Rediscovering Myst From a Ham’s Perspective

The Myst series of games from Cyan Worlds has always been one of my favorites. I love exploring the beautiful locations, the atmospheric music, and the puzzles.

In my younger days, I confess I almost always had to refer to walkthroughs to complete the game, and seldom actually understood the solutions fully.

But now, since studying for both Technician, General, and about halfway to the Amateur Extra License license test, replaying the games is much more enjoyable as I’m discovering the solutions are often based on real principals of electronics and mechanical engineering.

Take the Myst III Exile game which I just completed yesterday. In the age of Amateria, which is basically a giant pinball roller coaster, one of the first puzzles involves a sluice chute that has to be positioned using weights, and a movable support point. If not positioned correctly one end will be too high or too low, and the ice-sphere you need to roll across it will shatter.

In the end, you need 22 units of weight on one side (the side you can adjust is the left- you have no control over the weight on the right side) and to move the support point (otherwise known as the fulcrum) to the furthest left position of the three positions.

I started wondering afterwards, if there was a real-world formula for this puzzle solution. And there is!

I found this very enlightening article over on Engineer’s Edge that explains how to calculate how much weight is needed on one side to counterbalance the “lever” weight on the other if you know the length of each side of the balance beam.

Here’s the formula:

F x L = W x X

F = Downward Force on the right side
L = Length of the right side of the balance beam

W = Weight on the left side of the balance beam
X = Length of the left side of the balance beam

In the puzzle, F=11 (7 Wood Pieces of the Sphere + 1 Crystal Piece, which weighs as much as 4 wood pieces)

The total length of the balance beam is 3, and you can place it at positions 1, 1.5, or 2 units.

So in the left most fulcrum position ( L=2, X = 1, F = 11), we get this:

11 x 2 = ? x 1 — so W = 22

This comes out to 6 wood (6 x 1) and 1 metal (1 x 16) piece. (Metal weighs 4 x Crystal, Crystal weighs 4 x Wood)

That balances the beam at the correct position (equilibrium) for the ice-sphere to roll across both ends without shattering!

Today I just started replaying Myst IV Revelation, and the very first puzzle is very radio related. You’re looking at what is basically a MIXER/FILTER, where you are combining two signals (heterodyning), one external signal, and one from your local oscillator, to create a third signal matching or amplifying the signature of an age.

The controls represent AMPLITUDE, FREQUENCY, and PHASE.

This puzzle too, is much more entertaining now that I actually understand from an RF perspective what those terms mean.

It’s a pleasant surprise that the puzzles are not just meaningless, and there are known principals you can use to solve them.