Märklin uses magnetraction?

Marklin's version of magnetraction is different from the traditional Lionel system.  Like Lionel, Marklin rail is steel, so they can use a magnetic adhesion system as well.  Where as Lionel uses magnetic wheels, Marklin uses two small bar magnets mounted low in the frame and parallel to rails to "pull" the engine harder to the track.  They only Markin locomotives which use this system (to my knowledge) as the small "Kof" engines.

Stuart

I have 3 Williams GG1s with magnetic adhesion enhancement(magne-traction). I wonder if there are others as well.

The method of implementation of the magnets is interesting. I would imagine that the set up works best on their M-track with the sheet metal base versus their C- or K-track systems.  

busch makes a small locomotive for H0 what is using magnets as well

http://www.busch-model.com/onl..._id=en&spur_h0=1

Cor

Do it like the real ones do.  Weigh down the scale loco to scale weight.  That would be about 8,800 lbs for a 1:48 scale AC 6000, the largest loco in use today I think.  That would give you plenty of traction!  Of course, you would need a crane to place your loco and it would almost certainly collapse your rails and table.  I will just stick with rubber tires or multiple engines, like the real ones do.

The Marklin system reminds of what AFX H.O. race cars did, they have magnets very close to the bottom of the car chassis to help keep the car on the track better as the rails were made from a steel compatible with magnetic adhesion.

Lee Fritz

That's incorrect. From another post about scale weight...

WEIGHT

Using the formula given by WindUpGuy:

To find the scale weight of an O scale model - assuming it was built using exactly the same materials in perfect scale dimensions - you would divide the prototype weight by the scale factor cubed, i.e. 48^3 or 110592. 

So, 432,000 lb (max weight of an AC6000) / 110592 = 3.9 lbs not 8000 lbs.

Jerry

All I know is the original poster was able to get umlauts ( a vs ä  ) in the title.

I've seen that posted before by others more than once over the years, and just like then as well as now, I still don't buy it (and physics doesn't either).  It's kind of like that scene in the James Bond movie Thunderball where a full-size sedan is put in a crusher then you are supposed to believe that since it's now smaller, it's also lighter and therefore it can be put in the back of Oddjob's Ford Ranchero without collapsing the suspension & blowing out the tires. 

If you were able to take a prototype engine that is x- number of pounds in the real world and magically shrink it down to 1/48 scale, it would still be the same weight as it was in the 12" = 1' world, although significantly more dense (obviously) and weigh ever-so-slightly less since there would also less air occupying empty spaces in the loco (air has weight too).  So a locomotive that's 432,000 pounds would still be 432,000 lbs, no matter if it was half it's original size or if it were as small as a sugar cube.

Truth is that performance (adhesion, rolling resistance, stability) is what is factored when designing scale model and toy trains, not by "scaling down" what the weight would be if it were 12 inches = 1 foot.

Actually, the James Bond movie with Odd Job was Goldfinger.

Stuart

You're wrong John, we're not crushing a full size and weight locomotive down to a block the size of an O scale model.  A 48 (48x48x48) square foot cube of steel (real world size) would not weigh the same as a 1 (1x1x1) square foot cube of steel(O gauge size)or vice versa.

Jerry

Volume is cubic, length, width, and height are linear.   

The ratio of 1:48 applies to any linear dimension.   But the volume is 1 to 48 cubed, ie it goes out in 3 axis.  

Think how many inches in a cubic foot.   There are 12 inches in a foot (linear).   There 12x12 or 144 square inches in a square foot (area).   There are 1728 cubic inches in a cubic foot.

When you go from O scale to full size, you expand the volume by going in all 3 dimensions.

I did some research on this back some years and realized our freight cars tend to be much heavier proportionally than the prototypes.  

Jerry,

Wrong, it would still be the same weight regardless of it's size, if you don't add or subtract material from that steel cube, which was the point of my post.

The so-called academic "suggestion" of dividing the weight of something by the scale of something cubed is just that:  Academic.  It does not alter the laws of physics where in the real word object X that weighs Y will still be the same even if you can compress the size of something (while retaining all it's relative dimensional elements).  It will increase in density but it's mass will remain the same.

So John what you're telling me is that if I cast a driver for SP 4449 and use the same metal batch to cast an O scale wheel for an O scale 4449 they would both weigh the same?Neat trick, please explain how they would weigh the same. The O scale wheel would weigh close to what a scale wheel should weigh.

Jerry

John,

I think you are violating both the laws of physics and geometry.   However, you are entitled to  have that opinion.

I have posted a reply to this discussion on how much should scale trains weigh here:

https://ogrforum.com/t...78#45918751657218278

So tell me how an object becomes lighter without subtracting molecules from it.

No Jerry, because what you're doing there is making a completely different casting for a completely different object that uses less materials.  I'm saying is that if you were able to compress the size of that 12" = 1' driver you originally casted down to O scale size, it's weight would be the same.  The only other attribute other than the size difference, would be its density, which would be higher.

Making a completely different casting using less materials that results in a smaller object is completely different.

British 00/H0 Hornby had their "Magnadhesion" system back when, which was also a magnet low down in the locomotive frame. It was matched up with the older Hornby Super 4 steel track. Later nickel-silver track made it redundant, and some modelers removed it and replaced it with lead.

T-gauge trains (1/450 scale) also use magnatraction, like Lionel, using magnetic wheels and steel rail.

Stuart

OK, I think this is called "scale modelling". So what should a 1:48 scale model of a 425,000 lb. AC 6000 weigh? If its weight is proportionate to its size?

Rob, if that scale model was built 100% accurate with the same real-world materials as the prototype, it would weigh 3.84 pounds (425,000 / 48^3).

This works very well for calculating the weight of live steam locomotives that are accurate models of the real thing and use the same materials (at the same real-world densities) as the prototype.  It's not as accurate with O scale models that use different materials and internal construction from the prototype.

Hope that helps...

You are correct.

Well the rest of us are talking about making a scale model of an object using scaled down real world materials used in the original and in the same proportions. You're talking about magic, shrinking something down by somehow compressing it, none of us are saying that because in the reality that we exist it can't be done. We just want to know how much an exact copy of an object scaled down would weigh not some esoteric thought exercise. If you could magically scale it down why can't you also magically fix the weight issue too?

James, this is the very point I was trying to make but there always has to be that one that can't follow along.

Jerry

It is physically impossible to have a 1:48 scale model of an AC6000 weigh 8000 pounds as a substance with the required density simply does not exist on Earth. Such a substance does not even exist in our solar system. If you made a 1:48 scale model AC6000 out of solid iron, it would weigh only 25-30 pounds. 

Therefore, the hypothesis that scale weight is equivalent to the prototype weight divided by the scale dimension is wrong. 

To make a 1:48 scale AC6000 with a weight on Earth of 8000 pounds you have to use the core material from a star that is roughly 100 times the mass of our sun (Sol). 

Such stars do exist, but they are very rare.  For the curious https://en.wikipedia.org/wiki/..._massive_known_stars

Obviously the prototype weight divided by the cube of the dimension is correct. 

Hi John,

We went through this once before, didn't we?

A model is not an original object compressed down to a smaller size.

It is a completely different object, in most cases made out of different materials. The materials used have the same normal properties as any other earthly materials. They are simply fashioned into an object that "looks" like the original.

Now, if you ask, what would a 200 ton engine weigh if it were "compressed" down to O scale - that would be a totally different question and have nothing to do with modeling. It would still weigh 200 tons as you say.

It does bring up the question of how the surgical team, plus submarine, were miniaturized in the movie "Fantastic Voyage" before being injected into that guy's body. If they were just compressed and weighed the same, it would have been more than a little tough on the patient's body.

Jim

Scale weight should be proportioned like scale size.    You don't scale it down in size and not weight.

That is just plain silly.    It would be like saying you scale down the length to O scale size, but not width.    You would end up with a 10 ft wide model that is about 18 inches long!   

The whole meaning of "scale" is reducing the entire item proportionally.    Even if you used exactly the same materials it would be much less.  

And taking John's argument another step into fantasy, you scale down the molecules of each component material so you have the same number of molecules but they would all be smaller.    And since molecules have weight, wouldn't they also then weigh less proportionally?   

Phew, I think I need an adult beverage!!