Sunday, March 19, 2017

Portland to New York Using only Gravity (Part 3)

Portland to New York Using only Gravity (Part 3) Here we are at part 3 of a series of posts relating to an "underground brachistochone" where we wish to build a tunnel for a train propelled only by gravity from Portland, Oregon to New York, New York.

We left off in part 2 with the computation of the time to traverse the tunnel equal to $$\Delta t = \pi \sqrt{\frac{R^2 - r_0^2}{gR}}$$ Of course, this requires us to know the depth of the tunnel but we were able to compute the time as a function of a fraction of the earth's radius, $$R$$. What we want is a way to compute the depth of the tunnel, $$r$$ as a function of the angle we're at or vice versa.

To start this process, we recall, from part 2 $$$$\theta' = \frac{1}{r}\sqrt{\frac{\frac{k^2g}{R}(R^2-r^2)}{r^2-\frac{k^2g}{R}(R^2-r^2)}} \label{eq:tp}$$$$ and recalling our definition of $$r_0$$ stemming from solving for the conditions at the curve minimum $$$$\frac{k^2g}{R} = \frac{r_0^2}{R^2-r_0^2} \label{eq:r0}$$$$ Substuting \ref{eq:r0} into \ref{eq:tp} gives us, after a little bit of algebra, $$$$d\theta = \frac{1}{r}\sqrt{\frac {r_0^2(R^2-r^2)} {R^2(r^2-r_0^2)} }dr \label{eq:dt}$$$$ Moving some constants out of the way helps, sometimes, so lets do $$$$\frac{R}{r_0}d\theta = \frac{1}{r}\sqrt{\frac {R^2-r^2} {r^2-r_0^2} }dr \label{eq:dt2}$$$$ If you were to enter \ref{eq:dt2} into Wolfram-Alpha you would get a hot mess in return. Wolfram-Alpha is really, really smart but sometimes a little trickery is required to bend the integrand into something that it finds more recognizable. To that end, let $$u^2 = \frac{r^2-r_0^2}{R^2-r^2}$$ This is kind of strange but making $$u$$ equal to the upside-down-integrand kind of makes sense as the derivative of a square root is 1-over-the-square-root (multiplied by the derivative of the ... radicand ... it's been a while, is that right)? Solving for $$r^2$$, $$r^2 = \frac{u^2R^2 + r_0^2}{1+u^2}$$ Now we find $$dr$$ but instead of taking the square root and differentiating, we will differentiate implicitly. \eqalign { 2rdr &= {{2R^2u(1+u^2) - 2u(u^2R^2+r_0^2)} \over {(1+u^2)^2}}du \cr &= {{2u(R^2+r_0^2)} \over {(1+u^2)^2}}du \cr dr &= \frac{u}{r}{{(R^2+r_0^2)} \over {(1+u^2)^2}}du } This will get a little messy but hang in there. First off, notice that $$u=\sqrt{\frac{r^2 - r_0^2}{R^2-r^2}}$$ so $$\frac{1}{u}=\sqrt{\frac{R^2-r^2}{r^2 - r_0^2}}$$ which is the most of the original integrand. Working our way through this \eqalign { \frac{R}{r_0}\theta = \int\frac{1}{r}\sqrt{\frac {R^2-r^2} {r^2-r_0^2} }dr &= \int\frac{1}{r}\frac{1}{u}\frac{u}{r}\frac{R^2-r_0^2}{(1+u^2)^2}du \cr &= \int\frac{1}{r^2}\frac{R^2-r_0^2}{(1+u^2)^2}du \cr \text{(substituting for } r^2 \text{)} &= \int\frac{1+u^2}{(u^2R^2+r_0^2)}\frac{R^2-r_0^2}{(1+u^2)^2}du \cr &= \int\frac{R^2-r_0^2}{(u^2R^2+r_0^2)(1+u^2)}du \cr } That last integrand looks like a job for partial fractions. \eqalign { \frac{R^2-r_0^2}{(u^2R^2+r_0^2)(1+u^2)} &= \frac{A}{u^2R^2+r_0^2} + \frac{B}{1+u^2} \cr R^2-r_0^2 &= A(1+u^2) + B(u^2R^2+r_0^2) \cr &= u^2(A+BR^2) + A + Br_0^2 \cr \therefore A=R^2, B=-1 } Putting this back into the integral \eqalign{ \frac{R}{r_0}\theta &= \int\frac{R^2-r_0^2}{(u^2R^2+r_0^2)(1+u^2)}du = R^2\int\frac{du}{u^2R^2+r_0^2} - \int\frac{du}{1+u^2} \cr &= \frac{R}{r_0}\tan^{-1}\left(\frac{Ru}{r_0}\right) - \tan^{-1}u + C \cr \theta &= \tan^{-1}\left(\frac{R}{r_0} \sqrt{\frac{r^2 - r_0^2}{R^2-r^2}} \right) - \frac{r_0}{R}\tan^{-1}\sqrt{\frac{r^2 - r_0^2}{R^2-r^2}} + C\cr } Here we let $$\theta(r_0) = 0 \rightarrow C = 0$$. At the surface, when $$r = R$$, we can compute that $$\theta = \frac{\pi}{2} - \frac{r_0}{R}\frac{\pi}{2}$$ $$$$\theta = \frac{\pi}{2} (1 - \frac{r_0}{R}) \label{eq:thetaR}$$$$ The angle $$\theta(R)$$ in \ref{eq:thetaR} is the angle from where the curve is the deepest to the surface. So, given that the tunnel will travel from the surface down to the minimum then back up again, the total span, $$\Psi$$, of the tunnel is $$\Psi = 2\theta = \pi (1 - \frac{r_0}{R})$$ And solving for $$r_0$$ $$$$r_0 = R(1 - \frac{\Psi}{\pi}) \label{eq:r0x}$$$$ The earth's radius in kilometers is $$R=6371\text{km}$$. The flight distance from PDX to JFK is $$d=3949\text{km}$$. Since $$d=R\Psi$$ the required angle spanned by this distance is $$\Psi = \frac{d}{R} = \frac{3949}{6371} = 0.62$$. \eqalign { \Psi &= 0.62 \cr r_0 &= R(1 - \frac{\Psi}{\pi}) = 6371(1 - \frac{0.62}{\pi}) = 5114\text{ km} \cr } And from part 2 of this series, the time $$\Delta t$$ is given by (remember to adjust the units!) \eqalign { \Delta t &= \pi \sqrt{\frac{R^2 - r_0^2}{gR}} \cr &= \pi \sqrt{\frac{6371000^2 - 5114000^2}{(9.8)(6371000)}} \cr &\approx 25 \text{ minutes} } We can also ask about the maximum speed attained. Since we've declared that the sum of the kinetic and potential energy are zero and that this is true for the entire traversal of the path, we can deduce that the kinetic energy is at a maximum where the potential energy is at a minimum. This happens at $$r=r_0$$. We saw from part one of this series that \eqalign { |v| &= \sqrt{\frac{g}{R}(R^2-r^2)} \cr &= \sqrt{\frac{9.8}{6371000}(6371000^2 - 5114000^2)} \cr &\approx 4713 \frac{\text{m}}{\text{s}} \cr &\approx 16970 \frac{\text{km}}{\text{hr}} } So there you have it... How long is the tunnel? How many Gs are being generated at $$r=r_0$$. I'll leave those for the reader.

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