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@bigkid posted:

One note, people are using pounds to indicate mass, in the English system mass is weight (in lbs)/32 (something called slugs). In the Metric system mass is measured in KG, the weight is KG*9.8.

Doesn't really matter in terms of this discussion (which is a bit off the rails, pun intended, I think the original poster was just asking what the values would be in 1/48th scale if you scaled down weight and the power of prototype rolling stock). It is pretty easy to see why we can't really scale mass and power down, it is why transformers in the scale world had things like momentum and braking and the command control systems spend so much effort on making the train seem realistic in terms of getting to speed and braking. Open the throttle to 18v and watch how fast our trains move, even just the engine. Obviously most of us don't run mile long trains either, and even long trains in our domain on a big layout if you went full out would accelerate way faster than a train the same length in 1:1.

I disagree.  I think we did answer the question for the OP...  Take the weight and divide by 483 cubed (or multiply by 1/483). 
I also don't think you assertion about mass and power not scaling down is correct.  The reason we need things like momentum and braking is that most all modern locomotives don't have free rolling wheels (a.k.a. "back-drivable gears") and that traction motors in the real world are directly geared to the drive wheels and spin freely when the brakes and dynamic brakes are not being applied.  And for steam engines... well, I think the difference to our models is pretty obvious.

@Will Wilkin posted:

Thinking as Godzilla here, who doesn't do any homework, would not the volume of 1:48 scale be the inverse of 48 x 48 x 48?  If our models had the same density (weight per volume) as the protype, how much would these locomotives weigh at O-scale?  I don't know how to calculate it but I'm sure all the retired engineers here have already worked this problem out?  Would it be the cube root of 48?

Same question for power, or hopefully in similar proportions, either power scaled by weight (traditional measure) or, plausibly valid in modeling, power scaled by volume of the model?  How much power should these model engines have if scaled fo   r power?

Will,

There are some things that can’t be scaled, gravity and time are two of them.  As some have said the molecular structure can’t be scaled – that has to do with gravity – either.  Model airplanes have a similar problem in that they don’t fly in a scale atmosphere so their control surfaces have to deal with the physics of our planet.

I would guess if you want to find a ‘scale’ weight you could do that same thing you do for a scale length; multiply the real by the scale factor of you scale.  I don’t see the logic of this cube root stuff.  If you know the real weight (63,000 lbs for a boxcar) why would you need the cubic volume of the model?  Or for that matter the cubic number of the scale factor?

Tom Stoltz

in Maine

@Tom Stoltz posted:

Will,

There are some things that can’t be scaled, gravity and time are two of them.  As some have said the molecular structure can’t be scaled – that has to do with gravity – either.  Model airplanes have a similar problem in that they don’t fly in a scale atmosphere so their control surfaces have to deal with the physics of our planet.

Model airplanes follow the same rules of physics.  They fly because of the same lift, weight, thrust, and drag ratios of real airplanes.  The main difference is that typically the power to weight ratio of models is skewed massively in favor of the model.  You can do barrel rolls with a model 747 because of that fact.  This is so because the engines used favor power, and don't have to worry about efficiency and flight duration like a real aircraft does flying across the Atlantic consuming several tens of thousands of pounds of fuel.

I would guess if you want to find a ‘scale’ weight you could do that same thing you do for a scale length; multiply the real by the scale factor of you scale.  I don’t see the logic of this cube root stuff.  If you know the real weight (63,000 lbs for a boxcar) why would you need the cubic volume of the model?  Or for that matter the cubic number of the scale factor?

No one is advocating any "cube root stuff".  To scale the weight properly you need to divide all three dimensions by the scale factor e.g. 48 for O, 87 for HO, 64 for S, etc. That leads to dividing the weight by "48 cubed" for our O scale models.  The math works out as previously discussed.

@rplst8 posted:

Model airplanes follow the same rules of physics.  They fly because of the same lift, weight, thrust, and drag ratios of real airplanes. 



Hi rplst8,

Exactly, the physics of the real world rather than a scale world.

No one is advocating any "cube root stuff".  To scale the weight properly you need to divide all three dimensions by the scale factor e.g. 48 for O, 87 for HO, 64 for S, etc. That leads to dividing the weight by "48 cubed" for our O scale models.  The math works out as previously discussed.

This is the part I don’t agree with… or should I say, I don’t understand why (I’m denser than most, maybe that has something to do with it).  The volume of the model has nothing to do with the weight of the prototype.  If the boxcar weights 63,000 lbs, wouldn’t that be what you scale – the weight?  For the thought experiment, let’s assume a coach and a boxcar have the same prototype weight.  The models have very different 3 dimensional measurements, yet they should weight the same.  However the math of the volume would not support this.  But I do think scaling the weight (by the factors you suggest) would give you what you are looking for.  And surprisingly, the cars would be heavier than we could lift if the weight were scaled that way.

I’m probably on the wrong track, but I’m married so I’m used to that…

Tom Stoltz

in Maine

I'll try to explain this as best I can - I've used the same method to estimate the weight of live steam locomotives.  Since they are built using very similar materials making up the same components as their full-sized brethren, this method works very well.

Weight of an object is determined by the volume of the material it is made of.  Twice as much volume of a given material, twice as much weight, right?  If you Google "weight of concrete" you will see that concrete weighs 4050 lbs per cubic yard - the weight is given in relationship to a stated volume.

Now, when we build scale models, we use the scale factor to divide each dimension.  But, as we all know, it takes three dimensions to determine the volume of an object.  That is why the volume of a scale model is divided by the scale factor multiplied by itself 3 times - for height, width, and depth.

So, let's make a very simple example - we have an actual cubic yard of concrete - a block that measures 3' x 3' x 3'.  We decide we want to make a 1/3rd scale model of this cubic yard of concrete.  So we divide each dimension by 3, and wind up with a "model" that is 1' x 1' x 1'.  How much does our model weigh?

It's a 1/3rd scale model, so the scale factor is 3.  We multiply 3x3x3 for the height, width, and depth reduction, which equals 27.  We divide the weight of the "prototype" cubic yard of concrete - 4050 lbs - by the scale factor cubed - 27 - and find out our 1/3rd scale model should weigh 150lbs.

The material of the prototype and model is the same - the density, size of the atoms, etc don't change, just the volume, and the volume is reduced by the scale factor in all three dimensions.

If we built perfect O scale models using the exact same materials as the prototype, our trains would weigh 1/110592 as much as the real thing - 48x48x48 = 110592.  However, that isn't how models are generally built, so the weight may be greater or lighter than we calculate.

As I said in the beginning, I have found this useful in the live steam world.  Not sure how useful it is in the O gauge world.  But the math doesn't lie...

Hope that helps.

Last edited by WindupGuy
@WindupGuy posted:

I'll try to explain this as best I can -

Hi James,

Thank you for the explanation, I think I’m starting to see the logic behind using the cube of the scale factor for weight (volume).  I did Google the question ‘how to find scale weight of a model’.  There are a lot of interesting discussions for many different modeling groups, boats and planes to be exact.  Most agree with your explanation, that weight is a function of volume, as I now do also.  Some of the discussions did go into scaling power for those interested.

So what is the scaled weight of the 63,000 lb boxcar in 1/48th scale?  Would I be correct in assuming that if the weight of the model is near the calculated weight it would just be a coincidence?  And is that weight anywhere near the recommended NMRA weight?

Told you I was good at being wrong…

Tom Stoltz

in Maine

@WindupGuy posted:

I'll try to explain this as best I can - I've used the same method to estimate the weight of live steam locomotives.  Since they are built using very similar materials making up the same components as their full-sized brethren, this method works very well.

Weight of an object is determined by the volume of the material it is made of.  Twice as much volume of a given material, twice as much weight, right?  If you Google "weight of concrete" you will see that concrete weighs 4050 lbs per cubic yard - the weight is given in relationship to a stated volume.

Now, when we build scale models, we use the scale factor to divide each dimension.  But, as we all know, it takes three dimensions to determine the volume of an object.  That is why the volume of a scale model is divided by the scale factor multiplied by itself 3 times - for height, width, and depth.

So, let's make a very simple example - we have an actual cubic yard of concrete - a block that measures 3' x 3' x 3'.  We decide we want to make a 1/3rd scale model of this cubic yard of concrete.  So we divide each dimension by 3, and wind up with a "model" that is 1' x 1' x 1'.  How much does our model weigh?

It's a 1/3rd scale model, so the scale factor is 3.  We multiply 3x3x3 for the height, width, and depth reduction, which equals 27.  We divide the weight of the "prototype" cubic yard of concrete - 4050 lbs - by the scale factor cubed - 27 - and find out our 1/3rd scale model should weigh 150lbs.

The material of the prototype and model is the same - the density, size of the atoms, etc don't change, just the volume, and the volume is reduced by the scale factor in all three dimensions.

If we built perfect O scale models using the exact same materials as the prototype, our trains would weigh 1/110592 as much as the real thing - 48x48x48 = 110592. However, that isn't how models are generally built, so the weight may be greater or lighter than we calculate.

As I said in the beginning, I have found this useful in the live steam world.  Not sure how useful it is in the O gauge world.  But the math doesn't lie...

Hope that helps.

So, in O scale - 1/110,592 = .00000904224
Multiply .00000904224 X 410,000 lbs. (weight of and SD70/Dash 9, etc.) = 3.7073 lbs.

In HO scale - 1/658,503 = .00000151859
Multiply .00000151859 X 410,000 lbs. = .6226 lbs.

Last edited by Big Jim
@Tom Stoltz posted:

So what is the scaled weight of the 63,000 lb boxcar in 1/48th scale?  Would I be correct in assuming that if the weight of the model is near the calculated weight it would just be a coincidence?  And is that weight anywhere near the recommended NMRA weight?

The scaled weight of the 63,000 lbs boxcar (which my be the 64 thousand dollar question ) is 9.1 ounces.

The recommended weight from the NMRA is 5oz + 1oz per inch of length.  Assuming we're talking about a 40' boxcar, that works out to 10 inches scale, so a weight of 15 ounces.  Since a 63,000 lbs box car could probably carry about as much as it weighs give or take 10 tons, the NMRA recommendation would put it at about half full.

The fact that a model would scale out around 9.1 ounces (empty) would be in fact coincidence.  However, since we don't build our models with lead or gold, nor with Styrofoam, beryllium or carbon fiber, the range of densities of the different materials are not that far off from one another.  In addition, models tend to have "thicker walls" so to speak so even if they are made with a less dense material they're not all that far off... and definitely "in the ballpark" at least.

P.S. all scaled values are for O a.k.a. 1:48

Last edited by rplst8

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