# How much clearance does a car need when turning a corner?

I am contemplating buying a new car. However, the approach to the underground garage in my apartment has a 90 degree frustrating turn. Given the dimensions of the approach and car, what is the max turn circle for the car to fit the garage and turn?

given Ackerman steering and the overhanging front part of the car I believe you can use Pythagoreas' theorem to get R min and R max. delta R should be less than the shortest path in the pathway, ie 2.5m. unfortunately the result does not seem plausible. feedback would be greatly appreciated.

• Do you know the maximum wheel deflection? That's kinda important for this. – ratchet freak Jul 11 '15 at 16:39
• But if you have the max wheel deflection then the turn circle would also be given? What I'm looking for is the max turn circle that still would leave the car without scratches. – Misha Jul 11 '15 at 16:44
• What's the car's width? The "table" has it as 2120mm, but the drawing has it as 2200mm. – Wasabi Jul 11 '15 at 19:28
• For that matter, can you write down all the longitudinal dimensions? I can't read them. As I read them, the length is 5030mm, the distance between axes is 2900mm, the rear distance is 1248mm and the front distance would have to be 882mm, but I'm pretty sure that's not what's written down. What have I misread? – Wasabi Jul 11 '15 at 20:15
• Although I agree @EnergyNumbers' arguments, on my opinion these arguments extended with a small explanation, how the turning circle can be calcuated (formulas), could serve as a good quality answer. So I voted for leave open. – peterh - Reinstate Monica Jul 12 '15 at 7:22

To slightly generalize I'll reform the question slightly.

A ridged 2-D body (car) has a line $l$ that moves with it. The car can be linearly transformed as long as the instantaneous center of rotation lies along $l$ at least distance $R$ away from a point $c$ that also moves with the car.

In this case point $c$ lies in the center of the rear axle and $l$ lies on the rear axle.

Now imagine the car's domain is limited to a quarter plane with edges $A$ and $B$. It initially is placed against $A$, far from $B$ with $l$ perpendicular to $A$, and the goal is to translate the car so that it is against $B$ far from $A$ while minimizing the maximum distance from the closest edge.

($A$ and $B$ can be placed a inch away from the actual walls to prevent scratches and allow for non-idealized vehicle movement.)

## Reversals allowed

The solution is to advance the car along $A$ until it is an infinitesimal distance from $B$ (using an infinite turning radius to travel in a straight line) Then rotate about the tightest turning radius until in contact with $B$ Then rotate about the tightest turning radius on the opposite side until back in contact with $A$. This results in linear movement in the opposite direction but rotation in the same direction. These two steps can be repeated (infinitely) until $l$ is perpendicular to $B$ at which point it can advance away from $A$ in a straight line. From a macro perspective this looks like the car sliding along $A$ until it reaches $B$, then rotating while maintaining contact with both walls and finally advancing along $B$. This solution is independent of turning radius but involved infinite reversals.

## No reversals

Now lets further constrain our translations so that center of rotation must be further from $A$ and $B$ than $c$. (This removes the usefulness of backing up) Now the middle of the optimum strategy is obvious: turn at the maximum turning radius, but how do you minimize the distance to the wall approaching and exiting this strategy?

You remain in contact with the wall.

As you approach the wall and see you're just about to clear it, rather than continuing to turn you can gradually increase the turning radius to remain in contact with the wall. To remain in contact with the wall means the line between the contact point and the center of rotation is perpendicular to the wall.

From this we can get the position of the center of rotation while in the minimum turning radius portion of the turn.

$$D_{rear}=\sqrt{{O_{rear}}^2+(R_{min}+W)^2}$$ $$D_{front}=\sqrt{(O_{front}+WB)^2+(R_{min}+W)^2}$$

This point fully defines the most interesting portion of the turn allowing one to see if any obstacle on the other side would be struck. To clear:

$$\sqrt{(D_{rear}-b)^2+(D_{front}-a)^2} \leq R_{min}$$

Note that it makes a difference if you're going forward or backward. To see if you'd clear both directions you'd have to test with a and b reversed.

Indeed in the diagram above I've set $a=5.9m$ and $b=3.3m$. In this case it was because while the thick arc defined by drawing and equations above might be the most interesting portion of the curve, it is not going to be the limiting factor when $a$ and $b$ are not flipped. So we need to extend that curve.

The end points can be found using similar triangles, from there, the curve will just be an tangent exponential decay to a distance $W$ from the wall.

With these curves we can define a function $C$ to tell us whether the vehicle would clear an object placed at $(a,b)$:

$$C(a,b) = \begin{cases} \hfill \sqrt{(D_{rear}-b)^2+(D_{front}-a)^2} \leq R_{min} \hfill & \text{ if } a \leq a_{check} \text{ and } b \leq b_{check} \\ \hfill W+W_{rear}e^{\frac{(a_{check}-a)O_{rear}}{(R_{min}+W)W_{rear}}} \leq b \hfill & \text{ if } a > a_{check} \text{ and } b \leq b_{check} \\ \hfill W+W_{front}e^{\frac{(b_{check}-b)(O_{front}+WB)}{(R_{min}+W)W_{front}}} \leq a \hfill & \text{ if } a \leq a_{check} \text{ and } b > b_{check} \\ \hfill true \hfill & \text{ if } a > a_{check} \text{ and } b > b_{check} \\ \end{cases}$$

Where:

$$a_{check}=D_{front}-O_{rear}\frac{R_{min}}{D_{rear}}$$ $$b_{check}=D_{rear}-(O_{front}+WB)\frac{R_{min}}{D_{front}}$$ $$W_{front}=D_{front}-(R_{min}+W)\frac{R_{min}}{D_{rear}}-W$$ $$W_{rear}=D_{rear}-(R_{min}+W)\frac{R_{min}}{D_{front}}-W$$

Now to solve this system backwards to get the maximum $R_{min}$ that would allow passage requires making a few observations and assumptions. First we'll assume that you want to be able to drive around the corner in ether direction that means we'll swap $a$ and $b$ for whichever scenario is worse. If The front corner is further from the fixed axle than the rear corner (as is the case for all front steering vehicles I know of) then a < b is the tighter scenario.

Then one could use a numerical method to find the $R_{min}$ that gave equality for the second inequality. If $a\geq a_{check}$ then you're done. If not, then find the $R_{min}$ that gives equality for the first inequality.

## Glossary

• $W$ - Width of car
• $WB$ - Wheel Base
• $O_{front/rear}$ - Overhang of the front/rear
• $R_{min}$ - minimum distance between the center of rotation and the car
• $a$ - distance from outside wall to inside corner
• $b$ - distance from outside wall to inside corner

## Plugging in

With the numbers given It turns out that the maximum $R_{min}$ is slightly under $6.6m$

But you might have to fold the right mirror in.

• WOW- that is one elaborate answer. However, I cannot get the meaning of "The car can be linearly transformed as long as the instantaneous center of rotation lies along l at least distance R away from a point c that also moves with the car." furthermore, quarter plane - what is it? Finally, how did you arrive at the final equation? NB - I had a second look in the garage - this time with a measure. Turns out a- 3.3m, b=5.2m. – Misha Aug 2 '15 at 21:05
• The first quote describes the motion that ackerman steering allows in a rigorous manner. Basically, for every steering wheel position, the car would move in a circle about some center of rotation. That center of rotation always is in line with the rear axle, and the radius of that circle is no smaller than some certain distance.  A quarter plane is a 2D space that is bounded by two lines at right angles. A quadrant of a graph is an example of a quarter plane.  Diagrams to help explain are forthcoming.  I'll update with new numbers. – Rick Aug 3 '15 at 12:53
• Impressive - Most carmakers supply their info sheet with curb to curb turning diameter. Hence, I believe I add the width of the car to the minimum radius and multiply by 2. (1.67m (w) + 6.6)*2=16.5 m curb to curb turning radius (ie diameter). en.wikipedia.org/wiki/Turning_radius – Misha Aug 4 '15 at 21:47
• Now that was 2D and one obstacle - for those moving furniture up and down revolving staircases, narrow doorframes and halls- The even harder 3D version - How can you determine if the object will fit or not and also how to determine the optimal angulation of the object? – Misha Aug 4 '15 at 21:50
• @Misha That's actually a current research topic in computing (one that I studied in grad school at Berkeley). So while it's a very interesting topic, it's too broad to discuss here in detail here. One method I find interesting is to create a 6 dimensional space (three directional three rotational), project the obstacles through the space, then offset the surfaces according to the projected width of the object in the orientation corresponding to the rotational coordinate. Then any path that does not intersect this 6 dimensional geometry will work for moving the object through the obstacles. – Rick Aug 5 '15 at 11:47

Why not just take the car for a test drive and see if it can make the turn?

• This does not provide an answer to the question. To critique or request clarification from an author, leave a comment below their post. - From Review – Wasabi Jul 8 '17 at 0:17
• @Wasabi - I won't argue, as it doesn't explicitly answer the question as asked. But I do believe this answer is better than the accepted answer based on the wording of the question. If the question was about designing a turn in a new parking garage, or designing cars to accomodate tight garages then the accepted answer is far better than this one. But for a practical answer to a specific question by someone who wants to buy a car that will be able to make the turn in the garage, I believe the best engineering solution is to simply try it. Easy solution and guaranteed result. – Mark Jul 8 '17 at 1:44

On average allow a circle with a diameter of 13m (Radius 6.5m) for a carriage driveway.

• Please edit your answer with additional information such as explanations or sources for where you got this number. – Wasabi Jul 7 '17 at 17:40

Something important to consider is that if the corridor taking you down to the underground is narrower then the width of the way turning in, then there are certain sizes of cars that are able to go in but are unable to go out from the underground garage. So these cars can only go out in reverse.