Some more memories from Friesland. Binary Calculator.

A couple more images from our trip to Friesland. Simon’s binary calculator:


Doing math at a restaurant where we were celebrating his friend’s birthday:





The Leaning Tower of Lire

Also known as the Book-Stacking Problem. Simon had tried to build this tower at the Fries Museum where we visited a huge Escher exhibition (to the annoyance of the museum staff, to whom I had to explain that it was a serious math experiment and not just a kid dropping bricks), but it only worked with 4 blocks (possibly because the blocks were made of foam and weren’t rigid enough). He tried to stack the blocks on top of one another, shifting every next block first by one eighth, then by one sixth, next by one fourth, and next by one half – in the end, the top block would no longer be positioned above the bottom block.


He repeated the experiment at home, first doing some calculations and then using more rigid wooden blocks and managed to stack a tower of 6 blocks! (The top block still overlapped the bottom one by a bit though) :





The Pi Strip

Simon made a measuring tool to check the diameter of round objects: by wrapping the strip around them, he reads the Pi times the centimeters value, which basically gives him the diameter (as the circumference equals Pi times the diameter).

And here he is, measuring the diameters of Neva’s and Dad’s necks:

Simon’s Archimedean Solids Project

Simon is working on a project that will involve constructing the Archimedean solids from paper pieces that he programs in Processing (Java) and prints out. In the previous video, Simon worked out the distance between two points to measure the side length of a pentagon that has the radius of 1 (i.e. the distance between its adjacent vertices if the distance from its center to its vertices is 1). He first made a mistake in his calculation and got a result that would be true for a hexagon, not a pentagon. He then corrected himself and got the value that he thought he could use in the Processing code, but as it turned out, the ratio between the radius and the side length was still not right. We recorded a whole new video full of calculations and playing with the code, and achieved pretty neat results after Simon used the new value in the code, but still not good enough, as Simon wanted to have his pentagons to have the side length of 40 (to match the triangles and the squares he’d already made). Simon later found a solution using a different formula for a polygon with n sides (from trigonometry, defining the radius as the side length over (2sin times 180/n)) and succeeded in getting exactly the pentagons he wanted, with the side equalling 40. See the result here:

The winning formula:

If you are really into working out the calculations, feel free to check out our frantic attempts here:

Slowing down the time

The first thing Simon said this morning was: “Mom, do you know that if you keep moving, you get one quadrillionth of a second per second younger than if you just sit still?” Simon had heard this fact on VSauce the night before (and I think we watched another VSauce video earlier and also read in our Eine kleine Nachtphysik book about how in spacetime, we barely move through space and only move through time, and the faster one moves through space, the slower one moves through time – that’s why time stops for anyone moving through space at the speed of light.

This morning, Simon wanted to calculate how much less older one gets within a lifetime of 100 years if one keeps moving all the time. He first calculated how many seconds there are in a 100 years. He said he knew that there were 10! seconds in 6 weeks and went on from there:

And then, multiplied that value by one quadrillionth (see below). He ended up with approximately one three hundred thousandth of a second per 100 years – getting that much less older than if you sit still for a century. The value is very approximate, as Simon is not exactly sure whether the initial value of one quadrillionth per second is correct – it could be even less!




The Paradox of the Mathematical Cone

Simon showed me an interesting paradox that’s difficult to wrap my mind around. If you slice a cone (at a random height), the section is a circle. The chopped-off part (a small cone) also has a circle as its base. Are those circles equal? They are the same, because they result from the same cross section. Hence the difference between them is 0. Now imagine slicing the cone an infinite number of times. “The difference between the circles will come up an infinite number of times: zero times infinity”, – Simon explained. “But zero times infinity has no value (or has any value, it’s indeterminate). Zero times infinity is the same as infinity minus infinity, which means that it can be whatever you want. Riemann’s rearrangement theory makes this true.”