On the first point, the best information can be obtained from experiments made by Mr. Daniel Gooch during the gauge controversy. The results are very suitable for use in the present investigation, as the South Devon was to be a broad-gauge railway. Moreover, as the broad-gauge engine with which these experiments were tried was one of a class more powerful for their weight not only than the contemporary narrow-gauge engine, but also than the engines Mr. Brunel had experience of when he wrote his report three years previously, the results may be considered to represent very favourably the then existing case for the locomotives.

The engine employed in the experiments weighed, with its tender, about fifty tons. The maximum power it was capable of delivering by the pressure of steam in its cylinders was represented as a tractive force of 4,900 lbs. at a speed of 60 miles an hour, equivalent to 784 indicated horse-power; and at 40 miles an hour 5,200 lbs., equivalent to 555 indicated horse-power.

It is next to be considered how this power would, when running at the speeds mentioned, be employed in overcoming the elements of resistance. These are:—

(1) The working friction of the machinery.
(2) The rolling resistance of the engine and tender.
(3) The air resistance due to the engine frontage.

(4) The rolling resistance of the train.
(5) The air resistance on the portion of the train unprotected by the tender.
(6) The resistance due to gradient.

The following symbols and quantities may be conveniently made
use of to denote the various terms of the equation between force
and resistance.

Total available tractive force in lbs.F
Weight of engine and tender (superfluous load) in tons 50
Weight of train (useful load) in tonsW
The sum of the resistances of machinery, rolling resistance,
and air resistance of engine and tender
R
Rolling resistance of train in lbs. per tonK
GradientG
Speed in miles per hourV
Resistance of air (according to the received empirical formula)
=1
400
(frontage area) × V2
Frontage area of train in square feet63
Frontage area of portion of train unprotected by the tender,
in square feet
24

For a locomotive train therefore

F = R + WK + 24
400
V2 + (50 + W) 2240 G.

For a system that dispenses with the locomotive

Tractive force = WK +63
400
V2 + W 2240 G.