Clarification in Fundamentals of Vehicle Dynamics (Milliken)

[mark512]

I'm having a problem understanding the terms in Chapter 6 (steady-state cornering)

They start with a free body diagram of a bicycle model (front and rear axles are each represented by a single wheel) turning a corner in equilibrium at velocity V with radius r. With vehicle mass M, wheelbase L, and CG to front axle distance b and CG to rear axle distance c, the cornering forces on the front and rear are given as:

Fyf = M*(c/L)*(V^2/r)
Fyr = M*(b/L)*(V^2/r)

They then make a substitution because M*(c/L) and M*(b/L) "are the portions of the vehicle mass carried on the front/rear axles" and "are equal to" Wf/g and Wr/g. They continue using the terms Wf and Wr throughout the entire chapter.

The problem I'm having is that the list of symbols at the beginning of the book defines Wf as "the dynamic weight on the front axle" and Wr as "the dynamic weight on the rear axle" (and Wfs and Wrs as the respective static weights.) Within the chapter, Wf and Ws are periodically defined as "the load on" the respective axles.

From these definitions, it seems like Wf and Wr are the dynamic axle loads (i.e. affected by longitudinal weight transfer) but everything seems to be derived from the bicycle model sum of the moments around the CG, which to me, implies it should be the static weights because the CG position (and therefore its moment arm lengths to the front and rear axles) don't actually change?

Thanks in advance!
Mark

 

[GregLocock]

And why would you expect longitudinal weight transfer in steady state cornering?

Cheers

Greg Locock


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[mark512]

(I realized the author is Gillespie, not Milliken)

I guess the devil is in the details: *steady state* cornering - I assumed that meant they are not considering changes in the radius of the path the vehicle is taking (i.e. corner entry and exit), as opposed to they are also not considering longitudinal weight transfer due to acceleration/braking.

 

[mark512]

There is a part at the end of the steady-state cornering chapter on "Effect of Tractive Forces on Cornering" that re-does the bicycle model but this time with Fx's in addition to Fy's on the wheels. I think that's what threw me off, but I am now assuming they are referring to tractive forces like rolling resistance and drive torque necessary to maintain speed, where Fx is small compared to Fy and there is negligible longitudinal weight transfer.

Could accounting for longitudinal weight transfer due to acceleration and braking in a corner be as straightforward as adjusting for changes in Fz, camber, etc. due to the change in axle loading, or does the fact that the tire is producing larger forces in both the X and Y directions (and therefore experiencing simultaneous sideslip and longditudonal slip) throw a huge proverbial wrench into the analysis?

Thanks again,
Mark

 

[GregLocock]

The bicycle model is excellent for many things but it runs out of steam as you ramp the complexity up. You spend more time bodging the bicycle so it behaves like a 4 wheeler than if you started with a 4 wheeler. I think combined accelerations are a good example of that. Having said that combining an ax and an ay is probably doable in a bicycle model if that is as far as you need to go. I think your last para is a good approach.

The devil will be in your tire model, not the vehicle model.

Cheers

Greg Locock


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[SwinnyGG]

This would get you a close approximation, depending on the depth of complication in your tire model. But, you'd still be evaluating a steady-state condition, meaning constant acceleration in all directions. This would be useful only for an instantaneous point, and I don't think very applicable to any real situation.

 

[mark512]

That sounds good for now, thanks for your help guys!