The Physics of SUV Rollover Accidents

A 2001 PBS FrontLine program about SUV vehicle rollover accidents amazed me that no one, among the carmakers, among the legislators, or among the researchers, seemed to understand the very basic Physics involved! When I taught High School Physics, I gave problems comparable to that to the students!

This presentation will describe the physics (and math) of what occurs, for NON-physicist-types!

Sidebar regarding Centrifugal and Centripetal!

Virtually all non-Physicists regularly talk about Centrifugal force. That drives Physicists crazy, because there is actually no such thing! Circular motion can only occur for any massive object, if there is a CENTRAL force, or what is called a Centripetal force. A vehicle going around in fast donuts (circles)(kids, don't do this at home!) are able to do that because the tires have enough traction to provide the necessary Centripetal force to enable the vehicle to keep turning. Ah, but Newton told us that all (massive) objects try to go in straight lines unless a force is applied to them. Therefore, YOU, sitting IN that vehicle doing circles, are actually trying to go off on a straight line, at every instant! The vehicle keeps scooting sideways out from under you (due to the Centripetal force provided by the traction of the tires on the road). From YOUR point of view, where you tend to consider the seat of the vehicle to be a "reference frame" (a high-class term!), then you definitely feel that you are constantly "being thrown outward by Centrifugal force". Nope! Didn't happen! There are all sorts of things where we imagine that a Centrifugal force is existing, but in every one of those cases, the reality is that the effect is just due to Newton's INERTIA trying to cause an object to keep its current motion, like a speed, in whatever direction it is now pointed. So there really is no such thing as Centrifugal force, even though even Physicists occasionally mention it! More advanced: A constant velocity circular motion can be described in a coordinate plane as q (angular position) * r (radius of the circle as giving a location on that plane. In familiar x-y terms, we could say that is x = r * cos(q) and y = r * sin(q). (where q constantly increases at a constant rate in time, and we can specify that rate of change of q as w, an angular velocity) The position vector is then from the origin to a point on the circle at position-angle q. We could do a simple Calculus magic of Differentiation of this to get dx/dt = -w * r * sin(q) and dy/dt = w * r * cos(q). (dx/dt and dy/dt are the rates of changes of position along the x and y axes, sort of speeds). We could do that same Differentiation a second time to find that dx2/dt = - w 2 * r * cos(q) and that dy2/dt = - w2 * r * sin(q). (these are the rates of changes of the speeds along those axes, in other words the accelerations.) By comparing these acceleration values with the position values, we can see that they are identical, except for a constant factor of - w2. What this means is that this acceleration, required for the constant speed circular motion, is negative, meaning that it is opposite the direction of the position vector, which means it is always exactly toward the origin, in other words a CENTRAL acceleration. Newton told us that an acceleration times the mass involved equals the force, which means we are talking about a Central (Centripetal) Force for this circular motion. We even can see that it is exactly proportional to the square of the angular speed w. We have just strictly proven that circular motion can only occur under the influence of a central (Centripetal) force, and even the equation for calculating it! Below, you will see the v * v (square of velocity) in the equations, really just the same because v = r * w.

We show the tire-road frictional force as the red arrow to the right in the drawing. This represents a turn to the right. That force, acting on the vehicle, is just the simple F = ma (force equals mass times acceleration, as Newton said), expressed for a curved motion. For this example, we will consider a vehicle turning in a circular course, to the right, approximately like circling clockwise in an intersection, with a circle radius of 25 feet. Say the vehicle weighs 3200 pounds, and we will consider it moving at 20 mph. (20 mph is equal to about 29.3 feet/second). And the acceleration due to gravity is 32 feet/second/second. Now you know everything necessary!

With the force due to the tire-road friction being to the right, there is an effect, often called centrifugal force, that then also seems to exist to the left. It is actually just the condition of the vehicle wanting to go straight (in accordance with Newton's Laws) and so this apparent centrifugal force (to the left) is exactly equal in size with the (actual) centripetal force (to the right) making the vehicle turn the corner. Sorry about the confusing two words! The driver feels as though there is a force on him to the left. The reality is that the vehicle and seat are moving to the right underneath him, and the driver has that sensation (of centrifugal force) only because his body is trying to go straight.

The F = ma for a circular motion is in the form F = (w * v * v)/(g * r). [w = vehicle weight; v = velocity/speed; r = circle radius; and g = acceleration due to gravity]. That version of Newton's Law is very straightforward. (It's actually in the sidebar box above) This gives a centripetal force for our example: (F) = (3200 * 29.3 * 29.3)/(32 * 25) or around 3400 pounds!

That is the force that the tires traction must exert sideways on the vehicle to make it turn in the circle rather than going straight the way it would have normally wanted to go. (On an icy surface, nowhere near that much side force can be applied, and the vehicle slides straight instead of going around the turn. Now you know why!)

By the way, in nearly all vehicles, if you try to turn really fast circles to try to rollover a vehicle, the tires nearly always break free of the pavement before enough side force can be produced. But when a vehicle is traveling at speed, and a sudden and extreme side turn is made, the tire traction can exist long enough to cause a rollover. With switchback turning (zig-zagging) the tire traction can be plenty for such a rollover. And this is the source of the danger being considered here.

We have shown this force acting at one point of the vehicle, often called the center-of-gravity. Since the vehicle is considered here as one solid object, it can be mathematically treated as though all of its weight is at that one point. We are now going to change our position and look at the vehicle from the front, and we see the red spot that indicates the center-of-gravity.

In other words, the entire weight of the vehicle acts as though it was at that point/line, height. Viewing from the front, we're looking at the end of that line, so it looks like a point in our drawing.

Now, we have that all of the weight (mass) of the vehicle acts as though it is at the height of the center-of-gravity, and that there is a lateral force, the centripetal force being applied by the tires and road AT THE HEIGHT OF THE ROAD SURFACE on that weight, to the left in our drawing. This horizontal force does NOT act directly at/on the center-of-gravity directly because it is at the wrong height. Therefore, it not only acts to push the weight/center-of-gravity sideways but it also acts to try to twist the vehicle around the center-of-gravity. This twisting action is where the problem arises.

In basic Physics, forces can be analyzed for the separate effects of translation (straight-line motion) and rotation. This can be done equally correctly in different frames of reference. So far, we have been considering the force acting ON the vehicle as seen from the ground, and so we would draw our horizontal arrow TO THE LEFT in this drawing at a height of the bottom of the tires. This would represent the ACTUAL centripetal force acting on the vehicle by the road, and would be analyzable as to both translation and rotation effects. I think it may be easier to understand if we look at this in a different, but exactly equivalent way. Instead of saying that the centripetal force is acting to the left (in our drawing), at the height of the tires, we can say that there is an identical force to the right, acting ON the center-of-gravity. The effect is exactly the same, regarding the action to rotate/twist the vehicle, but I am hoping that the arrows and discussion might better show why this acts to roll the vehicle over sideways.

Here is the (3400 pound) sideways force that is acting on the vehicle, as seen from the front, and still acting on the center-of-gravity. In this drawing, you can see that if the (outer) rightmost tire is held from sliding sideways to the right (by friction with the roadway), then the force arrow will act to try to roll the vehicle over (to the right in the drawing). This all happens because the height of the center-of-gravity and the height of the frictional force with the road are at different heights, which then allows the twisting effect as a result.

Now, this does not mean that every turn results in a rollover! All (three or more-wheeled) vehicles are "meta-stable" because they have contacts with the roadway which are spaced apart sideways. In the case of conventional vehicles, these are the tires on the left and right sides of the vehicle. Where a bicycle does not have this advantage, and can easily be pushed over sideways (being naturally unstable), the sideways spacing of the tires on conventional vehicles, called the "track", makes for better stability regarding rolling over. Again, looking at this drawing, you can see that the force arrow shown would actually have to LIFT UP the weight of the vehicle (a little) in order to roll it over. Consider this drawing with the vehicle tilted up balanced on those rightmost two tires and you will see that the center-of-gravity has raised up, to be straight above the point of contact of the tires with the road. This necessity of "lifting" the vehicle weight up is what provides the meta-stability for normal driving. However, with ANY vehicle which has its center-of-gravity above the road, in other words, ALL actual vehicles, if sufficient lateral force is applied by the tire-road friction, the vehicle can rollover. We are going to proceed with the analysis of this now.

In physics, a force can be thought of as a "resultant force" which is a combination of two "component forces."

We know that the tire treads are going to somehow be involved, especially since we know that on icy roads, the vehicle does not turn. So we know that the sideways force must be being applied between the tire treads and the roadway.

Our (horizontal) force in the previous drawing can therefore be thought of as a combination of two separate component forces. One of the component forces is directly toward the tread of the tire, as shown here. This component is actually the portion of the force that actually acts to give the vehicle sideways force (through the tires' friction with the road) to make it turn. More to the point here, this component does NOT act to try to rotate/twist the vehicle.

This is the physics way of showing how the tire treads are able to transfer a sideways force to the vehicle, to make it possible to turn corners.

More Technically Presented: In actual Physics, this subject of rotational effects is described in terms of something called "moments". A moment is a Force acting on an "arm" to cause a torque or a tendency to rotate. In this case, there is one moment that is due to the weight of the SUV, and in our example that moment is a Vector aimed toward the FRONT of the vehicle (don't ask!) of W x r which has a magnitude W * r * cos(q). The moment that is due to the horizontal lateral force is a Vector aimed toward the REAR of the vehicle of F * r which has a magnitude F * r * cos(90° - q) or F * r * sin(q). In the situation of being just about to roll over, these two must be equal, or F * sin(q) = W * cos(q), or F * sin(q) / cos(q) = W, or F * tan(q) = W. q is the angle between the two red arrows shown in the two drawings above, and r is the actual (angled) distance between the center-of-gravity and the tire tread. If F is any greater than that amount, the vehicle will roll over. An equivalent description in Physics considers that F acts to rotate the vehicle (to the right in our drawings) at a height of the center-of gravity (which is r * sin(q); and the weight W acts to rotate the vehicle (to the left, around the outer, rightmost tire) at a horizontal distance of half the tire track (which is r * cos(q). You can see that this gives the exact same equations. Since the height of the center-of-gravity of a particular vehicle is known, as is its weight and its tire track width, F, the maximum sideways force before rollover is easily known for any vehicle. It would be totally incompetent if Ford Engineers, for example, did not know this value, as they testified in Court!

The other component force (of the horizontal lateral force discussed above) is at right angles to the component between the C-G and the tire tread, and so it is aimed upward and outward, as shown here. This component acts to roll the vehicle over, around that (outer) tire tread.

We are concerned about the extreme situation where the vehicle actually WOULD roll over sideways. From our front view, we then need to know the force necessary to act AT the center of gravity, AROUND the outermost tire treads, as shown in this drawing. Let's call this F_unknown. This force is upward at a (90 - q) angle, which depends on the geometry of the vehicle. You can also see that the magnitude of this component, F_unknown, is F_horizontal * sin(q). (We already know the value of F_horizontal)

You can see that this is the direction that a force would have to act in order to "roll" the vehicle over that tire's point of contact with the road. It represents the "torque" around that outer tire tread, due to the turning.

We're not quite done! This angled force component can now be considered to be a combination of a vertical and a horizontal force. If the vehicle is going to be about to roll over, the UPWARD part of it (shown as a yellow arrow) is what we are interested in and it must be at least exactly equal to the weight of the vehicle. That would then LIFT the weight of the vehicle up off the roadway. The (red) upward angled force gets a leverage advantage around the tire tread, and so it is able to have a lifting effect greater than its own strength, actually, equal to its strength divided by the cosine of q. If the vehicle is about to roll over, this lifting effect must be equal to the weight of the vehicle. F_unknown therefore equals Weight * cosine(q), (or Weight = F_unknown / cos(q). The cosine can never be more than 1.00, so this means that its leveraged (lifting) effect is always greater than its true force.

Standard geometry is then all that is needed to get actual numbers! It tells us that this angled (red) F_unknown must therefore be the weight of the vehicle times the cosine of the angle shown (q, between the red and yellow arrow).

For a rollover to begin to occur,

We learned F_horizontal up above, as ONLY depending on the speed and radius of the turn and the weight of the vehicle. The only part not yet known is the angle q.

The center of gravity is always above the ground. For a fairly tall vehicle like an SUV, it can commonly be 30" (or more) above the ground. Vehicle manufacturers used to divulge the height of the center-of-gravity of their vehicles, but they no longer do. For the vehicle illustrated, we believe that the center-of-gravity is probably around 34" above the road surface. Rather than using that value, we will again use a more generous (and more stable) 30" height for our example calculations. This is represented by the vertical green line shown here.

Now, look again at the force acting between the tire tread and the center-of-gravity of the vehicle (in red). This force is at an angle which has a horizontal component proportional in length to half the track of the vehicle. It also has a vertical component that is exactly proportional to the height of the center-of-gravity above the roadway. We KNOW both of these distances! Therefore, the angle at the CG (which we have been calling q), between the red and green lines, is the angle whose TANGENT is equal to (Height of the CG)/(half the track). In the case of this example, that vertical side is 30" and half-the-track is also 30". This means that the angle we have been describing has a tangent of 30/30 or 1.000, and so our angle q is 45°.

Since we now know the angle q in the triangles we have been dealing with, we can do the final calculations. Using the specific dimensions (tread track and center-of-gravity height) of ANY vehicle, it is therefore VERY easy to quickly calculate at what speed it would roll over! No "expert" could possibly get away with testifying in court that he does not know!

We have already determined that the actual lifting force on the center of gravity of the vehicle is the centripetal force times the tangent of that angle, in other words in our example, 3400 pounds times 1.000 or 3400 pounds vertical lifting effect!

Now, since the vehicle only weighs 3200 pounds, there is only 3200 pounds of gravitational force holding the vehicle down, and so this 3400 pound lifting force is enough to lift it up and roll it over. Our example vehicle would roll over in making that moderate turn at just 20 mph!

That's all there is to the calculations! If the resultant vertical lifting force is less than the weight of the vehicle, it will not roll over; if it is more, it is certain to roll over.

In our example, at just 20 mph, our vehicle would roll over when trying to turn in our 25-foot radius circle!

Actual SUVs are much heavier than we have used. However, the heavier weight increases the centripetal force proportionally, which increases the force that acts to roll the vehicle, EXACTLY in the same proportion that the necessary lifting force is increased for the heavier vehicle. The actual overall weight of the vehicle does not alter these results! It is not actually necessary to know the weight of the vehicle, and heavier vehicles are no more stable than identical lighter ones. The track-width, the center-of-gravity height, the turning radius, and the vehicle speed are the ONLY variables!

Long ago, Ralph Nader wrote a book "Unsafe at any Speed" about the poorly designed Corvair cars. It is hard to see how that is not even more true today about SUVs! Since some States now have Interstate highway speed limits of 75 mph, SUVs can commonly be traveling at 80 mph. The families inside them have NO idea of what a dangerous situation they are in! The worst part of all this is that all of the SUV manufacturers have always known this simple Physics! In order to sell the extremely popular SUV vehicles, they have chosen to ignore the amazing danger involved in driving them!

It is hard to see how the vehicle manufacturers will be able to financially respond to hundreds of thousands of lawsuits by people who will get injured inside SUVs. As of now, they simply are aggressively working at keeping the publicity of all those crashes as quiet as possible, in a way reminiscent of the way the tobacco companies insisted that cigarettes "had never been proven to be dangerous". In both cases, the manufacturers certainly DID know for many years how terribly dangerous their products are. And, in both cases, the companies involved are so huge that the American economy could not handle their going out of business. Therefore, we still have tobacco companies prospering (and now with total immunity from litigation and with strong support from all State governments due to their large incomes from the tobacco settlement. It is hard to see how the SUVs could result in anything different.

The really sad part is that countless thousands of people are going to die while the SUV manufacturers continue to deny all culpability. At least, with the Corvair, Mr. Nader got public outrage started after only a few hundred people died from their failings. This SUV situation will cause the deaths of far more people, and already has! But they are so popular, and the buyers and drivers of them BELIEVE they are safer inside them, that SUVs will continue to be extremely popular, until gasoline prices get so high that people start buying smaller vehicles.

(I realize that the above paragraphs are essentially an Editorial Opinion, and not usually heard from a Physicist, but it seems outrageous that so many people are dying just because giant corporations insist on selling popular products, even though they are extremely dangerous. If that is not a deja vu regarding cigarettes, I don't know what is!)

Back to science and Physics!

The point of much of this discussion is that it is relatively simple to calculate these things! It would not be necessary to keep doing road tests with vehicles constantly going up on two wheels, as the government and testing labs keep seeming to do.

The general form of the calculations above is vertical LIFT = centripetal force * Tangent(q). In our case, that was 3400 * (1.000) or 3400 pounds lift. The 3400 pounds of centripetal force was easy to calculate, using Newton's formula based somply on the speed and the radius of the turn. q is easy to calculate from two simple measurements of any vehicle. Why do "experts" claim to not understand such things, and even testify to such ignorance in Court?

The PBS FrontLine program included some "manufacturer's experts" claiming that widening the track by two inches would do miracles. That was clearly a lie! It is easy to do this math again to show that it would only make an incremental difference. In our example, widening the track by 2" makes our horizontal 30" into 31", while leaving the center of gravity height of 30" the same. It is easy to check that this geometrical change makes our 45 degree angle now become 44.06 degrees. The tangent of that angle is 0.9677. When put in our problem, we would now have 3400 * (0.9677) or 3290 pounds, STILL making the vehicle roll over! The dangerous lift is only minimally reduced, from 3400 to 3290 pounds! A two inch wider track would have a VERY slight benefit as to safety from rollovers. A benefit, true, but so small as to not provide significant extra safety for passengers. Actually, where our original vehicle could have made our circle at up to 19.3 mph while being on the verge of rolling over, the one with the two inch wider track would be able to do it at 19.6 mph instead, less than half a mph improvement!

Sports Cars

Recovery is Very Difficult, even for a Professional Driver

Let's look at how the forces we have been discussing change.

To start with, this doesn't look all that different.

But compare this drawing with the similar ones above. You can see that the center-of-gravity has gotten a little higher. At a 10° tilt, it is actually raised up by 5.2". And if the tire's low pressure allows it to deform horizontally, then it is less far horizontally from the tire footprint. We will estimate 2". These two effects combine in making that important q angle bigger. We would now have (vertical) 30 + 5.2 or 35.2" and (horizontally) 30 - 2 or 28". The tangent is now 35.2/28, or 1.257. Our 3400 pound horizontal centripetal force then would have the effect of 3400 * 1.257 or 4274 pounds of vertical lift!

The 3400 pound lifting/rolling effect has already gotten FAR greater. Even worse, there is now no realistic chance of recovery. Where the problem began with a lifting force just a little more than the vehicle weight, as soon as the vehicle starts tilting, the geometry gets catastrophically worse. An experienced driver would tend to immediately straighten the wheels out. In many vehicles, this can resolve the situation safely. But in tall vehicles such as SUVs, there is a tendency to rock the vehicle back the other way (inertia, momentum, etc) and the action of trying to recover tends to cause the SUV to roll over to the other side. So, even experienced drivers have great difficulty in recovering from incipient rollovers in tall vehicles. Imagine a less experienced soccer mom!

FYI: This is why test vehicles always have those "training wheels" arms on them, because once a rollover begins, it gets so much worse so fast that even professional drivers have great difficulty in recovering.

(Once a rollover has begun, the geometry actually gets even more complicated, involving the dynamic rotational inertia of the vehicle and the fact that the effective weight of the vehicle becomes less (even zero as it crosses above the tire tread), but this last calculation essentially shows how the problem gets far worse as the rollover proceeds. These complicating effects all act to make the rollover effect even worse. All of the earlier "level vehicle" calculations were exactly correct. They represent the most generous analysis regarding stability.)

Passengers and Luggage

Low Tire Pressures and Soft Suspensions

The other reason that the recommended low tire inflation pressures caused the accidents and deaths has to do with the normal way a tire functions. Each time a tire rotates, its sidewalls have to bend/flex/deform as that part is in contact with the road, as it must briefly support the weight of the vehicle. This causes the sidewalls of the tire to flex every single revolution. This flexing ALWAYS creates frictional heat within the sidewalls of the tires. When the tire pressure is low, this flexing is greatly exaggerated, resulting in much more internal heating of the sidewalls of the tires. This is why a tire that is extremely low on air pressure quickly blows out, because the sidewalls flex so much that they overheat and then fail, permitting the internal pressure of the tire to suddenly burst out.

When tires have abnormally low air pressure in them, on long trips of high-speed highway driving, they are especially susceptible to the sidewalls overheating in this way. For this reason, it is quite understandable that many tires failed and caused terrible accidents on those Ford vehicles and on all other SUV vehicles that recommend low tire pressures. This is actually an effect whether the vehicle is traveling straight or turning. If a particular tire had even a hint of a problem on its own, that situation would ensure that it would fail. So, whether or not the Firestone tires had any drastic flaw, even if they have a tiny additional inclination of a flaw as compared to other brands, the effect would have been tremendously magnified by the circumstances of the low recommended tire air pressures in those vehicles. The natural instability of the high center-of-gravity vehicles added to the problem, to cause the many rollovers once the vehicle got turned a little sideways.

Interestingly, the manufacturers cannot claim to not be aware of these matters! For example, Ford's SUV vehicles have wider track dimensions VERY much in line with these calculations! For example, their very tall Excursion (77.2" tall) has a MUCH wider track (64" compared to 56") as compared to their own Explorer Sport SUV, which is 68.4" tall. It is obvious from that that they are VERY aware of needing a wider track dimension for the even more unstable taller models. That means that they have always known what they were doing to begin with! And, assuming that their engineers were capable of the High School Physics presented here, they have been aware of the instabilities that are direct consequences of their decisions. This seems to imply an intentional knowledge that they were going to cause many injuries and deaths, which I see as far worse than just incompetence or unpredictable accidents.

Around 30 years ago, I owned a Cadillac that had a feature where the suspension was self-adjusting regarding vehicle height. The car had very soft suspension to start with, and when several people would get in the back seat, the car initially tilted where the headlights were then pointed somewhat skyward. This feature sensed this and switched on some small electric motors that pumped a little hydraulic fluid into the shock absorbers. You would hear these little motors run for a few seconds as the vehicle leveled itself back out.

Regarding SUVs, it seems to me something very obvious to add! If the vehicle speed is UNDER 30 MPH, a dashboard switch could be flipped to raise the vehicle to the high ground clearance that all SUVs now have, for doing off-road exploring. However, at all other times, the vehicle suspension would lower the entire SUV by maybe six inches vertically. That still leaves plenty of ground clearance for highway driving, but it importantly LOWERS THE CENTER OF GRAVITY BY SIX INCHES.

This would essentially eliminate the tendency of rolling over! You can do the math based on the equations presented above to see this! For our hypothetical vehicle discussed above, the CG is now at 24" high instead of 30". This makes our angle 38.6° rather than the 45° we had above. As a result, our vehicle could withstand a lateral (centripetal) force of 4,000 pounds instead of the 3,200 it could before. We have really increased the stability of the vehicle to rollovers by 25%, quite a large effect. It still would not eliminate ALL possibility of rollover accidents, because it still has an high CG, but a 25% improvement in stability would be a spectacular improvement.

And this is using only 30 year old technology, meaning virtually NO Research and Development costs, and relatively cheap application to vehicles!

How come none of the SUV manufacturers have ever used such an obvious idea?

Given how simple these calculations are, it is really disappointing that no one seems to be in a position to challenge the "engineering" of the carmakers! To watch in that PBS program, an "expert" representing a carmaker testifying that he just didn't know how great the benefit would be of widening the track two inches, was appalling! Tell them to get any high school Physics student to explain it to them! The calculations above show how simple those calculations are! There ARE no "complicated" calculations to do! This is all of it!

( http://mb-soft.com/public/othersci.html )

C Johnson, Physicist, Physics Degree from Univ of Chicago