The Classic Slide Rule

George,Y · 2007-05-30 01:21:55

The last post is updated and finishes all the usages of C and D.
The next post is going to illustrate CI and triangluar funtions.

Last edited by George,Y (2007-05-30 01:23:03)

George,Y · 2007-06-01 01:36:52

Sorry, but this post is an exercise post. Because I feel it necessary for you, a fan of the slide rule, to master the existing C and D scales before you forget the knowledge. So an exercise is necessary. And in this worksheet carefully designed by me, problems appeared earlier may have clues for later ones. So just finish it sequentially and you will master the whole of C and D scales.

I Simple Multiplication
2×2
2×3
2×4
2×1
1.2×4.5
1.2×4.5×1.7
1.8×1.05×3.2
1.04³×20 (suppose you have deposited 20 bucks in the bank for 3 years and the interest rate is 4%)

II Simple fractions
2×4/3
1×4/3
4/3
1/3×4
1/5×6
4/3×9/7
4/3×7/9
2×6/10
10/3×5/7

III Large results (use division by 10 wisely)
12×45
12×45×17
2×6
2×9
7/2×9
7/2×9/2

IV Small results (use multiplication by 10 wisely)
0.12×0.0045
1/3×5/7
1/3×4/7
1/3×2/7
1/5×1/9

George,Y · 2007-06-01 13:07:17

Uh, don't forget this
simulated slide rule

Clicking while pressing "Shift" will open a new window with the new page, rather than changing to the new page.

Last edited by George,Y (2007-06-02 02:01:15)

George,Y · 2007-06-02 01:28:16

This post on Trig functions and CI scale.

Open a new window with the virtual slide rule, and just look at the mid-scales in the sliding bar. Above the line there's a "T" symbol and below it an "S" symbol. Check the right end you will find that the T scale ends at 45 and that S ends at 90. Yes, T represents Tangent function, by degree. And S represent Sine function, by degree.

To be precise,

T=Arctan(C/10)
S=Arcsin(C/10)

or C= 10Tan(T)
C=10Sin(S)

So when C=10, T=45 and S=90
when C=1, T=arcsin0.1 and S=arcsin0.1, yes, the domain of arc funtions is incomplete. However there is a solution, which will be discussed later.

Sin(30)=0.5 10Sin(30)=5
Tan(30)=1.414/2=0.707 10Tan(30)=7.07, verify some values yourself.

Also you can note the scale value in red, backwards increasing- they are Cot and Cos.

The real problem comes when the given trig-function is not given by decimals, but fractions instead.

For example, Sinθ=3/5 and Tanθ=3/4. We don't have to transfer them to decimals.
Using the simple fraction calculation discussed earlier, we can get 4/5*10 by
C 5 10(right 1)
D 3 3/5*10
However this is an ineffective calculation because the S scale is a funtion of C instead of a function of D and the intermediate result 4/5*10 on D has little use.
Then we can change the roles C and D play.
C 3 3/5*10
D 5 10(right 1)
4/5*10 is got by 4*(10/5), however you can just think

It's simple algebra, having discussed in the previous page.
But it's of great convenience. Upper is 3, Lower is 5, then let 3 match 5 and 10 match the enlarged trig function value.
When C=10*3/5, S=Arcsin(3/5).
Check the Arctan(3/4) both 37°.

You may encounter 9/16 problem
but 9/1.6=9/16*10
so
C 9/1.6 9
D 1 16

This way you can calculate many values, however some values are just too small.

It doesn't matter because there is "ST" as a compensation. ST means "small tangents"
It calculates the arctangents on [0.01,0.1] So ST=Arctan(C/100)

The angles for ST just compensate T, [0.6, 5.8]U[5.7,45]
How about you have Sines of small value? almost equal to tangents of the same angle. So you may pretend to calculate arcsin by calculating arctan. In addition some slide rules provide sqrt(1-x[sup]2[/sup]) etc, by which you may get an accurate result after shifting sine value to tangent value.

Next, the CI thing and DI thing.
CI=10/C so you can inversions immediately from CI scale and C scale. (DI=10/D)
Another privilage is
CI[sub]1[/sub]D[sub]1[/sub]=CI[sub]2[/sub]D[sub]2[/sub]
This is from
D[sub]1[/sub]/C[sub]1[/sub]= D[sub]2[/sub]/C[sub]2[/sub]
eg.
CI 1 3
D 9 3
or
C 1 3
DI 9 3

You may find it simplier to operate, but I find backward scale and distant scales hard. They also has too large too small problems as CD scales. Almost the same problem
Now calculate 2*7 by both ways and you will discover the off-scale is the same severe.

Last edited by George,Y (2007-06-02 02:03:06)

George,Y · 2007-06-02 20:06:42

Next, LL scale.

Observing LL scale
Using LL scale

George,Y · 2007-06-02 20:55:57

This post is on what can we find only by looking at the LL scale

Here is a photo of a slide rule
click while pressing "shift" to get the pic in a new window

So you can observe the first side of the slide rule.
Note the bottom and the upper part. LL3 LL2 LL1 LL01 LL02 LL03
A little bit confusing.
But just drag the slide bar at the bottom of the window and check the right end of the slide rule.
The mark

And notice the x mark to the right of D scale.
Yes, exactly LL scales are exponent scales of D, note that D is already exponential. So e^D=e^(10^L). Note that LL3 scale increase dramatically.

Other than telling exponent of e and ln backwards,
having these 6 scales of exponent brings about two benefits.

One is that you can easily get the power of 10th or 1/10th easily.
Because e[sup]x[/sup]=e[sup]0.1x×10[/sup]= (e[sup]0.1x[/sup])[sup]10[/sup]
Check 2 on LL2 and the value on the scale above, the LL3 scale. 2^10≈1.03×10[sup]3[/sup]
Yes, if you are familiar with computer, 1KB=2^10Bytes=1024Bytes. 1030 is close to 1024, as far as the slide rule can tell.

Another benefit is the inversions
Because e^x=1/e^(-x) and so on, you can check inversions by two corresponding LL scales. LL3-LL03 LL2-LL02 LL1-LL01. The small scales of LL1 and LL01 can give very accurate inversions of numbers near 1. Suppose you are calculating how much you need to save under the interest rate at 3.5 to get 1000 next year. 1/1.035 is needed. By checking I can see it's 0.9656.

Additionally, just check the positions of 1.01 1.012 1.02 1.03 1.04 on LL1 and 1 1.2 2 3 4 on D. You will notice their positions are near, horizontally.
This is because e^x ≈ 1+x when x is small.
So e^.01x ≈1+.01x
The formula can be derived by derivative:
(e^x)'=e^x
e^(0+h) ≈e^0+(e^x)|'[sub]0[/sub]h- h is small.

Last edited by George,Y (2007-06-02 20:57:35)

brambuilt · 2007-06-18 11:13:34

I found this site that has brand new slide rules from the 70's. So if your wanting a nice new one they should have what you are looking for.
It's http://www.houseofsliderulers.com

Anthony.R.Brown · 2007-06-19 01:40:40

Is there a Condition! Known as SLIDE-RULE-NISM Where someone Loves! Slide Rules!........

A.R.B

George,Y · 2007-06-21 13:57:43

Hey, guys! Sorry for the delay of the last post on powers:

First, let me explain the basics:
Shifting Bases:

Suppose you have got a table of almost all of powers of 3,
..., 3^-0.1, 3^0, 3^0.1, 3^0.2,...
Now, however, you are to calculate a power of 9 instead of 3.

9^1.5

You cannot multiply
However, you know from the table of powers of 3 that 9=3^2
Then 9= (3[sup]2[/sup])[sup]1.5[/sup]=3[sup](2*1.5)[/sup]
Find 2*1.5 and by the table you can immediately get the answer, the power of 9

So with multiplication, a table of powers of a particular number can serve to find powers of any positive number.

That's how a slide rule works to help you guage a power.
LL scale, the power tables of e, e^0.1, e^0.01, e^-1, e^-0.1, e^-0.01 calculate more than powers of the only 5 e related bases.

Now suppose you are calculate 9^1.5 by the scale of e^x and multiplication on exponetial.

Simulated Slide Rule

Use the simulated slide rule (SSR) above
Flip the SSR, find LL3+ at the bottom, e to 20M (meaning 20,000,000)
LL3+=e^D
Now you should find D[sub]1[/sub]
which satisfy
e[sup]D1[/sup]=9

D D[sub]1[/sub]
LL3+ 9

and then find D[sub]2[/sub]=1.5D[sub]1[/sub] on D scale, by multiplication of slide rule.

then you can get e^D[sub]2[/sub]=9^1.5
D D[sub]2[/sub]
LL3+ e^D[sub]2[/sub]

Next topic is shifting LL scales, solving too large and too small problems.

Last edited by George,Y (2007-06-21 13:58:21)

George,Y · 2007-06-21 14:00:14

BTW 1.5D1 can be generated by sliding the middle bar, using C
C 1 1.5
D D[sub]1[/sub] 1.5D[sub]1[/sub]

Last edited by George,Y (2007-06-21 14:00:33)

George,Y · 2007-08-06 22:16:22

Okay, the final part.

Finding logs, roots and using LL shift.

C 1 3
LL3+ 3 27

b c
a d

if you follow the a-b-c-d sequence find 3 first and slide to find 1 to match 3 below, then 3 on the right, last 27 as the result of 3^3

But if you follow the sequence of a-b-d-c, you get log[sub]3[/sub]27 in the end.

And if you follow the sequence of d-c-b-a, you get 27[sup]1/3[/sup] in the end.

That's how simple logs and roots work. Just keep in mind that the D-value represents the exponent to whom e or other can be raised to generate the value on LL , and C-value-pairs represents the ratio between the two exponents on D.

George,Y · 2007-08-06 22:23:25

To illustrate the last sentence:

C: 1 3
D: 1.1 3.3
LL3+: 3 27

e[sup]1.1[/sup]=3 and e[sup]3.3[/sup]=27;

1.1/3.3=1/3

George,Y · 2007-08-16 01:31:28

Okay, I admit I am a Real procrastinator.

Here is the last demonstration for zealots in the slide rule:

Shifting the LL's

Click "Flip to the other side"

We have already learned how to use a slide rule to guage exponentials, roots and logarithms. But again like the multiplication, we have the "off-scale" problem because of the result might be too large or too small.

Just to think to calculate 2 raised to 3, or the cubic root of 8. The result for the former is 8, not in the scale LL2+ where 2 stays, because it's too large. (The largest value on LL2+ is e, approximately 2^1.44<2^3) And the result of the latter is 2, not in the same scale LL3+ where 8 stays.

To illustrate, let's focus on the 2^3.
Knowing that LL3+ is only LL2+^10 and use either formulae below:

Using the first formula:
LL2+: 2 <a>
C: 3<d> right 1(10)<c>
LL3+: 8<e> 1 M<b>

Logically a-b-c-d-e, and e is the result

Using the latter formula:
LL2+: 1.232<d> 2 <a>
C: 3<c> right 1(10)<b>
LL3+: 8<e>

a-b-c-d-e

BTW: e-d-c-b-a is the route to get the cubic root of 8

The last tip: for more accurate results for something like 3^7.8, you can split the exponent to a sum, like 7.8=2.8+5 and then multiplicate the splitted power, like 3[sup]2.8[/sup]×3[sup]5[/sup].

Hope you have a wonderful time with your own slide rule. Hope you love it!

The End

Last edited by George,Y (2007-08-18 02:36:49)

Anthony.R.Brown · 2007-08-18 02:04:19

A.R.B

As far as this Post is Concerned! it looks like SLIDE RULE´S!...................................................................................................

MathsIsFun · 2007-08-18 10:09:57

George, you have provided one of the best slide-rule resources on the web!

I wonder if we could develop our own virtual slide rule with a few modern twists? Like digital readout of the crosshairs, selectable conversions, ... any ideas?

uche · 2007-08-18 13:30:04

wow cool

Kargoneth · 2007-08-19 03:16:27

What an intriguing device... so simple, yet with proper knowledge and a precisely made slide, a very accurate and fast analog calculator. Many thanks, George.

(I like that internet version... you can pull the hairline slider off the device, lol)

George,Y · 2007-08-26 09:48:53

MathsIsFun wrote:

George, you have provided one of the best slide-rule resources on the web!
I wonder if we could develop our own virtual slide rule with a few modern twists? Like digital readout of the crosshairs, selectable conversions, ... any ideas?

Tks!

But well, it's a little bit hard to add those equipments on a tiny thin slide rule. For example, you can use a flexible electric resistant to record the position where the hairline is in. But it will be a larger device to tell how large the resistant is
+
|The starting position
|_______________________________
|The hairline
-
By dectecting the current magnitude from + to - under a certain voltage, we can determine how far the hairline is from the starting postion, then we have to transform the distance to real exponetial scales on the slide rule.

Perhaps there is a simplier way out of it. Is there a natural law that naturally changes exponetially? Then we may not employ the linear current-resistant phenomenon.

MathsIsFun · 2007-08-26 10:25:36

I don't know about making a real one, but maybe an internet version like the ones you found ... only better.

Take the best features of a slide rule and a calculator and merge them.

For example I made a slider for percentages here: Math Percentages. You can type in a number and then use a slider to get an answer.

Math Is Fun Forum

#26 2007-05-30 01:21:55

Re: The Classic Slide Rule

#27 2007-06-01 01:36:52

Re: The Classic Slide Rule

#28 2007-06-01 13:07:17

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#29 2007-06-02 01:28:16

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#30 2007-06-02 20:06:42

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#32 2007-06-18 11:13:34

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#33 2007-06-19 01:40:40

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#37 2007-08-06 22:23:25

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#38 2007-08-16 01:31:28

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#40 2007-08-18 10:09:57

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#41 2007-08-18 13:30:04

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#42 2007-08-19 03:16:27

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