Even if a hill or a road is too steep to climb, there is still a way to make progress, and that's by zig-zagging. Instead of going directly up the hill in the shortest route, it's possible to take an angled approach up the slope, increasing the path length, but making the climb angle less steep.
It is easier to outline this in a simplified diagram:
This triangular prism represents the face of a hill.
The angle directly up the hill is α and is shown in the pink triangle.
The angle of approach (i.e. the degree of zigzag, the deviation from the straight-up route) is ß, and is shown by the red and pink triangles combined.
The resultant angle (i.e. the actual angle of ascent) is δ and is shown by the blue triangle.
Each of the triangles is right-angled, so standard trigonometry functions can be applied (I haven't shown all the right angles in the diagram, but it is a regular triangular prism).
Considering each of these three angles in turn: the way to get to a simplified expression for δ is to express the three angles in the fewest numbers of lines. It's possible to express α, ß and δ in terms of the external dimensions of the prism (let's call them x, y and z) but this just leads to incompatible expressions that can't be simplified or combined.
α
ß
δ
The strategy here is to substitute for y and p in the expression for δ, and then to simplify.
Firstly, rearrange the expressions for α and ß to make y and p the subjects of those equations.
A very simple and elegant equation: the angle of ascent depends on how steep the hill is, and the amount by which you zigzag, and is completely independent of the size of the hill (i.e. none of the lengths are relevant in the calculation).
A few sanity checks:
If ß is zero, or close to zero, then δ approaches α - i.e. if you don't zigzag, then you approach the hill at its actual angle.
If ß approaches 90 degrees, then δ approaches zero - you hardly climb at all, but you'll need to travel much further to climb the hill. In fact, as ß tends towards 90 degrees, path length p tends to infinity.
If α increases, then δ increases for constant ß (something that was worth checking).
An interesting note:
At first glance, you may think that a path (or zigzag) angle of 45 degrees would reduce the angle of ascent by half (e.g. from 60 degrees to 30 degrees), simply because 45 is half of 90. However, this isn't the case. In order to get a reduction of a half, cos ß needs to equal 0.5. If cos ß = 0.5, then ß = 60 degrees. A much larger deviation from the straight-up angle is needed.
In conclusion
This question was first put to me when I was in high school (a few years ago now) and it's been nagging at me ever since. I'm pleased to have been able to solve it, and I'm pleased with how surprisingly simple the final expression is (previously, my 3-D geometry and logic weren't quite up to scratch, and I ended up going round in circles!).
It is easier to outline this in a simplified diagram:
This triangular prism represents the face of a hill.
The angle directly up the hill is α and is shown in the pink triangle.
The angle of approach (i.e. the degree of zigzag, the deviation from the straight-up route) is ß, and is shown by the red and pink triangles combined.
The resultant angle (i.e. the actual angle of ascent) is δ and is shown by the blue triangle.
Each of the triangles is right-angled, so standard trigonometry functions can be applied (I haven't shown all the right angles in the diagram, but it is a regular triangular prism).
Considering each of these three angles in turn: the way to get to a simplified expression for δ is to express the three angles in the fewest numbers of lines. It's possible to express α, ß and δ in terms of the external dimensions of the prism (let's call them x, y and z) but this just leads to incompatible expressions that can't be simplified or combined.
α
ß
δ
The strategy here is to substitute for y and p in the expression for δ, and then to simplify.
Firstly, rearrange the expressions for α and ß to make y and p the subjects of those equations.
A very simple and elegant equation: the angle of ascent depends on how steep the hill is, and the amount by which you zigzag, and is completely independent of the size of the hill (i.e. none of the lengths are relevant in the calculation).
A few sanity checks:
If ß is zero, or close to zero, then δ approaches α - i.e. if you don't zigzag, then you approach the hill at its actual angle.
If ß approaches 90 degrees, then δ approaches zero - you hardly climb at all, but you'll need to travel much further to climb the hill. In fact, as ß tends towards 90 degrees, path length p tends to infinity.
If α increases, then δ increases for constant ß (something that was worth checking).
An interesting note:
At first glance, you may think that a path (or zigzag) angle of 45 degrees would reduce the angle of ascent by half (e.g. from 60 degrees to 30 degrees), simply because 45 is half of 90. However, this isn't the case. In order to get a reduction of a half, cos ß needs to equal 0.5. If cos ß = 0.5, then ß = 60 degrees. A much larger deviation from the straight-up angle is needed.
In conclusion
This question was first put to me when I was in high school (a few years ago now) and it's been nagging at me ever since. I'm pleased to have been able to solve it, and I'm pleased with how surprisingly simple the final expression is (previously, my 3-D geometry and logic weren't quite up to scratch, and I ended up going round in circles!).
No comments:
Post a Comment