Dave Morgan wrote:I feel linkage geometry has a much greater effect than IC location....
Dave
As for the significance of the lower link being parallel, let's plug in some numbers and see what happens.
Suppose the lower link is horizontal with a rear attachment point directly below the axle centerline and 9 inches up from the strip surface.
The upper link's rear attachment point is directly above the axle centerline and 22 inches up from the strip surface. The front attachment point is 24 inches forward of the axle centerline and 15.76 inches above the strip surface.
These link lines intersect at 50 inches out and 9 inches up.
Now, we'll assume a 6000 pound force is acting horizontally at the rear tire patch to accelerate the car forward.
The lower link will be in compression. The upper link will be in tension. (For simplification, I'm lumping left and right links and tires together to make the problem two dimensional.) So, the horizontal forces acting at the rear link attachment points must balance. That is, the sum of the tire patch force and the upper link force must equal the magnitude of the lower link force.
In addition to the force balance, there must also be a moment (torque) balance. Taking moments about the tire patch, 9 times the lower link horizontal force must equal 22 times the upper link horizontal force.
It is possible, then, to solve for the horizontal forces at the links' rear attachment points. There would be 10154 pounds directed rearward at the lower link and 4154 pounds directed forward at the upper link.
Finally, there must be a vertical force balance. Since the lower link is horizontal, there is no lower link vertical force component. The upper link is angled down and is in tension, so there is a vertical force component downward which must be equal to the weight transfer experienced by the rear tire. The ratio of (22-15.76) to 24 would equal the ratio of this unknown force to 4154 pounds. This would be a downward vertical force of 1080 pounds.
The force acting at the rear tire patch would, then, have a horizontal component of 6000 pounds and an upward vertical component of 1080 pounds. The ratio of 1080 to 6000, when multiplied by 50, results in 9, meaning that the resultant force vector passes through the instant center located at 50 out and 9 up.
Now, suppose we retain the upper link and remove the horizontal lower link. A new link will be added whiich has a rear attachment point directly above the axle centerline and 100 inches above the strip surface. Yes, it's an absurd arrangement, but useful as an example. The front attachment point is 24 inches forward of the axle centerline and 56.32 inches above the strip surface.
Again, the link lines intersect at 50 inches out and 9 inches up.
The horizontal forces now are 1692.3 pounds at the upper link and 7692.3 pounds at the lower link. The upper link is exerting a downward force of 3080 pounds and the lower link is exerting an upward force of 2000 pounds.
The resultant tire patch force components remain 6000 pounds horizontal and 1080 pounds upward.
In other words, excluding strength and weight considerations, link angles and locations can take on any values and, so long as the instant center location is unchanged, launch characteristics remain unchanged.
Taking this exercise a step further, the instant center can be moved to any other point on the force line of action (which is also a line of constant percent antisquat) and the results will remain the same. Take it to 500 out and 90 up and see for yourself.
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