Gradients

Hey you lot Rebuilding a new layout at last. Woodland scenics have a 2 or 4 % incline kit. I will be running 6 coach trains. Do you think 4% is too steep, given Im only in a garage? Ta much Rob

Reply to
Rob Kemp
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Rob -

Yes. You will have enormous difficulty getting 6 coach trains up such a grade even on straight track. You should aim for 2% max on straight, and less on curves. If you can't fit this in you will probably have to settle for shorter trains.

Reply to
John Nuttall

Or redesign so that one track goes down and one up, giving twice the height difference in the same length.

MBQ

Reply to
manatbandq

Before planning my layout I did some tests. My 7-vehicle (one of the coaches is actually the "locomotive") Kato Nozomi 500 Shinkansen model was able to climb plain line on a 6% gradient, just about. 4% is utterly reliable.

My tests involved blue-tacking an oval of flexi-track to a flat dinner table and propping one end of the table up with various-height piles of books, to put the whole oval at whatever angle of slope I wanted to test. At 6% I found there was a risk of a tipping crash due to cant deficiency on the bottom end of the oval (which obviously had adverse cant in my test setup), but this particular model is a very fast mover (I've measured it at around 700mm/sec IIRC).

I have one US-made locomotive, which has more difficulty with longish trains up steep gradients, because (like most US models and apparently many UK ones) it lacks traction tyres.

Switches and crossings also cause extra drag, and opportunities to "catch", so avoid situations where a train has to be pulled up a steep hill at the same time as traversing pointwork.

I've settled on 4% as steepest gradient, basically because I want to cram as much as I can into the space available.

Reply to
Ian Jackson
Reply to
Just zis Guy, you know?

Also the train's weight and friction

A friend and I were privileged to visit the late Colonel Hare's famous

7mm scale Bromford and High Peak railway, with its American-style helix between the higher and lower levels. His heavy prize-winning B1 with a large Pittman motor had difficulty pulling a train that it took with ease on the rest of the layout.

Reply to
Christopher A. Lee
Reply to
Just zis Guy, you know?

Yes.

This engine was properly weighted, had a huge, powerful motor and had just won a trophy at Telford a week earlier.

One solution would be to use a banking engine (helper on the left side of the Atlantic) just like the real thing.

Another friend has an O-scale Lickey banker 0-10-0 built from an Eric Underhill (now JLRT) kit for this purpose but he ended up with a single level layout.

In OO you could use an engine with the front coupler removed.

The train engine stops on a short isolated section, the banker stops at the back of the train, power is restored to the isolated section and they both set off. Not having any load the banker will want to go faster than the train engine.

Similarly the train stops with the train engine just in front of a longer isolated section. Power is removed from this and the train sets off leaving the banker behind.

With DCC it's even easier.

Or you could double-head trains.

Reply to
Christopher A. Lee

John Nuttall wrote:

I've seen a couple of compact HO layouts at Canadian shows with a triple helix climbing up a circular mountain. The train disappears into a tunnel at the top, descends a (presumably steeper) helix inside the mountain, and re-emerges at the bottom level. The grade and curvature are quite sharp, but the train is a short one with a few freight cars, usually hauled by a Bo-Bo diesel loco.

Reply to
MartinS
Reply to
Just zis Guy, you know?

In OO I manage 1 in 30 3.33% new Hornby HST power car + 5 trailers + dummy power car gets up with no problem. Hornby 50 gets 6 coaches up no problem. That's on 3rd radius helix as well.

Reply to
Chris

Wasn't there one where the whole mountain rotated while the train climbed against the gradient? The result being you could observe the train climbing vertically but it stayed in front of the viewer until it finally went into a tunnel near the top. I saw it in the UK but have a feeling that it came from across Pond and has since gone back.

G.Harman

Reply to
damduck-egg

I reckon that the average train of 6 will struggle at steeper than

1:35 and certainly jib at restarting. I use rather more than a garage and have problems making incline approaches to be free of curves and points! Most tyred locos are 2 wheel drives and untyred 6 wheel so they are about equal in traction. I make my inclines with a 3m rising length and 2.5m falling (for an 80mm rise or fall). Just to be sure of good traction on the climb I use steel rail.
Reply to
Sailor

Yes, I think I remember seeing that one in Ottawa some years ago. The one I saw just last week at a local show had a stationary mountain.

Reply to
MartinS

"John Nuttall" wrote in news:4d79fc89$0$2445 $ snipped-for-privacy@news.optusnet.com.au:

Can someone please remind me how you convert ratios into percentages and vice versa (and I'll refrain from commenting on the sillyness of using "percentage" as a measure of gradient).

Reply to
Chris Wilson
Reply to
Just zis Guy, you know?

A percentage is a ratio. ;-) It's the ratio "per hundred". It (or the more precise permille) is the standard engineering spec everywhere except in the UK, where "percent" is for some reason a shibboleth.

OK, example:

3% means 3 foot rise/fall in 100 feet. Could be written as "3%, 3 in a hundred, 3:100, 3/100," etc.

The UK gradient specification is given as "1 foot in X feet", so 3% is 1 foot in 33-1/3rd feet, or 1 in 33, rounded off.

You can avoid all this by making a gradient tool. Materials: a 55cm piece of 1x2 lumber, a 1/4" bolt, a 9 inch level. Drill a hole for the bolt exactly 50cm from end. Thread the bolt in from below.

On piece of paper, make a table of gradients, using the fact that you have a 500mm run. Eg, 1 "1 in 250" is equivalent to "2 in 500". IOW, divide the run into 500, and that will give you the rise (in mm) of the trackboard over 500mm. Thus:

1 in 250: 2mm 1 in 200: 2.5mm 1 in 150: 3.3mm 1 in 100: 5mm 1 in 50: 10mm

and so on. Glue this table to the tool for ready reference.

Now measure the exposure of the bolt below the tool, set it on the track board with the bolt down hill, place the level on it, and raise and lower the track board until the tool is level. The track board will now be at the set gradient.

HTH Wolf K.

Reply to
Wolf K

$2445

What Wolf has not stated is that gradients on road and rail are measured as unit rise over linear distance travelled. The mathematical equivilent is unit rise per horizontal distance travelled (Tan theta). The difference is significant at the steeper values. Mathematically 1:1 is a slope of 45=B0 but in road/rail terms it is vertical! Clearly this is a non linear function such that % should not be applicable but expressing rise per 100 units travelled is convenient if expressed as percentage and makes the smaller figure into the shallowest rise (but the biggest fraction). Peter A Montarlot

Reply to
Sailor

$2445

I did not add that throughout Europe the road grads are marked off in % signs.

Reply to
Sailor

???????

I don't think 100% grade means "vertical" at all. It just means what it=20 says: 100 foot rise in 100 foot run. It's possible (but feasible only on =

roller coasters or in material delivery systems) to have grades in=20 excess of 100%.

IOW, "3% grade" does not mean "3/100 of the vertical slope". It's=20 possible that some (many?) people have a vague fuzzy notion that is what =

it means, but that's likely just because they haven't thought much=20 beyond "How do I calcuate how much space I need to run my track back and =

over itself?"

"Vertical" would be zero run over any (=3Dindeterminate) rise, IOW an=20 infinite grade.... Again, think roller coaster. Normally, we don't give=20 "grade" a direction or dimension, but on a roller coaster we could say=20 you can go from zero to positive to infinite to imaginary to infinite to =

negative and finally to zero grade again. And you get to pay for the=20 privilege!

Again, I think you've missed the point. Of course it's a non-linear=20 function. So why should that prevent us from expressing it as a percent? =

I don't follow your thinking here. The tangent (Rise/run) is a ratio,=20 which can be expressed in any way convenient to the user. The fact that=20 it's a non-linear function is neither here nor there.

Some background:

In N. America, by the mid-1800s the surveyors had adopted the 100-ft=20 "chain", and laid out the railway line in "stations" of one chain.=20 Elevations were calculated for each station, so it was handy to plot=20 grade (change in elevation) as so many feet per station =3D so many feet =

per 100 ft. So a grade of "X feet per hundred" became " X percent grade".=

In Europe, the metric system encouraged a superficially more precise=20 measure of gradient, the "pro mille". What in N. America is designated=20 as, say, 2.5% is Europe is designated 25,0 "pro mille" (sorry,I don't=20 have a character for that.)

No matter how it's designated, a grade/gradient is a ratio. The=20 conventions are historically explicable, and interesting as clues to how =

our ancestors discovered/invented/adapted the new technologies to their=20 needs.

Cheers, Wolf K.

PS: and let us now start a side bar on frog angles and turnout=20 numbers.... ;-)

Reply to
Wolf K

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