Since my last post was cycle related, I thought I would share the most fiendish problem regarding bikes that I have come across so far, courtesy of one my colleagues from Southampton (GR!). It has resurfaced with a colleague at Imperial (PB) and I fear I may in the spur of the moment given the wrong solution. Here is my attempt to right my wrongs.
Imagine you have a bike with its pedals in the vertical position. The bike is assumed to be magically balancing and not falling sidewise, whilst still being free to move backwards and forwards. You grab the bottom pedal and push it backwards. Will the bike move forwards or backwards?
Consider the diagram below:
The two important points on this diagram are: 1), the bottom of the rear wheel and 2) the bottom pedal.What the problem boils down to is the position of the bike as a function of the position of 2. The other quantities shown on the diagram are the angles φ and θ describing the positions of 1 and 2 relative to the bike itself, the radius of the back wheel ‘r’ and the length of the pedal cranks ‘k’.
Assuming that the bike travels in a straight line, the centre of the cranks (the bottom bracket) is at a point with coordinates (d,0), where d is the distance that the bike has travelled. I will define d such that it is positive if the bike is moving forward and that it has a value of 0 when the pedal and the bottom of the wheel are in the positions shown in the diagram. It is very straightforward to relate d to the angle φ:
this is simply because the wheel is not allowed to slip, therefore for each inch that the bike moves forward the wheel must rotate anti-clockwise. Notice that φ is implicitly defined to be positive when the wheel is rotating anti-clockwise.
Next we want to relate φ to θ. In order to do this, we need to know the gearing g of the bike. I define this as the number of turns the wheel makes for every turn the pedals make. So a high gearing is the one you use when speeding downhill and a low gearing when struggling to get up a steep mountain. With this definition in mind, it is clear that the two angles can be related by:
Now we know φ, all that is left to do is to write down the position of the pedal in cartesian coordinates. Since the position of the centre of the cranks is (d,0) and since the pedals are turning in a circle of radius k and defining φ as 0 when in the position shown in the diagram above we conclude that the position of the cranks can be written as:
we can use our other formulae to eliminate d for φ:
When we push the pedal backwards, we are making the x position of the pedal small and negative. The question boils down to whether this corresponds to φ being positive or negative. It is evident from the formula above that the answer must be “it depends” on the values of g,r and k. Let’s do a little bit more analysis to show exactly how. If the x position is small, then we can assume that φ is also small and that the sin can be approximated linearly. If we then measure distance in units of k and divide the whole problem through by k we find:
where g’ has been defined as . So there are two cases. If g’ is greater than 1, then a negative x corresponds to a negative φ. This means that moving the pedal backwards corresponds to the cranks and the wheel rotating clockwise and hence to the bike moving backwards. On the other hand, if g’ is smaller than 1, then a negative x corresponds to a positive φ. In this case, moving the pedal backward makes the wheel turn anti-clockwise and hence the bike moves forward. Since most normal bikes have a gearing g greater than 1, and since it would be hard to design a bike with cranks longer than the radius of the wheel, g’ is almost always greater than 1 therefore for a normal bike, if you push the pedals back, the bike moves back!