# A question for physicst



## rkapuaala (Jan 3, 2008)

I don't know if it is fatigue from all the moving I've been doing (by myself cause the wife is in Hawaii helping with my sons recovery) but this anomaly has been bothering me since I got involved in scale modeling. What I'm talking about is the weight of our models. 
When you think of it, even if a steam engine only weighed a ton (most steam engines weigh several tons, I know),,, if you scale that down it to 1:20.32 the model should still weigh around 100 lbs. Especially if you use brass and steel instead of plastic. 
Then it occurred me on that 40 minute trek back to Aptos to finish packing my belongings in boxes,,, there is gravity to consider and mass as well,,,, right? 
Wouldn't all those factors need to be scaled down too in order to calculate the accurate scale weight of a model? If so, what would a ton be equal to on our full scale,,,, scales? Am I getting derranged from packing too many boxes, is it only me that wonders about these kinds of questions???? 
Anyway, I know there are a few physicist on this list,,, and I'm hoping that the problem is intriguing enough to get your attention and possible answer to my question.


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## toddalin (Jan 4, 2008)

Posted By rkapuaala on 27 Jul 2010 12:05 PM 
I don't know if it is fatigue from all the moving I've been doing (by myself cause the wife is in Hawaii helping with my sons recovery) but this anomaly has been bothering me since I got involved in scale modeling. What I'm talking about is the weight of our models. 
When you think of it, even if a steam engine only weighed a ton (most steam engines weigh several tons, I know),,, if you scale that down it to 1:20.32 the model should still weigh around 100 lbs. Anyway, I know there are a few physicist on this list,,, and I'm hoping that the problem is intriguing enough to get your attention and possible answer to my question.



I'm no physicst (biologist/scientist actually), but I can see that your math is incorrect. You need to take the cube root of the weight to consider all three dimensions.

A ton scales to 0.2384 pound in 1:20.32 (i.e., 2,000# / 20.32 / 20.32 / 20.32 = 0.2384#)


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## Mr Ron (Sep 23, 2009)

Doesn't look like you are a mathematican either. The cube root of 2000 is 12.5992. I'm neither a mathematican or a physicist, but I stayed in a Holiday Inn Express.







I think if everything was scaled down, the motor would not be big enough to operate it.


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## Semper Vaporo (Jan 2, 2008)

Nope! Weight is a factor of the VOLUME, not the length or width or height individually. You divide by the CUBE of the scale (or divide by the scale 3 times in succession), not take the cube root of the weight.


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## markoles (Jan 2, 2008)

Are you using scale boiler plate? Steel has a different density than brass. Also, are you modeling all of the stuff that is inside your locomotive? In other words, are you doing a boiler with insulation and jacketing? Are you modeling the firebox, grates, all the boiler tubes, tube sheets, superheaters, plus the piping, etc..etc...ad nauseum. Plus all the water weight. To put it a different way, the Strasburg Railroad built several fake Thomas the tank engine replicas for the Day Out with Thomas events held at tourist railroads. That is essentially a model, scaled up, of a toy train, that gets shoved around by the railroad's diesel. Sure, it goes on the tracks and looks like a locomotive, but it is no where near the weight of a working steam locomotive. 

Also, I tend to think that you do not scale gravitational acceleration, since the gravity where your model will be is the same as where the 1:1 locomotive is.


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## Semper Vaporo (Jan 2, 2008)

A USRA Light Makado weighs about 292,000 lb, so a 1:32 scale model of it should weigh about 9 pounds. My Aster Mikes weighs about 12 pounds (sans tender). That is well within the range of possible values given the totally different materials between the prototype and the model.


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## Mr Ron (Sep 23, 2009)

You need to take the *cube root of the weight* to consider all three dimensions.

A ton scales to 0.2384 pound in 1:20.32 (i.e., 2,000# / 20.32 / 20.32 / 20.32 = 0.2384#)
*The cube root of 2000 = 12.599.*


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## msimpson (Jan 5, 2009)

20.3 x 20.3 x 20.3 = 8365.427. Therefore an accurate 1:20.3 model will have 1/8365.427 the volumne, mass, and wieght of the original. Thus a 254,000 lb. fully loaded Mikado and tender would scale down to about 30 lbs. A little 14 ton Shay would scale down to about three and one third pounds, somewhat more loaded. In our world things don't work exactly this way, because some things don't scale precisely, but not too far out of the ball park. 

When you scale something down, you reduce each of the three dimensions, and it is the product of the reduction which gives you the overall reduction in weight. Thus, a 7/8 model is 1/2571.353 of the original weight. Divide the prototype's weight by the cube of the scale. 

And actually, if you reduce Russell's 11 x 15 inch cylinders, for example, by a Roundhouse model's scale of 1:19, you get .579 x .789, very close to the .5625 (9/16) bore of the model and somewhat longer than the .625 stroke of the model. Scaled down originals will do work, but there are a lot of opportunities for fudging the results. 

Sorry, doddering off gain. (I got a degree in mathematics once, but I only use it -- and my fingers and toes -- to count my pocket change, now.) 

Regards, Mike 

Sorry, Mr. Ron, but the cube root of the weight is irrelevant. Otherwise, a model of an eight ton locomotive would weigh only twice what a model of a 1 ton locomotive would weigh. But even scaled down, the larger model will weigh 8 times what it's smaller cousin weighs. It's the weight divided by the cube (not the cube root) of the scale.

From another point of view, if you simply took the cube root of the weight, then the model would weigh the same, no matter whether the scale was 1:1 or 1:220. Yet, experience tells us that Z scale models are a bit lighter than the prototype.


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## Greg Elmassian (Jan 3, 2008)

I'm trying to stop laughing. (and I am a physicist) 

Numbers right, description wrong. 

take weight, divide by the cube of the scale = scale weight. 

weight = 1 ton = 2000 pounds.... divide by 29 cubed (so I'm standard gauge) = 0.082 scale pounds per prototype ton... 

I have a 70 ton capacity box car... that's 70 * 0.082 = 5.74 pounds ... and that's right. (fully loaded box car)... I tend to have my cars in the 3-4 pound range. 

Looking further on the web, a 60' box car had a light weight of 74,000 pounds and a max loaded weight of 286,000... that's 3 pounds and 11.7 pounds... 

All these numbers make sense... 

Greg


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## Semper Vaporo (Jan 2, 2008)

Posted By Greg Elmassian on 27 Jul 2010 01:49 PM 
I'm trying to stop laughing. (and I am a physicist) 

Numbers right, description wrong. 

take weight, divide by the cube of the scale = scale weight. 

weight = 1 ton = 2000 pounds.... divide by 29 cubed (so I'm standard gauge) = 0.082 scale pounds per prototype ton... 

I have a 70 ton capacity box car... that's 70 * 0.082 = 5.74 pounds ... and that's right. (fully loaded box car)... I tend to have my cars in the 3-4 pound range. 

Looking further on the web, a 60' box car had a light weight of 74,000 pounds and a max loaded weight of 286,000... that's 3 pounds and 11.7 pounds... 

All these numbers make sense... 

Greg 


Greg, Greg, Greg, Greg, Greg, Greg, Greg, Greg, Greg, .... tch tch tch! A Physicist! And to make such a monumental mistake!

I don't dissagree with most of your mathematics... but, honestly... how gross can you get...

"29 cubed (so I'm standard gauge)" ???

OH DEARY ME!

1:29 is NOT "Standard Gauge"... maybe close, but NOT standard gauge. 1:32 is the PROPER scale for 45mm track to represent "Standard Gauge".

I sure hope you don't design bridges, buildings or nuclear reactors!

It seems you have succumbed to the myth that the "G" in "G scale" represents the word "Garden"... it means "GOOFY" my dear man, GOOFY!



ACK! HALP! HALP! HALP!

GREG! PUT THAT FLAME THROWER DOWN!

SOMEBODY! HALP... A CROWD IS GATHERING ON MY FRONT LAWN WITH PITCHFORKS AND TORCHES! HALP!

PSZAPPPP!







(woo hoo, that wasn't a flame thrower, that were a laser blaster!)


*grins!*


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## Greg Elmassian (Jan 3, 2008)

I model "standard gauge" trains as opposed to "narrow gauge".... 

I did not say that the scale was appropriate of course. 

(I knew I was going to get flamed! I could feel the warmth building!) 

hahahaha! 

Anyway, it does look like we have the math right. Interesting how heavy a fully laden car would be. Wonder if we used scale grades and curves, how our locos would pull a "scale length and weight" train? 

OK, going to change out of the flame-retardant underwear! 

Greg


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## toddalin (Jan 4, 2008)

Posted By toddalin on 27 Jul 2010 12:12 PM 
Posted By rkapuaala on 27 Jul 2010 12:05 PM 
I don't know if it is fatigue from all the moving I've been doing (by myself cause the wife is in Hawaii helping with my sons recovery) but this anomaly has been bothering me since I got involved in scale modeling. What I'm talking about is the weight of our models. 
When you think of it, even if a steam engine only weighed a ton (most steam engines weigh several tons, I know),,, if you scale that down it to 1:20.32 the model should still weigh around 100 lbs. Anyway, I know there are a few physicist on this list,,, and I'm hoping that the problem is intriguing enough to get your attention and possible answer to my question.



I'm no physicst (biologist/scientist actually), but I can see that your math is incorrect. You need to take the cube root of the weight to consider all three dimensions.

A ton scales to 0.2384 pound in 1:20.32 (i.e., 2,000# / 20.32 / 20.32 / 20.32 = 0.2384#)


I understand the concept and perhaps the use of the "cube root of the weight" was a poor choice of words, but I was trying to keep it simple. The math is there, it's correct, and gives the proper value for the OP as well as the means for him to properly consider any scale.


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## RimfireJim (Mar 25, 2009)

This falls into a topic of engineering called model studies, part of dimensional analysis and similitude. The simple approach of using the cube root of the scale factor works in this case because of the simple relationship between weight and size. It doesn't work for more complex problems such as those encountered in fluid dynamics. That's why accurately scaled rudders don't work very well on scale boats, for example. When scale models are built for flow testing, dimensionless numbers (Reynolds number, Froude number) are used to get the proper relationship between the model and the full size object. These numbers relate all the parameters that affect the results. It's a very important principle in civil, marine, mechanical, and aeronautical engineering.


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## Mr Ron (Sep 23, 2009)

I really think scale can be applied to dimensions only. Applying scale to weight or speed is meaningless. It's like saying a dog's life is 1/7th of a human's life. It's just an approximation for perspective purposes only. If you scale down dimensions, the weight will not scale. The weight will be whatever the scaled down model weighs. The relationship will be like the dog example. Scale relates to dimensional properties only. That is what I believe and I'm sticking with it (unless you prove me wrong).


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## Semper Vaporo (Jan 2, 2008)

If you take a block of lead that is 1-ft by 1-ft by 1-ft, it is one cubic ft of lead. If you scale that block by 1/2 then the weight will change by the cube of the scale...cut it in half one way and it will weigh 1/2 of the cubic ft. Cut that block in half in the 2nd dimension and it will weight 1/2 of that half or 1/4 of the cubic ft. Cut that remaining block in half in the 3rd dimension and it will weight 1/2 of the 1/4 which is 1/8 of the original cubic ft. One half of one half of one half is one eighth which is the cube of 1:2 scale.

Weight does scale the same as the three dimension coordinates. A model may not scale in weight unless the exact same materials are used and ALL dimensions get scaled the same... component wall thicknesses (cab walls, pipes walls, boiler plate, sheathing, etc.) included! But since a 1/2 inch outside diameter pipe in the prototype would be reduced to 1/64 diameter in 1:32 scale and the prototype pipe wall thickness of .1 inch would reduce to 0.03125 thickness, it would behoove you to also reduce the steam pressure in a live steam model by the same amount or that "tin foil" pipe will burst!


Edit: Oops, typo... missed a zero.. that pipe thickness would be 0.003125 inches!


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## Greg Elmassian (Jan 3, 2008)

Using scale reduction of weight for models is a time-proven reality. Nothing to do with dogs. Dog's life is part of their unique biological makeup. 

Also, it's not the cube root of the scale, it's weight divided by the cube of the scale... scale to the 3rd power not scale to the 1/3 power... 

cube root of 29 is 3.07... not same as dividing the weight by 29 cubed. 

Regards, Greg


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## jgallaway81 (Jan 5, 2009)

First, just food for thought.... a fully loaded coal train from Shire Oaks, PA to (anywhere east) must go up and over the Allegheny Summit via Horseshoe Curve. These trains, totaling as many as 130 fully loaded coal hoppers, will easily reach as much as 19,000+ tons.

Now, having said that, I want to ask a different but related question. Does horsepower scale down? For that matter does tractive effort or adhesion scale down? Those 19,000ton trains will be lifted up and over the summit by (as many as) 2x 4,000hp road units (DASH-9's or SD-70M-2's) 2x 3,000hp helper-ahead units (SD-40E's & SD-40-2's) and 4x 3,000hp rear-end helpers (SD-40E's & SD-40-2's). This means it can take as much as 6,000hp + 8,000hp + 12,000hp, or 26,000hp to move that train over the summit. 


Tell me that wouldn't be cool to model? And keep in mind that is just ONE coal train!


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## rkapuaala (Jan 3, 2008)

Wow,,,, my head is swimming! I'm really going to have some things to ponder during the trip to Aptos this morning. Horse Power!!! What about time? Is my little Ruby right out of the box really traveling closer to scale speed? I mean if we scale down time, then a half mile a minute would be more like 130 feet in 2.95 seconds meaning that even at full throttled my Ruby is going too slow! 
If I lay an oval track in my backyard with a radius of 4 feet and two straight aways of 8 feet each then I've got approximately 41 feet of track meaning that in 2.95 seconds my little Ruby would have go around the track just over 3 times???? I don't think it would make it through the radius without derailing.


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## Torby (Jan 2, 2008)

Well, usually we run our trains way too fast, except me, I tend to run a little over slow.

Your Ruby is likely a narrow gauge engine, so let's figure it's 1/20. (We're not going to measure anything close enough for the 3 tenths to matter.) Your track, is 4 * 2 * pi + 16 ft. That's 41 ft, as you figured, or 822 scale ft. 

Now 10 miles / hour is 14 ft per second, or 56 seconds around your track.
15 miles/hour is 37 seconds around your track.
20 miles/hour is 28 seconds around your track.

Now Ruby is going to be a pretty slow loco. I expect 10 to 15 scale mph would be about the max.

I commonly run 40ft freight cars on my train, and I just keep in mind that one freight car/second is about 27 miles/hour. This bit of info suits any scale, so if I'm at a train show and somebody's string of 40ft freight cars are going by at 2/second, I know the train is going about 55 miles/hour.

And Greg is correct. The "scale weight" of something is the prototype weight / scale cubed, but who cares, as long as it's heavy enough not to fall off the track and the wheels roll smoothly?


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## toddalin (Jan 4, 2008)

Posted By rkapuaala on 28 Jul 2010 08:00 AM 
Wow,,,, my head is swimming! I'm really going to have some things to ponder during the trip to Aptos this morning. Horse Power!!! What about time? Is my little Ruby right out of the box really traveling closer to scale speed? I mean if we scale down time, then a half mile a minute would be more like 130 feet in 2.95 seconds meaning that even at full throttled my Ruby is going too slow! 
If I lay an oval track in my backyard with a radius of 4 feet and two straight aways of 8 feet each then I've got approximately 41 feet of track meaning that in 2.95 seconds my little Ruby would have go around the track just over 3 times???? I don't think it would make it through the radius without derailing. 


Time (perhaps the fourth dimension) is a constant in all scales and does not change. If a real train (1:1) covers a mile in a minute (60 mph), at 1:20.32 the distance is reduced to ~260 feet, and if the it model train is going 60 scale miles an hour (~260 feet per minute), it still takes a ~minute to cover the distance.


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## Larry Green (Jan 2, 2008)

Rick, perhaps some pineapple juice with rum will help.


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## Torby (Jan 2, 2008)

Time (perhaps the fourth dimension) is a constant in all scales and does not change.

Well, unless you're really heavy, really far away, or going really fast. They have to consider relativity when working with the GPS satellites as GPS relies on knowing the time exactly.


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## Andre Anderson (Jan 3, 2008)

Positively it is time for pineapple juice and rum, lots of rum


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## rkapuaala (Jan 3, 2008)

Is it ok to dring 3 or 4? Full scale portions of course..... I'm sick of moving,,,, I will never move again.


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## Larry Green (Jan 2, 2008)

How about a new 1:20 figure--you sitting on a moving box, pineapple and rum drink in hand? Sort of a commemorative item? 

Never gonna move again, eh? Maybe so, but you have kids, don't you? Helped move one daughter out three times before she stayed "gone". At least her sister had pity on me and moved into an apartment right from college. Good luck! 

Larry


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## rkapuaala (Jan 3, 2008)

Larry, did you have to remind me about the kids,,, not to mention the nieces and nephews!!!!!!!!!


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## Chucks_Trains (Jan 2, 2008)

Posted By jgallaway81 on 27 Jul 2010 10:42 PM 
First, just food for thought.... a fully loaded coal train from Shire Oaks, PA to (anywhere east) must go up and over the Allegheny Summit via Horseshoe Curve. These trains, totaling as many as 130 fully loaded coal hoppers, will easily reach as much as 19,000+ tons.

Now, having said that, I want to ask a different but related question. Does horsepower scale down? For that matter does tractive effort or adhesion scale down? Those 19,000ton trains will be lifted up and over the summit by (as many as) 2x 4,000hp road units (DASH-9's or SD-70M-2's) 2x 3,000hp helper-ahead units (SD-40E's & SD-40-2's) and 4x 3,000hp rear-end helpers (SD-40E's & SD-40-2's). This means it can take as much as 6,000hp + 8,000hp + 12,000hp, or 26,000hp to move that train over the summit. 


Tell me that wouldn't be cool to model? And keep in mind that is just ONE coal train! 

WOW!! Can't beleive that nobody tore into this one so I will. The drawbar pull needed to hold or move a train on a grade is 20 lbs per ton per grade percent. A 19,000 ton train not including the motive power will take somewhere near 380,000 lbs. of tractive force just to hold it on a 1% grade..from rolling downhill. Modern couplers are only rated at something like 385,000 lbs hence the pushers needed.

If that train is going slow like under 14 mph then the "modern" diesels don't really do much hence all the good old workhorse SD40's









All the horsepower that you figured isn't really a true indicator of the force required as that kind of heavy hard pulling is more in the tractive effort realm..unless that train is doing like over 20 mph.

All in all that train would definately be very cool to do in 1:29


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## jgallaway81 (Jan 5, 2009)

I will grant that coming up out of CP-C (Conemaugh), the train would be better served by mother/slug sets since a diesel electric engine can produce more power at low speeds than can be put to the rail head without slipping the drivers.

However, once underway, even a heavy 14k-16k ton train can reach 40-45 on the straight & level stretches between South Fork and Cassandra. Course, once we reach Cassandra, the speed drops very quickly as the train hits the steepest part of the West Slope. 


Now, coming down the East Slope, coal trains are restricted to a maximum speed of 15mph down hill. Once entirely on the east slope, it will take the combined effort off all locomotive units in dynamic braking combined with an 8-10 pound application on the train's air brakes to prevent a runaway. I'm not certain here if the locomotives can produce maximum braking effort at such slow speeds or not.

However, the DASH-9, SD-70M, GEVO and SD-70-M2 all win in dynamic braking ability over the old SD-40-2's.


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