pathology_doc

So over the holidays, I became the proud owner of a Pickett N600ES, a baby slide rule (which can be found mimicked electronically here) no more than 6 inches long. Lo and behold, what should I find but instructions as to how to get the best out of it.

What puzzled me are the instructions on page 9 on extending the use of the natural log scale.

"To find e^6.54 first divide 6.54 by 2.303, obtaining quotient 2 and remainder 1.934."

In other words, separate out e^2.303 (i.e. 10) raised to some integer power, leaving e raised to some power between 0 and 2.3 (and thus within the limits of the Ln function of the slide rule), then multiply the two together.

Okay, so I try it, and 6.54/2.303 = about 2.83 on this teeny tiny scale... But now what? We have the quotient 2, but what about that 0.83? How do we get 1.934 out of that?

Hang on, if we carry the two lots of 2.303 away to form 100 (i.e. (e^2.303)²)... and here's where I'm going wrong - I forgot I was dealing with lots of 2.303! When I multiply that fractional remainder out by 2.303 again, I get the remainder they were looking for.

Let's try... 0.83 * 2.303 = 1.95, not quite what they get (1.934) but they're using a 10 inch rule, and possibly one with a magnifying cursor. e^1.95 gives 7, multiply that by a hundred = 700. (On my 10 inch Pickett Microline 140, I get 1.935, which is pretty well bang on the money, but the Microline 140 doesn't have an Ln scale (though I suppose there is the LL3 scale at a pinch) and the diminutive 600ES does.)

The book gives 692. That's just over 1% out.

Let's try again with another of their examples, e^17.4.

17.4/2.303 is about 7.55, take the 7 away to form 10 million, 0.55x2.303=1.265 (on the small Pickett), e^1.265 = 3.54 and multiply by the 10 million. They say 3.60, about a 2% error there.

Not bad for a six inch slipstick...