2022-04-09
但凡汽车转弯，就必定会磨损轮胎，尤其是多排轮的大货车，但具体为什么会产生大量摩擦？背后的原理是什么呢？
Only when all of the circle centers of the four wheels point to the same point, the car can move forward without stretching and friction. This is the fundamental rule of moving forward.
But for some truck which has multiple wheel rows but only one row of wheel for control direction, whose backward wheels can't control direction, must gonna suffer friction, and retire early than other cars.
You must have seen such tire marks at the crossing, especially on these roads which have more multi-row wheel trucks. That's not because of braking, it's because of the original structure flaw of these trucks.
Here I want to solve several basic but essential questions about car moving and driving.
- What's the turning radius of a specific car?
- What's the safe distance between the car and road border when encountering an alpha-degree crossing?
- What should drivers do for standard turning and crossing roads?
Above shows a card turning sketch map:
- L means the length between the front and back wheels of the car.
- L' means where the car will go in the next time frame.
- r1 means the turning radius of the front wheel.
- r2 means the turning radius of the back wheel.
- d means the safe distance between wheels and the road brick.
- there also has an angle between r1 and r2, let's say it's x.
As we can see,
- r1 = L / sin(x)
- r2 = L / tan(x)
- d = L * sin(l / L), where l is the length that the back wheel between the road's front edge.
For example, let's assume we want to bypass a 90-degree intersection, the wheelbase is 3 meters, the back wheel is L meters from the road's front edge, and we turn the degree of the front wheels to 45.
then we can evaluate the b from the above formula:
b = L * sin(l / L)
b = 3 * sin(3 / 3)
b = 3 = wheelbase
So when the front wheel stands on the road's front edge, if we want to turn 90 degrees, the safe distance is the wheelbase, here are 3 meters.
Another extreme condition is the back wheel stands on the edge, in that situation, the safe distance "d=L*sin(0/L)" equals 0, and there got no risk, just in case the opposite road edge, which requires the road at least wider than 1.414 * L.
In a moderate example, let's assume the road's edge is in the center of two wheels.
b = L * sin(l / L)
b = 3 * sin(1/2)
b = 2.12