Rtr and Junior, That .637 ought to be .707 first off. As to what it means, it is a conversion based on average voltage of a sine wave. AC voltage is typically measured in RMS. Root-Mean-Square is calculated by measuring the voltage many times as it rises and falls through the sine wave, then squaring the values (ex: 3x3=9), the squared values are then averaged, and finally the square root of the average is is reported as the reading. The point of all the math is that this RMS value shows the amount of actual work the AC power can do. Many early true RMS meters used a heating element rather than do all the math. Essentially an AC voltage in RMS will produce the same amount of heat as that same voltage in DC. The 1.4141/.707 figures are only accurate on a pure sine wave, however, and as these are used on many cheap meters you can get screwy results if measuring the voltage on other waveforms, including track powered by modern transformers.

Stan, sorry for the crude drawing. It was only meant to show the possible path for electricity to flow. When the phase is going one way it has a path, in the other direction it does not. The point is that a bridge does nothing when used in the method used above in the post, only one of its diodes is being used.

Junior, Already having them is a good enough reason to use the bridges, I suppose, but it may be worth ordering a dozen diodes for a buck instead, and saving the bridges for something where they are needed.

On the last point, The switch machines won't care if they get half-wave power or what the peak voltage is. They only care about the average, or RMS voltage. The LED's on the other hand do care about peak voltage. Typical LEDs are intended to operate on a current of around 20mA, and most will fail in short order with currents much higher. Looking at a random LED data sheet shows a 30mA max forward current. That figure isn't an average, it's a maximum peak, and every time it is exceeded it does damage to the LED until the LED fails. if one optimized for 20mA at 16VDC, at the true peak to peak of the wave form from an 18VAC supply they would be pushing about 32mA through the LED. it's just for an instant, but all those instants 60 times a second do add up, and after some time, much less than it should last, the LED will fail.

JGL