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Mark Ortiz Automotive is a chassis consulting service primarily serving oval track and road racers. This newsletter is a free service intended to benefit racers and enthusiasts by offering useful insights into chassis engineering and answers to questions. Readers may mail questions to: 155 Wankel Dr., Kannapolis, NC 28083-8200; submit questions by phone at 704-933-8876; or submit questions by e-mail to: mortiz49@earthlink.net. Readers are invited to subscribe to this newsletter by e-mail. Just e-mail me and request to be added to the list.

TRAILING ARMS WITH TORSIONAL RUBBER SPRINGING

I wonder if you are familiar with torsion rubber suspension systems used on travel trailers and boat trailers. For instance see the Henschen dura torque axle system below:

http://www.inlandrv.com/tours/axle-plant/index.htm

There really is just a single trailing link with this system. Where is the roll center? Is it at the center of the axle tube? Is it essentially fixed?

Is the virtual swing arm length infinite?

Has anyone ever tried to use an axle like this on the rear of a fwd vehicle?

This system is a pure trailing arm system with no anti-roll bar. The front view swing arm length is infinite, or undefined. The side view instant center is at the pivot axis.

This creates a system with no camber change in ride, no camber recovery in roll, very little geometric anti-roll, and a lot of geometric anti-lift in braking.

The roll center is not at the pivot axis. It's at ground level in the unrolled condition, and slightly above ground level when cornering, i.e. with some roll and with more than 50% of the lateral force generated by the outside tire.

This particular design of trailer axle uses rubber in a combination of compression, shear, and torsion as the ride springing medium. The assembly includes a square-tube crossmember that mounts it to the frame. Each trailing arm has welded to it a piece of square bar that is inserted into the crossmember. The square bar is sized so that its cross-sectional width across points is slightly smaller than the crossmember tube's inside width across flats. In other words, the square bar is just small enough to fit inside the square tube and turn.

As assembled, the square bar sits at a 45 degree rotational angle inside the tube: its points are centered on the tube's flats. This leaves four roughly triangular cavities between the flats of the bar and the interior surface of the tube. These cavities are filled with rubber.

One might suppose this would be done by bonding custom extrusions to the metal, and also adding some metal retention system to keep the bar in the tube. But that's not what Henschen does. They buy round rubber bars or cords that, as purchased, would not go into the cavity. Then they squash them to the shape of the cavities in a hydraulic press, and freeze them with liquid nitrogen while they're still in the press. When they come out, they will hold their molded shape until they thaw. Then they will try to resume their cylindrical shape.

The rubber pieces and the trailing arm with the square bar welded on are assembled into the crossmember tube while the rubber is frozen. When the rubber thaws and tries to resume its former shape, it becomes a very tight interference fit in its cavity. Far as I can see from the website, this is deemed sufficient to retain the trailing arm to the crossmember with no additional hardware.

It would be possible to add a torsion bar connecting the square bars, to serve as an anti-roll bar, but this is not done. It would be possible to add hydraulic shocks, but that is not done either. The system relies on the internal hysteresis of the rubber for damping.

The result is a very simple suspension system, and a very compact one.

This exact design has not been used at the rear of a fwd vehicle, but certainly a trailing arm suspension with rubber springs has. That's what the original Mini had.

One problem with rubber as a spring material is that it cold flows under sustained load, causing the suspension to sag. The best application for rubber springing is a light trailer that sits unloaded between trips, or a light motorcycle or scooter that sits on its center stand with the suspension extended when not being ridden.

Probably the biggest car to use rubber-in-torsion springs was the Tucker. I'm told that owners of the few surviving Tuckers store their cars on blocks with the suspension at or near full droop to keep the cars from sagging. Sagging is less of a problem when the car is light compared to its cargo, the suspension is stiff, and longevity is not a high design priority. That's how the Mini got away with rubber springs.

With or without rubber springs, a suspension for a car needs proper dampers. Rubber does provide some internal damping, but a car suspension needs real shocks.

RIDE AND ROLL RATES WITH MONOSHOCKS

I was wondering if you might be keen to discuss and explain monoshock suspensions, with particular reference to Dallara F3 cars, and explain how one might go about calculating belleville stack rates, wheel rates, load transfer distributions (bearing in mind the oddity of one wheel going up and the other down at the front in roll) and so on. I am soon to race a 1998 Dallara in Monoposto, having been racing a conventional Reynard 883 for two years, and have been scratching my head as to how to apply the 'classical' equations for load transfer distributions and wheel rates etc. when using a single ride spring and belleville stacks.

The water is further muddied by the application of preload on ride springs and stacks, but I'm hoping to ignore them for now, as one ignores many dynamic aspects in such analysis. Is this wise?

Taking the last question first, no we cannot ignore preload, especially on the washer stacks. On the ride spring, it depends on whether the spring is preloaded at static ride height, or only at full droop.

If the ride spring is preloaded at static, the suspension is solid in ride (wheel rate approaching infinity, or undefined) until the preload is exceeded. If it is not preloaded at static, it provides a rate for the wheel pair that is equal to the spring rate times the square of the spring-to-wheel motion ratio. This is the number of pounds of load change for the wheel pair, per inch of ride travel. It corresponds to the sum of the two individual wheel rates in ride in a conventional suspension. So if you have an equation that calls for individual wheel rate in ride, you use half of that ride-spring-rate-times-square-of-motion-ratio quantity.

If the equation in question is for lateral load transfer, you use zero for the ride spring rate. The ride spring in a monoshock suspension acts only in ride and does not contribute to roll resistance or elastic lateral load transfer at all.

All the elastic roll resistance in a monoshock setup comes from the Belleville washer stacks. There is a stack on each side of the rocker where the shuttle passes through. Ordinarily, we use identical stacks on both sides of the rocker. The stacks act in parallel, so the rate of force change with respect to displacement at the shuttle is the sum of the rates of the individual stacks, if they are both active. They may, or may not, both be active. That's where preload comes in.

When both stacks are equally preloaded, there is a compressive load on each stack, and a reaction force from each stack trying to extend itself. But since these act on the shuttle in opposite directions, there is no net force trying to move the shuttle to either side.

Correspondingly, if there is an increase in extension force from one stack, and a decrease in extension force on the other, those force changes act additively, and the force trying to re-center the shuttle is the sum of the two.

That force acts on the wheels, through the pushrods and the rest of the suspension, at some motion ratio. The rate of force change with respect to displacement at the wheels is the rate at the shuttle, times the square of the motion ratio.

Like an anti-roll bar, the shuttle mechanism is an interconnective springing system. It generates force in response to a displacement difference between two wheels, i.e. an oppositional displacement of the pair. To define a motion ratio for such a system, we need to resolve the question of what we call an inch (or mm) of motion: is it an inch (or mm) at each wheel, meaning two inches (or millimeters) difference between the two, or is it half an inch (or mm) at each wheel, meaning one inch (or mm) difference between the two?

I prefer the former method, because it puts wheel rates for all modes, from all springing devices, in the same terms: force per unit of displacement per wheel. Using this method, the angular roll resistance is the wheel rate in roll, times the square of the track, times ½. That gives the angular rate in lb-in or N-mm per radian. To get lb-in or N-mm per degree, divide by 180/π or 57.3.

Most books use the method above to calculate the component of angular roll resistance due to the ride springs (again, this is zero for a monoshock suspension), and use a different formula for the component due to the anti-roll bar. The more common method for the bar component is as above, except the rate is taken as force per unit of displacement per wheel pair, and the angular rate from the bar is then as above, except with the multiplication by ½ omitted.

Both methods work fine, and give the same answer for total angular rate, provided you use the method that goes with your expression of rate.

The angular roll displacement is then the sprung mass, times the lateral acceleration, times the moment arm of the sprung mass c.g. about the roll axis, divided by the sum of the front and rear angular roll resistances. The front elastic lateral load transfer at that roll displacement is front angular roll resistance, times angular roll displacement, divided by front track.

If you are setting up a factory-built race car with a monoshock suspension, the factory will generally furnish a table stating the rates for various stacks. If you are building a car, or want to try stacks that aren't in the table, it helps to understand the basic rules governing the behavior of Belleville washer stacks.

For readers unfamiliar with Belleville washers, they are simply spring steel flat washers, dished a bit so they can serve as a short-travel compression spring. The dished washer has a concave, or cup, side, and a convex, or cone, side. If we stack a number of washers cup-to-cup and cone-to-cone, we have a compression spring with a useful amount of travel.

A Belleville washer is not a perfectly linear spring, but if we approximate its behavior as a constant rate, and call that k, and if we have a number of washers which we call N, we can state some rules about the rates of combinations of such washers.

If we stack N washers cup-to-cone, or nest them, they act in parallel and the rate is kN. If we stack them cup-to-cup and cone-to-cone, they act in series and the rate is k/N. If we have two stacks, each of an overall rate K, on each side of the rocker, and both are active, the rate of the assembly is 2K.

Both stacks are active as long as there is a load on both. Once the inside stack (the one toward the inside of the turn) unloads, it ceases to contribute to the rate, and the rate from that displacement on is K rather than 2K. If the stacks are not preloaded at all, only the outside one is ever active, and the rate is always K. If the stacks are loose on the shuttle at static, the rate is zero until the clearance on the outside stack is taken up, and then it's K.

It is possible to create stacks in which some of the washers are nested and some act singly. In that case we can have a rising-rate stack. If the portion with nested washers has a rate L and the portion with unnested washers has a rate M, then the whole stack has a rate of LM/(L+M). If we compress this stack far enough, the unnested washers bottom out (get squashed completely flat) and we are left with the nested ones still active. The rate then goes to L (possibly in a handbasket).

When we have variable-rate stacks, either rising-rate ones or ones where there is a chance of a preloaded stack unloading, we have to determine at what roll displacement we will get a rate change, and compare that to our predicted roll displacement using the rate we have at static. If this comparison shows that we will encounter a rate change, we have to work backwards and calculate at what lateral acceleration we encounter the rate change. We then calculate what load transfers we have at that lateral acceleration, using the angular roll resistance rates for the first part of the travel, and finally we calculate further load transfer for the additional increment of lateral acceleration, using the rate that applies for that increment.