With some high school physics, we can roughly determine a safe vehicle speed given weather conditions for various tires on various surfaces and at various speeds.
By Ted Mitchell
Published February 11, 2009
What does high school physics have to do with safe driving?
Here in the middle of winter, you can observe very different responses to snow, ice and slush on roads. Some people tend to drive as if nothing much was different, while others crawl down the road.
The police sometimes sum up a crash with the phrase "excessive speed for conditions". This is pretty vague. Can it be quantified?
Yes and no. Things like visibility, driver inattention and poor skill or judgment are not easily analyzed. But every vehicle is subject to the laws of physics.
The first step is to set a benchmark. What is safe? I theorize that adjusting your speed to a constant stopping distance on various surfaces will give a good estimate of what is a "safe speed for conditions".Say we are talking about a typical urban street with a speed limit of 50 km/h. This being southern Ontario, the enforced speed limit is higher than posted and a speed of 60 km/h is universally tolerated.
We need a braking force to slow us down: Ff = ma. On flat ground, that braking force comes from friction Ff = μFn. The coefficient of friction is designated my mu (μ). But Fn = mg, which is the weight of the car. Rearrange, and a = μg. That is, mass cancels, and deceleration is directly proportional to the coefficient of friction.
But what is μ? It turns out that this is not so easy to determine, and it varies a lot. For dry pavement and good tires, μ is approximately 0.8.
For constant acceleration, v12 - v22 = 2ad. v1 is starting speed, v2 is zero, so rearranging, braking distance db = v12/2μg. Make note of what this means: braking distance is proportional to the square of speed, and directly (inversely) proportional to traction.
For our example above, 60 km/h in real units is 16.7 m/s. So db = 16.72/(2)(0.8)(9.81) = 17.7 m.
This is just the braking distance, for the stopping distance we need to assume a decision-reaction time of about 1 second (relatively quick). So we need to add reaction distance dr = vt = 16.7 (1) = 16.7 m.
Then ds = dr + db.
Stopping distance = 16.7 + 17.7 = 34.4 m.
How much should you slow down to keep constant stopping distance? Because we now have reaction time to deal with, that question becomes a little cumbersome for high school math, so it helps to use a spreadsheet.
|Speed (km/h)||Conditions||Coefficient of friction||Reaction distance (m)||Braking distance (m)||Stopping distance (m)|
Now with snow μ = 0.3, what happens if you don't slow down?
Braking distance is now db = 16.72/(2)(0.3)(9.81) = 47.2m.
Stopping distance is now 16.7 + 47.2 = 63.9m. That is 6.6 car lengths past where you can stop on dry pavement.
|Speed (km/h)||Conditions||Coefficient of friction||Reaction Distance (m)||Braking distance (m)||Stopping distance (m)||Car lengths @ 4.5m|
Try out a few numbers and you'll find that the proportional slowing needed for a safe stopping distance is greater with increasing speed.
Several other factors come into play in the real world. Kinetic energy (mv2) decreases markedly at lower speeds, therefore crashes are less severe and less fatal if they happen at lower speeds - and of course, vice versa.
Then there's visibility, traffic congestion, pedestrians, and many important driver factors like inattention and decision - reaction time. Visual processing time (decision speed) is especially sensitive to age-related slowing even in drivers who have normal simple-task reaction times.
Pedestrians and cyclists are at least as affected by slippery conditions (sidewalks not shoveled) and cannot react to avoid collisions as well as usual. They also tend to be hindered by heavy clothing which limits vision and hearing.
Perhaps more important than stopping, control is more likely to be lost in low friction situations, so the vehicle can do more damage and lead to domino effect crashes.
Putting all this together, the prudent driver will treat slippery conditions with respect and slow down at least as much as the table above would suggest.
This analysis is not transferable to traction for acceleration, because it matters if there are two or four wheels driven. Situations exist where friction is marginal for being able to get a two wheel drive vehicle up a small hill or accelerate adequately where four wheel drive has no problem. However, once power is not applied to the wheels, it ceases to matter.
Avoiding a crash requires braking and control and only very rarely calls for acceleration, so it is not surprising that four/all wheel drive has no benefit for safety. Statistics, and a scan of ditches on slippery days, suggest the opposite.
Four wheel drive can be very handy but fails in the safety department. But there is a way to stack the deck in your favour for increased winter traction: snow tires. Not tires with big lugs, those are obsolete. Modern winter tires have a 'mountain snowflake' symbol and several key properties:
Compared with a good set of all-season tires, the coefficient of friction can be up to 25% higher on snow and ice.
Let's say we have a slippery surface with μ = 0.20 for all-season tires and μ = 0.25 for snow tires. Maybe that doesn't seem like much. But you're on the highway traveling 80 km/hr.
|Speed (km/h)||Conditions||Coefficient of friction||Reaction distance (m)||Braking distance (m)||Stopping distance (m)||car lengths @ 4.5m|
There is a difference in stopping distance of 25.2 m, or 5.6 car lengths.
Put another way, when you have come to a complete stop just in time with your snow tires, the car with all-seasons is still moving - but how fast?
Recall v12 - v22 = 2ad, a = μg. d = db ; braking distance with snow tires, μ = 0.20, and v2 is the unknown we want: the speed of the car without snow tires.
Turns out v2 = 35.7 km/h with all season tires at the point where the car with snow tires has stopped. If there's a stopped transport in front of you that would be bad. Convinced?
The cost of snow tires may seem expensive, but expect a set to last 5 years with moderate use. For most cars, $1500 will buy eight tires and four rims which will last 10 years, and save wear on your summer tires. $150 / year is cheap insurance.