Converting Prototype to HO Scale

[jkeaton]

If the local hobby shop doesn't have an HO scale ruler, then have them order one from Walthers - or order one yourself from Micro-Mark (www.micromark.com).

Jim

[dbduck]

should be able to find HO ruler at local hobby shop

[fishplate]

OH GREAT! Another bullet for the nitpickers. Now I'm going to get told that my drop in coal loads are incorrect scale ton weight.[B)]

[Anonymous]

Mass is a cubic measurement. It does NOT scale down linearly. Thus, by cubing the HO scale ratio 35/3048 (calculated exact using 3.5mm to the foot) one gets the coefficient of scale mass to be ~1.514e-6 (.000001514). You simply multiply this by the mass you want to scale down to HO.

Since mass is directly proportional to weight, a loaded 75 ton freight car would weigh (75*1.514e-6) tons * 2000 lbs/ton * 16 ounces/pound = 3.63 ounces, which is completely reasonable.

[Anonymous]

Where do you buy an HO ruler?

[ericboone]

QUOTE: Originally posted by CPPedler

As far as I know the calculation by 'ndbprr' is absolutely right for height,width and length but I have never found a formula for working out scale weight this is a little tricky because if something weighs 87 tons (U.K. measurements!!!) it can't possibly weigh 1ton in H.O. I've never yet found anyone who does know that answer. CPPedler

I was incorrect, thus the edit of this posting. lemscate has the correct end result. However, units of mass themselves are not cubic. The cubic ratio comes from the volume being scaled.

Mass(g) = Density(g/cm^3) x Volume(cm^3)
(These metric units are used just as an example. cm^3 means a cubic centimeter)

If your model is perfectly scaled, then, as previously discussed, the volume of the model will be the volume of the prototype times the scale ratio cubed. In HO scale, this would mean multiplying the prototype volume by 1 / 87.1 cubed or 1 / 660776.311 to obtain the model volume.

Assuming your model is made from the exact same materials, then the density will be the same for both the model and the prototype.

Looking back at the formula, it is easy to see that if the density of the model and the prototype are the same and the volume of the model is 1 / 660776.311 of the prototype, then the mass of the model must also be 1 / 660776.311 of the mass of the prototype.

[dehusman]

A handy way to judge the size of strip styrene is to look at the size in .010 of an inch.

A piece that is .010 x .040 is roughly the size of a 1x4. A .060 x .100 piece is a 6x10 roughly.

1/8 in is .125 so that is roughly a foot. 1/8 square balsa is roughly a foot square.

For fine measurements I use a dial caliper and a cheap calculator. A dial caliper is one of the best investments you could make. It and a #11 Xacto blade are probably my two most valuable ing tools.

Dave H.

[leighant]

The proportion for HO and modeling scales is generally given in LINEAR dimensions, ie, one dimension. HO is 1/87th actual LINEAR size (one dimension). For a real object 87 feet long, the HO model would be 1 foot long. I think you go that.

But if you figure out how big a one acre field is in actual feet, and build an HO model is HO scale feet, the model will not be 1/87th of a real acre, because AREA is two dimensional, length and width if it is rectangular, to make it easy. An HO scale area would be 1/87 x 1/87, or 1/7569 of real area. Or it would take 7569 HO acres to fill one real acre. Got that.

VOLUME is 3 dimensions, so you would have to "cube" the linear proportion, ie 1/87 x 1/87 x 1/87. That comes out to 1/658503. So if you had a cube 87 feet tall, 87 feet wide and 87 feet long, an HO model of it would be 1 foot on each side. 87 of them lined up end to end would run along one edge of the 87 foot cube, right?
7569 one-foot cubes would cover one side of the 87 foot cube.
But to fill the 87 foot cube would take 658503 one-foot cubes-- about 2/3rds of a million.

[CPPedler]

As far as I know the calculation by 'ndbprr' is absolutely right for height,width and length but I have never found a formula for working out scale weight this is a little tricky because if something weighs 87 tons (U.K. measurements!!!) it can't possibly weigh 1ton in H.O. I've never yet found anyone who does know that answer. CPPedler

[challenger3802]

Since HO is 3.5mm to the foot, here is an alternative way to get the right length from the prototype.

(And if that's not confusing enough I have OO gauge stock on my layout which uses 4mm to the foot)

Ian

[Anonymous]

Thank You.

Dividing by 87 just seemed too easy. I needed someone to tell me I am making it too hard.

I will pick up a ruler tonight.

[ndbprr]

Two ways:
1. You buy an HO Scale ruler that has dimensions in feet and inches in HO scale. Available at most hobby shops or from Walthers catalog.
2. You divide the dimension by 87 since HO is 1/87th full scale. So something 87' long in real life will be 1' long in HO scale. Something 43.5' will be 1/2' or 6".

[Anonymous]

The kids will be gone in ten years, the house will be paid iff and I will gut the basement and fill it with railroad tracks. So now I am researching the prototype line I want to model, and buying Kalmbach how to books. One great book is "Trackwork and Lineside Detail".

It gives the prototype dimensions for sign posts, cabinets, etc. Sooo, I go to thinking, how do you convert that to an HO scale?