Why would cyclists push a bike? In the case of a recent stage of Tirreno-Adriatic, there were three parts with a 27 percent gradient. Yes. That's pretty steep for a bike.
Surely you know I don't really report on the news. If you want more details about the actual race, check out VeloNews.com. For me, I am interested in this question from Mark Cavendish (@MarkCavendish) - oh, hat tip to Chris Hill (@Hillby258):
Now for the physics. I think there are three reasons why a slope would be too steep. For all of these cases, I am going to assume that it is a prolonged slope. This means that you can't just build up a large speed and zoom up the slope. If this was the case, you could go straight up a wall (which you can for a short time).
Limit Due to Human Power
Here is a diagram of a biker going up a hill.
First, a quick note on grades. What does a 30% grade mean? It means that if you travel a distance up the incline, the ratio of vertical to horizontal distance (times 100) would give you the grade. We usually represent the steepness of a slope with an angle, but this essentially does the same thing. I'm not sure of the international symbol for grade, so I will use r. In terms of the height (h) and horizontal distance (s), the grade would be:
Let's say that the rider is moving with some speed v and this speed is slow enough that air resistance isn't a significant factor. How much energy would it take to move up the slope at a constant speed? In this case, I could consider just the energy going into the change in gravitational potential energy of the rider plus bike. The change in energy would be:
I don't really care about the energy. I want to know the power needed to go up this slope. Power is defined as the change in energy over the change in time. But how long does it take to go up this incline? If the velocity is v, I can write the vertical speed as:
__Update: __I hate this gradient notation. It isn't very useful for calculations. So, the above equation has a mistake (I crossed it out). However, the vertical velocity is still h over the change in time. I added more useful expression for the vertical velocity (in terms of theta).
Now, I can solve for the change in time and use this to calculate the power.
__Update 2: __Ok, I had to fix this equation too. Again, I switched back to using theta for the slope angle instead of that silly gradient. I included a calculation to determine the angle from the gradient.
Let's just put in some values here. Suppose the bike plus rider has a mass of 75 kg with an average speed of 2 m/s. If the grade was 30, this would require a power of 441 Watts 422 Watts. That's some serious power. It's possible, but it would wear you out quickly. I don't have a good feel for the power of a cyclist, but I asked my brother who rides quite a bit. He said that on a 40k ride, he averages 280 Watts. He's not a wimp, so I would say that this 441 Watts 422 Watts is pretty tough.
Remember, this is the power with no friction and no air resistance. It would actually be even higher. How about a plot of power need for different grade slopes? __Note: __I replaced the previous graph with an updated graph to adjust for the power error above. The change wasn't huge.
From this, if you want a minimum speed of 1 m/s then a slope with a 40% grade would take a minimum of 300 watts. I think this is a little much. I would go for a speed of 2 m/s with a maximum power of maybe 300 watts. This would put the maximum grade at 20%.
__UPDATE: __Here is a possibly useful graph of human power limits posted on the International Human Powered Vehicle Association website.
I'm not sure where this data came from, but it looks reasonable. I would love to see the data this is based on (hopefully it wasn't just sketched up on a bar napkin during a heated discussion). Anyway, this seems to suggest that a top athlete could produce 0.4 horsepower - this is around 300 watts. So, my estimate doesn't seem so crazy.
Oh, Hat Tip to Eric Booth (again).
Limit Due to Center of Mass
Let me go ahead and say that I suspect the power limit above is going to be lower than the next two limits to the slope. There, I said it. I might be wrong though.
For a bike going up an incline, the center of mass has be horizontally in between the two support forces. In this case the support forces are the contact forces on the two tires. I'm not going to re-derive this center of mass stuff - if you want more details, check out my post on the mass of Darth Vader.
Here is a diagram of a biker going up a hill. In this case, I am assuming the biker is leaning forward as much as possible. This might put the center of mass of the bike plus the rider right over the handle bars.
This diagram is more complicated than it needs to be. That is my fault. I really didn't know what I wanted to draw until I started drawing it. Oh well, this will work. Here I have labeled a as the distance of the center of mass above the ground and c as the horizontal distance of the center of mass in front of the real wheel contact point.
If the red dot for the center of mass is to the left of the blue dot for the back wheel, the bike will tip over. If I measured the angle of the line from back wheel to center of mass, the angle of this line must be less than 90°. This angle is the sum of the slope angle (I will call θ) and the angle of the center of mass with respect to the wheels (which I will call α). Looking at the triangle I added, I can find α.
For the maximum slope, the sum of these angles would be 90°.
Bikes can be different sizes, so let me just guess that a is about 0.8 meters and and c is around 0.75 meters. This would put the maximum slope angle at around 43°. In terms of grade, this would be a 93.7%. Of course, I already calculated that this would probably too steep to ride up due to power constraints.
Limit Due to Friction
A bike is more complicated than a solid block. However, I am going to model the bike as a block anyway. In order for a bike to go up an incline at a constant speed, the net force must be zero (zero vector). Here is my bike-block.
Here I have picked the x-axis to be along the direction of the incline and the y-axis to be perpendicular to that. If the forces add up to the zero vector, they have to add up to zero in both the x- and y-directions.
Now, I will use the fairly standard model for static friction. It says that the frictional force is proportional to the force the surface pushes back perpendicular to the surface (we call this the normal force).
The μs is the coefficient of static friction. It is a value that depends on the types of materials interacting (in this case rubber and asphalt or cement). The less than or equals sign is there because the static friction force will only push as had as it needs to to prevent the two surfaces from slipping up to its maximum amount. Yes, it should be static friction and not kinetic. Static is used when the two surfaces don't slide relative to each other (that's what we have).
I want to use this model for friction to solve for the maximum angle before this thing slides.
I just need the coefficient of friction for a tire and a road interacting. Based on this I am going to guess a coefficient of 0.8. This would give a maximum incline of 38.7° (80% grade). Of course if the road is wet, this would go down to a coefficient maybe as low as 0.45. This would make the maximum slope angle at 24° (45% grade). These are all much higher than the power limit.
Really, the friction problem might be worse than this. The bike only uses the back wheel for moving forward, so it is the friction on the back wheel that matters. If the biker is leaning forward, the weight distribution might not be even on the two wheels. I will leave this estimation (combining the previous two limits) as an exercise for the reader.
