## Introduction

At least one of my other posts ** How Distance Reduces The Power of Radio Waves** requires an understanding of mathematical proportionality. In particular it requires an understanding of

**Inverse Proportion**. So I have written this article to try and clarify it. It gives an explanation of the two types of

**proportion**, otherwise known as

**variation**, which are:

**Direct Proportion**,**Inverse Proportion**otherwise known as**Indirect Proportion**.

## The Symbol Used To Represent Proportion

In math there is a symbol for proportionality thus:

**∝**

So if **y** is proportional to **x** it is shown as:

**y ∝ x**

## Direct Proportion

Anything that is in **direct proportion** to something else rises and falls in step with it, i.e. if one doubles in size so does the other.

So if **y ∝ x** then **2y ∝ 2x** and **y/2 ∝ x/2**.

### In the graphs below y ∝ x (a linear function)

**Fig. 1** above shows a chart with three graphs where the values on the **y** axis go up and down in ** direct proportion** to the values on the

**x**axis.

i.e. **y ∝ x**

or **y = kx** where **k** is a constant.

### Examples of Direct Proportion

My first three examples are from each of the graphs in **Fig. 1**. I’ve used the same value for **x**, i.e. **x = 3**, throughout. Each uses a different constant **k** thus:

**Example 1:**Let**k = 1**in this example, making**y = 1x**. So**y = (1 × 3) = 3**.**Example 2:**Let**k = 3**in this example, making**y = 3x**. So**y = (3 × 3) = 9**.**Example 3:**Let**k = 6**in this example, making**y = 6x**. So**y = (6 × 3) = 18**.

My next three examples, **4 to 6**, assume the value of **x** is doubled to **6**. **y**, being in proportion, is doubled too:

**Example 4:**When**k = 1**,**y = (1 × 6) = 6**. (Previously in**example 1**,**y**was**3**.)**Example 5:**When**k = 3**,**y = (3 × 6) = 18**. (Previously in**example 2**,**y**was**9**.)**Example 6:**When**k = 6**,**y = (6 × 6) = 36**. (Previously in**example 3**,**y**was**18**.)

In the examples above **y** rises and falls in step with **x** and so has gone up in ** direct proportion** to

**x**.

Since **y = kx**, **x** and **y** have a constant ratio **k** thus: **y/x = k**

### In the graph below y ∝ x^{2} (a non-linear function)

**Fig. 2** above shows a graph where the values on the **y** axis go up and down in ** direct proportion** to the

**of the values on the**

*squares***x**axis.

i.e. **y ∝ x ^{2}** (x squared)

or **y = kx ^{2}** where

**k**is a constant.

**k = 1** in this example making **y = 1x ^{2}**.

So if **x = 5** then **y = 1 × (5 × 5) = 25**.

Then if **x** is doubled from **5** to **10**, **y** is ** quadrupled** from

**25**to

**100**,

i.e. **y = 1 × (5 × 2) × (5 × 2)**.

So **y** now rises and falls in step with **x ^{2}** and is no longer in

**to just**

*direct proportion***x**but is now in

**to**

*direct proportion***x**.

^{2}**See other people’s explanations:**

## Inverse Proportion or Indirect Proportion

Anything that is in inverse proportion (or indirect proportion) to something else will fall when the other thing rises and rise when it falls, i.e. if one doubles in size the other halves.

So if **y ∝ 1/x** then **2y ∝ 2/x** and **y/2 ∝ 1/(2x)**.

### In the graph below y ∝ 1/x (a non-linear function)

**Fig. 3** above shows a graph where the values on the **y** axis go up and down in ** inverse proportion** to the values on the

**x**axis.

i.e. **y ∝ 1/x**

or **y = k/x** where **k** is a constant.

**k = 1** in this example making **y = 1/x**.

So if **x = 5**, **y = 1/5 = 0.2**

Then if **x** is doubled to **10**, **y** is halved to **0.1**,

i.e. **y = 1/(2 × 5) = 1/10 = 0.1**

So **y** falls in step with **x** rising, and rises in step with **x** falling since **y** is in ** inverse proportion** to

**x**.

### In the graph below y ∝ 1/x^{2} (a non-linear function)

**Fig. 4** above shows a graph where the values on the **y** axis go down in ** inverse proportion** to the

**of the values on the**

*squares***x**axis.

i.e. **y ∝ 1/x ^{2}**

or **y = k/x ^{2}** where

**k**is a constant.

**k = 1** in this example making **y = 1/x ^{2}**.

So if **x = 5**, **y = 1/(5×5) = 1/25 = 0.04**.

Then if **x** is doubled to **10**, **y** is ** quartered** to

**0.01**,

i.e. **y = 1/(2**×**5)**×**(2×5) = 1/(10****×10) = 1/100 = 0.01**.

So **y** falls in step with **x ^{2}** rising, and rises in step with

**x**falling since

^{2}**y**is no longer in

**to just**

*inverse proportion***x**but is now in

**to**

*inverse proportion***x**.

^{2}**See other people’s explanations:**

### Comparing The Inverse Proportion 1/x With 1/x^{2}

This is the same as comparing the variation of apparent size of objects seen at a selection of distances with their visual areas at those distances.

If I said an object’s apparent size (height and width) was ** inversely proportional** to its distance away

**(x)**, I would mean the further away it got the smaller it would look in both height and width

**(1/x)**. So if the distance was doubled its apparent size would halve

**(1/2x)**and if the distance was doubled again its apparent size would halve again

**(1/4x)**. It would then appear to be a quarter of its original size because its distance would be four times the original distance away.

This ** inverse proportionality** relating apparent size to distance away is supported by

**Casey Rule**, a programmer and composer, in his quote below.

An object’s apparent size within a field of view decreases linearly as it’s distance from the viewer increases. So, a circle that appears 1 inch in diameter when 5 meters from the camera will appear to be 5 inches when 1 meter from the camera.

by Casey Rule re Relationship between the size of the object and the distance to the camera

I have checked that **1/x** or **x ^{-1}** is a non-linear function. The only power of

**x**that is linear is

**x**. I cite InesWalston at Brainly. I checked this because Casey Rule states:

^{1}*An object’s apparent size within a field of view decreases linearly as it’s distance from the viewer increases.*And his use of the word linearly may be confusing. I think the diagram below might show what he means.

Now remembering all those formulae for calculating the areas of various shapes such as: rectangles, circles, triangles, parallelograms, etc. will make it obvious that the areas of those shapes are not inversely proportional to the observer’s distance from them as their sizes (heights and widths) are. Those areas are in fact inversely proportional to the ** square** of the distance away from the observer

**(1/x**.

^{2})### Comparing The Inverse Proportion Graphs In Figs. 3 & 4.

**Fig. 5** compares {**y** when it’s inversely proportional to **x**} with {**y** when it’s inversely proportional to **x ^{2}**}.

i.e. it compares **y = k/x** with **y = k/x ^{2}** when

**k = 1**.

You can clearly see from these graphs that the **1/x ^{2}** graph drops rapidly for small values of

**x**compared to the

**1/x**graph. This is good news for anyone concerned about radiated power as it usually results in the received power at any point being less than expected.

**NOTE:** The graphs in **Fig. 5** are based on the concept that the power levels in the real world graph **y = (k/x ^{2})** and surmised graph

**y = (k/x)**are the same when the unit of distance from the source is

**1**, according to the units used for

**x**, e.g.:

**1 metre**when**x**is measured in metres,**1 yard**when**x**is measured in yards,**1 foot**when**x**is measured in feet.

**NOTE:** In all the examples and discussions shown here **x, y** and **k** can have negative values from the mathematical viewpoint. However I am trying to aim the discussion towards the radiation of positive power and energy where negative values don’t usually play a part.

**Note:** The section below may be removed from this post and inserted in my future post ** How Distance Reduces The Power of Radio Waves**.

### How These Observations Relate To The Radiation of Energy

In physics the power of radiated signals containing energy, such as radio waves, light and sound, diminish in the same way as the apparent area of a distant object. Their power diminishes in a way that is * inversely proportional to the square* of the distance over which they are transmitted. That means doubling the distance reduces the power not to just a half but to

**of what it was, and doubling it again reduces it to**

*a half of a half which is a quarter***of what it was.**

*a quarter of a quarter which is a sixteenth*This “square law” makes all the difference to the power behind received energy. It reduces it to a lower level than you might expect.

If you were sent to this post while reading ** How Distance Reduces The Power of Radio Waves** you should now return to it and continue reading.

## Leave a Reply