I don't think so, but...
http://www.youtube.com/watch?v=rC8VzVmNPOI
Rotor
Jake: How often does the train go by? Elwood: So often you won't even notice ...
loathar wrote:No. Not unless you figure out a way to scale down the molecular hardness factor of copper.
Exactly what I was thinking, kinda tough to scale down the density of the copper.
Modeling B&O- Chessie Bob K. www.ssmrc.org
Depends on how much weight you put in your loco. Enough weight and that scale penny will flatten right out. Maybe you could research it and let us know.
Enjoy
Paul
bogp40 wrote: loathar wrote:No. Not unless you figure out a way to scale down the molecular hardness factor of copper.Exactly what I was thinking, kinda tough to scale down the density of the copper.
But by the brilliant U-Tube shown, it's not hard to scale down the density of the brain!
Teditor
Modeling the N&W freelanced at the height of their steam era in HO.
Daniel G.
If there are no dogs in heaven,then I want to go where they go.
Kurn wrote:No need to scale down the copper.Just weight the Big Boy 'til it weighs scale weight.Around 9500 lbs.
Frankly, I've wondered about this, too. Since weight is propotional to volume, the HO scale weight would be the weight of the prototype divided by 87 cubed (87 X 87 X 87), right? So the scale weight of UP's 844 would be it's actual weight (1,000,000 pounds) divided by 658,503 (87 cubed)... Could it be that 1.518 pounds is the scale weight of an HO scale 844? Correct me if I'm wrong in this thinking.
Anyway, I've also thought about a scale penny, a couple thousandths thick and I've imagined that the scale penny would be unaltered if run over by the 1.5 pound model loco. I've tried to test it, but the scale penny keeps falling off the track (just kidding).
I did try it in 1:1. The prototype 844 flattened a real penny to about 1/20 of its original thickness. It must be that molecular hardness thing that keeps it from happening in scale. I'm glad to know that.
Regarding having too much time to think, I run shays. Enough said?
Phil, I'm not a rocket scientist; they are my students.
The answer lies in simple Newtonian kinematics. The penny is deformed by the force applied by the weight of the locomotive. This is the mass of the locomotive times the gravitational acceleration. This doesn't scale because it's not a function of size but of mass.
For a given thickness of zinc with a copper jacket one could calculate the force required to deform it based on the tensile strength of the material. I'm a dynamic meteorologist and not a metallurgist, so I can't help with that. But once you know the required force, you go back to Newton's second law:
F = m * a (where in this case a = -g, or +9.8 m/s^2 or +32 ft/s^2).
So, assuming your metallurgist friend has given you F for an HO scale penny, m = F/a. Now you know the required mass for your HO locomotive.
I would wager that the mass you'd calculate is still so high, that in order to make the locomotive still be HO scale it would have to be made of a material so dense it would not exist on the Periodic Table.
Good luck!
Modeling the Rio Grande Southern First District circa 1938-1946 in HOn3.
Dave Vollmer wrote: The answer lies in simple Newtonian kinematics. The penny is deformed by the force applied by the weight of the locomotive. This is the mass of the locomotive times the gravitational acceleration. This doesn't scale because it's not a function of size but of mass.For a given thickness of zinc with a copper jacket one could calculate the force required to deform it based on the tensile strength of the material. I'm a dynamic meteorologist and not a metallurgist, so I can't help with that. But once you know the required force, you go back to Newton's second law:F = m * a (where in this case a = -g, or +9.8 m/s^2 or +32 ft/s^2).So, assuming your metallurgist friend has given you F for an HO scale penny, m = F/a. Now you know the required mass for your HO locomotive.I would wager that the mass you'd calculate is still so high, that in order to make the locmotive still be HO scale it would have to be made of a material so dense it would not exist on the Periodic Table.Good luck!
I would wager that the mass you'd calculate is still so high, that in order to make the locmotive still be HO scale it would have to be made of a material so dense it would not exist on the Periodic Table.
I must have missed something.......you said simple................
Don't forget the rolling effect of the wheel as it goes over the penny. Like the rollers in a steel mill, or just a plain rolling pin as it's making cookies... Probably has as much of an influence over the flattening as the weight. Otherwise, the tracks would be flattened, nevermind the ties... Ok, I know, the weight of the engine is distributed over many ties, but...
p.
-|----|- Peter D. Verheyen-|----|- verheyen@philobiblon.com -|----|- http://www.philobiblon.com/eisenbahn -|----|- http://papphausen.blogspot.com/-|----|- http://www.youtube.com/user/papphausen2
Humm... a scale penny, 1/87th of the size, Height, weight and thickness, would not be much copper. Some of the newer diesels can weight that 1.5 lbs. Humm... not much copper to deform...humm...
Tilden
P.S. I had some spare time.
loathar wrote:concretelackey-Simple version-An atom of copper is an atom of copper. It can't be scaled.(Yet...)
Yes, but wouldn't "our scaled" copper atom have to be 1/87th of the 1:1?
bogp40 wrote: loathar wrote:concretelackey-Simple version-An atom of copper is an atom of copper. It can't be scaled.(Yet...)Yes, but wouldn't "our scaled" copper atom have to be 1/87th of the 1:1?
See the part above again. An atom of copper CAN'T be scaled. It is what it is. It will have the same atomic bonds whether it's in a 1:1 penny or a 1:87 penny.(simple 9th grade chemistry)
concretelackey wrote: Dave Vollmer wrote: The answer lies in simple Newtonian kinematics. The penny is deformed by the force applied by the weight of the locomotive. This is the mass of the locomotive times the gravitational acceleration. This doesn't scale because it's not a function of size but of mass.For a given thickness of zinc with a copper jacket one could calculate the force required to deform it based on the tensile strength of the material. I'm a dynamic meteorologist and not a metallurgist, so I can't help with that. But once you know the required force, you go back to Newton's second law:F = m * a (where in this case a = -g, or +9.8 m/s^2 or +32 ft/s^2).So, assuming your metallurgist friend has given you F for an HO scale penny, m = F/a. Now you know the required mass for your HO locomotive.I would wager that the mass you'd calculate is still so high, that in order to make the locmotive still be HO scale it would have to be made of a material so dense it would not exist on the Periodic Table.Good luck!I must have missed something.......you said simple................
My PhD deals with computational fluid dynamics. Solid body physics is simple by comparison!
Dave
Just be glad you don't have to press "2" for English.
http://www.youtube.com/watch?v=zQ_ALEdDUB8
http://www.youtube.com/watch?v=6hqFS1GZL4s
http://s73.photobucket.com/user/steemtrayn/media/MovingcoalontheDCM.mp4.html?sort=3&o=27
Dave Vollmer wrote:I would wager that the mass you'd calculate is still so high, that in order to make the locomotive still be HO scale it would have to be made of a material so dense it would not exist on the Periodic Table.
Actually, such material does exist. It's known as the Bowser element.
Nelson
Ex-Southern 385 Being Hoisted
Dave Vollmer wrote:The answer lies in simple Newtonian kinematics. The penny is deformed by the force applied by the weight of the locomotive. This is the mass of the locomotive times the gravitational acceleration. This doesn't scale because it's not a function of size but of mass.For a given thickness of zinc with a copper jacket one could calculate the force required to deform it based on the tensile strength of the material. I'm a dynamic meteorologist and not a metallurgist, so I can't help with that. But once you know the required force, you go back to Newton's second law:F = m * a (where in this case a = -g, or +9.8 m/s^2 or +32 ft/s^2).So, assuming your metallurgist friend has given you F for an HO scale penny, m = F/a. Now you know the required mass for your HO locomotive.I would wager that the mass you'd calculate is still so high, that in order to make the locomotive still be HO scale it would have to be made of a material so dense it would not exist on the Periodic Table.Good luck!
WARNING! SCIENCE CONTENT!
Since gravity isn't scalable, the felt effect of it would be massively increased in HO. We measure gravity as an accelerative force, one normal G is 32 feet per second squared. In HO this scales out to 2784 feet per second squared.
If everybody is thinking alike, then nobody is really thinking.
http://photobucket.com/tandarailroad/
Short answer no. The HO penny would be 1/87th the thickness of a real one, and cover 1/(87 squared) the area of the 1:1 version. But your model loco of scale mass has only 1/(87 cubed) the mass so it misses on having the same effect by a factor of 87. The critical factor is force over area. If you want it to work, build a platform on your Big Boy and get your better half to balance on it. That might just do the trick. Otherwise try a plasticine penny.
Cheers, Ian the metallurgist.