6/12/2016 > Math

Similarly, let's call the position of the back wheel \((b_x,b_y)\) (since its \(y\) value is not constant).

To quantify the above observations (\(L\) is the length of the bicycle and \(A\) and \(B\) are constants): \[(f_x-b_x)^2+b_y^2=L^2\] \[\frac{db_y}{db_x}=\frac{-b_y}{f_x-b_x}\] The second equation is a little harder to grasp than the first. However, all that it is saying is that the direction of the tangent line is toward the front wheel. Really it is about similar triangles, with \(-b_y\) being the height and \(f_x-b_x\) being the width of one triangle as opposed to \(db_y\) and \(db_x\) for the "differential triangle" that is the slope.

Now solve for \(f_x\) in the first equation (I omitted the negative solution because I assume that \(f_x\) is positive) \[f_x=\sqrt{L^2-b_y^2}+b_x\] Plug it in to the second equation: \[\frac{db_y}{db_x}=\frac{-b_y}{\sqrt{L^2-b_y^2}+b_x-b_x}=\frac{-b_y}{\sqrt{L^2-b_y^2}}\] Happily, the \(b_x\) cancels. This isn't just a coincidence: the direction of the back wheel should be independant from its arbitrary location on the \(x\) axis.

And now we have a neat differential equation. We can solve this with separation of variables. \[\int-\frac{\sqrt{L^2-b_y^2}}{b_y}\,\,\,db_y=\int db_x\] Now that is a lovely integral! I spent awhile trying to crack this one, but considering what Wolfram-Alpha spit out, it would have take a lot longer than "awhile" (I actually felt guilty for asking the internet, so I did it myself): \[-\sqrt{L^2-b_y^2}+L\log{(L\sqrt{L^2-b_y^2}+L^2)}-L\log{(b_y)}+C=b_x\] Note that I combined the "C"s from both indefinate integrals into the one on the left side.

Now we can setup some initial conditions. At the start, if we assume that the back wheel starts on the \(y\) axis, the back wheel is at \((0,b_{yi})\) Solving for C: \[C=\sqrt{L^2-b_{yi}^2}-L\log{(L\sqrt{L^2-b_{yi}^2}+L^2)}+L\log{(b_{yi})}\] I made myself a Desmos graph of the situation. Feel free to play around with it. The red line is the path of the back wheel. The black line is the frame of the bicycle.

I actually tested this today. I was going for a bike ride and I thought to see if my solution looked remotely correct before I started. It looked pretty good: the back wheel never passed the centerline, leading me to believe that my implicit assumption about friction was correct.

Interestingly, it seems impossible to compute the resulting path with acceleration. Because the velocity cannot change instantaneously, the length of the bicycle cannot remain constant, hence breaking my equation. It's a little bit of a shame, but I guess I'll be happy with my current solution.

# The Bicycle Problem

You're on a bicycle. You are going diagonally across a road. When you cross the middle of the road,
you suddenly turn your front wheel so that it travels on the yellow midline. What is the path
of your back wheel?

My friend brought this problem up before physics one day. He was very excited. He mentioned tensors,
to which I, having no experience with the subject, merely waited patiently. Luckily, the solution to this problem requires
no tensors at all.
## The Solution

When I actually got to looking at this problem, a few things were pretty clear:- The distance from the back wheel to the front wheel is constant. Bicycles don't stretch.
- The back wheel is pulled toward the front wheel at all times. The force on the back wheel is provided by the frame, which is in the direction of the front wheel.

Similarly, let's call the position of the back wheel \((b_x,b_y)\) (since its \(y\) value is not constant).

To quantify the above observations (\(L\) is the length of the bicycle and \(A\) and \(B\) are constants): \[(f_x-b_x)^2+b_y^2=L^2\] \[\frac{db_y}{db_x}=\frac{-b_y}{f_x-b_x}\] The second equation is a little harder to grasp than the first. However, all that it is saying is that the direction of the tangent line is toward the front wheel. Really it is about similar triangles, with \(-b_y\) being the height and \(f_x-b_x\) being the width of one triangle as opposed to \(db_y\) and \(db_x\) for the "differential triangle" that is the slope.

Now solve for \(f_x\) in the first equation (I omitted the negative solution because I assume that \(f_x\) is positive) \[f_x=\sqrt{L^2-b_y^2}+b_x\] Plug it in to the second equation: \[\frac{db_y}{db_x}=\frac{-b_y}{\sqrt{L^2-b_y^2}+b_x-b_x}=\frac{-b_y}{\sqrt{L^2-b_y^2}}\] Happily, the \(b_x\) cancels. This isn't just a coincidence: the direction of the back wheel should be independant from its arbitrary location on the \(x\) axis.

And now we have a neat differential equation. We can solve this with separation of variables. \[\int-\frac{\sqrt{L^2-b_y^2}}{b_y}\,\,\,db_y=\int db_x\] Now that is a lovely integral! I spent awhile trying to crack this one, but considering what Wolfram-Alpha spit out, it would have take a lot longer than "awhile" (I actually felt guilty for asking the internet, so I did it myself): \[-\sqrt{L^2-b_y^2}+L\log{(L\sqrt{L^2-b_y^2}+L^2)}-L\log{(b_y)}+C=b_x\] Note that I combined the "C"s from both indefinate integrals into the one on the left side.

Now we can setup some initial conditions. At the start, if we assume that the back wheel starts on the \(y\) axis, the back wheel is at \((0,b_{yi})\) Solving for C: \[C=\sqrt{L^2-b_{yi}^2}-L\log{(L\sqrt{L^2-b_{yi}^2}+L^2)}+L\log{(b_{yi})}\] I made myself a Desmos graph of the situation. Feel free to play around with it. The red line is the path of the back wheel. The black line is the frame of the bicycle.

## Epilogue

This got me thinking. I assumed, early on in the problem, that the direction of motion was always in the direction of the applied force. Technically, the back wheel should have some sort of inertia that keeps it moving in the same direction. Thus the acceleration should have been in the direction of the applied force, per Newton's 2nd law. However, any effects of inertia are quickly brought to a halt by friction.I actually tested this today. I was going for a bike ride and I thought to see if my solution looked remotely correct before I started. It looked pretty good: the back wheel never passed the centerline, leading me to believe that my implicit assumption about friction was correct.

Interestingly, it seems impossible to compute the resulting path with acceleration. Because the velocity cannot change instantaneously, the length of the bicycle cannot remain constant, hence breaking my equation. It's a little bit of a shame, but I guess I'll be happy with my current solution.