This weekend I purchased a 1970s Schwinn Varsity, which this article calls "the single most significant American bicycle." I am in the process of converting it into a fixed gear bicycle, and I'll have photos and a description of that process soon. Today's post, however, is on some math with fixed gear bicycles. On fixed gear bikes, there is one speed, one chainring, and one cog. What makes a fixed gear different from a generic single speed is that the drivetrain is connected directly to the hub of the rear wheel without a winch mechanism permitting free turning of the wheel while the pedals remain stationary. You have to pedal continuously on a fixed gear bike as long as you're moving forward!
One of the most important decisions when making a fixed gear bike is choosing which gear you want to be stuck in forever. I chose to make this decision based on my desired leg revolutions per minute and desired speed. Since I'll be using the bike for getting around town and maybe a few longer rides, I want to be going about 18 mph when I'm pedaling at 90 revolutions per minute. Now, how do I figure out what gear ratio to use?
18 mph = 28,962 meters per hour = 482.7 meters per minute
The circumference of my bike wheel (700x23c size) will be 2.09858 meters. This means the rear wheel will need to turn 230.013 turns per minute. I want my pedals to be turning at 90 per minute. This means the gear ratio I'll need is 230.013/90 = 2.5557. Of course, chainrings and cogs come in limited sizes, and to minimize wear on the chain both should have an even number of teeth. I've decided on a 42x16 combination, which is actually a gear ratio of 42/16 = 2.625, corresponding to a speed of 18.5 mph at 90 rpm.
Now that we've done some basic arithmetic, how about a little number theory? As I was reading up on fixed gear bikes, there was much discussion on "skid patches." Some die-hards believe that adding a break to a fixed gear ruins its sleekness, and so they rely on skidding to stop their bike. Basically they lean forward to take weight off the back wheel, lock their legs, and then skid to stop. Since most people skid with their pedals in the same position every time, this causes skid patches to form on the tire at certain locations. The gear ratio determines how many skid patches you end up with. For example, if your gear ratio is 45x15, then every pedal stroke corresponds to 3 whole wheel revolutions, and you will have one skid patch. If your gear ratio is 42x16, as I plan mine to be, you will have 8 skid patches. How do you determine how many skid patches you'll have?
Theorem. Let a/b be the reduced gear ratio, meaning that a and b are relatively prime integers (In the above example, 42x16 reduces to 21/8). Then there are b skid patches.
Proof. First we show that there cannot be more than b skid patches. If a/b is the reduced gear ratio, then for every one pedal revolution there are a/b revolutions of the wheel. So for every b pedal revolutions there are a wheel revolutions. Thus, b pedal revolutions returns us to the original starting location on the wheel, since a is an integer. So there can't be more than b skid patches.
Now we show that there can't be fewer than b skid patches. Let's assume that there are fewer than b skid patches. Then there exist two different numbers of pedal revolutions that correspond to the same location on the wheel. Let's call these two different numbers of pedal revolutions m and n, with 0<n<m<b. If m and n pedal revolutions get us to the same spot on the wheel, then their difference, m-n pedal revolutions will get us to the same spot on the wheel. Now m-n pedal revolutions is equal to (m-n)(a/b) wheel revolutions, which must be an integer in our case. This means that b divides m-n or b divides a. But b can't divide m-n because m and n are both less than b, and b can't divide a because b and a are relatively prime. This is a contradiction, so after every pedal revolution [0,1,2...b-1] we must be at a different location on the wheel, and there must not be fewer than b skid patches.
There are no more or no fewer than b skid patches, so there must be exactly b skid patches.
Some fixed gear skidders are ambidextrous, and can skid with either their left or their right leg forward.
Theorem. Consider the case of an ambidextrous skidder. If the reduced gear ratio a/b has an even numerator, then there are b skid patches. If the reduced gear ratio has an odd numerator, then there are 2b skid patches.
Proof. Ambidextrous skidders can skid every half pedal revolution, which is equivalent to a situation where we have a single-side skidder using a front chainring half as large. The gear ratio for an ambidextrous skidder, then, is effectively (a/2)/b. If a is even, then this can simplify such that a and b remain integers, and as above there are b skid patches. If a is odd, the gear ratio could only be simplified to a/2b and according to the theorem above we have 2b skid patches.
One of the most important decisions when making a fixed gear bike is choosing which gear you want to be stuck in forever. I chose to make this decision based on my desired leg revolutions per minute and desired speed. Since I'll be using the bike for getting around town and maybe a few longer rides, I want to be going about 18 mph when I'm pedaling at 90 revolutions per minute. Now, how do I figure out what gear ratio to use?
18 mph = 28,962 meters per hour = 482.7 meters per minute
The circumference of my bike wheel (700x23c size) will be 2.09858 meters. This means the rear wheel will need to turn 230.013 turns per minute. I want my pedals to be turning at 90 per minute. This means the gear ratio I'll need is 230.013/90 = 2.5557. Of course, chainrings and cogs come in limited sizes, and to minimize wear on the chain both should have an even number of teeth. I've decided on a 42x16 combination, which is actually a gear ratio of 42/16 = 2.625, corresponding to a speed of 18.5 mph at 90 rpm.
Now that we've done some basic arithmetic, how about a little number theory? As I was reading up on fixed gear bikes, there was much discussion on "skid patches." Some die-hards believe that adding a break to a fixed gear ruins its sleekness, and so they rely on skidding to stop their bike. Basically they lean forward to take weight off the back wheel, lock their legs, and then skid to stop. Since most people skid with their pedals in the same position every time, this causes skid patches to form on the tire at certain locations. The gear ratio determines how many skid patches you end up with. For example, if your gear ratio is 45x15, then every pedal stroke corresponds to 3 whole wheel revolutions, and you will have one skid patch. If your gear ratio is 42x16, as I plan mine to be, you will have 8 skid patches. How do you determine how many skid patches you'll have?
Theorem. Let a/b be the reduced gear ratio, meaning that a and b are relatively prime integers (In the above example, 42x16 reduces to 21/8). Then there are b skid patches.
Proof. First we show that there cannot be more than b skid patches. If a/b is the reduced gear ratio, then for every one pedal revolution there are a/b revolutions of the wheel. So for every b pedal revolutions there are a wheel revolutions. Thus, b pedal revolutions returns us to the original starting location on the wheel, since a is an integer. So there can't be more than b skid patches.
Now we show that there can't be fewer than b skid patches. Let's assume that there are fewer than b skid patches. Then there exist two different numbers of pedal revolutions that correspond to the same location on the wheel. Let's call these two different numbers of pedal revolutions m and n, with 0<n<m<b. If m and n pedal revolutions get us to the same spot on the wheel, then their difference, m-n pedal revolutions will get us to the same spot on the wheel. Now m-n pedal revolutions is equal to (m-n)(a/b) wheel revolutions, which must be an integer in our case. This means that b divides m-n or b divides a. But b can't divide m-n because m and n are both less than b, and b can't divide a because b and a are relatively prime. This is a contradiction, so after every pedal revolution [0,1,2...b-1] we must be at a different location on the wheel, and there must not be fewer than b skid patches.
There are no more or no fewer than b skid patches, so there must be exactly b skid patches.
Some fixed gear skidders are ambidextrous, and can skid with either their left or their right leg forward.
Theorem. Consider the case of an ambidextrous skidder. If the reduced gear ratio a/b has an even numerator, then there are b skid patches. If the reduced gear ratio has an odd numerator, then there are 2b skid patches.
Proof. Ambidextrous skidders can skid every half pedal revolution, which is equivalent to a situation where we have a single-side skidder using a front chainring half as large. The gear ratio for an ambidextrous skidder, then, is effectively (a/2)/b. If a is even, then this can simplify such that a and b remain integers, and as above there are b skid patches. If a is odd, the gear ratio could only be simplified to a/2b and according to the theorem above we have 2b skid patches.
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