234 
OF THE PARTS OF 
[sect. VII. 
as negative in the calculation, when the radius bar is to be less than the parallel 
bar. 
This expression, which will be found to be of important use in rectifying and 
adjusting the motions of old engines, and indeed in all cases where it may be 
necessary or desirable to have the radius bar to work from a stated centre, was 
first investigated in the valuable treatise on ‘ Mechanics for Practical Men,’ by 
Messrs. James Hann and Isaac Dodds, (Newcastle, 1833,) a work in which mathe 
matical theory and useful practice are combined throughout, with remarkable 
clearness and simplicity. 
If the beam from the centre of motion to the point F be one and a half times 
the length of the stroke or b = 3 s; then, 
/3 . £-.\ s (3 5 — 2 c). f, ni ao & (3 ^ — 2 c) . 
r = (— + 2) — + C = 2-9142 _i L + C. 
'2 ' c c 
In any case, except when the radius bar C D, and parallel bar D G are of the 
same length, the deviation is increased by increasing the quantity of angular 
motion of the beam. Hence, beams having short parallel bars should be limited 
in the extent of angular movement; indeed the motion should not in general 
exceed 20°, and this is very nearly the case when the distance of the end F of the 
beam from its centre of motion A, is to the length of the stroke as 3 : 2; in 
such case the radius bar may be found as follows: 
493. Rule hi. To find the length of the radius bar, when the length of the 
beam from the centre of motion is to half the length of the stroke, as 3 to 2. 
From three times half the length of the stroke subtract twice the length of the 
parallel bar, and multiply the difference first by the half length of the stroke, and 
then by the number 2*914. Divide the product by the length of the parallel bar, 
and the quotient added to the length of the parallel bar will be the length of the 
radius bar. 
Example. Let the length of the stroke be 8 feet, its half 4 feet, and let 
the length of the parallel bar D G be 3 feet; then 3x4 — 2x3 = 6; and 
6 x 4 x 2*914 = 69*936, which divided by 3 gives 23*312 ; add to this the 
length of the parallel bar 3 feet, and we have 23*312 + 3 = 26*312 feet, for the 
radius bar C D. 
A short parallel bar is assumed in this example, to show the great length of 
radius bar required in such a case. 
The length of the links D B, G F, are from four to five-tenths of the length of 
the stroke, depending on convenience and space; but the longer they can be made, 
the less oblique strain will take place during the motion. The vertical distance 
t