Too much voltage!

A resistor is a very poor choice.  The way to do this is with some diodes.  Each diode pair drops the voltage about .7 volts.  To drop it 5 volts, you'd need seven or eight diode pairs.

A more compact way to do this is with bridge rectifiers, you get two diode pairs for each bridge.  Here's a diagram of two bridge rectifiers that are the same as four pairs of diodes.  I'd use 3 amp bridge rectifiers.

Thanks, Gunrunner. I'll do some bridge rectifier shopping.

You can also put some 5 volt Zener diodes end to end and do the job with a smaller package.

In the trolley I'd use 3a+

At the track I'd match it above the supply max

Postwar or can motor?

Postwar: mod at the track is likely easier.

Postwar. It's a Lionel #60. And i agree on the track mod.

@RoyBoy posted:

You can also put some 5 volt Zener diodes end to end and do the job with a smaller package.

The Zener diodes sound like a simpler solution. So I'll contact Mr. Zener, and see if he'll sell me some:-)

I found some 5.1v Zener diodes online that are 0.5w - 500mW. Would they work?

I seldom looked them let alone designed with them. They were either good or cooked and it showed visually with burnt or shattered glass.   A General purpose diode was always on hand. Oversized = cooler, packaged as a BR, cooler yet.  I choose overkill vz heat every time.

1/2 watt Zener diodes would be way too small.

Since the Zener diode will be dropping a lot more voltage, it's also going to be dissipating a lot more power.  The 5V Zener will be dropping the full motor current and dissipating 5V in the process.  Let's assume that the motor voltage peaks at 1A, that's 5 watts of power dissipation in the Zener diode.  If this is in the AC feed line, that's only happening half the time, the other half of the time it's dissipating maybe .7 watts.  So the average power dissipation is around 3 watts in each Zener.  Given the rule of thumb that the component rating should be twice the expected load, the Zener diode should be rated at 6 watts or more.

5W 5.1V Zener Diode

If using the Zener method, as noted earlier this is done in pairs.  The total voltage drop is the sum of the two diodes.  At any given time, one of the diodes is dropping the Zener voltage (e.g. 5.1V) and the other is dropping about 0.7V... in this case the sum being almost 6V.

If you only want to drop 5V (from 14 down to 9), you might consider a lower Zener voltage...such as a 4.7V, 4.3V, or even 3.9V.  All the same price, all sold by DigiKey (and others of course).  As GRJ points out choosing with suitable power-handling capability (Watts) is required.

Note that you can get MORE drop by adding more diodes.  But if you go with the pair of Zener diodes, you can't get LESS drop after the fact.  Hence another reason to consider something less than 5.1V.

Additionally, nothing stops you from mixing/matching a pair of 5 Watt Zener diodes with one or more pairs of inexpensive 3A diodes allowing fine-tuning of the voltage drop in ~0.7V steps.  If buying from DigiKey, you'll be paying way more for shipping than the parts...so minds as well toss some into the order.

Here's a recycled photo from an earlier OGR thread illustrating how a pair of Zener diodes can do the bulk of the voltage drop...and then be augmented with diodes and/or bridge rectifiers to fine-tune the voltage drop.

All of you guys, as usual, have been a tremendous help, as electronics isn't exactly my forte'. I'm assuming that most, if not all of you are MIT graduates in electronics design and applications. Myself, a retired toolmaker, is more at home rebuilding motors and fabrication. That is, as long as the motor doesn't have a short :-)

Thank you stan2004, as that diagram helped me considerably, because without it - - - well, let's just say I would have hooked them up differently.