Atmospheric pressure acting on a surface the size of a large dinner-plate exerts a force sufficient to propel a ten ton train! The *That’s Maths* column ( TM027 ) in the* Irish Times* this week is about the atmospheric railway.

For more than ten years from 1843 a train without a locomotive plied the 2.8 km route between Kingstown (now Dun Laoghaire) and Dalkey, near Dublin. Trains running every thirty minutes were propelled up the 1-in-110 gradient to Dalkey by atmospheric pressure. Returning trains coasted down to Kingstown under gravity.

A fifteen-inch cast-iron pipe was laid between the railway tracks and a piston in the pipe was linked through a slot to the front of the train. The slot was sealed by a greased leather flap to ensure that it was air-tight. The air in the section of pipe ahead of the train was exhausted by a steam-driven air pump in an engine house at the Dalkey terminus. With a partial vacuum ahead, the atmospheric pressure behind the piston drove the train forward.

**Rough calculations**

Available technical data are scant, but adequate for some back-of-an-envelope calculations to estimate average speeds and travel times. Pressure is force per unit area: *p* = *F/A*. The atmosphere presses on the piston from both ends; the nett force is the pressure difference multiplied by the area. The area of the piston-face follows from the diameter (diameter 0.38 m, radius 0.19 m): *A* = *π*(0.19)^2 = 0.1134 m^2. With a working vacuum of about 15 inches of mercury, or half an atmosphere, the pressure difference is 500 hPa. So the force is 5,670 newtons, or about 6 kiloNewtons (6 kN).

Using Newton’s Law, we can divide the force by the mass of the train to get the acceleration. Assuming a train of mass *M* = 10 tonnes, the acceleration comes to about about 0.6 metric units (0.6 metres per second per second). This may be compared to the 10 metric units of the acceleration *g* due to gravity but, with the small gradient, only about one percent of the gravitational acceleration acts along the pipe, so the nett acceleration a is about 0.5 metric units.

Now the standard equations of elementary mechanics come to the fore. Knowing the acceleration and the length of the line (*L* = 2800 metres) we can estimate the maximum speed and journey time. The formula *V*^2 = 2*aL* gives a terminal speed of 53 m/s or 190 km/h. However, we have neglected friction, which reduces this considerably. Typical running speeds were about 40 km/h, giving a journey time for the up-train of about four minutes.

*MgH*where the elevation H is 25 metres above sea-level. Supposing this to be converted completely to kinetic energy ½

*MV*^2, the terminal speed would be about 80 km/h. A mean speed of 40 km/h is an over-estimate as frictional losses have again been ignored. According to contemporary reports, the return journey under gravity was “very lady-like”, the average speed being about 30 km/h and the journey taking between five and six minutes.

**Prospects**

Although reasonably successful, the atmospheric system had several operational inconveniences and was abandoned after about ten years. But a system using similar pneumatic principles is running today in Jakarta and another, called Aeromovel, is proposed for construction in Brazil. So air-power, which seemed like a white elephant 160 years ago may again provide fast, clean and frequent urban transport.

**Sources**

Hadfield, Charles, 1967:* Atmospheric Railways: **a Victorian Adventure in Silent Speed. *David & Charles, Newton Abbot.

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