Front View Geometry and Roll Center Height 1

Given,

Tire radius = 300 mm; Track = 1600 mm

Using equation of straight line in yz plane, (z-z1) = (z2-z1/y2-y1)(y-y1),

From the coordinates given in the question,

Equation of line joining the UCA ball joint and subframe pivot is:

z = 0.1944y+404.58 ............ (I)

Likewise,

Equation of line joining the LCA ball joint and subframe pivot is:

z = -0.0566y+190.94 ............ (II)

Solving equation (I) and (II) gives the intersection point between them which is the coordinate of the instant center as:

0.1944y+404.58 = -0.0566y+190.94

Solving, y = -851.15

Substituting value of y in (I), we get, z = 239.12

Thus, Instant center location = (-851.15,239.12)

Now,

Equation of line joining tire center on ground and the IC is given by the equation of a straight line passing through origin i.e. z = slope * y.

Slope = (239.12-0)/(-851.15-0) = -0.28094

Then,

Equation of line joining tire center on ground and IC is:

z = -0.28094y ............ (III)

The kinematic roll center RC lies on the line given by (III).

We know, RC lies in the vehicle centerline for symmetric suspension.

So, y coordinate of RC, y_RC = track/2 = 1600/2 = 800 mm

Substituting y_RC = 800 in equation (III),

z_RC = -224.75

Thus, Roll center height = -224.75 mm (-ve sign indicates that RC lies 224.75 mm below the ground surface)