Tag Archives: puzzle

Puzzle Time – Rate Of Growth Of Shadow

Note: The draft of this article had been staring back at me since April 2022. I’m only now getting around to publishing it. Consistent, timely and regular publishing is a natural skill of mine.

Have you ever walked at night on a street that’s lit with street lights? As you pass the lamp post, you start to see a small shadow in front of you. Slowly, as you walk further, the shadow grows, and it keeps growing until it fades to the point where you can’t see it, the next lamp post’s illumination overpowers it or both.

I had this question pop in my mind about the rate at which such a shadow grows, and if is a constant speed or somehow accelerating as the person goes further and further away from the pole. To find it for myself, I spent some time figuring it out today; a lazy rainy Sunday afternoon.

Assuming that

  • A person 1.8m tall
  • Walks at 2m/s
  • Past a lamppost 9m tall

We’re asking,

  1. How fast is the head of the shadow moving away from the lamppost and
  2. How fast is the head of the shadow moving away from the person

Let’s draw some figures to better visualize the question.

To solve this problem, the first piece of information that we note is that the top of the pole, the bottom of the pole and the head of the shadow form a right angled triangle.

The top of the person’s head, their feet and the head of the shadow also form a right angled triangle.

Since we drew a line from the lamp to the person’s head extending all the way to the head of the shadow, the angles thus formed at the lamp and the person (θ) are the same. We conclude here that the smaller triangle is similar to the larger triangle.

Using the property of similar triangles, we get an equation of x in terms of y.

or x = 4y

Since we already know the rate of change of x, which is just the walking speed of the person, we can take derivative on both sides to get the rate of change of y.

So the shadow is growing at 0.5m/s relative to the walking person. Or, if we want to get rate at which the head of the shadow is moving away from the lamppost, we express z in terms of y and then apply derivative on both sides.

2.5m/s away from the lamppost or 0.5m/s faster than the person walking, which it why it seems to grow in length away from us the further we go from the lamppost.

In closing

Hope you enjoyed reading this silly thought, and perhaps even learned something. Thank you for reading!

Puzzle Time – Distance To Horizon

A couple of months ago I was on a flight and had a thought: How far is the horizon at this very moment?

I got my notebook and pen out and tried to solve this using only the knowledge I had, since there was no internet anyway. Of course, many assumptions were made as were convenient for the calculation.

I planned on checking my answer once I landed. If you want to attempt it, stop reading here and come back later when you have a solution. Unfortunately for me, I made a little error subtracting between two big numbers and got a completely wrong answer.


I assumed the following

  • Perfect visibility, no degradation due to atmosphere
  • Earth is a perfect sphere with a radius of 6400km
  • Aeroplane flying at an altitude of 10km


Thanks to the assumptions, the problem reduces to primary school geometry problem and is trivial to solve once visualized clearly on a piece of paper.

Using Pythagoras Theorem we know how to calculate the third side of a right angled triangle give the first two. We also know that a tangent to a circle (the line of sight from an aeroplane) meets the circle in a way that it is perpendicular to the radius drawn from the point of contact.

Combining the two pieces of knowledge, the problem can be trivially arranged as follows

(R + r)² = R² + Dh²


  • R is the radius of the earth (in Kilometers)
  • r is the distance of the aeroplane from the surface of the earth (in Kilometers)
  • Dh is the distance to the horizon

Rearranging, we get

Dh² = (R + r)² – R²

Expanding the (a + b)² equation

Dh² = R² + 2Rr + r² – R²

Dh² = 2Rr + r²

Taking square root on both sides

Dh = √(2Rr + r²)

Substituting the values

Dh = √(2*6400*10 + 10²)

Dh = √(128000 + 100)

Dh = √(128100)

Dh = 357.9

Distance to horizon is about 358 kilometers when the aeroplane is at an altitude of 10 kilometers. Now looking at the map while sitting in a flying aeroplane is a lot more insightful as I can guess which cities I’m looking at outside the window.

That’s it for this little post. Thank you for reading!