Series resister for downhill run?

Yes, series resistors to slow a train going down hill was common.  I have used some of Lionel's prewar variable resistors to do this, as the resistance required may be different for different engines. 

Usually, what goes down must come up, and visa versa. Unless, of course, you have a continuous loop where your trains are always going in the same direction on the same section of track. IMO, a layout gets boring quickly without the ability to reverse direction with some sort of return loop or one of its' variants.

Lionel made use of the four channels on a ZW to set different voltages on tracks sections by isolating that section with the center rail insulating pin and powering it with a different handle set at the desired voltage.

E.g.: A variable, B uphill, D downhill and C accessories. Can also be done with multiple transformers. Then you can adjust up and down easily for different engines and train lengths.

Thanks David.  Do you have model numbers for those parts so I can see if someone online specs or has some for sale?

If you know the specs, that would be awesome.

Paul

The layout has two loops with interconnecting turnouts.  There is no room for a reversing loop on the layout.  The one siding (a dead end) would only be accessible from one direction, so 90% of the time, the direction will be clock-wise to be able to stop, backup cars into the siding (simulate unloading/loading) and then forward again.  All this in on the inner loop.  Only the outer-loop has the elevator portion and is the one that would need the resistor on the downhill section of track.  I could use two isolated track sections (one up, one down) and toggle switches to switch the resistors into the power feed for the section that was on the downhill portion at that time.   Multiple transformers (or a multi-throttle one) is not an option at this time as last time I looked they would cost more than this entire layout. LOL.

Paul

L M G T F Y

And eBay...

Thanks Rob.

Paul

Lionel recommended using a 10ohm,10watt resistor for preserving voltage to the e-unit. (from the 1954 Operators Manual)

So, perhaps that would be sufficient to control the downhill track voltage. Here is a 10ohm, 25watt available from Digi-Key. $6.14 + regular mail shipping.

It would be cheaper and much more versatile to use the similarly spec'd Lionel Rheostats.

That's about what I calculated too.  Between 5 and 10 ohms. I found something similar but rated at 25watts. Better safety margin.  I just wasn't sure about the exact value.  This is an old 1960 2-4-2 engine. Transformer is rated a total of 45 watts and 8 to 15VAC. Works great still.  

I set up the outer loop in a temporary setup with the expected grades and except for the downhill portions it worked fine pulling 6 cars. Little slow on the uphill grade, but there is a sharp O-27 90deg turn at the bottom of the downhill run(s).  

So that section of track's center rail will be isolated and get the resistor in series with it.  I thinking some double-pole/double-throw switches in the wiring to those sections to be able to switch the resistor in or out out of the feed line depending on which way around the loop the train is heading. 

Paul

One more item, I heard that Radio-Shack is closing their doors, so it might be worth a visit to get all those switches and other connectors while they have them.

Paul

CEO of the SBRRR (Spare Bed-Room Rail Road)

I don't know if they're closing, but there is a looming bankruptcy, that's not good.

Paul, in the 1950's, I used such a system.  For the resistors, I removed one strand from a length of #18 wire, and put about a foot of it into the line to the downhill portion.  I actually used a shorter length for the level portions, so that I could apply full power to the uphills.  I used a SPDT switch to directly connect to thone of the other of the grades.  Would get about 2 months of a wire before it got brittle from the heat and broke, but replacement was easy.

A short on the line would really heat that strand up although it was in air with nothing to ignite, I would hold current to below the transformer breaker opening point, so one should use in-line breaker.

Try this ebay link for a 100 watt, 10 ohm wire wound, aluminum shell with cooling fins resistor for $2.79, with free shipping

http://www.ebay.com/itm/1-2-4-...;hash=item1e89f27e9b

Fill in other ohms if 10 ohms is not good for you

It is hard to beat ebay free shipping from China for small electrical parts if you can wait a week or so.

charlie

Thanks all for the replies and ideas.  

Choo Choo Charlie, I think I will order some tomorrow.  A couple of different ratings so I can test which works best.

At the moment, the "hill" with about a 4% grade (up and down) is set up on the bench for testing prior to actually constructing the layout.  So far, so good. 

Paul

10 ohms might be a bit high.  Figure 1 volt of drop for each amp you're pushing through a 1 ohm resistor.

Note that resistor has to be bolted to a LARGE heatsink with some thermal compound to achieve anything close to it's rated power dissipation.

So, grj, can diode pairs be used to drop the voltage on the downhill section?

Unless there is a direct short, I don't see the resistor dissipating 100W.  More like 10 to 20 watts. The most the transformer will deliver is 45 watts.  

Watts Law: P=I^2 * R 

Do plan for shorts.  

Inline circuit breaker for those.

That would certainly work.  It has the added benefit that the voltage doesn't change based on the load.  I'm not a fan of resistors for these kinds of applications.

If someone could enlighten me on the use of a diode pair for this?

If you connect pairs of diodes back to back and put this chain in series with the track feed, each diode pair will drop about .6 - .7 volts.  If you want to drop 5 volts, you'd have perhaps 7 diode pairs. 

Here's a post on voltage dropping with diodes.

http://www.jcstudiosinc.com/Voltage-Dropper-Diodes

For example, eBay has 6A bridges at 5 for $2.32 (free shipping from Asia). 

http://www.ebay.com/itm/5pcs-K...;hash=item51a6659068

If your layout runs on 45 Watts, 6 Amp diodes ought to be good enough. Or you can buy individual 6A diodes for about 15 cents each (free shipping).  Note that 5 bridges = 20 diodes so about 12 cents each.

http://www.ebay.com/itm/20PCS-...;hash=item4ae2f40d62

As Dale explains in his write-up, a bridge conveniently packages 2 pairs of of diodes saving time on soldering/assembly.  So with, say, 3 bridges you get 6 pairs of diodes each pair providing ~0.7V AC voltage drop...plus the convenience of multiple taps (~0.7V steps) so you can conveniently "fine-tune" the voltage drop for your application.

Multiple taps make it possible to gradually step down the voltage at the expense of wiring up multiple blocks on the grade.  Or the taps could be arranged to provide no drop on the uphills, a small drop when flat, a bigger drop on downhills.

Gee Stan, with 45 watts, I think 3A diodes should be sufficient.  OTOH, going with 6A certainly gives you a margin for safety, and they're cheap enough.

Agreed.  The nice thing about those 6A bridges is the mounting hole in the center of the package.  I'm not sure I've seen a mounting hole on lower Amp bridges. In additional to making for more tidy assembly, it makes it convenient to fasten to a metal plate (heat-sink) albeit not needed for a 45 Watt layout.

They are a nice convenient package, as as you say, easier to wire as well, less wires.

Paul,

There you have it. Thank you John and Stan.

You can control the downhill or the entire loop with a series of these mounted near the transformer.

The 12 position Euro style terminal strip with single diodes would allow you to make all of the connections without soldering and select the desired voltage as in Dale's diagram by connecting a wire in the desired terminal.

I hope that helps.

Should there not be a breaker in the line equal to max amperage rating of the diodes?

I would note that one should not wire this as a bridge rectifier:  Only 2 terminals will be used.