Hello list

I am happy to report that Tony Brewis appears to have deciphered part of the
rule problem.

I attach his recent mail, at the bottom of this message, for information and
would invite any comment.

I have today scanned the actual rule and it has copied very well.  I am in
the process of manipulating the image to make it a manageable size and I
will then post it on my web site so that you can all see the actual thing.
(it currently sits at two images of 6.5Mb in size each)

Regards

Phil Clifford

[log in to unmask]

http://freespace.virgin.net/kathryn.c/index.htm

Tony Brewis wrote :-

I think I have "cracked the code" of the slide rule gauge points listed in
the "Pumping Engines" table. The G.P. for 24 inch diameter, by the way,
should be 106, not 406 as listed. The figures given enable an engineer to
calculate the gallons per hour delivered by a pump if he knows the piston
diameter in inches, the piston stroke length in feet and the number of
strokes the piston makes per minute.

The Gauge Points are given to three significant figures, as this is the
practical accuracy when using a slide rule. Powers of ten are left to be
understood - the engineer has to do a back-of-the-envelope calculation to
decide where to put the decimal point.

For the cases where only two digits are given, a third digit zero may be
left to be understood. In using a slide rule, it would not be necessary to
quote this.  I think in fact the real value of the G.P for 3 inch diameter
is 16.5. for 4 inch 29.2, for 5 inch 45.7, and so on. The G.P. for 21 inch
diameter is 810, and for 23 it is 970, followed by 1060 for 24 inch, up to
1650 for 30 inches.

It works like this:

If the area of the piston is     A  (square feet);
The piston stroke is            L    (feet)
And the piston makes    N   strokes per minute,

Then the volume swept by the piston per minute is A x  L x N.

If the pump were 100 percent efficient, this would be the volume of water
pumped per minute.  However, we have to assume it is less than perfect, and
I believe Mr Routledge had assumed an efficiency "E" of 90 percent ( =
0.90 ).

To get the volume per hour, we multiply by 60. Also, the number of gallons
in a cubic foot is 6.23, if I remember rightly, so the gallons per hour is
given by:

A x L x N x E x 60 x 6.23.   (Equation 1)

Now, we have expressed "A" in square feet, whereas the pump piston diameters
are usually expressed in inches.

If the piston diameter is D inches, then it is d feet, where   d = D/12.

The area in square feet  is   ( pi  x d x d) divided by 4.

The constant pi/4 is   0.7854.

Lo! and behold, we see this listed ( times ten) in the table above, under
Gauge Points of a Circle, under Area.

So, substituting for "A" in Equation (1), and changing the order somewhat,
we get:

( 0.7854 x D/12 x D/12 x 60 x 6.23 x 0.90 ) x N x L

Let us look at a couple of examples.

If D = 12 inches. the figure in brackets becomes

0.7854 x 1 x 1 x 60 x 6.23 x 0.90  =  264.2, which to three significant
figures is 264

The slide rule gauge point given in your table is 264.

If D = 24 inches, the figure in brackets becomes

0.7854 x 2 x 2 x 60 x 6.23 x 0.90 = 1056.9, which to three significant
figures is 1060

The slide rule gauge point given in your table should be 106.

If D = 30, the figure in brackets becomes

0.7854 x 2.5 x 2.5 x 60 x 6.23 x 0.90 = 1651, which to three significant
figures is 1650

The slide rule gauge point given in your table is 165. (Remember, you work
out powers of ten in a rough calculation ).

So, knowing the pump piston diameter (and hence its gauge point), the stroke
length and the number of strokes per minute, you can work out the output in
gallons per minute just by multiplying three numbers together -- the figures
in brackets have been worked out for you by Mr Routledge.

What the numbers are on the remaining tables, I have yet to tackle!

Thanks for the challenge!

Regards,

Tony Brewis



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