The Trials and Tribulations of Tribology
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How's This Good For Games?
An obvious application of the Coulomb dry friction model is for travel over surfaces. You may have a game that requires a character to travel over various types of terrain. By specifying different coefficients of friction for different types of terrain (asphalt, grass, ice, and so on), you can simulate movement over this terrain in a realistic, and even more importantly, a physically consistent manner.
Many games simulate friction simply by reducing the velocity by a percentage based on the surface type. This may seem at first to be the same thing as the dry friction model described above. However, it differs from it in many critical ways. By adjusting the velocity directly, you eliminate the side effects of applying the friction as a force. These side effects are what make objects in the physical simulation behave the way players expect them to behave. These small breakdowns in the simulation make it glaringly apparent that the world is fake. Perhaps an example would help here.
The Adventures of Sara Craft
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Figure 3. You can
control how much force Sara must exert on the box before it
moves |
Say I'm creating an adventure game starring a beautiful woman named Sara running around a dangerous, mystical temple in a stunning cocktail dress. To escape from the temple, Sara must manipulate a series of wooden boxes to activate various switches embedded in the floor. (Don't blame me, my producer came up with the concept.)
Sara pushes the boxes around by applying a horizontal force to the objects. If I do not consider friction at all, then once the boxes are sliding they will slide all around the room, bouncing off the walls forever. Clearly something needs to be done. So, I simply reduce the velocity of the object as it slides around. This can be made to look pretty good. However, there is still a problem.
If you have ever pushed a box really hard, particularly if your point of contact is near the top of the box, the box will sometimes tip over before it starts sliding. In fact, if you throw a box across the room, once it hits the floor it will tumble all over the place instead of simply sliding to a halt. People are used to these facts. They live with them every day. If your world does not address these behaviors, it will not feel right.
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Figure 2. Forces exerted on a box
as it verges on tipping over. |
But why does the box tip over? Well, guess what, it is all about friction. Take a look at the box in Figure 2. Sara will be applying a force, F, to the box h units above the ground. What I'm looking for is a state for the system where the box is about to tip over at point A. I can apply the principles of statics to solve this problem. (If you are not familiar with statics, check out the For Futher Info section at the end of this column.) For an object to be in static equilibrium, the sum of all forces and the sum of all moments in the body must equal zero.
When the box is about to tip over, there is only a reaction to the ground at point A. The support on the other side has no reaction to the ground. Therefore, we can state the equilibrium equations. Let me start with the sum of forces.
The sum of horizontal forces consists only of F and f, and they directly oppose each other. In the vertical direction, the weight W and normal force N are also equal and opposite. The sum of moments however, is a bit more complicated. You may remember from physics that the moment of a force about a point P is
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Figure 4. Sara tips
the box over instead of sliding it
away. |
where D is the perpendicular distance from the point P to the line of action of the force F. Forces are sliding vectors, meaning that they act equally along their entire line of action. Let's look back at the drawing in Figure 2. When the object is about to tip over, it makes sense to look at the sum of moments about the point A. There are two moments being applied about point A. The force Sara is applying, F, and the force of the weight of the object, W.
At the point of equilibrium where the box is about to slip,
So, I can substitute leaving
If Sara applies the force at a point (0.5d)/ µs units high or higher on the box, the box is going to tip over before it starts sliding. What's even more interesting is the fact that the equation above states that the value for h is not dependent on anything other than the dimensions of the box and the coefficient of static friction. The magnitude of the force F does not matter at all. It may seem that if Sara pushes harder, the box would be more likely to tip. Statics proves that this is not the case.
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