pathology_doc: Ginny Weasley (film) clutching Riddle's diary: Ginny/Horcrux OTP (Default)
Dear Sun/Hemmi,

I know it's at least half a century too late for this rant, and probably more. I know there's absolutely nothing I can do now except curse the shade of Jiro Hemmi and all who worked under him, but what the hell were you thinking when you drafted the design of the Model 152 and didn't include a common log scale on an otherwise incredibly useful slide rule?

OTOH, you do get bonus points for the Teledyne-Post 44BA-47O: the extra precision between 4 and 5 on the C and D scales (keeping the 0.02 divisions when just about everyone else with a 10 inch scale reverts to 0.05) is tops. If I had to start up slide rule manufacture again, I'd go with this model modified to the Rietz scale set. There's more than enough room on the front for the L scale if you move the manufacturer's logo back where Post's used to be.

(They did, to their credit, rectify the omission on the 153. And as a bonus, they put on an extra scale to allow calculation of hyperbolic functions.)
pathology_doc: Ginny Weasley (film) clutching Riddle's diary: Ginny/Horcrux OTP (Default)
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 e6.54 first divide 6.54 by 2.303, obtaining quotient 2 and remainder 1.934."

In other words, separate out e2.303 (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. (e2.303)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. e1.95 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, e17.4.

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), e1.265 = 3.54 and multiply by the 10 million. They say 3.60, about a 2% error there.

Not bad for a six inch slipstick...


pathology_doc: Ginny Weasley (film) clutching Riddle's diary: Ginny/Horcrux OTP (Default)

March 2015

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