-zero) of the moment from the virtual force by the same component of the direction of the reaction moment times the moment arm length.

8. Knowing the force through the pushrod, we can use the same procedure above to find the moments around the bellcrank's axis of rotation, solving for the spring force at the end.

9. Because we applied a "virtual force" to the wheel, the force we see in the spring can be divided by the magnitude of the original "virtual force," and this ratio should not only be the ratio in the forces, but also the ratio in movement (say instead of a virtual force, we applied an infinitesimally small displacement).

Any help is appreciated! Can anyone see what I'm doing wrong or suggest a better way to solve the problem? If this is clear as mud, let me know and I'll try to post some pictures.

Thanks!

-Kerry

 

[KLoux]

Making this post has got me thinking...

First, my last step in the previous post is wrong - we need to apply the principle of virtual work, so force dot distance at the wheel must be the same as the force dot distance at the spring. That means the ratio of the forces is the inverse of the ratio of the displacements.

Also, I think I have found the cause of the problem here (but I'm still working on the solution). The moments are not just r cross F, because this does not take into account the portion of the force that is parallel to the axis of rotation. In other words, we need to project the forces onto a plane that has the axis of rotation as its normal.

I think I can figure the rest out on my own, but if anyone is interested to see what I come up with, let me know and I'll make a follow up post.

Thanks,

Kerry