Physics of Automobile Rollovers

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Introduction

The National Highway Traffic Safety Administration (NHTSA) of the Department of Transportation of the United States government recently promulgated rollover resistance ratings (http://www.nhtsa.dot.gov/cars/testing/ncap) for automobiles. A parameter called the Static Stability Factor (SSF) is assigned to each vehicle. It is defined as one-half the track width divided by the height of the center of gravity. It is called "static" because SSF is essentially the tangent of the slope angle for an vehicle to just roll over while sitting on the slope. Another, slightly less static, way to look at SSF is "equal to the lateral acceleration in g's at which rollover begins in the most simplified rollover analysis of a vehicle represented by a rigid body without suspension movement or tire deflections" (Taken from http://www.nhtsa.dot.gov/cars/rules/rulings/Rollover/Chapt05.html.) Note the words "at which rollover begins".

This diagram from that web site shows the lateral acceleration vector a to the right and the gravity acceleration vector g downward. The vehicle will just start rotating about the pivot point at the right tire when $\tan \theta =\frac{a}{g}.$

One needs SSF to be as large as possible to make the lateral acceleration required to cause a rollover to be large.

In a real situation a vehicle is usually moving when it rolls over, not standing still on a slope. We consider two types of idealized moving rollover situations:

1. A vehicle is moving (sliding) sideways and the wheels strike a solid obstacle that provides a pivot point for a possible rollover. Rollover occurs when the ensuing rotation causes the force of gravity vector to pass through the pivot point.
2. A vehicle is moving, without slipping, around a circular curve at a constant speed high enough to just cause rollover. Rollover occurs when the force of gravity vector passes through the pivot point.

It should be emphasized that the effects of suspension movement, tire movement or electronic/mechanical stability control may be very important in the rollover tendency for an vehicle. Suspension and tire movements would likely increase the tendency for rollovers, while electronic/mechanical stability control should make it less likely that a vehicle would get in a situation where rollovers occur. These analyses do not account for such and, thus, can only compare vehicles as rigid unintelligent bodies.

Rollover Physics for a vehicle Sliding Sideways

Consider a vehicle sliding sideways until both side wheels strike a solid obstacle, such as a curb. The curb provides a pivot point for the car to rotate. The half track $\frac{t}{2}$ and the height $h$ of the center of gravity are shown in this diagram at the point when the vehicle strikes a curb at sideways speed v.

Here the vehicle is at the critical point for rollover. If the speed is greater than zero at this point, rollover will occur. If the sideways speed just goes to zero at this point in the rotation, the vehicle is just on the verge of rollover (the critical point), The height of the center of gravity at the critical point has increased to $r=\sqrt{{\left(\frac{1}{2}t\right)}^2+h^2},$ by Pythagorus' theorem as the hypotenuse of a right triangle with sides $\frac{t}{2}$ and $h$ .

Energy conservation requires that: energy before hitting the curb = energy at the critical point, where energy = kinetic energy $\left(\frac{1}{2}m{v}^2\right)$ + gravitational potential energy (mg × height above surface):

Therefore, at the critical point where $v=0$ :

$\frac{1}{2}m{v}^2+mgh=mgr=mg\sqrt{{\left(\frac{t}{2}\right)}^2+h^2}=mgh\sqrt{s^2+1},$ where $s=\mathrm{SSF}=\frac{\frac{t}{2}}{h}.$

Therefore, the initial speed that just produces the critical point is: $v^2=2g(\sqrt{{\left(\frac{t}{2}\right)}^2+h^2}-h).$

This is the square of the initial speed, when hitting the curb, that yields zero speed at the critical point, the condition to just begin a rollover; call this initial speed the critical speed. Thus, we see that the critical initial speed depends not only on the SSF, but also on $h$ or $\frac{t}{2}$ . In fact, it is better to not even consider SSF, but both $h$ and $\frac{t}{2}$ instead.

For $s=0$ (i.e., $t=0$ ): $v^2=0$ .

For $s=1$ (i.e., $h=\frac{t}{2}$ ): $v^2=2gh(\sqrt{2}-1)=0.828gh=0.414gt$ .

For $s=\infty$ (i.e., $h=0$ ): $v^2=gt$ .

Most vehicles were rated at $1.0<s<1.5$ in the NHTS ratings (http://www.nhtsa.dot.gov/cars/rules/rulings/roll_resistance/appendix1.html). (It is interesting to note that the incidence of rollovers is about five times greater for $s=1.0$ vehicles as it is for $s=1.5$ vehicles.)

Of course, energy is not strictly conserved, so the actual critical initial rollover speed will be larger than this calculated speed. However, rollovers often occur after the vehicle has left the road bed and is on the downward slope beside the road bed, which would decrease the critical initial rollover speed.

	This is a plot of the critical-point initial speed $v$ in miles/hour versus $h$ in feet for different values of $t$ . (Bottom to top: $t=0.5,1,2,3,4,5$ feet). One wants $v$ to be as large as possible. Therefore, one wants $h$ to be as small as possible.
	This is a plot of the critical-point initial speed $v$ in miles/hour versus $t$ in feet for different values of $h$ . (Bottom to top: $h=5,4,3,2,1,0.5$ feet). One wants $v$ to be as large as possible. Therefore, one wants $t$ to be as large as possible.
	This is a 3-D plot of the critical-point initial speed $v$ in miles/hour versus $h$ and $t$ in feet. One wants $h$ to be as small as possible and $t$ to be as large as possible. Note that it is more effective for producing the desirable high $v$ to decrease $h$ (center of gravity) than to increase $t$ (track).

Classifying Automobiles for Rollover Safety

A better way to rate vehicles for sideways rollovers is by calculating critical-point initial speed

$v=\sqrt{2g(\sqrt{{\left(\frac{t}{2}\right)}^2+h^2}-h)}$ from $t$ and $h$ , instead of by static stability factor $s=\mathrm{SSF}=\frac{\frac{t}{2}}{h}$ . Critical-point initial speed also has the advantage of being a value with units (miles/hour) to which people can relate. For example

(In case these appear to the reader to be small speeds for a vehicle, remember that they are sidways slipping speeds, not forward speeds.)

Values for s for various vehicles are given in http://www.nhtsa.dot.gov/cars/rules/rulings/Rollover/Appendix.html, but the track and center-of-gravity values are not given there. I was able to find track/tread values for some of the listed vehicles on the Internet. (Ford Motor Company did not list the track for its vehicles.) I used them and the SSF, I calculated the center-of-gravity height. (It is not clear on the web pages what is meant by "track width" and "tread width"; It appears that some or all that use the term "track" really mean "tread". In every case I have assumed that they meant the width to the outside edges of the tires; so I have subtracted off 6" for the approximate width of the tire.) Then I calculated the critical-point initial sideways rollover speed, as shown in the following table, sorted by increasing SSF:

This shows the critical-point initial rollover sideways speed plotted against the static stability factor (SSF). Note that, at high SSF, the two are essentially equivalent measures of relative rollover tendency, but at low SSF they differ considerably in the rankings. For example, the Chevrolet Suburban is fourth from lowest in SSF rankings, but lowest in critical-speed rankings. Just above the Suburban, four vehicles have their rankings reversed when one uses critical speed instead of SSF. There are three reversals of two vehicles at higher SSF. I believe that this shows convincingly that critical-point initial speed should be used, rather than SSF, in ranking vehicles for rollover tendency, since it embodies the physics better and does not always agree with SSF rankings.

Of course, in a real sideways rollover, there will not be a solid curb providing a fixed pivot point. Instead, the pivot point may actually be moving, say in soft earth or pavement friction. Then the critical speed will be higher than the calculated value for a fixed pivot point. Nevertheless, the critical speed calculated for a fixed pivot point is a good compartive measure of the rollover hazard for different vehicles. Most rollovers occur as sideways rollovers (http://www.nhtsa.dot.gov/hot/rollover: "Most rollover crashes occur when a vehicle runs off the road and is tripped by a ditch, curb, soft soil, or other object causing it to rollover."). There is usually a forward speed, as well as the sideways speed that causes the rollover, which greatly increases the likelihood of damage to the vehicle and its occupants during rollover.

Rollover Physics for an Automobile in a Circular Path

Consider an vehicle moving at initial speed $v$ when the driver suddenly turns the steering wheel such that the vehicle's wheels move in a circle of radius $r$ with no slipping.

The centripetal force acting toward the center of the circle is provided by the friction of the road bed with the tires and causes ( $f=ma$ ) a centripetal acceleration toward the center of the circle with value $a=\frac{v^2}{r}$ . This force acts at the places where the tires touch the road bed, but the acceleration acts through the center of gravity, as does the gravitation force w (weight) downward. There is also a normal force n of the road on the tires. The forces f and n act at all four tires, although they are shown at only one tire in this diagram. The force of gravity is $w=mg$ , where the acceleration of gravity is $g=32.2\frac{\mathrm{ft}}{{\sec }^2}$ . ${\theta }_c=\arctan \left(\frac{h}{\frac{t}{2}}\right)$ is the initial angle between the vehicle floor and the vector to the center of mass from the pivot point.

The frictional force $f$ depends on the coefficient of static friction $\mu$ between the tires and the road; "static" because the tires are not slipping. The usual assumption is that $f\le \mu n$ ; that is, the frictional force can take on any value from 0 to $\mu n$ , depending on what is needed to provide the centripetal acceleration $a$ .

We treat the two inner tires as one tire and the two outer tires as one tire. Since, relative to the earth, the vehicle will rotate around the point where the outer tire touches the earth, we take that point as the point about which to calculate torques. We call the force doublet ( $f,n$ ) at the inner tire ( $f_i,n_i$ ) and the force doublet at the outer tire ( $f_o,n_o$ ) .

Balancing forces:

$\Sigma f_x=ma=\frac{m{v}^2}{r}=\frac{w{v}^2}{\mathrm{gr}}=f_i+f_o\le \mu (n_i+n_o)$

$\Sigma f_y=n_i+n_o-w=0$

The more relevant situation for rollover tendency is when the center of gravity is over the point of contact of the wheel that is touching the road bed. Then the only force causing a torque about the point of contact is $f_c$ . At this point any centripetal acceration value would cause the vehicle to roll over. Call this situation the "critical point" for having a rollover. This critical-point rollover angle is given by $\tan {\theta }_p=\frac{\frac{t}{2}}{h}=s=\mathrm{SSF}$ . This is another reason that SSF was chosen by the NHTSA as the parameter to compare vehicles for rollover tendency. One wants this angle to be as large as possible; i.e., one wants large $t$ and small $h$ .

In the diagram above the height of the center of gravity above the road surface is $c=\sqrt{{\left(\frac{t}{2}\right)}^2+h^2}$ .

At any intermediate angle between ( $0^{°}$ → ${\theta }_p$ ) the centripetal acceleration required to balance the torques lies between ( $s·g$ →0):

$\tau =ma(h\cos \theta +\frac{t}{2}\sin \theta )+w(h\sin \theta -\frac{t}{2}\cos \theta )\Rightarrow a=g\frac{s-\tan \theta }{1+s\tan \theta }$ . Thus, $a=0$ at $\theta ={\theta }_p$ .

In the equation above it appears that the centripetal acceleration $a=\frac{v^2}{r}$ is constant, but that is not true. Energy is assumed conserved, so $v^2$ must decrease to $v^2-2g(\sqrt{{\left(\frac{t}{2}\right)}^2+h^2}-h)$ as the center of mass rises from initially $h$ to $\sqrt{{\left(\frac{t}{2}\right)}^2+h^2}$ at the critical point of rollover. Also, since the wheels are locked to move in a circle of radius $r$ , the radius of the center of mass changes from initially $r$ to $r+\frac{t}{2}$ at the critical point of rollover. As functions of $\theta$ , the angle between the vehicle floor and the road bed, the torque/weight as a function of $\theta$ is

$\frac{\tau \left(\theta \right)}{w}=\frac{v^2-2g[h(\cos \theta -1)+\frac{t}{2}\sin \theta ]}{g[r+\frac{t}{2}(1-\cos \theta )+h\sin \theta ]}(h\cos \theta +\frac{t}{2}\sin \theta )+h\sin \theta -\frac{t}{2}\cos \theta$ .
(For the derivation of this equation, see torque.)

This shows the rollover torque as a function of angle for different values of $h\text{and}t$ up to complete rollover on the side ( $\theta =90°$ ) at the critical initial circular speed and $r=40$ ft. Any speed slightly above the critical initial circular speed will cause a rollover. The tip-over angle, given by $\tan {\theta }_p=\frac{\frac{t}{2}}{h}=s=\mathrm{SSF}$ , is shown for each case by an arrow head. The critical initial circular speeds for these cases are: $h=$ 2 ft, $t=$ 2 ft ( $s=$ 0.5) 17.3 mph $h=$ 2 ft, $t=$ 4 ft ( $s=$ 1.0) 24.5 mph $h=$ 2 ft, $t=$ 6 ft ( $s=$ 1.5) 30.0 mph $h=$ 2 ft, $t=$ 8 ft ( $s=$ 2.0) 34.6 mph

This shows the rollover torque as a function of angle for circular speed for 50% above the critical speed and $r=40$ ft. These graphs make it very clear, that once the initial circular speed is above the critical speed and, thus, the torque is above zero at zero angle, the rollover is certain since the non-zero torque causes an angular acceleration which then moves the vehicle to a larger angle where the torque is even larger.

If we assume that the system does not lose any energy to the environment during the course of going from initiating the rollover to the critical point of the rollover, the energy-conservation analysis carries through exactly as above for a sideways-slipping rollover when hitting a curb.

The energy-conservation initial critical speed is, as above: $v_{\mathrm{ec}}=\sqrt{2g(\sqrt{{\left(\frac{t}{2}\right)}^2+h^2}-h)}$

For the two examples given above:

When the torque is zero at an angle $\theta$ , then the initial speed required is given by

The critical circular initial speed for barely initiating ( $\theta =0°,\tau =0$ ) a rollover is $v_{\mathrm{ir}}=\sqrt{r·s·g}$ .

Equating the two critical initial speeds for energy conservation and barely initiating rollover, we get the following for the critical turning radius required for barely inititating rollover in circular motion of the type described herein:

$r=\frac{4h}{t}(\sqrt{{\left(\frac{t}{2}\right)}^2+h^2}-h).$

These radii are much smaller than any vehicle can steer. Larger, more realistic radii, give a larger critical speed for initiating a rollover, which definitely would cause a complete rollover, once the inner wheels leave the road bed.

For the two examples, a realistic radius of 40 feet yields:

For this kind of rollover, s is a good measure of the tendency to roll over. However, most rollovers are not of this kind (http://www.nhtsa.dot.gov/hot/rollover). Some rollovers may be a combination of sideways sliding and circular motion.

Conclusion

If a single measure of rollover-tendency ranking of vehicles is used, it should be critical-point initial speed,

$v=\sqrt{2g(\sqrt{{\left(\frac{t}{2}\right)}^2+h^2}-h)}$ , rather than static stability factor, $s=\mathrm{SSF}=\frac{\frac{t}{2}}{h}$ . This is strictly for sideways rollovers, not for circular-motion rollovers; however, that is the case for most rollovers according to NHTSA.

I suspect that a reasonable fraction of rollovers involve some circular motion, as the driver tries to maneuver out of trouble, as well as sidewise motion. Circular-motion rollover tendency depends monotonically on SSF. So, perhaps one should average the rankings for critical-point initial speed and SSF.

The rankings of the vehicles given above for the three ways to rank are as follows:

In all calculations above, the effects of suspension movement, tire movement or electronic/mechanical stability control were neglected. Suspension and tire movements would likely increase the tendency for rollovers, while electronic/mechanical stability control should make it less likely that a vehicle would get in a situation where rollovers occur.

Auto makers need to be required to publish the height of the center of mass for each automobile made, so that one can easily calculate the SSF. Better, yet they should be required to publish the SSF.

An industrial economist, Joe Kimmel, has developed a formula for predicting the probability of rollovers in accidents given the track width $t$ , height $h$ and weight $w$ of an auto. The formula is given at http://ads.usatoday.com/money/consumer/autos.nayti698.htm. There is a slight error in the formula as given there. The correct formula for the % probability of rollover is:

P=0.091 $(550000\frac{h}{t·w}-90).$

The values for 189 model year 2001 autos are given at:

http://cgi1.usatoday.com/money/consumer/autos/mauto695.htm.

Rollover risks for some 2001 luxury AWD automobiles are: