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.
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
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.
The Hornby A3 will pull 13 coaches up 2% even with a slight curve. The
9F will pull 50 trucks up 2% with ease. It will depend largely on the
mass of the engine, I reckon. I would not use the Woodland inclines,
though, I'd make the inclines with the baseboard because the track fixes
more securely that way. My experience with Woodland inclines has not
left me with any conviction that they beat the other methods. Even
carving the inclines from insulation board is better, IMO, because the
result will be less geometrically precise.
It's very easy to make inclines from insulation board carved to shape
with a sharp pallette knife and a Surform.
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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.
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.
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.
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.
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.
"John Nuttall" wrote in news:4d79fc89$0$2445
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).
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.
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.
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
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).
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 =
"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
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.
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
PS: and let us now start a side bar on frog angles and turnout=20