O-gauge Lima / Bachman scale question

Hello All,

I believe some Lima offerings in O gauge use differing scales, such as the coaches being 1/48 whilst locos are 1/43.

At shows, certain Lima outline offerings turn up, such as a French diesel I saw at Telford.

Does anyone know what scale was used for these models?

Also, does anyone know what is the linear scale used for the Bachman On30 narrow gauge model line?

Thanks in advance for any help

Pete.

Reply to
Peter Principle
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"Peter Principle" wrote

I think it's all to one scale (1:48), but some UK liveried items are actually continental prototypes and may appear to be overscale compared with the coaches.

John.

Reply to
John Turner

=>Also, does anyone know what is the linear scale used =>for the Bachman On30 narrow gauge model line?

1:48

Runs on 16.5mm track. Use Peco's On30 track if you want correct tie size and spacing. Or fudge it with standard gauge OO/HO track.

Reply to
Wolf Kirchmeir

I think the Lima were a hybrid scale, something like 1:45.

Those are 1/48, being intended for the US market.

It's a bit of a mix and match though, because the prototype was

3' gauge instead of 2'6". Although the little Porter was available for a variety of inductrial gauges including 2'6". But eg the the Colorado stuff was 3'.
Reply to
Christopher A. Lee

There are a number of different European "0" scales; 1:43.5, 1:45 and

23mm/metre - for obvious reasons they don't use 7mm/foot.
Reply to
Gregory Procter

"Gregory Procter" wrote

Strange then that they use 3.5mm : 1 foot for HO scale.

John.

Reply to
John Turner

No they don't - HO scale in Europe is 1:87 ratio. 3.5mm:1 foot is not

1:87. (nor is it 1:87.1 as used in the USa)

Regards, Greg.P.

Reply to
Gregory Procter

In message , Gregory Procter writes

You're right. It's 1:87.08571429, which to 1 decimal place is 1:87.1

Reply to
John Sullivan

That's dependent on what degree of compromise you're prepared to accept in scaling your models. I don't see any point in converting a metric prototype dimension to an Imperial measurement and then dividing by an Imperial:Metric ratio to get a metric model dimension which fractionally doesn't match the accepted and official metric scale. Once I know the scale dimensions I wish to model, I make the adjustments to suit the materials and tools available and to suit my modelling skills. I certainly wouldn't use the US rounded scaling factor, because that introduces an error before I make my own modelling compromises.

Regards, Greg.P.

Reply to
Gregory Procter

Oh yes - it is very close to 1:87. Try it on your Calculator!

304.8/3.5 = 87.085714

David.

Reply to
David F.

1:76.2 is "very close" to 1:87 depending on your perspective - why use other than the exact scale conversion.
Reply to
Gregory Procter

So your models of 1m diameter wheels will measure 11.4942528736--- mm or do you use something other than exact scale? Perhaps an 11.5mm wheel, ie scale of 1:86.96, or even an 11mm wheel, a scale of 1:90.96 (and I know they might be worn wheels, that's not the point)

The difference between 1:87 and 1:87.1 is not significant in modelling, even P87 track and wheels are compatible whichever scale you use. An 80ft car is 0.28mm different in length, can you model well enough to tell? Keith

Make friends in the hobby. Visit Garratt photos for the big steam lovers.

Reply to
Keith Norgrove

Hmmm, my wheels measure 11.5mm, 11.08mm and 10.5mm depending on the situation.

Yes there is a significant difference: - I start with a prototypical dimension - I apply my scale factor - I _then_ adjust to suit any factors I decide are appropriate.

You say the ".1" is not significant. That immediately brings forth two questions:

- why include it?

- when does a figure become significant?

-1:76(.2)?

-HO by MOROP standards is 1:87 with 1:80 as an alternative. I'm informed by some modellers that the difference isn't significant! That extra figure can affect rounding of a converted dimension when one works to the nearest 0.5mm. When I fit my wheels to, for example, a Roco wagon I have to take into account that Roco have already fudged the design to suit 11mm wheels, Märklin to suit

10.5mm wheels. Correct diameter 11.5mm wheels won't fit without fudging other dimensions.

A length over buffer beam and a length over buffers rounding can make a difference that is visible in a model.

Regards, Greg.P.

Reply to
Gregory Procter

In message , Gregory Procter writes

Thank goodness I model in OO. I don't have any of these problems.

Reply to
John Sullivan

Sure, the rails are on the correct side of each other and the gauge variation isn't significant ;-)

Reply to
Gregory Procter

Some of us are obsessed with building everything to the finest detail, some of use want to buy and run rtr trains. Nothing wrong with either.

Reply to
MartinS

Some of us wander around in the middle ground. I buy rtr trains and I build the models I do build to the best of my ability.

Reply to
Gregory Procter

We sure do. rtr locos with a few details added, throw in a little kitbashing of buildings, rolling stock and locos and that suits many of us to a tee.

-- Cheers Roger T.

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of the Great Eastern Railway

Reply to
Roger T.

So where do they get 1:43.5 and 23mm/m from then? Both of those are

7mm/ft rounded off as you so like to do but indistinguishable in practice except to those who can see the effect of 0.28mm in a 280mm long coach. 1:45 is the 'corrected' scale to correctly fit 32mm gauge. So, IMHO, there are two european 'O' scales and several ways of expressing them. Keith

Make friends in the hobby. Visit Garratt photos for the big steam lovers.

Reply to
Keith Norgrove

They come from dividing the prototype gauge by the model gauge - why would Europeans use English feet when converting a prototype metric measurement to a model metric measurement? I fell foul of this situation years ago in converting Württemberg prototype drawings in "fuss" (feet) to scale size using 3.5mm as the conversion factor. They are a quite different measurement!

Yes, there are at least two European continent 'O' scales and considerable arguement as to which is "correct".

Regards, Greg.P.

Reply to
Gregory Procter

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