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I was wondering how many kilometers/meters I would save if I always drove on the inside edge of the curves between my home and workplace, compared to driving in the middle of the lane.

If the lanes are approximately $3 m$ wide and I estimate that over $100 km$ I drive an average of:

  • 60x 90-degree curves with a length of $10 m$.
  • 20x traffic circles with a driving length of $45 m$ (half circle).
  • 30x 45-degree curves at $1000 m$.
  • 30x 30-degree curves at $250 m$.

How many meters would I save on a distance of $100 km$?

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2 Answers 2

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This is a cute problem because the answer turns out not to depend on how far you drive, only on how much you turn.

Suppose the road follows a curve whose turning radius in the middle of the lane is $r$ meters, and it turns for a total of $\theta$ degrees. If you drove in the middle of the lane, you would have travelled a distance $2\pi r\cdot\theta/360$ meters, because the whole circle is $2\pi r$ meters long but you've only covered a fraction $\theta/360$ of it. On the inside of the lane, you travel $2\pi(r-1.5)\theta/360$ meters, assuming the distance from the middle of the lane to the inside is $1.5$ meters. The difference is $2\pi\cdot1.5\cdot\theta/360 = \pi\theta/120$ meters, independent of $r$.

(This is the same as the distance you would travel if you went $\theta$ degrees around a circle of radius $1.5$ meters. This fact may or may not seem obvious once you think about it more.)

Now that you have multiple curves in the road, your shortest path depends on how they're arranged, but the simplest assumption is that there are enough straight stretches in the road between curves that you can position yourself on the inner side of each turn without significantly affecting your distance travelled. In that case, all we have to do is add up the sixty $90^\circ$ curves, twenty $180^\circ$ turns, thirty $45^\circ$ curves, and thirty $30^\circ$ curves, to get a total turning of $11250^\circ$. This corresponds to a saving of $11250\pi/120 \approx 294$ meters.

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  • $\begingroup$ Nice one - however, there is a light mistake in your calculation, you forgot a $pi$ along the road : $2\pi\cdot1.5\cdot\theta/360 = \pi\cdot\theta/120$ meters, which gives you in the end a saving of ~294,52 meters $\endgroup$
    – S3MP
    Commented Aug 21, 2012 at 11:46
  • $\begingroup$ @S3MP, thanks for the correction. $\endgroup$
    – user856
    Commented Aug 21, 2012 at 11:50
  • $\begingroup$ I'd just like to note that I wish I could pick 2 answers as the accepted answer. Both have good points and both answer the question equally well in my opinion. I chose this one, because of the interesting approach. $\endgroup$ Commented Aug 23, 2012 at 9:54
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If you admit that all your curves are derived from perfect circles, every curve length $c$ is related to the radius $r$ of the circle and the angle in degrees $\alpha$ by the formula: $$ c= r \cdot {{2\pi \cdot \alpha}\over{360}} $$ which gives you the radius of every circle as $$ r= c \cdot {{360}\over{2\pi \cdot \alpha}} $$

Applying this formula gives you 4 different radius: ~6.37m, ~14.32m, ~1273.24m and ~477.46m

Then two questions:

  • how wide is your car? 2 meters?
  • by how much would you reduce the radius of a circle if you drive on the inside edge?

It depends on which side of the road you are driving: suppose you're not British (ie you drive on the right side of the road).

  • If the curve bends to the right, you will not reduce the radius by more than 0.5 meter.
  • If the curve bends to the left, and you drive on the opposite lane to reduce the length of the curve (which I would not recommend, except if you are a cyclist during the Tour de France), you will reduce the radius of your circle by a maximum of 3.5 meters.

Assuming a 3.5 meters gain for the radius of each curve (ie all your curves are bending to the left, which is highly improbable but gives you the maximum you can gain), you would gain ~687.22m for the 140 curves you listed in your question. So less than 700 meters over a 100km drive...

Assuming half of the curve on the right, half on the left (equally distributed among your 4 classes), you would only gain ~392.70m.

You would also want to consider the extra distance you have to drive between two opposite curves (right then left or left then right) compared to a situation where you stay right in the middle of your lane... but that's an other story...

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  • $\begingroup$ I'd just like to note that I wish I could choose 2 answers as accepted, as both have good points and both answer the question equally well. $\endgroup$ Commented Aug 23, 2012 at 9:55
  • $\begingroup$ @MichaelFrey don't worry - I'm not offended at all and you're right: Rahul's approach is far better as it goes to the root of the problem: it's not about distance but about angle and I didn't cover it ;) $\endgroup$
    – S3MP
    Commented Aug 23, 2012 at 10:26

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