1873 - Old Forge Coal Mines

INSTECTORS OF MINES. 59
no proper rule for calculating this connection of the lever. Here are inserted
the action of 6 different dimensions of levers
Thus :—A lever 20 inches long from centre of fulcrum to where the
weight is suspended, and 2 inches long from fulcrum to centre of valve,
the weight of the lever being 4 lbs. 9 oz., require the weight to counterpoise
it. To obtain this test suspend weights from the hole which takes
the fulcrum pin and rests the pivot, which acts upon the valve as the centre
of motion.
Test No. 1 lever, 20 inches by 2-inch fulcrum, by 4 lbs. 9 oz., takes 19 lbs. to poise it.
« 2 " 20 " 3 " " 4 lbs. 9 oz., " 12 " "
" 3 « 24 "2 " " 5 lbs., " 28 " "
« 4 " 24 " 2*4 " " 5 lbs., " 21 " "
« 5 " 24 " 3 " " 5 lbs., " 16 " "
6 « 24 " Sy 2
" " 5 ^s., " lS^S "
The popular rule for calculating the action of the lever upon the valve
is so far from the actual truth that it is not recommended for adoption, viz :
Divide the whole length of the lever by the length of the fulcrum to valve,
and that quotient by half the weight of the lever. A nearer approximation
is to divide the whole length of lever by length of fulcrum to valve,
and quotient by the square root of the weight of the lever.
Example on No. 4 lever:—Thus, 24-^-2.5=9.6. Then the square root
of the weight of lever.
Here, ,\ 5 lbs=2.236X 9.0=21.4656 lbs. This gives nearly £ pound too
much. The same on No. 3 test would bring the action of the lever l£ lbs.
too little.
Having substituted the above formula there will be but little difficulty in
adjusting the load the safety valve has to bear.
Example 2 :—Suppose it is required to blow off the steam at 50 lbs. per
square inch, and we select No. 2 test lever for that purpose, what weight
will be required to be hung on the end of the lever ?
Here, lever "20 inches long- by 2 inches fulcrum, by 4 lb. 9 oz.=19 lbs.,
then deduct 19 pounds from 50 pounds=31 lbs., which=the required weight.
Select No. 3, and blow off at 60 lbs. per square inch, what then will be
the result ? This, 24 by 2 inch fulcrum by 5 lbs.=28 ;
hence deduct 28 from
60, which=32 pounds, the required weight upon the lever's end.
A lever 24 inches long, with 2 inches fulcrum, weighs 5 lbs., actuated by
28 lbs., riding over a valve 3 inches in diameter ; valve and pivot being 4
lb. weight and 32 lb. weight to be on the end of the lever, require the total
pressure on the valve, likewise the pressure in pounds per square inch.
Here, according to formula:—Valve No. 3 suppose to be 3 inches in diameter.
Hence 32X. 7854=1. 0686 or 7 square inches.
Whence 28X7=196 and 32X7 plus 4 lbs., weight of valve,=228.
Therefore 196-j-228=424 lbs.; total pressure upon the valve and 424-f-
7=60 pounds per square inch upon the boiler=4 atmospheres.
Again, suppose you graduate the test lever, No. 3, it being 22 inches long
from valve to the end where the weight is suspended.
Thus, 22 inches divided by 5^=4 divisions, require the weight when removed
to each division alternately.
Here, by rule of three:— 22:60::5i:15 lbs.=first division; 22:60::11:30
lbs.=second division ; 22:60::16^:45 fbs.=third division. Then 22:60::22:
60=60 lbs. lever end.
In resolving these values the learner will not fail to be instructed in this
most essential requisite.
When a spring balance is applied to the lever of a safety valve the distance
between fulcrum and valve equals the diameter of valve in inches