We next decide at which part of the diagram its expansion curve and the test curve shall touch, and in this example we have chosen that it shall be at line 10.
We have now to find what pressure the length of line 10 represents on the scale of the indicator spring, which in this case we will suppose to be 25 lbs., the line measuring 25⁄30 of an inch, and a 30 lb. spring having been used to draw the diagram. Next multiply the pressure (25 lbs.) by the number of the line (10), and divide the product (250) by the number of each of the other lines in succession, and the quotient will be the pressures to be represented by the lines.
For example, for line 9 we have that 250 divided by 9 gives 27.7, hence line 9 must be long or high enough to represent a pressure of 27.7 lbs. above a perfect vacuum, or in this case 27.7⁄30 of an inch.
For line 8 we have that 250 divided by 8 gives 31.25 lbs., hence line 8 must be high enough to represent a pressure of 31.25 lbs. above a perfect vacuum.
The atmospheric line is, in this case, of no other service than to form a guide wherefrom to mark in the line of no pressure, or of perfect vacuum.
Now take the case of line 5, and 250 divided by 5 gives 50, hence the height of line 5 must represent a pressure above vacuum of 50 lbs.
Having carried this out for all the lines from line 10 to line 1, we draw in the true expansion curve, which will touch the tops of all the lines.
Fig. 3370.