Difference between revisions of "Logarithmic Scale and the Slide Rule"

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Now, say we want to do the addition <math>4+3</math>, how would you use the two ruler to do such an addition? Well we will place <math>S-0</math> over <math>B-4</math> and the number opposite <math>S-4</math> on <math>B</math> is the answer which is <math>7</math>.
 
Now, say we want to do the addition <math>4+3</math>, how would you use the two ruler to do such an addition? Well we will place <math>S-0</math> over <math>B-4</math> and the number opposite <math>S-4</math> on <math>B</math> is the answer which is <math>7</math>.
[[Image:Demo1.png|center|1000px]]
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[[Image:Demo1.png|center|800px]]
  
 
What about <math>5+7</math>?
 
What about <math>5+7</math>?
  
[[Image:Demo2.png|center|1000px]]
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[[Image:Demo2.png|center|800px]]
  
 
The number opposite <math>S-7</math> went beyond scale <math>B</math>. What should we do then? Well we can extend the B scale like shown below.
 
The number opposite <math>S-7</math> went beyond scale <math>B</math>. What should we do then? Well we can extend the B scale like shown below.
  
[[Image:Demo3.png|center|1000px]]  
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[[Image:Demo3.png|center|800px]]  
  
 
Well then the answer is 12. But suppose our original scales are 10 inches, making it longer will force the rule to be inconveniently cumbersome. So what is the next solution? The more observant of you will find that <math>S-0</math> is over <math>B-5</math> and <math>S-(1)0</math> is over <math>B-(1)5</math>. If S-0 were over B-7, then S-(1)0 would be over B-(1)7. This should not come at a surprise since whatever single digit number <math>x</math> you add to the number 10 would end up <math>(1)x</math>. Hence the right hand digit does not change. All we have to pay attention the left hand digit. Hence for the previous case, we can actually place <math>S-(1)0</math> over <math>B-5</math> and the number opposite <math>S-7</math> is thus the right hand digit of the answer. Keeping track of the left hand side of the digit, we get the answer 12 as shown below.
 
Well then the answer is 12. But suppose our original scales are 10 inches, making it longer will force the rule to be inconveniently cumbersome. So what is the next solution? The more observant of you will find that <math>S-0</math> is over <math>B-5</math> and <math>S-(1)0</math> is over <math>B-(1)5</math>. If S-0 were over B-7, then S-(1)0 would be over B-(1)7. This should not come at a surprise since whatever single digit number <math>x</math> you add to the number 10 would end up <math>(1)x</math>. Hence the right hand digit does not change. All we have to pay attention the left hand digit. Hence for the previous case, we can actually place <math>S-(1)0</math> over <math>B-5</math> and the number opposite <math>S-7</math> is thus the right hand digit of the answer. Keeping track of the left hand side of the digit, we get the answer 12 as shown below.
  
[[Image:Demo4.png|center|1000px]]
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[[Image:Demo4.png|center|800px]]
  
 
What about <math>24+58</math>? Then we need a more finely divided scale with ten more division between each consecutive numbers. What we are going to do is to compute <math>2.4+5.8</math> and them move the decimal place afterward. Refer to the diagram below. Place S-0 over B-2.4 and the number opposite S-5.8 is 8.2. Move decimal point to the right, we have 82 which is the real answer. Now, you should realize that the same computation is conducted even if the desired calculation is <math>240+580</math> or <math>2400+5800</math>. All we have to do is to pay attention to the decimal place. Even though there is no limits to the subdivision, there is a limit to how finely we con read the subdivisions. Not only that, markings on the ruler have perceptible thickness themselves so there is actually limits to how many subdivisions we could put on the ruler. Therefore, we are condemned to inexactness if you desire to calculate <math>18578+8473954</math> in which case we have to approximate.   
 
What about <math>24+58</math>? Then we need a more finely divided scale with ten more division between each consecutive numbers. What we are going to do is to compute <math>2.4+5.8</math> and them move the decimal place afterward. Refer to the diagram below. Place S-0 over B-2.4 and the number opposite S-5.8 is 8.2. Move decimal point to the right, we have 82 which is the real answer. Now, you should realize that the same computation is conducted even if the desired calculation is <math>240+580</math> or <math>2400+5800</math>. All we have to do is to pay attention to the decimal place. Even though there is no limits to the subdivision, there is a limit to how finely we con read the subdivisions. Not only that, markings on the ruler have perceptible thickness themselves so there is actually limits to how many subdivisions we could put on the ruler. Therefore, we are condemned to inexactness if you desire to calculate <math>18578+8473954</math> in which case we have to approximate.   
  
[[Image:Demo6.png|center|1000px]]
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[[Image:Demo6.png|center|800px]]
  
 
The reverse process is subtraction. I need picture but my powerpoint in the server which cannot be accessed not...God I hate this power outage.  
 
The reverse process is subtraction. I need picture but my powerpoint in the server which cannot be accessed not...God I hate this power outage.  
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Use <math>5.78 \times 9.849 \times 10^5</math>. Place <math>C-10</math> over <math>D-9.85</math>, read the number opposite <math>C-5.78</math> which is approximately <math>5.69</math>. Now, take note that slide rule does not keep track of the decimal place so we have to keep track of it. Since <math>5.78 \approx 6</math> and <math>9.849 \approx 10</math>, the answer is a little less than <math>60</math>. Therefore the answer <math>56.9</math>. In addition, we have to multiply <math>56.9</math> with <math>10^5</math> to get the answer for <math>578 \times 9849</math> is approximately <math>5690000</math>. Use a calculator to check the answer, we have 5692722 which is not that far away from our approximated answer.  
 
Use <math>5.78 \times 9.849 \times 10^5</math>. Place <math>C-10</math> over <math>D-9.85</math>, read the number opposite <math>C-5.78</math> which is approximately <math>5.69</math>. Now, take note that slide rule does not keep track of the decimal place so we have to keep track of it. Since <math>5.78 \approx 6</math> and <math>9.849 \approx 10</math>, the answer is a little less than <math>60</math>. Therefore the answer <math>56.9</math>. In addition, we have to multiply <math>56.9</math> with <math>10^5</math> to get the answer for <math>578 \times 9849</math> is approximately <math>5690000</math>. Use a calculator to check the answer, we have 5692722 which is not that far away from our approximated answer.  
  
[[Image:Logdemo1.png|center|1400px]]
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[[Image:Logdemo1 5.png|center]]
 
[[Image:Logdemo1 5.png|center]]
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We first look for <math>\frac {1}{9.849}</math>.  
 
We first look for <math>\frac {1}{9.849}</math>.  
[[Image:Logdemo2.png|center|1400px]]
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[[Image:Logdemo2.png|center|1000px]]
 
Read the number opposite to <math>CI-9.849</math> on the <math>C</math> scale which is <math>1.011</math>. Keeping track of the decimal, the answer is <math>1.011 \times 10^{-1}</math>
 
Read the number opposite to <math>CI-9.849</math> on the <math>C</math> scale which is <math>1.011</math>. Keeping track of the decimal, the answer is <math>1.011 \times 10^{-1}</math>
 
[[Image:Logdemo2 5.png|center]]
 
[[Image:Logdemo2 5.png|center]]
  
 
Now we do the operation <math>5.78 \times 1.011 \times 10^{-1}</math>
 
Now we do the operation <math>5.78 \times 1.011 \times 10^{-1}</math>
[[Image:Logdemo3.png|center|1400px]]
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[[Image:Logdemo3.png|center|1000px]]
 
Place <math>C-1</math> over <math>D-5.78</math> and read the number opposite <math>C-1.011</math> which is <math>5.86</math>. Keeping track of the decimal, the answer is approximately <math>0.586</math>. The answer from a calculator which gives the answer <math>0.58686...</math>, we have the answer pretty close.  
 
Place <math>C-1</math> over <math>D-5.78</math> and read the number opposite <math>C-1.011</math> which is <math>5.86</math>. Keeping track of the decimal, the answer is approximately <math>0.586</math>. The answer from a calculator which gives the answer <math>0.58686...</math>, we have the answer pretty close.  
  
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To do squaring, we need a new scale that corresponds to the old C and D scale.  
 
To do squaring, we need a new scale that corresponds to the old C and D scale.  
  
[[Image:Sqr1.png|center|1450px]]
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[[Image:Sqr1.png|center|1000px]]
  
 
After we have fixed those points, we can then recalibrate the B scale on a logarithmic scale. Then we can do squaring by choosing number on the C scale and read the number on the B scale.  
 
After we have fixed those points, we can then recalibrate the B scale on a logarithmic scale. Then we can do squaring by choosing number on the C scale and read the number on the B scale.  
  
[[Image:Sqr02.png|center|1450px]]
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[[Image:Sqr02.png|center|1000px]]
  
 
[[Image:Sqr025.png|center]]
 
[[Image:Sqr025.png|center]]
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Example: <math>\sqrt {4850}</math>.
 
Example: <math>\sqrt {4850}</math>.
  
[[Image:Sqrt01.png|center|1400px]]
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[[Image:Sqrt01.png|center|1000px]]
  
 
[[Image:Sqrt015.png|center]]
 
[[Image:Sqrt015.png|center]]
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Similar to squaring, we need a new cube scale that corresponds the old C and D scales. Hence, we have the new K scale.
 
Similar to squaring, we need a new cube scale that corresponds the old C and D scales. Hence, we have the new K scale.
  
[[Image:Cube01.png|center|1400px]]
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[[Image:Cube01.png|center|1000px]]
  
 
Then we recalibrate the scale according to the logarithmic scale and hence we have the new K scale.  
 
Then we recalibrate the scale according to the logarithmic scale and hence we have the new K scale.  
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Example: <math>783^3</math>
 
Example: <math>783^3</math>
  
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[[Image:Democube25.png|center]]
 
[[Image:Democube25.png|center]]
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Example: <math>\sqrt [3] {783}</math>
 
Example: <math>\sqrt [3] {783}</math>
  
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[[Image:Democube3.png|center|1000px]]
  
 
[[Image:Democube35.png|center]]
 
[[Image:Democube35.png|center]]

Revision as of 14:31, 29 June 2010

Inprogress.png
150 Extra Engineers
150 extra engineers thumb.jpg
Field: Algebra
Image Created By: IBM
Website: Global Nerdy --- Tech Evangelist Joey deVilla on software development, tech news and other nerdy stuff

150 Extra Engineers

This was a picture of an IBM advertisement back in 1953.


Basic Description

This was a picture of an IBM advertisement back in 1953 during an age when "computer" referred to human who did calculations and computation solely. The advertisement boasted that


"An IBM Electronic Calculator speeds through thousands of intricate computations so quickly that on many complex problems it's just like having 150 Extra Engineers. No longer must valuable engineering personnel...now in critical shortage...spend priceless creative time at routine repetitive figuring. Thousands of IBM Electronic Business Machines...vital to our nation's defense...are at work for science, industry and armed forces, in laboratories, factories and offices, helping to meet urgent demands of greater production."


Doesn't one just want to go back to that age when engineers were in "critical shortage" and "If you had a degree, you had a job. If you didn't have a job it's because you didn't want one." Oh well...


Notice that in the picture, in addition to the lack of women engineers back in those days, the male engineers almost all have receding hairlines (possibly due to constant overwork and the disproportional allocations of oxygen and other nutrients between brain cells and scalp skin). It was also typical of the male engineer to be defined by a standard uniform: "white shirt, narrow tie, pocket protector and slide rule." I need to do footnote here. But I don't know how. Yeah! Pocket protector! I did not know what it is until I saw the picture in the Scientific American Article, When Slide Rule Ruled, by Cliff Stoll, who is a brilliant engineer, physicist and educator, and whose TED talk is both hilarious and inspiring. However, it is neither Cliff Stoll, nor the pocket protector that I want to talk about. (It is just a thing that, despite making the engineers look very geeky and very attractive in author's opinion, protected their shirt pockets from being worn out so quickly due to the numerous engineering essentials they had in there.) Rather, it is the ruler looking stuff that is the engineer's hand that I want to talk about. The Slide Rule.


Born into the digital and automatic age, hardly anyone of our generation ever gets to know anything that is analog or manual. Not only that, we hate to have anything to do with stuff that is analog. We have been cultured to base our life and happiness on gadgets that allow us to access the world and all the information with a finger tip. We don't realize that in fact, things that are analog laid the foundation for our modern society and ushered us into the digital age. Slide Rule is one of "those analog things". In the pre-computer age, it was one of those ingenious tools that enabled engineers, mathematicians and physicists to do calculations, and because of it, we have witnessed the erections of skyscrapers, harnessing of hydroelectric power, building of subways, advancement in the aeronautics, beyond that, it helped us won the WWII, sent astronauts into the space (as a matter of fact, a Picket 600-T Dual base Log Log slide rule was carried by the Aopollo crew to the space and moon should the on-board computers fail) and eventually, hastened its own demise.

A More Mathematical Explanation

Note: understanding of this explanation requires: *Algebra

Logarithm and logarithmic Scale

This section is not to preach you on logarithms since you are we [...]

Logarithm and logarithmic Scale

This section is not to preach you on logarithms since you are well familiar with its properties. Rather, I will simply introduce logarithmic scale which is less mentioned and understood but widely used and essential to the construction of a slide rule.

We are familiar with the linear scale, namely the equally divided scale, from the number 0 to whichever number as shown in scale A. We then use the identity x=log_22^x to relabel scale A to produce scale B. Then what we do to transform the linear scale to logarithmic scale is simply to take away the log_2, turning scale B to scale C. Calculating the numbers on scale C, we end up with scale D.

Sliderule001.png


Now, notice that the given numbers are equally divided but the differences between consecutive numbers are not the same. Our goal is to produce a logarithmic scale that has all the integers from 1 to 8. For example, if we want the number 3 on scale D, there will be a number x on scale A that corresponds to the number 3 on scale D. We have <template>AlignEquals




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About the Creator of this Image

International Business Machines (IBM) is a multinational computer, technology and IT consulting corporation headquartered in Armonk, North Castle, New York, United States. IBM is the world's fourth largest technology company and the second most valuable by global brand (after Coca-Cola). IBM is one of the few information technology companies with a continuous history dating back to the 19th century. IBM manufactures and sells computer hardware and software (with a focus on the latter), and offers infrastructure services, hosting services, and consulting services in areas ranging from mainframe computers to nanotechnology.


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