Exhibit 1: This video:

It shows a London Underground Central Line train arriving and then departing from Bethnal Green station. It showed some pretty sharp acceleration as it left. How many amps of traction current was it pulling to do that? Let's find out.

First we need to know about the train. It was a 92 Stock train, and we can learn about that here:

http://www.tfl.gov.uk/assets/downloads/foi/Rolling_stock_Data_Sheet_2nd_Edition.pdfUseful information is that the train is 132 m long and weighs 170 tonnes unladen.

From the video we see that it starts moving at 0m55s and the back of the train disappears into the tunnel at 1m10s. Making the reasonable assumption that the front of the train was at the tunnel mouth when it started, that is 132 m in 15 seconds.

Using our high school physics formula for rectilinear motion, we have:

s = 0.5 * (u + v) * t

where s is distance travelled, u is initial velocity, v is final velocity and t is time. Given an initial velocity of zero and making an assumption of constant acceleration we can find:

v = 2 * s / t = 2 * 132 / 15 = 17.6 m/s (~ 40 mph)

That's how fast the train was going when it vanished into the tunnel. Next, we can calculate the energy the train has at that speed. Again from physics:

E = 0.5 * M * v^2

Assuming the train was unladen (it looked pretty empty), we have:

E = 0.5 * 170 000 * 17.6^2 = 26 300 000 J = 26.3 MJ

From the energy we can get the power. The energy has to be imparted to the train in 15 s, so the power is given by:

W = E / t = 26.3 / 15 = 1.75 MW

That gets us very close to the traction current. We know that the train is supplied at a nominal 630 V DC, so:

I = W / V = 1 750 000 / 630 = 2 800 A

That's quite a lot of current.

We can see from some of the pictures on this page what size cables are used to carry that current:

http://www.strategicrail.co.uk/case-studies/survey-and-design/power-engineering/Nice thick cables and heavy bolts.