I hate to admit it, but I also had parking issues. Especially when you live in the city, searching for a place for your car can be a torture, because spaces often seem too small for your car. But is it really so? How much space does it have to be in a row of cars to fit in your car? Mathematics provides an answer to this question: as long as the space is slightly longer than your car, you can maneuver your way into it by skilfully driving back and forth. However, in some circumstances, this can take a lot of effort and a lot of sweat.
I’m rarely lucky enough to get the car into a small gap the first time and have to go back and forth a few times until I get it straight enough. I prefer to park elegantly in one go: put the car in reverse, pull it into the cleared area with a sharp turn, and then move forward slightly, so that the car is centered in the intended space. In this case, the parking lot should be a little longer than our car. But how much, for such a maneuver? Surprisingly, the minimum distance does not depend on the width of the car, although everyone knows the feeling of not being able to park a car that is too wide.
Park your car in the coolest way possible
The problem of finding the smallest possible parking space for a perfect maneuver can be solved with a bit of simple engineering. The best way to do this is to start at the end point and look back – let’s say you’ve parked right between two cars and want to get out of the gap. To do this, first put it in reverse and drive in a straight line to the car behind you. Then engage first gear, turn the steering wheel hard to the left and exit the space. Ideally, you’d barely clean the car in front of you.
This presents us with an engineering problem: we want to calculate the length of the pink segment c from fig. To do this, you need to know the turning radius of the car (here it is indicated by a dotted line). radius s Corresponds to the distance between the center point M of the rim and the right front tire E. So does the wheelbase The The (distance between the front and rear axle) of our car plays an important role: the Pythagorean theorem can be used to calculate the distance between the center M of the circle described by the maximum steering angle and the right rear wheel F: MF = √ (s 2 – The 2).
The distance (green part) between center point M and the right front end of car A can now be calculated, again using the Pythagorean theorem. To do this, you need to know the front overhang K (the distance between the front wheel and the front end of the body) of the vehicle. So we get: MA = √ (MF2 + (K + The)2) = √ ( s 2 – The 2 + (K + The)2 ). This length corresponds to the radius of the solid circle, and thus also to the distance between the outer rear corner G of the vehicle ahead and the central point M.
We are now only two steps away from finding the length c that we are looking for. First you need to draw a straight line parallel to the dock from G to MF. We denote by K the point of intersection of the two lines.
Since the car in front of us is not necessarily as wide as ours, the K does not necessarily have to match the rear left wheel. If the segment length GK is calculated now, the minimum possible distance can be calculated c Using all the quantities we already know. GK = √ (MG2 – Member of the Knesset2 ) = √ (MA2 – Member of the Knesset2). Now MK is simply the length of the MF segment minus the width B The car ahead: MK = MF – B. If we now use all the quantities already known, we get the following formula: GK = √ (s 2 – The 2 + ( K + The )2– (√ ( s 2 – The 2) – B)2). If now they pose The And the Kyou get the length you need c:
We now have a working formula for finding the smallest possible parking space to park one car at a time. All you need are some characteristics of your vehicle and the width of the vehicle you want to park behind.
For example, if I wanted to park my first car, a 1990 Volkswagen Golf Series II, behind another Golf of the same type, I would need the following information: The turning circle is about 10.5 meters, so s = 5.25 metres. move The The front overhang is about 2.5m long K About 0.88 meters long, width B About 1.68 meters and the total length of the car is about 4 meters.
Substituting these numbers into the formula, we get c = 1.53m: The space should be 1.5m wider than your car. The distance between two cars must be at least 5.53 meters for the second golf series to fit into the parking space. On the other hand, if you want to stand behind the 2.2m wide Hummer H1, you need about 30cm more space.
How does it fit into such a small space?
But some people like a challenge: They don’t want to park neatly in one go, but they do want to squeeze into the smallest parking spot possible. Mathematics offers a solution for that, too. One can formally prove that one can park a car (at least in theory) anywhere, no matter how small it is, as long as it is longer than the car – no matter how long it is: it can only be 30 centimeters or nanometers. The only prerequisite is to be able to drive a car with extreme precision.
How do you get to a standing spot of a certain length with the fewest maneuvers possible? To solve this problem, you can first refer to the above result.
In fact, at the first stage, you can proceed in the same way: if possible, pass the car in front of you in order to get as far as possible into the parking lot. To do this, the formula obtained above can be used, but instead of calculating the volume of the area cwe can now solve an equation B. In this case, B It no longer corresponds to the width of the car in front, but to the maximum offset that can be obtained by maneuvering to the right.
Once you’re done, you’ll find yourself in the parking lot, albeit not in an ideal position: you’re lined up parallel to other cars, but still leaning slightly to the left. This means that the position of the car needs to be corrected. To do this, turn the steering wheel to the right, travel in an arc to the middle of the space, and then turn left to stop straight again. [si uppone in questo caso che l’angolo dela sterzata verso destra si esattamente uguale a quello della successiva sterzata a sinistra, NdT]. Now the question is: what is the distance w Does the right cover up with these maneuvers?
To calculate it, you follow the center point of the car’s rear axle: the radius s From the circle traveling with the steering can be calculated starting from the turning radius [sempre indicato con r, NdT]: s = √ ( s 2 – The 2) – number/ 2, where n is the rear track, that is, the distance between the two rear wheels [il raggio percorso dal punto centrale dell’asse posteriore è pari a raggio percorso dalla ruota esterna diminuita di metà della carreggiata posteriore, NdT].
Referring to the figure below, note s and distance cheight w It can be calculated using the Pythagorean theorem w ⁄ 2 = r’ – √ ( s 2 – c 2/ 4).
Again, we can use specific numbers to get an idea of the size of car parks. If I want to park my second series Volkswagen Golf (track number = 1.44 m, which results in s = 3.9m) in a space only 40cm longer than my car, the time required can be very long: in the first maneuver I can move a maximum of 0.32m to the right. But inside the space, I have less freedom of movement: according to the previous formula for w, I can move a maximum of one centimeter to the right for each maneuver. So if I park another Volkswagen Golf 1.7 meters wide in front of me, I have to go back and forth 138 times!
With a slightly larger parking space, 80 cm longer than my car, the situation is even better: in the first maneuver I get a maximum of 71 cm. Each successive correction brings about 4 centimeters: in this case, “only” 25 maneuvers are required.
I guess I’d rather pay attention to the larger parking spaces I can fit the train in neatly.
(The original of this article was Published in Spektrum der Wisseschaft on November 25, 2022. Translation and editing by Le Scienze. Reproduction is authorized, all rights reserved.)
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