A Mathematical Problem - 5 comments

Before we begin, this post is a little odd. Especially so for it being Friday night, when other bloggers will be twisting and gyrating as the music takes control of them, deep within subterranean clubs - or else reclining on DFS sofas, watching the tail end of Newsnight Review with sophisticated friends, knocking back white wine by the bucketload.

I'd class myself as a militant pedestrian, having no truck with cyclists, car drivers, or taxis. If the journey'll take less than half-an-hour, I'll walk. I also work in central London, where the slow movement of traffic and weight of numbers has given pedestrians a healthy contempt for the rule of the road: cross where you like, whenever you see a gap.

The rather sad point I'm getting to is that these factors allow a pedestrian to consider the optimum route for crossing a road, to minimise the time and distance travelled. Assume one wishes to cross from point P on the left-hand side, to Q on the right-hand side, thereafter continuing until the end of that road, R. For the cautious pedestrian with a lot of time on their hands, the usual plan is to wait for a gap in traffic at least large enough for several carnival floats to pass, cross from P to Q at an angle of 90 degrees to the pavement, turn one's body to the left, and proceed from Q to R. They spurn the chance to take up Pythagoras' advice and walk directly from P to R, along the hypotenuse. Assuming a combined road-and-kerb width of 2.5 metres, setting off at an angle of 85 degrees from the pavement, and continuing until you reach the other side, will let you complete the triangle in a mere 28.68 metres, 2.39 metres less than if you had walked along the other two sides of the triangle, a saving of 7.69% and perhaps 1.5 seconds of your life.

Was it worth doing? Spending all that time in the road just to ensure you reach your destination before your rival? Soon the inquiring mind turns to the question of the optimum crossing angle for reducing the distance walked. Isn't it obvious what it is? Put it this way, there aren't many possibilities. All the same, can you prove it mathematically?

All distances can be expressed in terms of the angle at which you leave the pavement (0° ≤ θ < 90°) and the width of the kerb-plus-road (W). I'll also refer to the proportion of journey saved by walking the smart, dangerous way, as Z. With a bit of fiddling about, you come up with:

Z = 1 - ( sec θ / ( 1 + tan θ)) ... or ...
Z = 1 - 1 / ( cos θ . ( 1 + tan θ))

The higher the better, clearly. You want to find the angle θ where Z is at a maximum - that is to say, where the rate-of-change is zero, and the rate-of-(rate-of-change) is negative (slope has had a positive gradient, now flattening). To do this, you need to differentiate the above function, to give you a new function that gives you the rate of change of the the proportion of journey saved, not the actual value. Then, set that equal to zero, and you should be able to rearrange to get a formula for the optimum θ itself.

Well, that's as far as I got - my ill-remembered Maths lessons, a couple of books, the Wikipedia, and even this have failed to give me something that works. It was addictive, but I had to give up at 10.45 - it, basically, being a waste of my time.

Of course I know what the optimum angle is: (a) it's pretty obvious when you think about it, and (b) you can graph the above function in Excel to get a good idea, then focus in. The question is also academic, given that the purpose of your journey is actually to reach your destination R, not be marooned on the kerbside partway along the road, albeit with a smug grin on your face, knowing you've made an efficiency saving of 29.89% on your reduced journey. All the same, it would have been nice to know that I'd managed it properly.

For the sake of regular readers who may be a little concerned, let me assure them that I don't intend to publish any future trigonometrical teasers for A-level pupils. I have plenty more politics to post about, and normal service will be resumed...

Update: fixed the Z formula - duh.