formula

Compressors: Formula for Maximum Volume

Suppose we have an audio signal which peaks at L decibels. We apply a compressor with a threshold T (with T being smaller than L, otherwise the compressor will not spring into action) and ratio r. How does this effect the maximum volume of the audio signal? Let’s derive a formula for that. Remember that the compressor leaves the parts of the signal that are below the threshold unchanged and dampens the excess volume (threshold to signal level) by the ratio we set. So the dynamic range from the threshold to the peak, which is L – T, is compressed to (L – T) / r. Hence, the peak volume after compression is:

L’ = T + (L – T) / r

For example, suppose our mix peaks at L = – 2 dB. We compress it using a threshold of T = – 10 dB and a ratio r = 2:1. The maximum volume after compression is:

L’ = – 10 dB + ( – 2 dB – (- 10 dB) ) / 2 = – 10 dB + 8 dB / 2 = – 6 dB

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Overtones – What They Are And How To Compute Them

In theory, hitting the middle C on a piano should produce a sound wave with a frequency of 523.25 Hz and nothing else. However, running the resulting audio through a spectrum analyzer, it becomes obvious that there’s much more going on. This is true for all other instruments, from tubas to trumpets, basoons to flutes, contrabasses to violins. Play any note and you’ll get a package of sound waves at different frequencies rather than just one.

First of all: why is that? Let’s focus on stringed instruments. When you plug the string, it goes into its most basic vibration mode: it moves up and down as a whole at a certain frequency f. This is the so called first harmonic (or fundamental). But shortly after that, the nature of the vibration changes and the string enters a second mode: while one half of the string moves up, the other half moves down. This happens naturally and is just part of the string’s dynamics. In this mode, called the second harmonic, the vibration accelerates to a frequency of 2 * f. The story continues in this fashion as other modes of vibration appear: the third harmonic at a frequency 3 * f, the fourth harmonic at 4 * f, and so on.

String Vibrating Modes Overtones

A note is determined by the frequency. As already stated, the middle C on the piano should produce a sound wave with a frequency of 523.25 Hz. And indeed it does produce said sound wave, but it is only the first harmonic. As the string continues to vibrate, all the other harmonics follow, producing overtones. In the picture below you can see which notes you’ll get when playing a C (overtone series):

(The marked notes are only approximates. Taken from http://legacy.earlham.edu)

Quite the package! And note that the major chord is fully included within the first four overtones. So it’s buy a note, get a chord free. And unless you digitally produce a note, there’s no avoiding it. You might wonder why it is that we don’t seem to perceive the additional notes. Well, we do and we don’t. We don’t perceive the overtones consciously because the amplitude, and thus volume, of each harmonic is smaller then the amplitude of the previous one (however, this is a rule of thumb and exceptions are possible, any instrument will emphasize some overtones in particular). But I can assure you that when listening to a digitally produced note, you’ll feel that something’s missing. It will sound bland and cold. So unconsciously, we do perceive and desire the overtones.

If you’re not interested in mathematics, feel free to stop reading now (I hope you enjoyed the post so far). For all others: let’s get down to some mathematical business. The frequency of a note, or rather of its first harmonic, can be computed via:

(1) f(n) = 440 * 2n/12

With n = 0 being the chamber pitch and each step of n one half-tone. For example, from the chamber pitch (note A) to the middle C there are n = 3 half-tone steps (A#, B, C). So the frequency of the middle C is:

f(3) = 440 * 23/12 = 523.25 Hz

As expected. Given a fundamental frequency f = F, corresponding to a half-step-value of n = N, the freqency of the k-th harmonic is just:

(2) f(k) = k * F = k * 440 * 2N/12

Equating (1) and (2), we get a relationship that enables us to identify the musical pitch of any overtone:

440 * 2n/12 = k * 440 * 2N/12

2n/12 = k * 2N/12

n/12 * ln(2) = ln(k) + N/12 * ln(2)

n/12 = ln(k)/ln(2) + N/12

(3) n – N = 12 * ln(k) / ln(2) ≈ 17.31 * ln(k)

The equation results in this table:

k

n – N (rounded)

1 0
2 12
3 19
4 24
5 28

And so on. How does this tell us where the overtones are? Read it like this:

  • The first harmonic (k = 1) is zero half-steps from the fundamental (n-N = 0). So far, so duh.
  • The second harmonic (k = 2) is twelve half-steps, or one octave, from the fundamental (n-N = 12).
  • The third harmonic (k = 3) is nineteen half-steps, or one octave and a quint, from the fundamental (n-N = 19).
  • The fourth harmonic (k = 4) is twenty-four half-steps, or two octaves, from the fundamental (n-N = 24).
  • The fifth harmonic (k = 5) is twenty-wight half-steps, or two octaves and a third, from the fundamental (n-N = 28).

So indeed the formula produces the correct overtone series for any note. And for any note the same is true: The second overtone is exactly one octave higher, the third harmonic one octave and a quint higher, and so on. The corresponding major chord is always contained within the first five harmonics.

Wavelength (And: Why Is The Sky Blue?)

A very important type of length is wavelength, usually symbolized by the Greek letter λ (in m). It is defined as the distance from crest to crest (one complete cycle) and can easily be calculated for any wave by dividing the speed of the wave c (in m/s) by its frequency f (in Hz):

λ = c / f

What are typical wavelengths for sound? At room temperature, sound travels with a speed of c = 343 m/s. The chamber pitch has a frequency of f = 440 Hz. According to the equation, the corresponding wavelength is:

λ = 343 / 440 ≈ 0.8 m ≈ 2.6 ft

Are you surprised? I bet most people would greatly underestimate this value. Bass sounds are even longer than that. The lowest tone on a four-string bass guitar has a frequency of f = 41.2 Hz, which leads to the wavelength:

λ = 343 / 41.2 ≈ 8.3 m ≈ 27 ft

So the wave coming from the open E string of a bass guitar doesn’t even fit in a common room. In the case of light, the situation is very different. As noted in the introduction, the wavelength of light ranges between 4000 A (violet light) and 7000 A (red light), which is just below the size of a bacterium.

Wavelength plays an important role in explaining why the sky is blue. When light collides with a particle, parts of it are deflected while the rest continues along the initial path. This phenomenon is known as scattering. The smaller the wavelength of the light, the stronger the effect. This means that scattering is particularly pronounced for violet and blue light.

Unless you are looking directly at the Sun, all the light you see when looking at the sky is scattered light coming from the particles in the atmosphere. Since blue light tends to scatter so easily, the sky ends up in just this color. But why not violet? This is a legitimate question. After all, due to its smaller wavelength, violet light is even more willing to scatter. While this is true, it is also important to note that the sun’s rays don’t contain all the colors in the same ratio. In particular, they carry much less violet than blue light. On top of that, our eyes are less sensitive to violet light.

(This is an excerpt from my Kindle book: Physics! In Quantities and Examples)

Distribution of E-Book Sales on Amazon

For e-books on Amazon the relationship between the daily sales rate s and the rank r is approximately given by:

s = 100,000 / r

Such an inverse proportional relationship between a ranked quantity and the rank is called a Zipf distribution. So a book on rank r = 10,000 can be expected to sell s = 100,000 / 10,000 = 10 copies per day. As of November 2013, there are about 2.4 million e-books available on Amazon’s US store (talk about a tough competition). In this post we’ll answer two questions. The first one is: how many e-books are sold on Amazon each day? To answer that, we need to add the daily sales rate from r = 1 to r = 2,400,000.

s = 100,000 · ( 1/1 + 1/2 + … + 1/2,400,000 )

We can evaluate that using the approximation formula for harmonic sums:

1/1 + 1/2 + 1/3 + … + 1/r ≈ ln(r) + 0.58

Thus we get:

s ≈ 100,000 · ( ln(2,400,000) + 0.58 ) ≈ 1.5 million

That’s a lot of e-books! And a lot of saved trees for that matter. The second question: What percentage of the e-book sales come from the top 100 books? Have a guess before reading on. Let’s calculate the total daily sales for the top 100 e-books:

s ≈ 100,000 · ( ln(100) + 0.58 ) ≈ 0.5 million

So the top 100 e-books already make up one-third of all sales while the other 2,399,900 e-books have to share the remaining two-thirds. The cake is very unevenly distributed.

This was a slightly altered excerpt from More Great Formulas Explained, available on Amazon for Kindle. For more posts on the ebook market go to my E-Book Market and Sales Analysis Pool.

How To Calculate Maximum Car Speed + Examples (Mercedes C-180, Bugatti Veyron)

How do you determine the maximum possible speed your car can go? Well, one rather straight-forward option is to just get into your car, go on the Autobahn and push down the pedal until the needle stops moving. The problem with this option is that there’s not always an Autobahn nearby. So we need to find another way.

Luckily, physics can help us out here. You probably know that whenever a body is moving at constant speed, there must be a balance of forces in play. The force that is aiming to accelerate the object is exactly balanced by the force that wants to decelerate it. Our first job is to find out what forces we are dealing with.

Obvious candidates for the retarding forces are ground friction and air resistance. However, in our case looking at the latter is sufficient since at high speeds, air resistance becomes the dominating factor. This makes things considerably easier for us. So how can we calculate air resistance?

To compute air resistance we need to know several inputs. One of these is the air density D (in kg/m³), which at sea level has the value D = 1.25 kg/m³. We also need to know the projected area A (in m²) of the car, which is just the product of width times height. Of course there’s also the dependence on the velocity v (in m/s) relative to the air. The formula for the drag force is:

F = 0.5 · c · D · A · v²

with c (dimensionless) being the drag coefficient. This is the one quantity in this formula that is tough to determine. You probably don’t know this value for your car and there’s a good chance you will never find it out even if you try. In general, you want to have this value as low as possible.

On ecomodder.com you can find a table of drag coefficients for many common modern car models. Excluding prototype models, the drag coefficient in this list ranges between c = 0.25 for the Honda Insight to c = 0.58 for the Jeep Wrangler TJ Soft Top. The average value is c = 0.33. In first approximation you can estimate your car’s drag coefficient by placing it in this range depending on how streamlined it looks compared to the average car.

With the equation: power equals force times speed, we can use the above formula to find out how much power (in W) we need to provide to counter the air resistance at a certain speed:

P = F · v = 0.5 · c · D · A · v³

Of course we can also reverse this equation. Given that our car is able to provide a certain amount of power P, this is the maximum speed v we can achieve:

v = ( 2 · P / (c · D · A) )1/3

From the formula we can see that the top speed grows with the third root of the car’s power, meaning that when we increase the power eightfold, the maximum speed doubles. So even a slight increase in top speed has to be bought with a significant increase in energy output.

Note the we have to input the power in the standard physical unit watt rather than the often used unit horsepower. Luckily the conversion is very easy, just multiply horsepower with 746 to get to watt.

Let’s put the formula to the test.

—————————

I drive a ten year old Mercedes C180 Compressor. According the Mercedes-Benz homepage, its drag coefficient is c = 0.29 and its power P = 143 HP ≈ 106,680 W. Its width and height is w = 1.77 m and h = 1.45 m respectively. What is the maximum possible speed?

First we need the projected area of the car:

A = 1.77 m · 1.45 m ≈ 2.57 m²

Now we can use the formula:

v = ( 2 · 106,680 / (0.29 · 1.25 · 2.57) )1/3

v ≈ 61.2 m/s ≈ 220.3 km/h ≈ 136.6 mph

From my experience on the Autobahn, this seems to be very realistic. You can reach 200 Km/h quite well, but the acceleration is already noticeably lower at this point.

If you ever get the chance to visit Germany, make sure to rent a ridiculously fast sports car (you can rent a Porsche 911 Carrera for as little as 200 $ per day) and find a nice section on the Autobahn with unlimited speed. But remember: unless you’re overtaking, always use the right lane. The left lanes are reserved for overtaking. Never overtake on the right side, nobody will expect you there. And make sure to check the rear-view mirror often. You might think you’re going fast, but there’s always someone going even faster. Let them pass. Last but not least, stay focused and keep your eyes on the road. Traffic jams can appear out of nowhere and you don’t want to end up in the back of a truck at these speeds.

—————————

The fastest production car at the present time is the Bugatti Veyron Super Sport. Is has a drag coefficient of c = 0.35, width w = 2 m, height h = 1.19 m and power P = 1200 HP = 895,200 W. Let’s calculate its maximum possible speed:

v = ( 2 · 895,200 / (0.35 · 1.25 · 2 · 1.19) )1/3

v ≈ 119.8 m/s ≈ 431.3 km/h ≈ 267.4 mph

Does this seem unreasonably high? It does. But the car has actually been recorded going 431 Km/h, so we are right on target. If you’d like to purchase this car, make sure you have 4,000,000 $ in your bank account.

—————————

This was an excerpt from the ebook More Great Formulas Explained.

Check out my BEST OF for more interesting physics articles.

Sources:

http://ecomodder.com/wiki/index.php/Vehicle_Coefficient_of_Drag_List

http://www.mercedes-benz.de/content/germany/mpc/mpc_germany_website/de/home_mpc/passengercars/home/_used_cars/technical_data.0006.html

http://www.carfolio.com/specifications/models/car/?car=218999

Estimating Temperature Using Cricket Chirps

I stumbled upon a truly great formula on the GLOBE scientists’ blog. It allows you to compute the ambient air temperature from the number of cricket chirps in a fixed time interval and this with a surprising accuracy. The idea actually quite old, it dates back 1898 when the physicist Dalbear first analyzed this relationship, and has been revived from time to time ever since.

Here’s how it works: count the number of chirps N over 13 seconds. Add 40 to that and you got the outside temperature T in Fahrenheit.

T = N + 40

From the picture below you can see that the fit is really good. The error seems to be plus / minus 6 % at most in the range from 50 to 80 °F.

What is Torque? – A Short and Simple Explanation

Often times when doing physics we simply say “a force is acting on a body” without specifying which point of the body it is acting on. This is basically point-mass physics. We ignore the fact that the object has a complex three-dimensional shape and assume it to be a single point having a certain mass. Sometimes this is sufficient, other times we need to go beyond that. And this is where the concept of torque comes in.

Let’s define what is meant by torque. Assume a force F (in N) is acting on a body at a distance r (in m) from the axis of rotation. This distance is called the lever arm. Take a look at the image below for an example of such a set up.

Great Formulas II_html_m7f65e4ec

(Taken from sdsu-physics.org)

Relevant for the rotation of the body is only the force component perpendicular to the lever arm, which we will denote by F’. If given the angle Φ between the force and the lever arm (as shown in the image), we can easily compute the relevant force component by:

F’ = F · sin(Φ)

For example, if the total force is F = 50 N and it acts at an angle of Φ = 45° to the lever arm, only the the component F’ = 50 N · sin(45°) ≈ 35 N will work to rotate the body. So you can see that sometimes it makes sense to break a force down into its components. But this shouldn’t be cause for any worries, with the above formula it can be done quickly and painlessly.

With this out of the way, we can define what torque is in one simple sentence: Torque T (in Nm) is the product of the lever arm r and the force F’ acting perpendicular to it. In form of an equation the definition looks like this:

T = r · F’

In quantitative terms we can interpret torque as a measure of rotational push. If there’s a force acting at a large distance from the axis of rotation, the rotational push will be strong. However, if one and the same force is acting very close to said axis, we will see hardly any rotation. So when it comes to rotation, force is just one part of the picture. We also need to take into consideration where the force is applied.

Let’s compute a few values before going to the extremely useful law of the lever.

—————————

We’ll have a look at the wrench from the image. Suppose the wrench is r = 0.2 m long. What’s the resulting torque when applying a force of F = 80 N at an angle of Φ = 70° relative to the lever arm?

To answer the question, we first need to find the component of the force perpendicular to the lever arm.

F’ = 80 N · sin(70°) ≈ 75.18 N

Now onto the torque:

T = 0.2 m · 75.18 N ≈ 15.04 Nm

—————————

If this amount of torque is not sufficient to turn the nut, how could we increase that? Well, we could increase the force F and at the same time make sure that it is applied at a 90° angle to the wrench. Let’s assume that as a measure of last resort, you apply the force by standing on the wrench. Then the force perpendicular to the lever arm is just your gravitational pull:

F’ = F = m · g

Assuming a mass of m = 75 kg, we get:

F’ = 75 kg · 9.81 m/s² = 735.75 N

With this not very elegant, but certainly effective technique, we are able to increase the torque to:

T = 0.2 m · 735.75 N = 147.15 Nm

That should do the trick. If it doesn’t, there’s still one option left and that is using a longer wrench. With a longer wrench you can apply the force at a greater distance to the axis of rotation. And with r increased, the torque T is increased by the same factor.

—————————

This was an excerpt from my Kindle ebook More Great Formulas Explained.

Check out my BEST OF for more interesting physics articles.

Computing the Surface Area of a Person – Mosteller Formula

While doing research for my new book “More Great Formulas Explained”, I came across a neat formula that can be used to calculate the surface area of a person. It goes by the name Mosteller formula and requires two inputs: the mass m (in kg) and the height h (in cm). The surface area S (in m²) is proportional to the square root of m times h:

S = sqrt (m * h / 3600)

For example, a person with the weight m = 75 kg and height h = 175 cm can be expected to have the body surface area S = 1.91 m². A note for American readers: you can use this table to easily convert the height in feet / inches to centimeters.

What’s the use of this? In my book I needed to know this quantity to compute heat loss. According to Newton’s law of cooling, the heat loss rate P (in Watt = Joules per second) is proportional to the surface area S and the temperature difference ΔT (in °C or K):

P = a * S *ΔT

with a being the so called heat transfer coefficient. For calm air it has the value a = 10 W/(m² * K). A person’s body temperature is around 37 °C. So the m = 75 kg and h = 175 cm person from above would lose this amount of heat every second at an air temperature of 20 °C:

P = 10 W/(m² * K) * 1.91 m² * 17 °C = 325 Watt

That is of course assuming the person is naked, clothing will reduce this value significantly. So the surface area formula indeed is useful.

Sports: Elo-Rating and Win Probability / Carlsen vs. Anand

The Elo rating system is a commonly applied tool to judge the performance of players in sports. Most associate it with chess, but it is in use in other sports as well, for example American Major League Baseball and Basketball. A nice thing about the rating system is that you can easily estimate the win probabilities given the difference in Elo rating of the opponents.

ELO

So if the difference is zero (opponents of equal strength), the win probability is obviously 0.5 = 50 %. If one player has a 200 Elo points advantage, his win probability is about 76 %. In case you prefer a table over a graph, you can find the corresponding values here. And for those interested in formulas, I found that the Boltzmann function models the values extremely well (r stands for Elo rating, p for the win probability):

p = 1 – 1 / ( 1 + exp ( 0.00583 * r  – 0.0505) )

For example, plugging in r = 200 leads to p = 0.75 = 75 %, close enough. For a very rough linear approximation this formula will do:

p = 0.5 + 0.001 * r

By the way, what sports event is happening right now in India? Right, the World Chess Championship. At the moment Anand and Carlsen are locked in a bitter battle to death (ok, a little too dramatic, but still very interesting) and I just finished watching their brilliant third match, which was yet another draw. Here’s the current top ten for chess:

Rank Name Title Country Rating Games B-Year
 1  Carlsen, Magnus  g  NOR  2870  0  1990
 2  Aronian, Levon  g  ARM  2801  6  1982
 3  Kramnik, Vladimir  g  RUS  2793  9  1975
 4  Nakamura, Hikaru  g  USA  2786  17  1987
 5  Grischuk, Alexander  g  RUS  2785  17  1983
 6  Caruana, Fabiano  g  ITA  2782  25  1992
 7  Gelfand, Boris  g  ISR  2777  11  1968
 8  Anand, Viswanathan  g  IND  2775  0  1969
 9  Topalov, Veselin  g  BUL  2774  6  1975
 10  Mamedyarov, Shakhriyar  g  AZE  2757  11  1985

As you can see, Carlsen is currently the leading chess player with an Elo rating of 2870 and Anand is on rank eight with a rating of 2775. The difference is 2870 – 2775 = 95 points, which translates into a win probability of about 63 % for Carlsen and 37 % for Anand. If you have to bet, bet on Carlsen, but it’s going to be a close one. Both have shown a great performance so far, though it seemed that in the third match Anand did miss some opportunities. I’m looking forward to tomorrow’s match.

Increase Views per Visit by Linking Within your Blog

One of the most basic and useful performance indicator for blogs is the average number of views per visit. If it is high, that means visitors stick around to explore the blog after reading a post. They value the blog for being well-written and informative. But in the fast paced, content saturated online world, achieving a lot of views per visit is not easy.

You can help out a little by making exploring your blog easier for readers. A good way to do this is to link within your blog, that is, to provide internal links. Keep in mind though that random links won’t help much. If you link one of your blog post to another, they should be connected in a meaningful way, for example by covering the same topic or giving relevant additional information to what a visitor just read.

Being mathematically curious, I wanted to find a way to judge what impact such internal links have on the overall views per visit. Assume you start with no internal links and observe a current number views per visitor of x. Now you add n internal links in your blog, which has in total a number of m entries. Given that the probability for a visitor to make use of an internal link is p, what will the overall number of views per visit change to? Yesterday night I derived a formula for that:

x’ = x + (n / m) · (1 / (1-p) – 1)

For example, my blog (which has as of now very few internal links) has an average of x = 2.3 views per visit and m = 42 entries. If I were to add n = 30 internal links and assuming a reader makes use of an internal link with the probability p = 20 % = 0.2, this should theoretically change into:

x’ = 2.3 + (30 / 42) · (1 / 0.8 – 1) = 2.5 views per visit

A solid 9 % increase in views per visit and this just by providing visitors a simple way to explore. So make sure to go over your blog and connect articles that are relevant to each other. The higher the relevancy of the links, the higher the probability that readers will end up using them. For example, if I only added n = 10 internal links instead of thirty, but had them at such a level of relevancy that the probability of them being used increases to p = 40 % = 0.4, I would end up with the same overall views per visit:

x’ = 2.3 + (10 / 42) · (1 / 0.6 – 1) = 2.5 views per visit

So it’s about relevancy as much as it is about amount. And in the spirit of not spamming, I’d prefer adding a few high-relevancy internal links that a lot low-relevancy ones.

If you’d like to know more on how to optimize your blog, check out: Setting the Order for your WordPress Blog Posts and Keywords: How To Use Them Properly On a Website or Blog.

Computing and Tracking the Amazon Sales Rank

The webpage http://www.novelrank.com/ provides a very neat simple way to track the sales rank of any book on Amazon. This service is completely free.

The sales rank is computed from the sales rate. The more a book sells per day, the lower the rank will be.  Here’s an approximate formula, taken from: http://www.edwardwrobertson.com/2013/02/a-quick-way-to-calculate-amazon-sales.html.

100,000 / rank = sales per day

So if a book is on rank 50,000, it sells about twice a day. As far as I know, a borrow counts as a sale and a free download as one third of a sale.

I use novelrank to track my ebooks. This is what the output looks like (launch of “Great Formulas Explained”):

novelrankgreatformulas

Indeed a neat tool to see how a book is performing. Note that the tracking starts on the day you add it, dates before that are not shown.

As you can see, during the period when no sale is made the sales rank increases more or less linearly at about # 50,000 per day. The average rank during this time can be calculated by the formula: final minus initial rank divided by 2. When a sale is made, the rank makes a discontinuous jump to a lower value.

Mathematics of Explosions

When a strong explosion takes place, a shock wave forms that propagates in a spherical manner away from the source of the explosion. The shock front separates the air mass that is heated and compressed due to the explosion from the undisturbed air. In the picture below you can see the shock sphere that resulted from the explosion of Trinity, the first atomic bomb ever detonated.

Great Formulas_html_m67b54715

Using the concept of similarity solutions, the physicists Taylor and Sedov derived a simple formula that describes how the radius r (in m) of such a shock sphere grows with time t (in s). To apply it, we need to know two additional quantities: the energy of the explosion E (in J) and the density of the surrounding air D (in kg/m3). Here’s the formula:

r = 0.93 · (E / D)0.2 · t0.4

Let’s apply this formula for the Trinity blast.

———————-

In the explosion of the Trinity the amount of energy that was released was about 20 kilotons of TNT or:

E = 84 TJ = 84,000,000,000,000 J

Just to put that into perspective: in 2007 all of the households in Canada combined used about 1.4 TJ in energy. If you were able to convert the energy released in the Trinity explosion one-to-one into useable energy, you could power Canada for 60 years.

But back to the formula. The density of air at sea-level and lower heights is about D = 1.25 kg/m3. So the radius of the sphere approximately followed this law:

r = 542 · t0.4

After one second (t = 1), the shock front traveled 542 m. So the initial velocity was 542 m/s ≈ 1950 km/h ≈ 1210 mph. After ten seconds (t = 10), the shock front already covered a distance of about 1360 m ≈ 0.85 miles.

How long did it take the shock front to reach people two miles from the detonation? Two miles are approximately 3200 m. So we can set up this equation:

3200 = 542 · t0.4

We divide by 542:

5.90 t0.4

Then take both sides to the power of 2.5:

t 85 s ≈ 1 and 1/2 minutes

———————-

Let’s look at how the different parameters in the formula impact the radius of the shock sphere:

  • If you increase the time sixfold, the radius of the sphere doubles. So if it reached 0.85 miles after ten seconds, it will have reached 1.7 miles after 60 seconds. Note that this means that the speed of the shock front continuously decreases.

For the other two parameters, it will be more informative to look at the initial speed v (in m/s) rather the radius of the sphere at a certain time. As you noticed in the example, we get the initial speed by setting t = 1, leading to this formula:

v = 0.93 · (E / D)0.2

  • If you increase the energy of the detonation 35-fold, the initial speed of the shock front doubles. So for an atomic blast of 20 kt · 35 = 700 kt, the initial speed would be approximately 542 m /s · 2 = 1084 m/s.

  • The density behaves in the exact opposite way. If you increase it 35-fold, the initial speed halves. So if the test were conducted at an altitude of about 20 miles (where the density is only one thirty-fifth of its value on the ground), the shock wave would propagate at 1084 m/s

Another field in which the Taylor-Sedov formula is commonly applied is astrophysics, where it is used to model Supernova explosions. Since the energy released in such explosions dwarfs all atomic blasts and the surrounding density in space is very low, the initial expansion rate is extremely high.

This was an excerpt from the ebook “Great Formulas Explained – Physics, Mathematics, Economics”, released yesterday and available here: http://www.amazon.com/dp/B00G807Y00. You can take another quick look at the physics of shock waves here: Mach Cone.

Physics (And The Formula That Got Me Hooked)

A long time ago, in my teen years, this was the formula that got me hooked on physics. Why? I can’t say for sure. I guess I was very surprised that you could calculate something like this so easily. So with some nostalgia, I present another great formula from the field of physics. It will be a continuation of and a last section on energy.

To heat something, you need a certain amount of energy E (in J). How much exactly? To compute this we require three inputs: the mass m (in kg) of the object we want to heat, the temperature difference T (in °C) between initial and final state and the so called specific heat c (in J per kg °C) of the material that is heated. The relationship is quite simple:

E = c · m · T

If you double any of the input quantities, the energy required for heating will double as well. A very helpful addition to problems involving heating is this formula:

E = P · t

with P (in watt = W = J/s) being the power of the device that delivers heat and t (in s) the duration of the heat delivery.

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The specific heat of water is c = 4200 J per kg °C. How much energy do you need to heat m = 1 kg of water from room temperature (20 °C) to its boiling point (100 °C)? Note that the temperature difference between initial and final state is T = 80 °C. So we have all the quantities we need.

E = 4200 · 1 · 80 = 336,000 J

Additional question: How long will it take a water heater with an output of 2000 W to accomplish this? Let’s set up an equation for this using the second formula:

336,000 = 2000 · t

t ≈ 168 s ≈ 3 minutes

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We put m = 1 kg of water (c = 4200 J per kg °C) in one container and m = 1 kg of sand (c = 290 J per kg °C) in another next to it. This will serve as an artificial beach. Using a heater we add 10,000 J of heat to each container. By what temperature will the water and the sand be raised?

Let’s turn to the water. From the given data and the great formula we can set up this equation:

10,000 = 4200 · 1 · T

T ≈ 2.4 °C

So the water temperature will be raised by 2.4 °C. What about the sand? It also receives 10,000 J.

10,000 = 290 · 1 · T

T ≈ 34.5 °C

So sand (or any ground in general) will heat up much stronger than water. In other words: the temperature of ground reacts quite strongly to changes in energy input while water is rather sluggish. This explains why the climate near oceans is milder than inland, that is, why the summers are less hot and the winters less cold. The water efficiently dampens the changes in temperature.

It also explains the land-sea-breeze phenomenon (seen in the image below). During the day, the sun’s energy will cause the ground to be hotter than the water. The air above the ground rises, leading to cooler air flowing from the ocean to the land. At night, due to the lack of the sun’s power, the situation reverses. The ground cools off quickly and now it’s the air above the water that rises.

Image
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I hope this formula got you hooked as well. It’s simple, useful and can explain quite a lot of physics at the same time. It doesn’t get any better than this. Now it’s time to leave the concept of energy and turn to other topics.

This was an excerpt from my Kindle ebook: Great Formulas Explained – Physics, Mathematics, Economics. For another interesting physics quicky, check out: Intensity (or: How Much Power Will Burst Your Eardrums?).

How much habitable land is there on earth per person?

What is the total area of habitable land on Earth? And how much habitable land does that leave one person? We’ll use the value r = 6400 km as the radius of Earth. According to the corresponding formula for spheres, the surface area of Earth is:

S = 4 * π * (6400 km)^2 ≈ 515 million square km

Since about 30 % of Earth’s surface is land, this means that the total area of land is 0.3 * 515 ≈ 155 million square km, about half of which is habitable for humans. With roughly 7 billion people alive today, we can conclude that there is 0.011 square km habitable land available per person. This corresponds to a square with 100 m ≈ 330 ft length and width.

Average Size of Web Pages plus Prediction

Using data from websiteoptimization.com I plotted the development of web page sizes over the years. I also included the exponential fit:
Image

As you can see, the 1/2 MB mark was cracked in 2009 and the 1 MB mark was cracked in 2012. Despite the seemingly random fluctuations, an exponential trend is clearly visible. The power 0.3 indicates that the web page sizes doubles about every 2.3 years. Assuming this exponential trend continues we will have these average sizes in the coming years:

2013 – ca. 1600 kb
2014 – ca. 2100 kb
2015 – ca. 2900 kb

So the 2 MB will probably be cracked in 2014 and in 2015 we will already be close to the 3 MB mark. Of course the trend is bound to flat out, but at this point there’s no telling when it will happen.

If you like more Internet analysis, check out The Internet since 1998 in Numbers.

Missile Accuracy (CEP) – Excerpt from “Statistical Snacks”

An important quantity when comparing missiles is the CEP (Circular Error Probable). It is defined as the radius of the circle in which 50 % of the fired missiles land. The smaller it is, the better the accuracy of the missile. The German V2 rockets for example had a CEP of about 17 km. So there was a 50/50 chance of a V2 landing within 17 km of its target. Targeting smaller cities or even complexes was next to impossible with this accuracy, one could only aim for a general area in which it would land rather randomly.

Today’s missiles are significantly more accurate. The latest version of China’s DF-21 has a CEP about 40 m, allowing the accurate targeting of small complexes or large buildings, while CEP of the American made Hellfire is as low as 4 m, enabling precision strikes on small buildings or even tanks.

Assuming the impacts are normally distributed, one can derive a formula for the probability of striking a circular target of Radius R using a missile with a given CEP:

p = 1 – exp( -0.41 · R² / CEP² )

This quantity is also called the “single shot kill probability” (SSKP). Let’s include some numerical values. Assume a small complex with the dimensions 100 m by 100 m is targeted with a missile having a CEP of 150 m. Converting the rectangular area into a circle of equal area gives us a radius of about 56 m. Thus the SSKP is:

p = 1 – exp( -0.41 · 56² / 150² ) = 0.056 = 5.6 %

So the chances of hitting the target are relatively low. But the lack in accuracy can be compensated by firing several missiles in succession. What is the chance of at least one missile hitting the target if ten missiles are fired? First we look at the odds of all missiles missing the target and answer the question from that. One missile misses with 0.944 probability, the chance of having this event occur ten times in a row is:

p(all miss) = 0.94410 = 0.562

Thus the chance of at least one hit is:

p(at least one hit) = 1 – 0.562 = 0.438 = 43.8 %

Still not great considering that a single missile easily costs 10000 $ upwards. How many missiles of this kind must be fired at the complex to have a 90 % chance at a hit? A 90 % chance at a hit means that the chance of all missiles missing is 10 %. So we can turn the above formula for p(all miss) into an equation by inserting p(all miss) = 0.1 and leaving the number of missiles n undetermined:

0.1 = 0.944n

All that’s left is doing the algebra. Applying the natural logarithm to both sides and solving for n results in:

n = ln(0.1) / ln(0.944) = 40

So forty missiles with a CEP of 150 m are required to have a 90 % chance at hitting the complex. As you can verify by doing the appropriate calculations, three DF-21 missiles would have achieved the same result.

Liked the excerpt? Get the book “Statistical Snacks” by Metin Bektas here: http://www.amazon.com/Statistical-Snacks-ebook/dp/B00DWJZ9Z2. For more excerpts see The Probability of Becoming a Homicide Victim and How To Use the Expected Value.