I've kind of gone over this point before, but I've made more sense of it to myself now.
Under high lateral load, the tire - being the non-rigid structure that it is - deflects in a couple different directions and moves laterally under the wheel. If you have your own racecar and put a camera watching the tire footprints you can see it pretty clearly, even on a FSAE car.
This gives rise to a non-zero overturning moment at the contact patch (not to be confused with the total overturning moment at the hub, which also comes from lateral force and puts substantial radial load into wheel bearings). Alternatively you could think of this as the vertical reaction force vector moving laterally with the tire. I believe in the RCE article they just assumed a purely lateral and vertical force, at a fixed position.In any event, you'd think that adding that moment would change how the forces are resolved through the suspension links, and thus how those lines of action act about the CG. The greater the deflection of the tire, the greater this effect. While I had realized that tire Mx played into total load transfer, I didn't know how it was distributed.. be it as a geometric or elastic effect. I'd think it would be purely geometric, at this point in my understanding.
As an interesting consequence, bolting on two different tire sets, even of the same size, could result in different true roll centers (force application points, whatever you'd like to call them) when out on the track.
3 comments:
Hi, I think I've probably figured out the reason why others have ignored the pushrod in the force based roll centre calculation! If we consider one corner of the car, then no roll motion just means there is no further change of length in the spring, ie no force change in the push rod if there is no body roll!
That's irrelevant though, isn't it? Unless the wheel is off the ground, there's always load in the pushrod.
Think of the car now in equilibrium on the ground the ground. And there is now an extra horizontal force on the sprung mass. We can always think about how the suspension react to the extra force and superimpose that onto the original equilibrium solution. (or you can think of it as we wipe out the forces in the original equilibrium and calculate the equilibrium solutions of the changes)
according to SAE definition: "The point in the transverse vertical plane through any pair of wheel centers at which lateral forces may be applied to the sprung mass without producing suspension roll"
Therefore if that extra force causes no body roll, ie, no deflection in the srping, then the solution we superimpose on the original equilibrium solution will have no force in the push rod!
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