When a double four-bar linkage steering system is properly designed, a vehicle's turning radius is the same for left and right-hand turns.

** DAN FRITZINGER**Mechanical Engineer

Grabill, Ind.

The double four-bar linkage is one type of steering arrangement that engineers often use on simple vehicles, such as go-karts, trailers, and toy cars. In this arrangement, there are two four-bar linkages, whose bases are located on the centerline of the front axles. Typically, the two arms located at the center of the steering arrangement (the steering levers) drive the tie rods, spindle levers, and, ultimately, the front axles to cause the steering action.

The guiding principle behind proper steering-geometry design is that when a vehicle is turning, centerlines drawn through both front axles must intersect the same point on a line extending through the rear-axle centerline. If this is accomplished, all four wheels will pivot about the same turning point.

One of the two four-bar linkages drives the inner wheel and the other drives the outer wheel. The two four-bar linkages are mirror images of each other .

When the steering system is properly designed, the vehicle's turning radius based on calculations for the will be the same as the length of the vehicle's turning on calculations for the outer linkage. In other words, and outer linkage will both cause the vehicle to assume turning radius. This makes all wheels operate in harmony a smooth-turning vehicle.

To do the analysis, establish values for the length of the base of the four-bar linkage, *R*1, steering-lever length, *R*2, tie-rod length, *R*3, spindle-lever length, *R*4, the angle formed by *R*1 and *R*2 (Θ2), the angle that *R*2 is rotated through to cause the vehicle to turn, Δ and the centerline distance between the front and rear axles, *M*.

Next, solve for the values of one of the four-bar linkages when oriented such that the vehicle is traveling forward:

*Z*2 = *R*1 ^{2 }+ *R*2 ^{2 }2*R*1*R*2cos(Θ 2) *Y *= cos

^{1 }((Z

^{2 }

*R*3

^{2 }

*R*4

^{2 })/(2R3R4))

Φ= cos

^{1 }((Z

^{2 }+

*R*4

^{2 }

*R*3

^{2 })/(2ZR4))

β= cos

^{1 }((Z

^{2 }+

*R*1

^{2 }

*R*2

^{2 })/(2ZR1))

Θ

^{5 }= 180° Φβ

The next step is to rotate the four-bar linkages by the value of Δ and solve for the turning radii based on the inner and outer linkages. We'll work with the inner linkage first. Assume a new value for Θ2 by subtracting the value of Δ then calculate the new values of *Z*, Φ, and β. Next solve for Θ6 using:

Θ^{6 }= Θ5 + β+Φ 180°

Once Θ6 is known, solve for *D*1, which, along with *M*, forms two sides of a right triangle.

*D*1 = *M*/tan(Θ6)

Now solve for the inner linkage's turning radius, *Rt,I *using

*Rt,I *= *R*1 + *D*1

Next, solve for the outer linkage. First, add the value of Δ to the value of Θ2, solve for *Z*, , and as before, and then solve for Θ7 with

Θ^{7 }= 180° Θ5 β Φ

Next, solve for *D*2. As with *D*1 for the inner linkage, *D*2 and *M *form the two sides of a right triangle.

*D*2 = *M*/tan(Θ7)

Finally, solve for the outer link-age's turning radius, *Rt,O *using

*Rt,O *= *D*2 *R*1

If the steering system has been properly designed, the values for the inner and outer turning radii will be equal. In my spreadsheet, I have a cell that subtracts one turning radius from the other. If the answer in the cell is near or at zero, the steering system has been properly designed. A graphical layout on a board or with some sort of CAD package is recommended as a final check.