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.
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 = 2aL 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.We can also estimate the maximum speed on the downhill journey by invoking energy conservation. At Dalkey, the train has gained potential energy MgH where the elevation H is 25 metres above sea-level. Supposing this to be converted completely to kinetic energy ½MV^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.
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.
Hadfield, Charles, 1967: Atmospheric Railways: a Victorian Adventure in Silent Speed. David & Charles, Newton Abbot.