How can I calculate side force on my car?

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!

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 http://en.wikipedia.org/wiki/Drag_equation. Enjoy.

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: http://www.arraysolutions.com/Products/windloads.htm, 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.

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: http://www.google.ca/search?hl=en&safe=off&q=155.7+inches+x+48.4+inches&btnG=Search&meta=&aq=f&oq= Density of air: http://en.wikipedia.org/wiki/Density_of_air Force of wind is tricky because we only know its velocity. http://www.google.ca/url?sa=t&source=web&ct=res&cd=4&ved=0CBYQFjAD&url=http%3A%2F%2Fwww.greatrix.co.uk%2FWindForce.doc&ei=KWrqSpXpJMPVlAeTxoCABQ&usg=AFQjCNETjsSSVwDILPw7HIOPUZDSZ3HbeQ&sig2=wPD5ydttmbaiq8E0weKnsA 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 (http://www.google.ca/search?hl=en&safe=off&q=80+kph+in+m%2Fs&btnG=Search&meta=&aq=f&oq=) 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: http://en.wikipedia.org/wiki/Gravity#Earth.27s_gravity Mass of Miata: http://en.wikipedia.org/wiki/Mazda_MX-5 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: http://en.wikipedia.org/wiki/Friction#Coefficient_of_friction 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: http://www.google.ca/search?hl=en&safe=off&q=135+kph+in+m%2Fs&btnG=Search&meta=&aq=f&oq= 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.

Thanks NetZapper - that's pretty much what I was looking for.

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?

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.

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.

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 http://en.wikipedia.org/wiki/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?