How was this discovered? The Gregorian calendar – the one we use in modern 21st century western society – repeats itself every 400 years. Therefore, to calculate how many times the thirteenth falls on each day of the week, we only have to examine 400 years, or 4800 months. A bloke by the name of B.H. Brown did this by hand in 1933, but now we have computers. The program to do this is completely trivial once you have a calendar algorithm to follow the Gregorian calendar like all modern computer languages have built-in. Trivial or not, I couldn’t resist…let’s calculate how many times the 13th falls on each day of the week, shall we?

13th Histogram Calculator

When you click “GO!”, the calculator will start from today, January 13, 2012, and count which weekdays fall on the thirteenth of the month over the next 400 years until January 13, 2412. Here’s a hint: it goes up to 688.

GO!Reset

Exciting, huh? There was only one Friday the 13th in 2011, in May, so I've been waiting months to post this.

To the geometrymobile!

This is the composite photo, taken by Letian Wang in Beijing, China, on December 10, 2011, that originally sparked my interest. You can see that the radius of the Earth’s shadow is slightly greater than the Moon’s diameter.

First of all we need to define what we mean by “shadow”.

For the purposes of this post and subsequent calculations, we will only be dealing with the umbra, the dark bit where no direct sunlight hits the moon. We will be ignoring the indirect sunlight that is refracted through the Earth’s atmosphere during an eclipse, bathing the lunar surface in reddish hue.

Okay, let’s do some math!

Here we can see the relationship between the radius of the umbra to all the distances involved. Do I even have to mention that the diagram is not to scale?

If we draw two more lines, we get two congruent right triangles.

Since we know they are congruent, we know their sides are proportional and can write the following equation:

…which we can solve for the radius of the umbra:

When we plug in the values for the radii of the sun and Earth, we get:

Over time, with the two elliptical orbits involved, both the numerator and the denominator of our distance ratio vary. The ratio of the distance from the Earth to the Moon (d_moon) to the distance from the sun to the Earth (d_earth) is at its maximum when the Moon is at its apogee and the Earth is at its perigee (we are ignoring the likelihood of these two extremes coinciding, of course).

The ratio is at its minimum when the Moon is at its perigee and the Earth is at its apogee.

When we plug these back into our bigger equation, we discover that the radius of the Earth’s shadow at the distance of the moon varies from 4479 km to 4735 km, or from 2.578 to 2.725 moon radii.

To visualize this, let’s look at the minimum and maximum shadow sizes compared to the Moon.

Not a very big difference, I think you’ll agree. We did, however, answer our question.

At least now astrophotographer and artist Laurent Laveder will know the range of sizes for his hoops for his incredible lunar eclipse photography.

I did not draw this car, René van Belzen did.

If we stop to think about it – and we should, as it’s good practice – two features of the plot of our “minimum speed to keep the rear window dry given the angle, Ɵ” equation can be deduced. Firstly, when Ɵ is 90º, the minimum speed will be zero, since the rain – which we are assuming, for simplicity, is falling straight down – won’t hit it even if we are stopped. Secondly, as Ɵ approaches 0º, the minimum speed to keep the rear window dry will shoot off to infinity. Note that we shall also be ignoring near-speed-of-light Einsteinian physics, as the Newtonian model is more than sufficient for the speeds I drive. So we can picture our graph dropping down from infinity and then making its way down to the x-axis, maintaining a negative slope the entire time, until it hits 90º on the x-axis.

The time it takes the rain to fall the vertical height of our window is going to be the height divided by the terminal velocity of the rain. And the time it takes the car to move the horizontal length of our window is going to be the length divided by the velocity of the car. When those two times are equal, the car is traveling at the minimum velocity to keep the window dry.

Okay, so we have our equation, but we have too many variables. We need to find a way to write h in terms of l, and we need a rain velocity.

Simple trigonometry tells us that the tangent of the angle in a right triangle is equal to the opposite side divided by the adjacent side, so…

Thus we can replace h in our previous equation, giving us:

The length cancels out, and we get our final equation:

There are two things that I like about this. Neither the height nor length of the window really matter, just as my wife’s brain intuited instantly. The second is that of course the tangent operation is involved, since it zips off to infinity, yet calms down near the x-axis. Since we need it to zip off to positive infinity near the y-axis, it must be inverted. What I love about math and physics is just how intuitive it can become. My mathematician readers were probably already thinking about the inverse tangent function when I discussed the general shape of the data, particularly given that an angle – and thus trigonometry – was involved. Let’s continue!

Wait, before we continue, let me go off on a bit of a tangent (oh dear, trigonometry jokes are a bad sine)… After I finished writing this post and was lying in bed trying not to think about angles and hypotenuses, it occurred to me that the tangent only has to be inverted because of my arbitrary definition of where we start measuring the angle. If we define 0º as being the vector pointing straight down, the direction the rain is falling, and measure the angle away from the movement of the car, then we get a plot that starts at 0 mph at 0º and goes up to ∞ mph at 90º, just like the normal, unmodified tangent function does! But wait, this mathematical model gets even better! What happens if you increase the angle past 90º (sort of like a spoiler)? The tangent function flips back to negative infinity and comes up to 0 mph at 180º, which still accurately models the speed the car would have to travel to keep the rear window dry, since it would have to be going very fast in reverse, and then less fast as the angle approaches the vertical 180º! This is exactly the kind of real world application of a mathematical principle that is so often lacking in our education system. Anyway, I thought that was cool. Back to our regularly scheduled post…

We need a velocity for the rain. What is the terminal velocity for a rain drop? I asked the internet, and the internet said that a small drop of light stratisform rain falls at about 4.6 mph, and the largest possible drop – they can’t be bigger than 5 mm or they break up (awesome high speed video!) – from a huge thunderhead falls at up to 20 mph. We’re going for a minimum speed to guarantee a dry rear window, so we’ll use the 20 mph figure.

When we substitute 20 mph for our rain velocity and convert from radians to degrees, our final equation looks like this:

And now we’re ready to plot it! Drumroll please…

Exactly what we expected, isn’t it?

Examples

This Wolsely Hornet has a rear window angle of 57º. That means that, if it goes more than 13 mph in even the harshest thunderstorm (of absolutely vertical rain), no rain will hit its rear window. Isn’t that incredible? It seems slow to my internal physics engine.

This Peugeot 203 Limousine has a 35º rear window, so it will need to accelerate up to 28.5 mph in order to keep it dry. As you would expect, if you have a 45º window, the velocity you need to travel is 20 mph, exactly equal to the rain’s velocity.

And lastly, this Porsche has a 16º rear window, and thus it will need to zip up to 70 mph to keep dry. Obviously this will be no problem for the Porsche, but you can see how the velocities are increasing very rapidly as we approach the horizontal, due to the very nature of the tangent function.

My curiosity is satisfactorily quenched on this topic, and I hope you learned something. Math is all around us and lets us know things about the world through mere thought and some pen and paper. Isn’t that fascinating?

Leaves in the park.

The solar photons with longer wavelengths add a romance to sunset photography.

A stone path.

A flowery gate.

A cow posing for a silhouette.

A country path leading into the town of Colindres.

Striped flower pot.

Some gorgeously colored flowers…or are they leaves?

A long gate.

The pattern on the ground from the gate in the previous photo.

Cinderblocks.

Sunshine on tree bark, with a distant horse.

Three horses.

A lone flower.

Two old houses, one in the foreground and one in the background.

Time to head back into town, past Nora’s daycare and the town church.

One through seven.

Another angle.

One through five, stacked.

This is my favorite one. I got the depth of field I wanted very well, but it’s a shame I was too careless and lazy to remove the clutter from the background.

How can you place the arithmetical signs ‘+’ and ‘-’ between the consecutive numbers 123456789 so that the end result is 100? So, for example, you could go…..

12+34+56-7-89 , but that would make 6, so that doesn’t work.

Try to come up with the method that uses the least number of symbols.

My first thought was, “It would be so easy to write a program to try all the possibilities!” So I did. Thirty minutes and 41 lines of javascript code later, I had this:

Charting Number of Solutions

Once I had the code to quickly generate the solutions for any arbitrary goal value, I decided to see if there was a pattern in the number of solutions between the goal values from 1 to 100. There isn't.

Click to see large.

The Code

It's just a matter of changing the operation in each slot between nothing, + and -, and when you roll back to nothing, you increment the next slot. Javascript's built in eval() function came in very handy to calculate the value of each equation.

function test() {
  function next(slots, operations) {
    for (var i = 0; i < slots.length; i++) {
      if (!slots[i]) {
        slots[i] = operations[0];
        return slots;
      }
      else {
        var index = operations.indexOf(slots[i]);
        if (index + 1 < operations.length) {
          slots[i] = operations[index + 1];
          return slots;
        }
        else
          delete slots[i];
      }
    }
    return null;
  }
  function makeEquation(slots) {
    equation = ['1'];
    for (var i = 0; i < slots.length; i++) {
      if (slots[i])
        equation.push(' ' + slots[i] + ' ');
      equation.push(i + 2);
    }
    return equation.join('');
  }
  var resultsList = document.getElementById('resultsList');
  if (resultsList)
    resultsList.remove();
  var operations = ['+', '-'];
  var slots = new Array(document.getElementById('numDigits').value - 1);
  var goalValue = document.getElementById('goalValue').value;
  while (slots != null) {
    var equation = makeEquation(slots);
    if (eval(equation) == goalValue)
      printResult(equation + ' = ' + goalValue);
    slots = next(slots, operations);
  }
}

If we write down the equation for the ratio of the picture area to the border area (the full frame size – the picture’s area), we get this:

If we set our ratio to Phi and solve the equation, using the quadratic formula, we get that our doubled border width, x, is:

What does this give us? Now, for any length and width of picture, we can calculate how wide to make the border around it such that the ratio of the area of the picture to the area of the border is Phi…or any number for that matter!

Picture Border Ratio Calculator

Using the calculator below, if you input the length and width of your picture, and the area ratio (picture / border), it will tell you how large to make your border. And below the calculator is a rendering of the values you enter. Note that the default values in the calculator are for a golden rectangle and an area ratio of Phi. Looks pretty nice, doesn’t it? Go ahead, try out other picture sizes!

Wasilla

Wasilla, Alaska, is 1130 km from Russia.

How high above the Earth’s surface do you have to be to see a point 1130 km away along the Earth’s surface? If we start plugging numbers into my Distance To Horizon Calculator, we see that the answer is 101425.88 meters. You have to go 101 km (63 miles!) above Wasilla before you can see Russia. You can knock 104 meters off of that if you like to account for Wasilla’s elevation above sea level.

Just how high is 101 km? This graphic from Wikipedia puts it in perspective very well.

0.381 km – The top of the Empire State Building
8.848 km – The top of Mount Everest
10.668 km – Jet airliner cruising altitude
100 km – The Kármán Line that marks the border of “outer space”
101 km – Sarah Palin’s house
185 km – Space Shuttle operation

So an astronaut in the Space Shuttle passing directly over Wasilla could definitely see Russia as well, but no one down in the atmosphere that allows aeronautical flight could.

Juneau

Juneau, Alaska, is 2011 km from Russia.

To see Russia from Juneau, you’d have to go up 330,715.1 meters. That’s almost the 350 km altitude of the International Space Station.

My point is that, by necessity of both map folding and general human cognitive orientation, we all generally reason about the ground we live on as if it were two dimensional. When discussing line of sight, however, the curvature of the Earth comes into play even at very short distances. The millions of US voters that “sort of know that Alaska is up there to the ‘left’ of Canada” might actually believe that Palin waves goodnight to the Russians across the water every night. If you know someone like that, send them to this web page.

Disclaimer: these calculations have disregarded topography and the curvature of space-time, both of which would have a negligible effect.

At this time of year, these passion flowers are blooming vigorously here. Later, they will turn into passion fruit.

The name of the fruit comes from Christians seeing something so beautiful and being unable to appreciate it without relating it to Jesus somehow. I have to wonder if their interpretation of this flower had anything to do with the word stigmata referring to both the pollen receptors on a flower and the crucifixion wounds. From what I can tell the word comes from the ancient Greek word for tattoo! Go figure…

Thanks to sgazzetti for helping me identify this flower last year.

Phun is exactly the kind of computer “game” that my brain adores. I used to play hours and hours of The Incredible Machine, which is a very similar concept. For me, at least, Phun is seriously addictive. It’s like crack for Newtonian physics lovers.

The windows version has been out for a while, but they just released the Mac version about week ago, and I’ve been like a kitten in a tub full of catnip ever since.

Oh, and it also lets you play with particles that mimic liquid water. Here are two of the more interesting creations I’ve made. The first video had to be sped up a lot because my computer couldn’t render all those water particles very fast. For the second video, I added some sound effects. It’d be cool if Phun did sounds by itself.

The above video is also available on YouTube, Yahoo, Metacafe, Google, Revver, DailyMotion, Blip.tv, Veoh, Crackle, Stupid Videos, Sclipo and Viddler.

The above video is also available on YouTube, Metacafe, Google, Revver, DailyMotion, Blip.tv, Crackle, Stupid Videos, Sclipo and Viddler.

And, of course, there are thousands more videos like this on you know what site.