Sunbeam Tiger Steering Analysis
I have owned my Tiger for 31 years and during that time there have been many complaints about the Tiger's steering. Most of the complaints have come from the factory manual which depicts the problem quite clearly: the inside wheel turns less than the outside wheel, producing what I would say is classic oversteer. The manual shows that the outside wheel turns about 30 degrees when the inside wheel turns only about 26 degrees. However, in all my years, I have found no analysis of why this was so. I decided todetermine why and then could anything be done about it. My model is simple. I used an offset crank and slider mechanism for the analysis. The model comes from "Kinematic Analysis of Mechanisms", McGraw-Hill, 1969. Figure 1, below, shows the notation I used to develop the ovelall model, while figures 2 and three show left and right steering. The method I used was to build the math model so that if I moved the left steering arm a degree at a time, I could calculate the amount that the rack displaced. I reversed this for determining the right side arm movement. I calculated the correct steering arm angle based on ackerman theory. The Tiger's steering arm dimensions come from the manual. I bought two MGB steering arms and measured the length and steering arm angle if installed on the Tiger. And I designed an arm while I was waiting for the MGB arms to arrive.
The measurements I used are: 1) Wheelbase = 86 inches 2) Distance between ball joints = 45 inches 3) Rack length between end sockets = 25.25 inches 4) Rack location ahead of ball joints = 10 inches 5) MGB effective arm length = 5.815 inches (center of lower ball joint to tierod balljoint) 6) MGB steering arm angle whe installed = 14.5 degrees 7) Tiger effective arm length = 5.25 inches 8) Tiger steering arm angle when installed = 11 degrees 9) Special steering arm length = 5.8 inches 10) Special steering arm angle when installed = 14.6 degrees
Using the nomenclature of figures 1 and 2, the math model I developed for the left wheel is:
Xb^2 + (2 * R * COS(phi)) * Xb + + (Yb^2 + R^2 - S^2 - 2 *R * Yb * SIN(phi)) = 0
The equation is in the quadratic equation form and is solvable for Xb. Xb is the rack displacement required when the angle phi is input.
The right side wheel is a little more complicated because it is the angle we are looking for. Using the geometry as depicted in figures 1 and 3 the model develops as follows:
S^2 = Xb^2 + R^2 + Yb^2 - 2 * R * Xb * COS(phi) - 2 * R * Yb * SIN(phi) letting S^2 - Xb^2 - R^2 - Yb^2 K = ----------------------------------- - 2 * R * Xb Yb Z = --- XB then the equation reduces to: K = COS(phi) + Z * SIN(phi) remember that COS(phi) = SQRT(1 - Sin^2(phi)) and substituting and reducing, we find (Z^2 + 1) * SIN^2(phi) + (-2 * K * Z) * SIN(phi) + (K^2 - 1) = 0 which is again of the quadratic equation form with SIN(phi) as the variable.
The two equations were set up in Excel and solved for each input angle. The various steering arm lengths and angles were used to calculate the values in Table 1. The Perfect Ackerman angle was calculated using simple, by comparison, trig relationships. That exercise is left to you, the reader.
What can we surmise from the data. An improvement can be made albeit, only a small improvement. Use of the MGB arms require some metal work to install. The mounting hole spacing has to be opened up a small amount, about 0.020 inches to fit, and one boss requires machining to remove material equal to the thickness of the caliper mount flange. Alternately, a bushing could be used to space the arm outward, but then longer bolts are required. Spacing outwardly also changes the effective steering arm angle, reducing the overall effectiveness. The maximum benefit gained, at full lock of 26 degrees, is only 2 degrees. This is not enough of a gain for me - another 7 degrees is needed!
Use of the arm I designed produces similar results.
So what can be done? If we stick with the rack in its present location, nothing better than the arms above. We cannot move the rack back because it would interfere with the motor damper and it would hit the first crossmember mounting location. The only solution I see is to design a new crossmember, and I am doing that. The new design will use tubular a-arms, Heim fittings for rod ends, and will be able to relocate the rack down and back. I am trying to not modify any Tiger parts in the process. And everything will be a bolt on. I will post the design when finished.
Some have said that the perfect steering can be achieved by placing the rack directly in line with the steering rod ends. Not so! If a drag link, as shown in Figure 4, is used with the correct steering arm angle, then this would be reasonably correct. As a drag link moves during turning, it makes a skewed trapezoid to maintain all of the geometry. The center of the rack is pinned in place and only the tie rods move in concert with the steering arms. I ran many cases trying different steeing arm angles and different rack locations to determine if a best solution exists. Nope.
|Inside Wheel Input||Outside Wheel Output|