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Building a railroad system in my "Man Cave"...

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Building a railroad system in my "Man Cave"...
Posted by RobertB. on Monday, October 18, 2010 10:49 AM

I have converted my 2 car garage into a Man Cave and I am planning on incorporating my old H.O. train set to run around the Man Cave walls. I plan on running the rails on several different levels (different heights) and need advice as to prevent gravity conflicts from keeping my engine from pulling the cars up the hills. One of the routes I plan on using is a 76 1/2" to 64 1/4" drop in a 10' metal stud wall. And would like to keep the main attraction no higher than 48" above floor level. Any advice at to what pitch or curves allowed to run the engine up the different levels. If the main display of the town/city is going to be at 48" above floor for maximum viewing (and design work), what pitch per foot (inch rise per foot) would facilitate or allow the engine to pull the cars up the hills without stalling? Or, is my run in the wall too steep of a slope to cause the engine to derail (crash) when it reaches the end of the wall and turn to go around viable obstructions outside the wall?  How sharp of a turn am I allowed to make to tour around the city or to tour around the walls of the Man Cave for the Mountain Scenery?

Much help will be much appreciated before I waste money designing the Scenery and landscapes.

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Posted by jrbernier on Monday, October 18, 2010 11:27 AM

Robert,

  A lot is going to depend on your engines and how many cars they are pulling.  A 2% grade rises 1" every 50".  That 12+" rise you mentioned will take about 25'!  I think you will need to supply some more information.

Jim

Modeling BNSF  and Milwaukee Road in SW Wisconsin

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Posted by Bob W on Monday, October 18, 2010 11:48 AM

Would that not be 50'-0 + ?

Bob W [FL]

If it ain't broke.... Fix it till it is !

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Posted by selector on Monday, October 18, 2010 12:11 PM

Yes, it would be 50 feet (50').  That's darned near as long as my folded loop main track!  In that space, I rise only 8", and that means a 3% grade which I now (having been subjected to my own folly for four years) no longer recommend to anybody.  It is too limiting on the lengths of trains and too hard on drive train components, including the rods on steamers.

Settle for whatever height is accomplished via a 2.2% grade, just like the B&O did 150 years ago, and which was adopted as the 'maximum standard grade' for railroads in N. America.

Crandell

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Posted by odave on Monday, October 18, 2010 2:46 PM

To your question on curves,  how sharp you can go depends on several factors related to the length of the equipment you plan on running (i.e. 40' boxcars vs 89' autoracks).  The Layout Design Special Interest group has a handy rule of thumb  that explains what to expect with different curve radii.  Check it out at this link.

Note that curves on a grade will "feel" sharper to the equipment, so you may want to make those broader.

One way some folks handle elevation changes is with a helix.  A helix is a "corkscrew" construction that has the track spiral up gradually through a circle.  There are some challenges with helices - they can be tricky to build and maintain, and eat up a lot of space.  It's something to look into, anyway.

 

--O'Dave
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Posted by galaxy on Monday, October 18, 2010 3:44 PM

I would stick to a 2% grade and avoid curves, OR allow for wide radius curves.

Like Crandell I found out too quickly a {4%} steep grade was way to steep for any of my locos to reasonably pull a length of train cars. I reduced to 3% hoping for better, but as Crandell pointed out, still to steep. I scrapped that layout completely and started again.

I now only stick to 1 or 2% grades at the most, even though it menat I didn;t get quite what I thought I wanted...

A 2% grade or less is Easier on me AND my locos.

-G .

Just my thoughts, ideas, opinions and experiences. Others may vary.

 HO and N Scale.

After long and careful thought, they have convinced me. I have come to the conclusion that they are right. The aliens did it.

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Posted by RobertB. on Tuesday, October 19, 2010 5:31 AM

Exactly how are the Curve radius calculated?

Example:

Curve Radius
2X - 12"
3X - 18"
4X - 24"
5X - 30"

Is this the distance from the far right side of the track to the far left side of the track on the opposite sides of the curve? Or center to center of the track from right side of curve to left side of curve?

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Posted by RobertB. on Tuesday, October 19, 2010 5:48 AM

Jim,

 

I am still in the design stage of my idea. I am trying to decide how I want to setup the lay out to make the elevation changes to facilitate the opposite sides of the room and change in the different levels considering the heights of the walk space and viewing space around the tracks.  I have one wall where I am planning on starting the track 76-1/2" above floor entering the wall and slopping down to 64-1/4" height exiting the opposite end of the wall 10' in length and coming out of the wall and turning around an object 26-1/2" deep and then turning a 90 degree turn 47" away. Like a question mark offset and turn. Then after the offset and turn (question mark style turn/offset), running a straight run approximately 106" away before making any other turns or height changes... Then this area will be in the occupied walking space so I will have to manipulate other obstructions and levels.

 

Like I said, I'm still in the design stage...

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Posted by RobertB. on Tuesday, October 19, 2010 5:58 AM

So Jim, if I am dropping 8-1/4"(inches) in a hill 10'(feet) long, what size of a slope is that considered? Is that too steep of a slope to make my question mark offset - 90 degree turn previously mentioned?

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Posted by Anonymous on Tuesday, October 19, 2010 7:39 AM

That´d be a slope of a little more than 7 % - way to steep! For a drop of 8 3/4" you need about 24 feet of track, which equals just about 3 %! You should not exceed this value.

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Posted by galaxy on Tuesday, October 19, 2010 8:46 AM

RobertB.

So Jim, if I am dropping 8-1/4"(inches) in a hill 10'(feet) long, what size of a slope is that considered? Is that too steep of a slope to make my question mark offset - 90 degree turn previously mentioned?

Whoa Doggies!

I thought that was kinda what you meant in the OP {Original Post}, but wasn't totally clear to me. A drop of 8.25 inches in a 120 inch space is WAY TO STEEP unless your trains only will go downhill s l o w l y so as not to derail. AS was mentioned that's about a 7-7.5% grade and any engine/trains will have trouble climbing that grade!

Again, stick with only 2 percent or less grade and you'll not be sorry. {2 inches or less in a 100 inch rise}

Push for a steep grade and it will give you grief over and over and you will end up tearing it out for a smoother lower grade like the 2% or lesser. Ask me how I know.....I scrapped a whole layout that had both 4% and 3% grades and they failed me had to tear everything out and develop new track plan. My trains AND I are happier with 2% or less !!!

{Trust us}

-G .

Just my thoughts, ideas, opinions and experiences. Others may vary.

 HO and N Scale.

After long and careful thought, they have convinced me. I have come to the conclusion that they are right. The aliens did it.

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Posted by odave on Tuesday, October 19, 2010 9:04 AM

RobertB.

[Curve radius]...Is this the distance from the far right side of the track to the far left side of the track on the opposite sides of the curve? Or center to center of the track from right side of curve to left side of curve?

If I'm reading this correctly, you are talking about the diameter of the circle made by curving the track all the way (or half-way) around.  Radius is 1/2 that (Diameter = 2 * Radius)

The radius is the distance from the center point of the circle out to the edge.   The center of the track (between the rails) is the "edge" of the circle.

So a curve with a radius of 18" will produce a circle of track that is 36" wide, measured from track center to track center.

--O'Dave
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Posted by Paulus Jas on Tuesday, October 19, 2010 2:57 PM

hi

If you look at the drawing at the upper left; you can easily calculate the needed width.

Two times the radius plus about 5 inches extra space on both sides (10 in total)

if you are running 89 feet modern cars, the length of that car is about one foot (12 inch) in HO.

The required radius need to be 2.5 times as large at least.

The width of your table will be  2 x 2.5 x 12 + 10 = 70 inch, almost 6 feet.

And you must be able to walk around the table and to reach in.

Paul

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Posted by selector on Tuesday, October 19, 2010 3:26 PM

A circle has a diameter of X in units of length.  A radius is half that.  We can't practically deal with circles of track, especially in the real world, so we use arcs of track, degrees of curvature, and radius of curvature.  Each diameter length comprises two radii; a radius is halfway across the diameter.  As stated above, the radius is really measured from the midpoint of any one diameter line, and extends out to the rim of the circle.  On our railroads, the radius pivot point can be placed anywhere, with a radius made any length we choose, and even begin and end at any length we need so that the next tangent segments, the straight track, is oriented in the direction we need them to go.

Generally, if we have wide curves, those with radii greater than about 30", we tend to need deeper surfaces for our railroads, and that can cause problems during construction and later during operation and maintenance because we have to reach so far into it.  Hard on the back, our elbows often are hard on fragile items delicately oriented and situated, and so on.  So having a large 'table' top layout with curves in the range of 26'-36" would mean almost impossible reaches to the center of the area to be built and scenicked.   It is one of the great Cardinal Sins of model railroad layouts....building a framework that forces us to use ladders and creepers just to get it erected, let alone enjoy its use later on.

Crandell

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Posted by RobertB. on Tuesday, October 19, 2010 6:01 PM

Galaxy/Jim,

So, if I am running my route through a 10' wall and the top of the hill (the entrance into the wall) is 76.75" above floor, how low should I be having the exit end of the lower end of the hill to allow only a 1% to 2% slope? To put it simply, how low should the bottom of the hill be than the top of the hill on a 10' long hill? As a scale to measure future slopes, how low (in inches) should the bottom of a one foot hill be? For up hill climbing or for downhill free rolling hills?  I can tell you when I get home tonight, how many current cars my current engine will be pulling. I am at the public library right now.

Thanks for all your help in designing my future Man Cave City/Mountain layout.

One other thing I just thought of, how do I know how powerful my current engine is and how many cars I can add to my train in the near furure?

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Posted by galaxy on Tuesday, October 19, 2010 9:09 PM

If you have a 76.75" high entrance, the lower end exit should be no lower than about 74.5", allowing a 2.25" drop in 120" of straight downhill means the grade is about 2%. A one percent grade would mean you downhill exit would be at about 75.76" or 75.5".

Remember a 2% grade means a 2" rise over a length of 100" {8 feet 4inches}. A 1% grade is one inch rise over 100 inches.

Adding a curve to that and you effectively increase the grade percentage. There is a "formula" for figuring it out I will have to see if I can dig it up if someone doesn't post it here.

The engine pulling cars thing: It depends on the engine and the cars. SOme engines {regardless of price} are better pullers than others. There is no hard and fast rule here. AS far as cars pulled, that would also depend on the cars. If you go with the National Model Railroaders Association's {NMRA} recommended weight for cars of 5 ounces you may not pull as many as cars that incorrectly only weigh 3 ounces.

DCC is something you can look into. With DCC you consist engines together programed at the same speed to pull longer trains. It can be done with DC, but it is harder and you have to have engines that move at nearly the same speed to consist together so they don't play "push me pull me" and derail each other.

Hope this helps you

-G .

Just my thoughts, ideas, opinions and experiences. Others may vary.

 HO and N Scale.

After long and careful thought, they have convinced me. I have come to the conclusion that they are right. The aliens did it.

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Posted by RobertB. on Tuesday, October 19, 2010 11:50 PM

What is the formula for calculating the grade % per foot? How far do I have drop the grade in a foot to maintain the 2% grade?

 

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Posted by selector on Wednesday, October 20, 2010 12:01 AM

If you have an incline 10' long and it dips, or rises, at a rate of 1.5% (one and one-half percent),  then at the end of the dip or rise, you will have a differential in height of only 1.5".  Not much.  Some of us swallow hard and impose double that rate, or 3%, on our trains and in that perfect 10' incline we will now have a whopping 3" of height or fall.  Again....whoppeee.....I guess.

This is a serious problem, and which causes a lot of us to seriously consider the type of layout and track plan we can cram into a given area and still have something that is both likable and functional...it actually works for reasonable trains.  I have 3% grades and my engines work hard on shorter trains.  But with a half circle grade that runs about 20 feet, I manage to get near 8.5" in height at the top, including having the space hog transitions into and out of the grade.  Trains need a gradual change from level to grade at either end, and they eat into whatever length you have to work with...otherwise your engines nosedive into the rails or come out of them.

Crandell

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Posted by HarryHotspur on Wednesday, October 20, 2010 1:28 AM

RobertB.

What is the formula for calculating the grade % per foot? How far do I have drop the grade in a foot to maintain the 2% grade?

It's just a simple percentage.  If the length of the run is 10' (120") and the drop (or rise) is 6", then you have a 5% grade (6/120=.05).  Just remember to use the same units of measure, i.e. inches or feet, in both parts of the equation.

- Harry

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Posted by Paulus Jas on Wednesday, October 20, 2010 4:07 AM

hi

i'll give you 3 formula's

1)   percentage = ( height (in inches) X 100)   /   (  length (in feet) x 12 )

2)  height (in inches) = percentage x length (in feet) x 12  /   100

3)  length ( in feet) = ( height ( in inches) x 100 )   /   ( percentage x 12 )    +   2 feet

the last 2 feet are needed for vertical easements, a gradual change between the horizontal and the grade ( one feet for the bottom, one for the top of the grade)

i think the best start is to make a drawing of your room and trackplan, and put in on here.

Believe previous followers of this thread, anything more then 3% is asking for trouble; depends on train length, length of the grade and much more. 

paul

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Posted by cudaken on Wednesday, October 20, 2010 3:22 PM

 Robert, Hum Man Cave I like and love mine.

 First, why 76 inch tall. Could you be trying to get a car or truck in to the man cave? If it is I have a plan that I drew up for my layout, but have not done yet. If that is what you are wanting to do, let me know.

 Now are you looking for the elevation for the scenic value? I as well have fought the battle of the grades, and with my current good sizes layout (G shaped 19 X 13) there is no way I could get the trains up 6 inches and back down again with out one side being a roller coaster. Plus I like long trains, norm for me is 20 to 40 car's.

 But, I have a plan for when I do build a new bench. It will be two levels but built to look like one! Lets see if I can explain this.

          Level 2 ______________ 

                     Mountain Side goes here

                     Apx 12 to 18 inches between upper and lower

                                                    Level 1    __________________

 The way I inversion my next layout it will be a 23 X 20 foot around the Man Cave layout. Level 1 will have 4 main lines, some industrial sidings, team tracks small engine yard with a turntable and will be in a valley. Upper level will have two main lines that look down into the valley (you will be able to see both upper and lower level at the same time). It as well will have a small engine yard with a turntable.

 Now, both upper and lower can be ran as separate lines, but there will be one line (set of rails) that connect the 2 levels. With only having one line to worry about I can say use the 18 inch height, and make the connecting line 792 inches for a 2.27 percent grade. Only going up 16 inches would net a 2.02 grade.  With Strong Doubled Headed Engines, you could pull 40 cars up that kind of grade. I all so have a turntable on both levels (there are other ways to turn engines) so I can turn the engines and not back a train down the grade.

 Plus, you can still have smaller grades for visual interest on both levels!

 I know this has been explained, but here is how I do my math.

 I want to go up 4 inches and I have 114 inches to do it in. Take the 4 and divide it by 114 = .035. That is a 3.5 percent grade. (boy I hope I am doing the math right!)

 If I want to go up 3 inches and still have 114 inches it will be a 2.63 grade.

 Other term used is radius. Radius in a nut shell is how wide (Inches) it takes to make a 180 degree turn.

 36 inches is a 18 inch radius.

 48 inches is a 24 inch radius.

 60 inches is a 30 inch radius.

 Now, lets say you only need to turn the train 90 degrees, it could still be a 18 inch radius you just stop where you want. But if it did not stop, it would still take 36 inches to make a 180 degree turn.

 For the nick pickers, yes I know it should be 360 degrees but when I was learning 180 was simpler.

 Robert, I hope I was of some help to get you thinking about different approaches to achieve your goals and dreams. Day dreaming is a big part of this, and most hobbies.                         

 All so drop by Elliot's Dinner in the General Discussion Section and I will buy you a Beer It is a chat room where trains and none train things are talked about. Some great modelers hang out there as well.

 Looking forward to seeing what you think, Cuda Ken

 

I hate Rust

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Posted by Richard Kucharski on Friday, October 22, 2010 12:30 AM

The easiest way I found for ballpark grades is 1" per 100" (8' roughly) = 1 degree.  2" per 100" = 2 degrees, etc., etc.  Don't go steeper then 3 degrees.

Rich

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Posted by HaroldA on Friday, October 22, 2010 5:41 AM

Like someone said, the formula is 'Grade=rise/run/'  What I did was to create a chart of various grades and then used those dimensions in the initial planning of the 12' x 4' extension I added this spring.  I still had to make some adjustments when I actually built it but having the information saved me some real headaches down the road. 

I would also recommend you get a copy of the John Armstrong book "Track Planning for Realistic Operation.'  To me it is still the gold standard for good track design.

 

There's never time to do it right, but always time to do it over.....

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Posted by RobertB. on Sunday, October 24, 2010 1:55 PM

I've started buying the flex tracks and the automatic diversion track to make the train pull into town. Still need to get the stationary diversion track to make it return to the scenic route around the "Man Cave" mountain tour route and some more flex tracks. I calculated 150' of Mountain Tour Route tracks and with a price of $4.25 per 3' track, I need to budget for the tour route. Plus, I still need to cut the tour route bed sections and build the hanger supports. Part of my planned route will run through two of the metal stud walls. Several sections of the route will be suspended above head and parts will be run along a couple of the walls.

One thought my daughter had made was to run guard rails along the suspended rails and the wall mounted rails to prevent the trains from derailing and falling off the tracks. What would you guys thing?

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Posted by galaxy on Sunday, October 24, 2010 2:43 PM

RobertB.

... I calculated 150' of Mountain Tour Route tracks and with a price of $4.25 per 3' track, I need to budget for the tour route....

...One thought my daughter had made was to run guard rails along the suspended rails and the wall mounted rails to prevent the trains from derailing and falling off the tracks. What would you guys thing?

Here is a link to code 100 flex track thats cheaper:

http://www.wholesaletrains.com/Detail.asp?Scale=HO&Item=150TRACK100&ID=200806687

or code 83 track if you are using it:

http://www.wholesaletrains.com/Detail.asp?Scale=HO&Item=150TRACK83&ID=20014766

 

Also anything that keeps a train from derailing onto the floor is a good thing. Listen to your daughter!

-G .

Just my thoughts, ideas, opinions and experiences. Others may vary.

 HO and N Scale.

After long and careful thought, they have convinced me. I have come to the conclusion that they are right. The aliens did it.

  • Member since
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Posted by RobertB. on Sunday, October 24, 2010 2:57 PM

The reason for the 76" height is so the tracks are going to run over head for human traffic, run downhill (1 degree) and offset back around a toilet area, clear a 42" width room, make a 90 degree turn and run over a window curtain rod, make another 90, run back across the opposite side of the office inside the opposite wall, exit the wall after dropping the required distance and maintaining the height for walkway passage, turn 90 again over my "Man Cave" bed area, run to the outside wall of my work shop area, turn 90 and run along the mountain scenes and to the town dropping the required distance for viewing and maneuvering around the city to pick up passengers and towns' people. Then after the train enters the city area, the electric automatic diversion track (activated by a reed switch and magnet in the engine and relay setup)  will turn the train into town to the train depot, stop for passengers. Circle around the town and finally return to the Mountain Scenic Route...

The tracks that will be running outside the walls will be hung on trapeze type mountings (above walkway clearances) after it exits the last wall until it reaches the outside wall of the shop area and then along the wall to the town.

You might want to look at my facebook page and look at my "Home Automation" pictures and look at my relay cabinet setup: http://www.facebook.com/godswindowcomputerservices

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Posted by RobertB. on Sunday, October 31, 2010 9:52 PM

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