ROLE OF A ROLL CENTER

(Mystery solved or mystery continues?)

Introduction

Basic definition of roll center is the point about which the sprung mass rolls under the influence of centrifugal force. It is also a point through which the lateral forces transmitted from Tire’s contact patch acts upon a chassis.

There has been many theories on “Force through a roll center” or in a better way “Force dissipation from Tire contact patch (TCP) to Sprung mass (SM)”.

Some of them are- 1.Lateral force acting at the Roll Center-given in Tune To Win. 2. Lateral force acting at Instantaneous Center IC-RCVD. 3. Concept of Kinematic Roll Center 4. Force Based Roll Center 5. Force Application Point FAP- Mitchell’s paper 6. Concepts given in “Suspension Geometry And Computation” by J.C.Dixon 6. N-line concept

Out of so many articles, research papers, theories by numerous engineers, everything made more confused. The one that we could closely justify is the N-line concept. It is very simple, interesting and at-least I could justify it to some extent.

The N-line concept and the one given in Dixon closely resemble each other except for some differences.

N-LINE CONCEPT

N-Line is simply a No-motion line.

Force Analysis using N-line concept

Image by ‘Z’ in Forum

Steps-

1. Draw vector Fg + Fi=R from COG. Here, Fg=mg i.e. weight and Fi is the inertial force

2. Assume/get the coefficient of friction of one tire. 3. Let this be u which is equal to tan(theta) where theta is the angle that resultant of tire force acts

with normal 4. Let this force be Fl 5. Get the intersection point of Fl and R 6. From this point draw Fr to Right wheel print center. 7. Fl and Fr are the resultant forces acting on Left tire and Right tire. 8. Now, resolve Fl and Fr along and perpendicular to n-line-fig(b) 9. Let Fl.ca be the force along n-line and Fl.sd be the force perpendicular to n-line 10. Now, consider Fl.ca and Fr.ca

Resolve them into horizontal and vertical component namely Fl.ca.y and Fl.ca.j similarly with right side-fig(d)

11. Net jacking force=Fl.ca.j-Fr.ca.j

REASONS AND JUSTIFICATIONS

(This part is more about our interpretation of the n-line concept)

• Part 1- Significance of all force

1. Fg=mg :- the weight of the car.

2. Fi :- inertia force i.e. centrifugal force.

3. Fg+Fi :- vector sum of both the forces i.e. R

4. u=tan(theta) :- coefficient of friction of one tire

5. Fl and Fr :- Components of R acting on that tire at their respective conditions i.e. u(tan theta)

6. Fl.sd and Fr.sd :- Force vertical to n-line that is taken up by spring-damper.

7. Fl.ca and Fr.ca :- Force along the n-line that is taken up by Control Arms CA.

8. Fl.ca.y and Fr.ca.y :- Lateral components of Fl.ca and Fr.ca

9. Fl.ca.j and Fr.ca.j :- Vertical components of Fl.ca and Fr.ca

10. Fj=jacking force :- Fj =Fl.ca.j - Fr.ca.j

• Part 2-Reasoning

1. Why to consider n-line? N-line is basically a no motion line. Consider following analogy-

Consider a circular plate with a rigid link on one side. Case A- If the force is applid perpendicular to the rigid link and along the diameter, then there will be no-motion of a circular plate. Case B- If a force is applied at any direction, then there will be a translation motion of a circular plate in upward/downward direction.

Ø The rigid link here is analogous to Control Arms and the circular plate is analogous to upright/wheel.

Ø Hence any force on upright (the circular plate) along the n-line will not translate it but, force applied in a direction other than n-line, translates the upright in upward/downward direction.

Hence, n-line is a no motion line.

>…………………………< >…………………………<

2. Why a force perpendicular to n-line is taken up by Spring-Damper?

Consider the following case-

Make four links viz. Upright, Upper control arm, lower control arm and Chassis. Pin them and make chassis link fix.

Now, apply a force along the n-line, there will be no vertical movement of upright link.

But if we now, apply force in any direction, there will be a vertical movement of that link.

What does this conclude?

This concludes that, if any force is not acting along the n-line, there will be a vertical movement of upright, that will provide movement to push/pull rod,which will actuate the rocker and hence Spring-damper. Hence this will be taken up by spring Damper.

>…………………………< >…………………………<

3. Why a force along the n-line is considered to be taken up by control arms? In this case, if the force is applied along the n-line, there will be no vertical movement of upright link. So where is the force gone? This force is dissipated into the control arms, so they are under stress and finally goes to sprung mass/chassis.

CONCLUSION

1. Forces do not act at a particular point rather it acts along its line of action. 2. Roll center migration should not be a headache because unless the jacking force is under

control (if you want so) the roll center migration is irrelevant. 3. The more the inclination of n-line, the more will be the jacking force. So migration of n-line

should be controlled. 4. Forces acting along the n-line are taken up by the control arms. 5. Forces perpendicular to n-line are taken up by Spring-Damper.

DOUBTS

1. In this concept, what constitues a roll couple? Which force acts along with the

Centrifugal force and what is the moment arm in this case?

2. How to resolve Fl.sd and Fr.sd? Does this entire force is taken up by spring-damper.

3. What is the idea behind the intersection of R and angle theta that we get from tire’s coefficient of friction?

4. Why coefficient of friction of only one tire is considered?

5. Why in Tune To Win, Carroll Smith has entirely a different approach? He used to incline

the lateral force along the roll center?

6. So many concepts, which one to follow? Or more specifically, which one is more closer to rwality?

By-VD Team -Mrudula Amar Warde -Prateek Tripathi -Siddharth Sharma