I turned my phone into a lab instrument for Istanbul's metro.

A tilt error \(\delta\theta\) puts a slice of gravity into the measured acceleration:

$$\delta a = g\,\sin\delta\theta \;\approx\; g\,\delta\theta, \qquad \delta\theta = 0.1^\circ \;\Rightarrow\; \delta a \approx 0.017\ \mathrm{m/s^2} \;\Rightarrow\; \delta v \approx 1.5\ \mathrm{m/s}\ \text{after } 90\ \mathrm{s}$$

So the error is modelled as an offset plus a slow drift, \(e(t) = c_0 + c_1 t\), and taken out of the integral:

$$v(t) = A(t) - c_0\,t - \tfrac{1}{2}\,c_1\,t^2, \qquad A(t) = \int_0^{t} a_{\mathrm{m}}(\tau)\,\mathrm{d}\tau$$

Two facts are known for every run of length \(T\) between stations \(L\) apart: the train is stopped at both ends, and it covers the track distance. They fix \(c_0\) and \(c_1\) exactly:

$$v(T) = 0, \qquad \int_0^{T} v(t)\,\mathrm{d}t = L \quad\Longrightarrow\quad \begin{bmatrix} T & \tfrac{1}{2}T^2 \\[2pt] \tfrac{1}{2}T^2 & \tfrac{1}{6}T^3 \end{bmatrix} \begin{bmatrix} c_0 \\ c_1 \end{bmatrix} = \begin{bmatrix} A(T) \\ \int_0^{T} A\,\mathrm{d}t - L \end{bmatrix}$$

\(|c_1|\,T\) — how much the error had to change over the run — grades it: under \(0.15\ \mathrm{m/s^2}\) is good, under \(0.6\) rough, above that the phone moved and the run is thrown out. On the run below it is 0.082.

The gyroscope measures rotation \(\boldsymbol{\omega}\) in the phone's own axes. Projected on the measured "up" \(\hat{\mathbf{g}}\), and minus the bias \(b\) read while the train stands still, it is the train's turn rate. Integrated with speed, it draws the plan:

$$\psi(t) = \int_0^{t} \big(\boldsymbol{\omega}\cdot\hat{\mathbf{g}} - b\big)\,\mathrm{d}\tau, \qquad x(t) = \int_0^{t} v\cos\psi\,\mathrm{d}\tau, \qquad y(t) = \int_0^{t} v\sin\psi\,\mathrm{d}\tau$$

Height comes from hydrostatics, read only in the calm air of a platform:

$$\Delta h = -\frac{\Delta p}{\rho\,g}, \qquad \rho \approx 1.2\ \mathrm{kg/m^3} \;\Rightarrow\; 1\ \mathrm{m} \leftrightarrow 11.8\ \mathrm{Pa}$$

Air rushing past a moving train has dynamic pressure \(q\). By Bernoulli, faster air means lower static pressure, so the car's pressure should dip in proportion to \(q\):

$$q = \tfrac{1}{2}\,\rho\,v^2, \qquad p_{\mathrm{car}}(t) - p_{\mathrm{line}}(t) \;\approx\; -\,C\,q(t)$$

where \(p_{\mathrm{line}}\) is the straight line between the two platform readings. At the line's top speed, \(q\) is already 296 Pa — the same as 25 m of height. That is why the slope between stations can't be read from a barometer on a moving train.

The field of a long straight current \(I\) at distance \(r\):

$$B = \frac{\mu_0\, I}{2\pi\, r}, \qquad \mu_0 = 4\pi\times10^{-7}\ \mathrm{T\,m/A}$$

The supply voltage \(U\) is fixed, so the current follows the power the motors deliver, and power is force times speed:

$$I = \frac{P}{U} = \frac{F\,v}{U}, \qquad F \approx m\,a \quad\Longrightarrow\quad B \;\propto\; a\,v$$

The typical peak while pulling is 158 µT — more than three times Earth's field. Taking the cables \(r \approx 1\) m away (an assumption; the real geometry is unknown), that is a current of order

$$I = \frac{2\pi\, r\, B}{\mu_0} \approx \frac{2\pi \times 1\ \mathrm{m} \times 158\ \mu\mathrm{T}}{4\pi\times10^{-7}\ \mathrm{T\,m/A}} \approx 800\ \mathrm{A}$$

Between two standstills at stations A and B, everything the motors put in, plus the height lost, must leave through the brakes and drag:

$$W_{\mathrm{brakes}} + W_{\mathrm{drag}} = W_{\mathrm{motors}} + m\,g\,(h_A - h_B)$$

On the steepest downhill that height alone is 398 J per kg of train. If the slope is even, most of it lands on the final stop, on top of the motion energy \(\tfrac{1}{2}v^2\) at the moment braking starts.

Ride comfort follows ISO 2631-1: each axis is filtered with a frequency weighting — \(W_d\) horizontal, \(W_k\) vertical — then combined:

$$a_w = \sqrt{\frac{1}{T}\int_0^{T} a_W^2(t)\,\mathrm{d}t}, \qquad a_v = \sqrt{a_{w,x}^2 + a_{w,y}^2 + a_{w,z}^2}$$