How can I calculate side force on my car?
October 29, 2009 8:22 PM Subscribe
Physics question - involving cars, air, momentum, force, etc... Basically - how can I calculate the side force from a crosswind on my car at any given speed?
(Yes, I googled as much as I could and couldn't come up with a formula for this.)
First off, I drive a 2001 Mazda Miata. 155.7 inches long by 48.4 inches tall. Considering the shape of the car, let's assume 155x35 inches as a side profile.
I'm not sure if the front dimensions are important, but I can measure the front end if necessary. For the sake of the math, I think we should disregard the front end of the car - I don't know what the drag ratio and force necessary to push through the air are on this car.
I guess my real question is how do I come up with a figure for how much force a cross wind applies to my car? And, at any given speed, how does that translate into lateral acceleration? (I understand I may be asking the wrong question here.) I know that a 10 mph crosswind will have a different effect when the car is traveling at 5 mph versus 50 mph.
This question comes out of the fact that while driving to work this morning I almost couldn't change lanes due to an 85 mph forward speed coupled with a 50 mph cross wind. I know that at a 10 mph forward speed the effect of that side wind would be lessened, I just want to understand the math behind why.
I apologize in advance for being unable to phrase this question properly. :) Perhaps part of this question is meant to ask "how do I phrase this properly such that I can get a solution to the question?"
(Yes, I googled as much as I could and couldn't come up with a formula for this.)
First off, I drive a 2001 Mazda Miata. 155.7 inches long by 48.4 inches tall. Considering the shape of the car, let's assume 155x35 inches as a side profile.
I'm not sure if the front dimensions are important, but I can measure the front end if necessary. For the sake of the math, I think we should disregard the front end of the car - I don't know what the drag ratio and force necessary to push through the air are on this car.
I guess my real question is how do I come up with a figure for how much force a cross wind applies to my car? And, at any given speed, how does that translate into lateral acceleration? (I understand I may be asking the wrong question here.) I know that a 10 mph crosswind will have a different effect when the car is traveling at 5 mph versus 50 mph.
This question comes out of the fact that while driving to work this morning I almost couldn't change lanes due to an 85 mph forward speed coupled with a 50 mph cross wind. I know that at a 10 mph forward speed the effect of that side wind would be lessened, I just want to understand the math behind why.
I apologize in advance for being unable to phrase this question properly. :) Perhaps part of this question is meant to ask "how do I phrase this properly such that I can get a solution to the question?"
Response by poster: Ah, okay, thanks JackFlash. I can understand your point. Assuming the wind was constant, over the same distance I would be blown to the side at 5 mph as much as I would at 50 mph, it would just take more time (not necessarily more distance) due to a slower rate of travel and assuming I don't correct for it. Am I correct, or close to correct?
posted by krisak at 8:42 PM on October 29, 2009
posted by krisak at 8:42 PM on October 29, 2009
My experience on motorbikes is the same as JackFlash's point about your intuition: a crosswind is just as forceful at 5km/h as at 100km/h, a strong gust will knock over an off-balance motorcyclist just as easily in the parking lot as on the highway.
The only difference is how much it'll hurt when you land.
posted by Fiasco da Gama at 8:51 PM on October 29, 2009
The only difference is how much it'll hurt when you land.
posted by Fiasco da Gama at 8:51 PM on October 29, 2009
Phrasing:
I am traveling 130 kph (~80 mph) (not to be a units nazi, but metric is usually standard for science like this) and the wind is blowing at 80 kph (~50 mph) perpendicular to my path. How will this affect my driving? How would it if I were travelling at 20 kph (~10 mph)?
Time to work on the answer...
posted by battlebison at 8:52 PM on October 29, 2009
I am traveling 130 kph (~80 mph) (not to be a units nazi, but metric is usually standard for science like this) and the wind is blowing at 80 kph (~50 mph) perpendicular to my path. How will this affect my driving? How would it if I were travelling at 20 kph (~10 mph)?
Time to work on the answer...
posted by battlebison at 8:52 PM on October 29, 2009
85mph on a 2001 Miata with a 50mph cross wind? Hope there was no one else around.
The force exerted by the crosswind will be (approximately) the same. My guess is that, at high speeds, the coefficient of friction between the tires and the road is lower due to aerodynamic lift, making it more difficult to exert enough force to move sideways. If the road was curving slightly as you tried to change lanes, the effect would be even more pronounced.
posted by Behemoth at 8:52 PM on October 29, 2009
The force exerted by the crosswind will be (approximately) the same. My guess is that, at high speeds, the coefficient of friction between the tires and the road is lower due to aerodynamic lift, making it more difficult to exert enough force to move sideways. If the road was curving slightly as you tried to change lanes, the effect would be even more pronounced.
posted by Behemoth at 8:52 PM on October 29, 2009
Response by poster: Behemoth - highway was fairly clear, and I have really good tires, so some of the risk was mitigated. If I'd felt that there was real danger I would've slow down (I don't mind taking myself out of the gene pool, but I do try to avoid involving anyone else in my destruction).
Let's assume two things:
1. Neutral lift. (The car has decent downforce at speed.) So let's assume the same pressure on the tires at 50 as at 5.
2. No curves in the road (at that point there actually weren't.) The most we can assume is a crappy road surface where my tires are only touching 50% of the normal contact patch.
posted by krisak at 8:57 PM on October 29, 2009
Let's assume two things:
1. Neutral lift. (The car has decent downforce at speed.) So let's assume the same pressure on the tires at 50 as at 5.
2. No curves in the road (at that point there actually weren't.) The most we can assume is a crappy road surface where my tires are only touching 50% of the normal contact patch.
posted by krisak at 8:57 PM on October 29, 2009
Best answer: Here's another way to think about it. Assume that a gust of wind from the right requires you to steer 10 degrees to the right to enable you to move down straight down the road. Let's say that it takes you 1 second to react to that gust of wind while you hold the steering wheel straight ahead which causes you to drift left. At 5 miles per hour with the steering wheel straight ahead and a 10 degree angle from the wind you will move about 1.3 feet to the left in error. At 50 mph with the same 10 degree angle from the wind you will move 10 ten times as far to the left or 13 feet.
posted by JackFlash at 9:01 PM on October 29, 2009
posted by JackFlash at 9:01 PM on October 29, 2009
Sure, but keep in mind a 50mph crosswind provides some degree of lift as well. If we assume no lift, however, then JackFlash's explanation makes the most sense to me.
posted by Behemoth at 9:02 PM on October 29, 2009
posted by Behemoth at 9:02 PM on October 29, 2009
Response by poster: Ah, okay. So, JackFlash, does that mean that it's a linear equation?
posted by krisak at 9:04 PM on October 29, 2009
posted by krisak at 9:04 PM on October 29, 2009
Best answer: I would call this a fluid mechanics / aerodynamics problem (rather than a physics problem, i.e., this calls for an engineer).
For a very Q&D answer, I would suggest using the aerodynamic drag equation. Imagine your car is traveling 50 mph, sideways, on a frozen lake. Now apply the drag equation.
Enjoy.
posted by coffeefilter at 9:05 PM on October 29, 2009 [1 favorite]
For a very Q&D answer, I would suggest using the aerodynamic drag equation. Imagine your car is traveling 50 mph, sideways, on a frozen lake. Now apply the drag equation.
Enjoy.
posted by coffeefilter at 9:05 PM on October 29, 2009 [1 favorite]
Response by poster: Coffee - that's funny, I almost flagged it as a fluid dynamics question. (I may not know all of the math, but I understand that differences in the disciplines.)
Thanks for the link. I may actually sit down and do the math (with some in/valid assumptions) on this.
posted by krisak at 9:11 PM on October 29, 2009
Thanks for the link. I may actually sit down and do the math (with some in/valid assumptions) on this.
posted by krisak at 9:11 PM on October 29, 2009
Best answer: The folks above are correct. Your forward velocity does not change the force of the side wind. They are two separate wind vectors. The force exerted by one is independent of the force exerted by the other.
[There is a concept from sailing called apparent wind. This is the vector sum of "wind" created by your forward motion and the real wind generated by atmospheric conditions. But that's mostly irrelevant, as you can analyze apparent wind by splitting out headwind vector and the crosswind vector and working on them separately.]
Anyway, the formula for wind pressure on a flat surface is: P = .00256 * V2, where V is wind velocity in mph and P is pressure in pounds per square foot.
It's then straightforward to apply your known car's side area for an approximation of the total force you're receiving from the wind.
posted by Netzapper at 9:49 PM on October 29, 2009 [1 favorite]
[There is a concept from sailing called apparent wind. This is the vector sum of "wind" created by your forward motion and the real wind generated by atmospheric conditions. But that's mostly irrelevant, as you can analyze apparent wind by splitting out headwind vector and the crosswind vector and working on them separately.]
Anyway, the formula for wind pressure on a flat surface is: P = .00256 * V2, where V is wind velocity in mph and P is pressure in pounds per square foot.
It's then straightforward to apply your known car's side area for an approximation of the total force you're receiving from the wind.
posted by Netzapper at 9:49 PM on October 29, 2009 [1 favorite]
Response by poster: Thanks NetZapper - that's pretty much what I was looking for.
posted by krisak at 9:53 PM on October 29, 2009
posted by krisak at 9:53 PM on October 29, 2009
Best answer: The drag equation requires velocity relative to the fluid. Your initial velocity in the lateral direction is 0, since you are only driving forwards. It will tell you how much extra force is required to drive forward, but for lateral we need something different.
Fortunately, this is all classical mechanics! Nothing too fancy here if we want to keep it simple. We need to know the force of the wind against the side of your car. I don't know anything about this sort of thing, but I'll give it a shot. I'll also try to show my steps for anyone who wants to know how this is done/get some (possibly wrong) help for grade 12 physics class.
Surface area of the Miata: 4.86 m^2
Density of air: ~1.3 kg/m^3
Force of wind is tricky because we only know its velocity. This source gives me this: The force of wind hitting something is basically the amount of momentum used each second = Volume of air hitting the car per second x Density x Velocity. Interestingly, we know the surface area of the air that is hitting your car, but the wind speed will actually give us the depth of air hitting that area per second, giving us the volume.
Volume of air hitting car per second: 4.86 m^2 x 22.2 m/s (80 kph in m/s) x 1 s = 107.9 m^3
Mass of air hitting car per second: V x p (rho, density) = 107.9 m^3 x 1.3 kg/m^3 = 140.3 kg
Momentum of air hitting car per second: m x V = 140.3 kg x 22.2 m/s = 3114.66 N (m kg / s^2)
This momentum is also the force of the air on the side of your car. For those keeping track of units, it is per second squared because of the "per second" way we've been calculating everything.
Ways this wind could affect you:
Firstly, it could cause your car to slip sideways on the asphalt. It would do this by overcoming the frictional force of your car. For this, we'll use frictional constants.
Acceleration due to gravity: 9.8 m/s^2
Mass of Miata: ~990 kg
Normal force of miata (force of the ground opposing its gravitational force): g x m = 9702 N
Coefficient of static friction between rubber and concrete: ~1
Frictional force = μ x Fn = 9702 N. Much more than the force of wind.
I don't really know why I did this part. I guess it just proves why your car does not go flying in the parking lot on a windy day?
The real implication is that this is the force on the side of your car. If you are changing lanes at 135 kph and changing lanes at an angle of, say, 10 degrees, we can use vector units to determine how much this will slow you down. We'll also assume that the surface area of your car does not change while doing this.
Angle between the heading of your car and being perpendicular to the road: 80 degrees.
Velocity going forward/10 degrees sideways: 37.5 m/s
Horizontal velocity = 37.5 m/s x cos(80) = 6.51 m/s
That is our desired velocity. To reach it, you must overcome the force on the side of your vehicle. This is why it was so difficult for you to change lanes.
This is a really long and contrived answer, and I hope it was helpful. I need to sleep, but maybe someone can pick up where I left off! Or tell me why I'm wrong.
posted by battlebison at 9:56 PM on October 29, 2009 [1 favorite]
Fortunately, this is all classical mechanics! Nothing too fancy here if we want to keep it simple. We need to know the force of the wind against the side of your car. I don't know anything about this sort of thing, but I'll give it a shot. I'll also try to show my steps for anyone who wants to know how this is done/get some (possibly wrong) help for grade 12 physics class.
Surface area of the Miata: 4.86 m^2
Density of air: ~1.3 kg/m^3
Force of wind is tricky because we only know its velocity. This source gives me this: The force of wind hitting something is basically the amount of momentum used each second = Volume of air hitting the car per second x Density x Velocity. Interestingly, we know the surface area of the air that is hitting your car, but the wind speed will actually give us the depth of air hitting that area per second, giving us the volume.
Volume of air hitting car per second: 4.86 m^2 x 22.2 m/s (80 kph in m/s) x 1 s = 107.9 m^3
Mass of air hitting car per second: V x p (rho, density) = 107.9 m^3 x 1.3 kg/m^3 = 140.3 kg
Momentum of air hitting car per second: m x V = 140.3 kg x 22.2 m/s = 3114.66 N (m kg / s^2)
This momentum is also the force of the air on the side of your car. For those keeping track of units, it is per second squared because of the "per second" way we've been calculating everything.
Ways this wind could affect you:
Firstly, it could cause your car to slip sideways on the asphalt. It would do this by overcoming the frictional force of your car. For this, we'll use frictional constants.
Acceleration due to gravity: 9.8 m/s^2
Mass of Miata: ~990 kg
Normal force of miata (force of the ground opposing its gravitational force): g x m = 9702 N
Coefficient of static friction between rubber and concrete: ~1
Frictional force = μ x Fn = 9702 N. Much more than the force of wind.
I don't really know why I did this part. I guess it just proves why your car does not go flying in the parking lot on a windy day?
The real implication is that this is the force on the side of your car. If you are changing lanes at 135 kph and changing lanes at an angle of, say, 10 degrees, we can use vector units to determine how much this will slow you down. We'll also assume that the surface area of your car does not change while doing this.
Angle between the heading of your car and being perpendicular to the road: 80 degrees.
Velocity going forward/10 degrees sideways: 37.5 m/s
Horizontal velocity = 37.5 m/s x cos(80) = 6.51 m/s
That is our desired velocity. To reach it, you must overcome the force on the side of your vehicle. This is why it was so difficult for you to change lanes.
This is a really long and contrived answer, and I hope it was helpful. I need to sleep, but maybe someone can pick up where I left off! Or tell me why I'm wrong.
posted by battlebison at 9:56 PM on October 29, 2009 [1 favorite]
Response by poster: Wow, thanks battle bison - that's exactly what I was looking for - the math behind why my car moved the way it did. I appreciate the explanation, and on Monday will have a fun discussion with my cube mate (he's a physics major).
posted by krisak at 10:12 PM on October 29, 2009
posted by krisak at 10:12 PM on October 29, 2009
One note - if you are going very fast, it would be easier for your car to enter a skidding state (by slamming on the brakes, turning too fast etc), in which case the effective coefficient of friction of the tires would be lower. So in a skid, wind would have a greater effect on the car.
posted by Earl the Polliwog at 10:40 PM on October 29, 2009
posted by Earl the Polliwog at 10:40 PM on October 29, 2009
Response by poster: Earl - interesting, but I wouldn't have entered a skid state. (Well, I hope not.)
But you do make a good point about the wind having a greater affect on the car in a skid state at 80 than at 20.
posted by krisak at 10:50 PM on October 29, 2009
But you do make a good point about the wind having a greater affect on the car in a skid state at 80 than at 20.
posted by krisak at 10:50 PM on October 29, 2009
1. OK, BB, here's why you're wrong. :) Your use of classical mechanics assumes a complete momentum exchange. That does not happen. Aerodynamic Drag is the way to go. Think about it. Still air, frozen lake, car moving sideways at 50 mph. Aerodynamic Drag Equation.
2. I thought about it some more. You know, your forward speed MIGHT ACTUALLY MATTER! Contrary to the basic physics common sense. (This is why common real world problems are solved by engineers, not physicists, and engineers have to be licensed. Unlike physicists.)
3. Here's why: when the relative wind (sum of side+fwd) is hitting your car at an ANGLE, you may be generating HORIZONTAL LIFT. Due to your, ahem, extreme forward speed, you may have achieved a high rate of laminar airflow diagonally across the Miata. I.e., it was a wing rolled 90 degrees and flying sideways. Maybe!
posted by coffeefilter at 1:44 AM on October 30, 2009 [1 favorite]
2. I thought about it some more. You know, your forward speed MIGHT ACTUALLY MATTER! Contrary to the basic physics common sense. (This is why common real world problems are solved by engineers, not physicists, and engineers have to be licensed. Unlike physicists.)
3. Here's why: when the relative wind (sum of side+fwd) is hitting your car at an ANGLE, you may be generating HORIZONTAL LIFT. Due to your, ahem, extreme forward speed, you may have achieved a high rate of laminar airflow diagonally across the Miata. I.e., it was a wing rolled 90 degrees and flying sideways. Maybe!
posted by coffeefilter at 1:44 AM on October 30, 2009 [1 favorite]
Ugh. All the "speed doesn't matter" answers are assuming a perfectly 90 degree crosswind, which is a ridiculous assumption. Without that assumption, the cross-section and shape of your car may be totally different depending on angle, and the sideways force versus backwards/forwards will be totally different as well. This is like the old physics joke that starts out with "assume a cow is a sphere..." Sure it lets you get an answer instead of saying "it depends", but not a useful one.
Bicycling magazine interviewed the guy in charge of Zipp wheels, and he talked about exactly this problem: tuning his wheels so that they performed the best under real-world crosswinds. If I remember correctly, they found that about 30 degrees was the most common angle.
posted by Dr.Enormous at 2:28 AM on October 30, 2009
Bicycling magazine interviewed the guy in charge of Zipp wheels, and he talked about exactly this problem: tuning his wheels so that they performed the best under real-world crosswinds. If I remember correctly, they found that about 30 degrees was the most common angle.
posted by Dr.Enormous at 2:28 AM on October 30, 2009
Response by poster: Enormous, this question is really based on a lot of assumptions. And I understand we're not going to be able to get a real world answer, I'm mostly looking to understand the mechanics of the situation.
posted by krisak at 8:11 AM on October 30, 2009
posted by krisak at 8:11 AM on October 30, 2009
Those who are talking about assumptions: you're absolutely right. A real car would not behave the same as our hypothetical one. But there is no way we can account for lift without some very specific measurements. There is no way we will know the exact surface area of the car without wasting all of krisak's time.
For physics questions like these, if we were an insurance company trying to figure out why exactly krisak crashed his car on that day, we would do this work. We would likely run computer simulations as doing all of this by hand is far too tedious. There are even things that we can account for, like rolling resistance, but don't because our life is easier that way.
Anyways, drag equation: I'm assuming that drag constant exists because not all of the air that hits the car is converted to momentum. As we all know, air can compress and be redirected. That would change the number I provided in my solution to something a little more realistic.
One thing I've noticed from this thread and my searching to find an answer is how much stuff I found that used the imperial system! Pounds per square foot and acceleration in terms of miles per hour and the like... is that standard for the United States? When you take introductory high school physics, do you record velocities in feet per second and force in pound-force?
posted by battlebison at 8:39 AM on October 30, 2009
For physics questions like these, if we were an insurance company trying to figure out why exactly krisak crashed his car on that day, we would do this work. We would likely run computer simulations as doing all of this by hand is far too tedious. There are even things that we can account for, like rolling resistance, but don't because our life is easier that way.
Anyways, drag equation: I'm assuming that drag constant exists because not all of the air that hits the car is converted to momentum. As we all know, air can compress and be redirected. That would change the number I provided in my solution to something a little more realistic.
One thing I've noticed from this thread and my searching to find an answer is how much stuff I found that used the imperial system! Pounds per square foot and acceleration in terms of miles per hour and the like... is that standard for the United States? When you take introductory high school physics, do you record velocities in feet per second and force in pound-force?
posted by battlebison at 8:39 AM on October 30, 2009
One thing I've noticed from this thread and my searching to find an answer is how much stuff I found that used the imperial system! Pounds per square foot and acceleration in terms of miles per hour and the like... is that standard for the United States? When you take introductory high school physics, do you record velocities in feet per second and force in pound-force?
The physics is independent of the units used. You get the same answers either way. Calculators and Google make it trivially easy to convert from one to the other. Unless you are talking exclusively to scientists it is preferred to use the units most familiar to the audience. In the U.S., for better or worse, that continues to be imperial units. In a science class students would be more likely to use the metric system, but sometimes they use both.
posted by JackFlash at 9:03 AM on October 30, 2009
The physics is independent of the units used. You get the same answers either way. Calculators and Google make it trivially easy to convert from one to the other. Unless you are talking exclusively to scientists it is preferred to use the units most familiar to the audience. In the U.S., for better or worse, that continues to be imperial units. In a science class students would be more likely to use the metric system, but sometimes they use both.
posted by JackFlash at 9:03 AM on October 30, 2009
As some wag said, every problem has a beautifully simple answer that's wrong.
This is not a simple problem. First, ignore the car's shape. (Or, as we used to say in school, assume a spherical car.) It's travelling along at speed v1, and there's a cross-wind v2. Now these are both vectors. To find the force on the car, you need to find the relative wind, which is the "vector sum" v2 - v1. (If the car is moving with velocity v1, the wind it feels is -v1.)
The vector v2 - v1 does not point in the "backwards" direction -- in other words, if you were in the car the wind would seem to coming at you from one side, not from directly in front.
Now let's go back to the car. What we have is an irregular shape with wind hitting it at an angle. There is no simple formula -- or any formula -- for the force exerted by the wind in this situation. In fact the force won't even be in line with the relative wind (the component at right angles to the wind is called lift). That's what makes crosswinds hard to drive in, small variations and gusts make the relative wind direction keep changing, so the side force on your car keeps changing.
You can do some back-of-the-envelope calculations to estimate the force involved. Fluid dynamic forces can usually be represented as 1/2 * (air density) * (speed)^2 * C * S, where C is the drag coefficient and S is the characteristic dimension (cross-sectional area) of the body. Drag coefficients for objects like cars tend to be around 0.5.
posted by phliar at 4:54 PM on October 30, 2009
This is not a simple problem. First, ignore the car's shape. (Or, as we used to say in school, assume a spherical car.) It's travelling along at speed v1, and there's a cross-wind v2. Now these are both vectors. To find the force on the car, you need to find the relative wind, which is the "vector sum" v2 - v1. (If the car is moving with velocity v1, the wind it feels is -v1.)
The vector v2 - v1 does not point in the "backwards" direction -- in other words, if you were in the car the wind would seem to coming at you from one side, not from directly in front.
Now let's go back to the car. What we have is an irregular shape with wind hitting it at an angle. There is no simple formula -- or any formula -- for the force exerted by the wind in this situation. In fact the force won't even be in line with the relative wind (the component at right angles to the wind is called lift). That's what makes crosswinds hard to drive in, small variations and gusts make the relative wind direction keep changing, so the side force on your car keeps changing.
You can do some back-of-the-envelope calculations to estimate the force involved. Fluid dynamic forces can usually be represented as 1/2 * (air density) * (speed)^2 * C * S, where C is the drag coefficient and S is the characteristic dimension (cross-sectional area) of the body. Drag coefficients for objects like cars tend to be around 0.5.
posted by phliar at 4:54 PM on October 30, 2009
I have a couple questions to all those who criticised the OP's initial assumption that the car speed has an influence on the force exerted by crosswinds:
a. car aerodynamics: isn't it that the faster the car goes, the higher the aerodynamic forces pushing it on the ground?
b. I think the angular momentum of 4 wheels spinning at a speed such as the car travels at 85mph should also be taken into account. Don't they act more or less like a gyroscope?
(these two effects might very well be absolutely negligible, IANAPhysicist, and most of the Physics I took was quickly forgotten)
in short: the force exerted on the car by the wind is of course the same, on the other hand couldn't the perceived "sway" actually be different between, say, 8 and 80mph?
posted by _dario at 5:59 PM on October 30, 2009
a. car aerodynamics: isn't it that the faster the car goes, the higher the aerodynamic forces pushing it on the ground?
b. I think the angular momentum of 4 wheels spinning at a speed such as the car travels at 85mph should also be taken into account. Don't they act more or less like a gyroscope?
(these two effects might very well be absolutely negligible, IANAPhysicist, and most of the Physics I took was quickly forgotten)
in short: the force exerted on the car by the wind is of course the same, on the other hand couldn't the perceived "sway" actually be different between, say, 8 and 80mph?
posted by _dario at 5:59 PM on October 30, 2009
This thread is closed to new comments.
I think your intuition is wrong. To a first order approximation, the force exerted by the crosswind is the same, whether you are traveling 5 mph or 50 mph.
When you drive with a crosswind, especially a gusty wind, you must continually adjust your steering to compensate for the sidewards force. This is a constant trial and error feedback in order to keep the vehicle moving straight down the road. The difference is that any error in your unconscious correction at 50 mph is much more immediate than at 5 mph. At 5 mph you have plenty of time to adjust. At 50 mph you might be in the ditch.
posted by JackFlash at 8:39 PM on October 29, 2009