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Pythagorean Method Update

Hello,

just a final comment about triangle solving using the Pythagorean method on the slide rule, that I proposed a few years ago.

When one of the catheters of the Egyptian triangle is nearly equal in length to the hypotenuse, the other catheter will be short, AND calculating its length accurately will then become difficult.

An example:

Consider the case where the length of the hypotenuse is 5 and the length of one catheter is 4.9.

By the proposed method, we should first divide 5 by 4.9, square the result, and then subtract 1.

We get (5/4)^2-1 or approximately 1.04 -1 = 0.04 with only ONE significant digit remaining.

Then, after taking the square root of 0.04 and multiplying by 4.9, we get 0.98 as the length of the hypotenuse, while the correct length actually is 0.995. The error here is greater than one percent.

The solution:

Using the CONJUGATE RULE, we rewrite (5^2-4.9^2)^½ as ((5-4.9)(5+4.9))^½ in order to arrive at (0.1 x 9.9)^½ = 0.995 as expected with no degradation of accuracy.

In cases like this, we may have to use paper and pen for additions and subtractions, only then deviating from the proposed "pen and paperless" method.

Cheers from

Kenneth J.

Sweden

What scale is used to calculate tangent in the range 85°-89.9990°, greater than 84°?

At that range, the tan function zooms off to infinity, so there is no scale on the slide rule.

Hello, I am a student from Korea. Thank you very much to make these wonderful pages. I want to share this with my friends but it is hard to share because it is written in English. Can I translate your “Slide Rule Calculations by Example” to Korean and pot it to my blog and website? I will leave an original link.

Sure, go ahead, sounds like a good idea!

New slide rule tricks

Hello Derek,

this web page should be a very interesting place for any slide rule enthusiasts still around. A big THANK YOU from an old-timer!

Also, here are two brand new(?) nice tricks to try out on your slipstick:

How do you perform Pythagorean operations when your sliderule has no such scale?

Try this:

Example 1

Find (1 – 0.5^2)^½

Set the cursor line at 5 on the D scale. Under the cursor line on the A scale, read the square 0.5^2 = 0.25.

The ”ones-complement” of that number is 1- 0.25 = 0.75 and you don't even have to calculate it.

Just read the A-scale backwards from A100 (representing the number 1.00 in this case) to the cursor line. That distance is obviously 0.75.

Therefore, set the cursor line at A75 (representing the number 0.75) and read the square root 0.866 on the D scale under the cursor line.

Answer: (1 – 0.5^2)^½ = 0.866

Example 2

How do you solve all Pythagorean triangles on the slide rule without any pen-and-paper operations whatsoever?

Try this on the Egyptian triangle to find the hypothenuse when the sides are 3 and 4:

Divide 4 by 3 on the C and D scales and find 1.33... on the D scale opposite to C1 on the slide.

Read the square of 1.33... (approximately 1.78) on the A scale opposite to B1. Add 1 mentally and set B1 at 2.78 on the A scale.

The square root of that number is 1.66.. on the D scale opposite to C1.

Move the cursor line to C3 in order to multiply 1.66.. by 3 and find the answer 5 under the cursor line on the D scale.

Answer: 3 x (1 + (4/3)^2)^½ = 5

Example 3

Try this on the Egyptian triangle to find the third side when the hypothenuse is 5 and one side is 4:

Set C4 opposite to D5 to divide 5 by 4. Read the square of that result as (1.25)^2 = 1.56...

(set the cursor line momentarily at C 1.25 in order to find the more accurate estimate (1.25)^2 = 1.563 )

Subtract 1 mentally and set the cursor line at 0.563 (actually 56.3) on the A scale.

Set CI 4 at the cursor line in order to multiply by 4, and read the result 3 on the D scale opposite to C1.

Answer: 4 x ((5/4)^2 -1)^½ = 3

Example 4

The same as in example 3, except that this time the hypothenuse is 5 and one side is 3.

Set C3 opposite to D5 to divide 5 by 3. Read the square of that result as (1,66.)^2 = 2.78 on the A scale.

Subtract 1 mentally and set the cursor line at 1.78 on the A scale.

Set CI 3 at the cursor line in order to multiply by 3 and read the result 4 on the D scale opposite to D10.

Answer: 3 x ((5/3)^2 -1)^½ = 4

In the last two examples we must divide 5 by either 3 or 4 in order to avoid a negative result after the subsequent subtraction of 1.

Dividing 3 by 5 or 4 by 5 doesn't work, since the following subtraction by 1 gives a negative number, and the subsequent square root operation fails.

However, example 2 works fine if we first divide 3 by 4, since the subsequent addition of 1 still produces a positive number.

- - -

Has no one ever stumbled on these very simple tricks before? That's hard to believe, but I have searched the internet eagerly on this subject several times, always without success. Still, I'm quite sure this is nothing new after all. It was just not reported. Now it is.

------------------------------------------------------------------------------------------------

Oops!

This was sent from

Kenneth J.

Sweden

Are your slide rule simulators open source? If so, how can I access it? I am asking on behalf of a computer museum.

Sure go ahead, but be warned that the photographs would still be copyright by the Pickett company, which originally made the slide rules.

What scale is used to calculate tangent in the range 85°-89.9990°, greater than 84°?

The tangent value increases too rapidly in that range, so it would be inaccurate on a slide rule. You would have to add 90 degrees to your value, take the tangent of that, then take the reciprocal to get the tangent.

QUESTION: I'm happened to meet your antiquark.com website when I searched slide rule. I tried your virtual slide rule page, it is very interesting. But when I tried to calculate 2.3 × 3.4 ( Simple Multiplication (uses C and D scales) ) I found my result is 6.5. Actually it should be 7.8. I don't know why I get the wrong result. I followed your steps. Could you read the attachment ( screenshot ) to give me some hints? Thanks for your website and your help.

ANSWER: Actually at that point in the scale you're in the "fine grained" part of the scale. It's not 2.3 you're looking at, but 1.23. The 2.3 value will be about 5cm to the right of that. However what you have described is a common mistake!

Where to buy them?

QUESTION: Hi, where can I purchase the Pickett n909-es

ANSWER: Usually the “go to” place to buy sliderules is eBay. There are some dedicated slide rule sellers on the internet outside of eBay, but they seem to charge a lot more.

QUESTION: I wonder if you could help me with a concept. For the past month I’ve been trying to learn the slide rule. Haven’t touched one in almost 45 years probably, but it’s fun to relearn it all. I’ve got several rules, and am working not with a K&E 4080-3 log log trig rule using the KE manual. (I’ve become quite hooked on this, even joined the Oughtred society!) I’m on the log log scales now. This rule has LL1, 2, and 3 and LL01, LL02, LL03. I’m doing pretty well figuring it out, but one portion of the manual (I’m determined to get through it and do every problem) has me confused after staring at it for two days. The author is trying to get us to visualize the LL scales laid out end to end, with three LL scales attached, and directly opposite them the three LL0 scales strung end to end, and between them a long C scale. So far so good. Here’s a copy of the offending page. I am lost as to why the example he gives would require 4 hypothetical end-to-end C scales rather than three. Can you perhaps give me a clue as to what he’s getting at? I’m also confused about the statement relating to the 4 hypothetical C scale coverage, from.1 to 1, 1 to 10, 10 to 100, and 100 to 1000. Why? Thank you very much.

ANSWER: I will try to answer your question to the best of my ability! The main concept of LL scales is that they aren't "cyclical." (for lack of a better term). The regular scales of a slide rule are "cyclical" in that if you were to extend them, they would basically form a repetitive scale. This is one of the amazing properties of the logarithm. So you can get a lot done without multiple log scales. For example, you don't need C, C2, C3 etc. A simple C scale will do. However Log-Log scales don't have that property. As a result, you actually need a very long slide rule to have a usable Log-Log scale. And the way of making a long scale short, is to break it down into pieces. On my Pickett N3, there are four scales, and they allow calculations up to about 20,000. Also, take a look at the following link on my site: http://www.antiquark.com/sliderule/sim/sr-calcs-by-example.html#mozTocId398379 It explains how to use LL scales, and the technique needed to switch to other scales. Switching scales is based on a property of exponents. I myself have found that running through examples, gives an intuitive understanding of slide rule operations.

#slide rules #log log #log log scale #LL0

Scanning Technique

QUESTION: how did you scan the images for your rules?

ANSWER: Hmmm... lets see if I remember... what I did for scanning was: 1. Try to adjust the slide rule parts so the slider and framing part are both properly aligned. 2. Remove the cursor. 3. Scan the slider and frame both together. That way, they will be (in relation to the scanner) at the identical height off the glass, identical angle, and with the same color balance. I found that the focus was good with this technique, even though the rules themselves were a few mm off the surface of the glass due to the end connectors. 4. Open the scan in Gimp/Photoshop, then do a minor image rotation to get the rule perfectly horizontal. 5. Chop the image apart to get separate slider and framing parts. 6. Due to the slider being scanned together with the frame, the ends will have been obscured by the end connectors. 7. Run another scan on the slider alone, and copy and paste the end pieces into the slider image. 8. Scan the cursor by itself-- only one side is necessary to fill in for both sides of the ruler. Hope that helps!

#slide rule

Accuracy and Precision

Dear Derek:

I found, and have enjoyed, your tutorial website, and I have been teaching my grandson, now an architectural student, in slide rule use. His teacher insists that his students employ slide rules, rather than calculators, for their work!! Apparently, the students will better learn what they are doing, limited to 1% accuracy (damn good, for when Washington Roebling used his slide rule when designing the Brooklyn Bridge, he still insisted on a 4x safety factor - later reduced to 3x because of a stress recalculation made by the cable wire manufacturer.)

Your exponential example 23 terms the calculation error ( made on a 5", crowded Pickett scale, as being due to scale inaccuracy. I disagree, the error is due to lack of precision in setting and in reading.

You might enjoy reading my 20 year old paper, now published only by a slide rule sale website, the paper being on the topic of accuracy and precision. (Here is the link).

Sincerely, A.M.

Derek’s response:

Thanks for the complements! Yes, in hindsight, I would tend to agree that a lot of error is due to setting and reading. I understand that after a lot of practice, people could develop a high degree of discrimination with the fine markings, and would be able to produce calculations of greater accuracy. I'll take a look at that paper you mentioned... Cheers, Derek

Welcome to the Auxiliary Site

Hello to my new tumblr-based site. The old site, now 10 years old, is creaking at the seams. It’s getting quite painful to even edit. So here you will be able to ask questions and post comments as I gradually re-add the various slide rules mentioned on the old page.

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