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Wiring a Helix for DCC

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Wiring a Helix for DCC
Posted by railandsail on Thursday, June 07, 2018 1:06 PM

I've seen a few references that suggest it is NOT necessary to run the bus-lines along the track route in a spiral helix, but rather just run the track feeders lines down one of the vertical legs (post) to the bus wires at the bottom. I thought this was a good idea.

My question is how many of these vertical connections should be made?

My helix is a double track affair with radii of 32.5” and 29.5”. I have plans to stagger the rail joints, and to solder all the rail joints,...code 100 Atlas rail.

Considering this sort of relatively small radius/circumference, I'm thinking I need only one feeder set of wires for each elevation/loop of the helix. And perhaps these feeders wires might be sized a bit larger than normal??

 

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Posted by BigDaddy on Thursday, June 07, 2018 2:35 PM

The circumfernce is 15 and 17 feet.  2*Pi*radius  Doesn't sound so small, when people talk about feeders every 6 feet.  I never built a helix, so it's all theoretical to me.

 

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Posted by nealknows on Thursday, June 07, 2018 2:52 PM

My helix has the radius you have. I put 2 sets of drops on each track and level. Call it overkill or whatever, but it lets me sleep at night. I did not solder all the rails just in case there is a major track issue, the helix does not have to be ripped apart. Mine was custom made and would cost me a small fortune if I had to replace it. 

In addition, I have all of my drops with 20 ga wire going to a track buss from the auto reverser. Works like a charm!

Good luck!

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Posted by carl425 on Thursday, June 07, 2018 3:33 PM

Code 100 vs 83 improves your odds of getting away with one set of feeders per loop, but I'd probably still do at least two sets 180° apart.

I used 2/lap on my 24" radius code 83 helix.

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Posted by PennCentral99 on Thursday, June 07, 2018 3:59 PM

My helix is 22" radius and has 4 levels. The levels are supported by 4 posts (lets say they're positioned at N-S-E-W. I ran the bus up the East and West posts and feeders to the rails. I used sections of flex track, which basically, ends up having a feeder every other joint.

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Posted by rrinker on Thursday, June 07, 2018 6:49 PM

 At least 2 - that makes it about 8 feet apart for the outer track. 4 verticals would probably be better.

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Posted by ATLANTIC CENTRAL on Sunday, June 10, 2018 7:19 AM

railandsail

I've seen a few references that suggest it is NOT necessary to run the bus-lines along the track route in a spiral helix, but rather just run the track feeders lines down one of the vertical legs (post) to the bus wires at the bottom. I thought this was a good idea.

My question is how many of these vertical connections should be made?

My helix is a double track affair with radii of 32.5” and 29.5”. I have plans to stagger the rail joints, and to solder all the rail joints,...code 100 Atlas rail.

Considering this sort of relatively small radius/circumference, I'm thinking I need only one feeder set of wires for each elevation/loop of the helix. And perhaps these feeders wires might be sized a bit larger than normal??

 

 

Will the rail joints be soldered?

I know lots of guys with large DCC layouts without all this "every 6-8 feet" feeder business and their layouts work fine. Why? Refer to my first question - their answer is yes.

But in a worst case, two bus locations 180 degrees apart, tapping each level should be more than enough.

I run DC, advanced cab control with base station radio throttles, detection and signaling. All my rail joints are soldered within each "block", many of which are 20' - 25' long, each has only one feed which must run thru a current detector.  No voltage drop issues.

Sheldon

 

    

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Posted by gregc on Sunday, June 10, 2018 7:40 AM

Nickle-silver track has a resistance of 1.45 millohm/in (0.057 ohm/m) which is the equivalent of #22g wire.

32" radius amounts to 17' of track per loop.    The track length fed from one feeder/loop is 8.5', reached from two separate points and ~0.15 ohm or 0.15 V drop at 1A.

greg - Philadelphia & Reading / Reading

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Posted by ROBERT PETRICK on Sunday, June 10, 2018 8:08 AM

gregc

Nickle-silver track has a resistance of 1.45 millohm/in (0.057 ohm/m) which is the equivalent of #22g wire.

32" radius amounts to 17' of track per loop.    The track length fed from one feeder/loop is 8.5', reached from two separate points and ~0.15 ohm or 0.15 V drop at 1A.

Just a quick math question . . .

If the feeder spacing is 8.5 feet, doesn't that mean that the load (engine) is always less than 4.25 feet from the feeder? So, why isn't the resistance of 0.00145 ohms per inch multiplied by 4.25 feet (51 inches)?

Just asking.

Robert 

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Posted by ATLANTIC CENTRAL on Sunday, June 10, 2018 8:19 AM

ROBERT PETRICK

 

 
gregc

Nickle-silver track has a resistance of 1.45 millohm/in (0.057 ohm/m) which is the equivalent of #22g wire.

32" radius amounts to 17' of track per loop.    The track length fed from one feeder/loop is 8.5', reached from two separate points and ~0.15 ohm or 0.15 V drop at 1A.

 

 

Just a quick math question . . .

If the feeder spacing is 8.5 feet, doesn't that mean that the load (engine) is always less than 4.25 feet from the feeder? So, why isn't the resistance of 0.00145 ohms per inch multiplied by 4.25 feet (51 inches)?

Just asking.

Robert 

 

Exactly. And none of our locos draw anything near 1 amp these days.

So the REAL voltage drop is minuscule.

Sheldon

    

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Posted by BMMECNYC on Sunday, June 10, 2018 8:28 AM

ROBERT PETRICK

 

 
gregc

Nickle-silver track has a resistance of 1.45 millohm/in (0.057 ohm/m) which is the equivalent of #22g wire.

32" radius amounts to 17' of track per loop.    The track length fed from one feeder/loop is 8.5', reached from two separate points and ~0.15 ohm or 0.15 V drop at 1A.

 

 

Just a quick math question . . .

If the feeder spacing is 8.5 feet, doesn't that mean that the load (engine) is always less than 4.25 feet from the feeder? So, why isn't the resistance of 0.00145 ohms per inch multiplied by 4.25 feet (51 inches)?

Just asking.

Robert 

 

Greg,

The postulated scenario states one set of feeders in 17' feet of track.  => Furthest distance a locomotive can be is 8.5 feet. 

Sheldon,

Most locomotives now draw 1/3 amp or so.  Is the OP running modern production equipment...  

That 1/3 amp everyone throws around is the steady state draw of the locomotive measure current draw on a 30" curve with a 3% grade (or whatever this ends up being).

You are also assuming one locomotive will pull the train up this grade.

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Posted by floridaflyer on Sunday, June 10, 2018 8:29 AM

What Robert and Sheldon said

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Posted by BMMECNYC on Sunday, June 10, 2018 8:32 AM

floridaflyer

What Robert and Sheldon said

 

Except the feeder spacing isnt every 8.5 feet, its every 17 feet.

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Posted by ROBERT PETRICK on Sunday, June 10, 2018 8:38 AM

BMMECNYC

 

floridaflyer

What Robert and Sheldon said

 

Except the feeder spacing isnt every 8.5 feet, its every 17 feet.

Okay. I thought we were talking about two feeders per loop. Sorry for adding to the confusion. Dang. Embarrassed

Robert

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Posted by gregc on Sunday, June 10, 2018 8:39 AM

yes the feeder spacing is 17', but the furthest distance the current needs to flow is half that distance as BMMECNYC stated.

greg - Philadelphia & Reading / Reading

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Posted by floridaflyer on Sunday, June 10, 2018 8:46 AM

I think Greg's comment of 'reached from two seperate points' clouded the issue. With one drop only, the 8.5 distance  and voltage drop is correct. Robert's conclusion,as well as mine, and maybe Sheldon's, was that there were two seperate points where there is a drop. Rereading Greg's post, he did say one drop.

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Posted by BMMECNYC on Sunday, June 10, 2018 8:56 AM

Since we are on the topic of the helix, I would suggest making the helix its own power district, protected by a circuit breaker separate from other layout circuit breakers.  

What I do know is it is easier to start both model train and real train on level ground than on a grade. 

What I dont know is 1) if the OP has tested the ability of his locomotives to start a train on his planned grade and 2)  will the locomotives will stay or slide backwards when power is secured.

Rule 108: In case of doubt or uncertainty, the safe course must be taken.
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Posted by ATLANTIC CENTRAL on Sunday, June 10, 2018 9:39 AM

BMMECNYC

 

 
ROBERT PETRICK

 

 
gregc

Nickle-silver track has a resistance of 1.45 millohm/in (0.057 ohm/m) which is the equivalent of #22g wire.

32" radius amounts to 17' of track per loop.    The track length fed from one feeder/loop is 8.5', reached from two separate points and ~0.15 ohm or 0.15 V drop at 1A.

 

 

Just a quick math question . . .

If the feeder spacing is 8.5 feet, doesn't that mean that the load (engine) is always less than 4.25 feet from the feeder? So, why isn't the resistance of 0.00145 ohms per inch multiplied by 4.25 feet (51 inches)?

Just asking.

Robert 

 

 

 

Greg,

The postulated scenario states one set of feeders in 17' feet of track.  => Furthest distance a locomotive can be is 8.5 feet. 

Sheldon,

Most locomotives now draw 1/3 amp or so.  Is the OP running modern production equipment...  

That 1/3 amp everyone throws around is the steady state draw of the locomotive measure current draw on a 30" curve with a 3% grade (or whatever this ends up being).

You are also assuming one locomotive will pull the train up this grade.

 

The OP's layout plan, posted on this forum for months now as he works out the details, is small with relatively short staging tracks, which likely equals short trains. I suspect one loco will handle most trains up that helix.

I don't personally use DCC, but I have been closely involved with the design, building and wiring of two large, basement filling DCC layouts, and the conversion of several DC layouts to DCC.

In the DC to DCC conversions, no addtional feeders were added. Existing block feeders were simply connected together into districts and fed. They all work just fine.

I think Paul3 and a few others on this forum with suitable experiance will support my view that this multiple feeder thing is way over kill. Especially in todays world of low current motors.

And again, the solution to a lot of these problems is soldered rail joints.

When I was a child, my father set up a large (5' x 18') "Christmas Garden" every year using TruScale Ready Track. Even for a 10 week Christmas display, he soldered all the rail joints, and then unsoldered them for dis-assembly.

That was back in the 60's with current hog open frame motors. He wired the layout with surplus telephone wire and fed each block/loop with only one feeder. Those trains ran fine.

Sheldon

 

    

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Posted by carl425 on Sunday, June 10, 2018 9:42 AM

ATLANTIC CENTRAL
Exactly. And none of our locos draw anything near 1 amp these days. So the REAL voltage drop is minuscule.

Yeah but...

With the small current draw of today's locomotives, the issue in feeder size and spacing is no longer the voltage drop when running the locomotives.  It is passing the "quarter test".  You want the circuit breaker to trip whenever there is a short on the track.  With feeders too far apart a short will not trip the breaker.

See "The Great Feeder Experiment" at http://www.wiringfordcc.com/track.htm

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Posted by gregc on Sunday, June 10, 2018 12:19 PM

carl425
With feeders too far apart a short will not trip the breaker.

See "The Great Feeder Experiment" at http://www.wiringfordcc.com/track.htm

i looked at The Great Feeder Experiment and have a hard time believing the results based on the description.

My understanding is he connected a 5A booster to a piece of 3ft flex track through

  • 25' of 14g
  • feeders of various gauge (18-26) and lengths (0.5 - 6ft)

am i correct?

based on wire resistance reported in the American Wire Gauge, i calculate the worst case wire resistance of 0.616 ohm for the 6 ft 26g case, total resistance for both paths.

if the booster supplies 15V, a short should result in 24A.

a more simplistic case is that 31ft of 26g wire has a resistance of 1.3 ohms or 2.6 ohms.

what am i missing?

i strung 54 ft of 30g wire, measured 5.8 ohm (5.6 according to the wire table) and 1+A bulb on my PowerCab lit up.

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Posted by BMMECNYC on Sunday, June 10, 2018 12:20 PM
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Posted by BMMECNYC on Sunday, June 10, 2018 12:29 PM

gregc
25' of 14g feeders of various gauge (18-26) and lengths (0.5 - 6ft) am i correct?

Yes

gregc
what am i missing?

I cant tell by your post.

The table provided indicates a Go/No go test.  The booster either tripped or it didnt.

Also for calculating the booster's circuit breaker tripping, you need to factor in the wire out to the track and the wire back to the booster.  

So its actually 50 ft of 14 AWG out and back + 1ft-12ft of feeders.

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Posted by BMMECNYC on Sunday, June 10, 2018 12:31 PM

gregc
if the booster supplies 15V, a short should result in 24A.

Where are you getting that number?

Rule 108: In case of doubt or uncertainty, the safe course must be taken.
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Posted by gregc on Sunday, June 10, 2018 12:33 PM

BMMECNYC
Also for calculating the booster's circuit breaker tripping, you need to factor in the wire out to the track and the wire back to the booster. 

gregc
i calculate the worst case wire resistance of 0.616 ohm for the 6 ft 26g case, total resistance for both paths.

greg - Philadelphia & Reading / Reading

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Posted by gregc on Sunday, June 10, 2018 12:34 PM

BMMECNYC
 
gregc
if the booster supplies 15V, a short should result in 24A.

Where are you getting that number?

24 A = 15 V / 0.61 Ohm

greg - Philadelphia & Reading / Reading

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Posted by BMMECNYC on Sunday, June 10, 2018 12:39 PM

gregc

 

 
BMMECNYC
 
gregc
if the booster supplies 15V, a short should result in 24A.

Where are you getting that number?

 

 

24 A = 15 V / 0.61 Ohm

 

Is the booster in question actually capable of suppling 24A to the short in question?

Did you calculate the resistance of the track? There is an additional 6 (ish) feet of rail involved in the test.

Also I think the AWG chart assumes pure copper wire, which most (smaller) wire you buy is not (this is also stated on the wiring for dcc website somewhere)

 http://www.wiringfordcc.com/trakwire.htm

 

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Posted by BMMECNYC on Sunday, June 10, 2018 12:46 PM

Greg,

I guess it might be also that there is resistance at each conection that is not accounted for in your calculation (wire to booster, wire to smaller wire, small wire to rail).

 

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Posted by rrinker on Sunday, June 10, 2018 12:55 PM

gregc

 

 
carl425
With feeders too far apart a short will not trip the breaker.

See "The Great Feeder Experiment" at http://www.wiringfordcc.com/track.htm

 

i looked at The Great Feeder Experiment and have a hard time believing the results based on the description.

My understanding is he connected a 5A booster to a piece of 3ft flex track through

  • 25' of 14g
  • feeders of various gauge (18-26) and lengths (0.5 - 6ft)

am i correct?

based on wire resistance reported in the American Wire Gauge, i calculate the worst case wire resistance of 0.616 ohm for the 6 ft 26g case, total resistance for both paths.

if the booster supplies 15V, a short should result in 24A.

a more simplistic case is that 31ft of 26g wire has a resistance of 1.3 ohms or 2.6 ohms.

what am i missing?

i strung 54 ft of 30g wire, measured 5.8 ohm (5.6 according to the wire table) and 1+A bulb on my PowerCab lit up.

 

 Did you measure the voltage at the far end? If the resistance was 5.8 ohms, with a 1 amp load on the far end the voltage should drop by 5.8 volts. FAR too much, that's more than 1/3, the equivalent of running the train at 2/3 speed vs full throttle, and definitely noticeable. That's why we don;t use long runs of #30 wire for powering the layout.

 It all depends on how the layout is used. If you only ever run one train by yourself, you'll probably never have even a 1 amp load. Lots of operatores, depends on how the layout is divided and if there are bottleneck points that are all the same power districts, you can easily exceed 1 or 2 amps of load in a single spot.

 Note sure where that 24 amps comes from. If the wire resistance id 5.8 ohms, then a dead short is 15v/5.8 or 2.6 amps. A 3 amp capable system wouldn't trip. And you are assuming a 0 ohm short. Even clipping a pair of leads to the wires int's really 0 ohms. If you set a quarter on the rails, NOT press down, which is the proper way to do the test, say you had 4 ohms of resistance and ZERO ohms in the wire, so absolutely no loss, ideal wire, ideal track. A 4 ohm load at 15 volts is 3.75 amps. Not enough to trip a 5 amp system. But 3.75 amps through a 4 ohm load is nearly 60 watts of heat. A 60 watt light bulb easily melts plastics. THAT'S why you use heavier than you might need wire, and use circuit breakers instead of feeding 5, 8 or 10 amps right to the rails, at least in smaller scales. You're unlikely even on a busy layout to have the 6-12 (or even more) locos it would take to draw 3 amps, but even that relatively high resistance short would exceed the 3 amp setting and cause the power to cut off.

                                 --Randy

 

 


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Visit my web site at www.readingeastpenn.com for construction updates, DCC Info, and more.

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Posted by BMMECNYC on Sunday, June 10, 2018 1:01 PM

This calculator https://www.cirris.com/learning-center/calculators/133-wire-resistance-calculator-table

 

says that 26 AWG wire 12ft long is .49 ohms +.126 ohms = .616 ohms of wire resistance.

Rail resistance: https://www.w8ji.com/track_resistance.htm

 

0.00145 ohm per inch code 100

.00145 x 70" (36" piece of flex track, soldered connections about half inch from end of rail, plus quarter would not be exacty at end of rail)

.1015 ohm

.616 ohm + .1015 ohm = .7175 ohm

+2 unknowns:

resistance of connections, at least one of the three is a screw terminal

resistance of the 5/8ths inch piece of material used to generate the short circuit.

 https://dccwiki.com/Rail_Size

this gives slightly different data for rail resistance.

 

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Posted by gregc on Sunday, June 10, 2018 1:04 PM

BMMECNYC
I guess it might be also that there is resistance at each conection that is not accounted for in your calculation (wire to booster, wire to smaller wire, small wire to rail).

he said the booster outputs 14.2V.    A resistance of 2.84 Ohm results in 5A.

if there is 0.61 ohm of resistance in 50 ft of 14g and 12 ft of 26g, you're suggesting that the majority of the resistance is in the connections (2.2 Ohm = 2.8 - 0.6), not in the wire.

greg - Philadelphia & Reading / Reading

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