Tony Thompson & Scale Weights


Paul Hillman
 

Tony,

I understand you have a Phd in metallurgy? Maybe you could answer, or
comment on a "dumb ( ? )" question I've often thought about, but never
tried or researched?

Say, in HO scale, 87.1:1, we physically scale-down the prototype
dimensions by that ratio. What if the car were built of the exact same
materials as the prototype, would the weight scale-down by 87.1 also?

A 60,000 lb. car would then weigh 689 lbs. in HO?

Paul Hillman


Tim O'Connor
 

A 60,000 lb. car would then weigh 689 lbs. in HO?
Paul Hillman
Paul, objects have 3 dimensions, so 87.1 x 87.1 x 87.1
equals (approx) 660776, so a 200,000 lb loaded prototype
car in should weigh 4.84 ounces in HO scale. The NMRA
recommendations are in that ballpark.

Tim O.


Anthony Thompson <thompson@...>
 

Paul Hillman wrote:
Say, in HO scale, 87.1:1, we physically scale-down the prototype
dimensions by that ratio. What if the car were built of the exact same
materials as the prototype, would the weight scale-down by 87.1 also?
No, it would scale by the volume, which goes as the cube root of the scale. That would make it roughly 1.5 to 2 ounces.

Tony Thompson Editor, Signature Press, Berkeley, CA
2906 Forest Ave., Berkeley, CA 94705 www.signaturepress.com
(510) 540-6538; fax, (510) 540-1937; e-mail, thompson@signaturepress.com
Publishers of books on railroad history


Paul Hillman
 

Tony & Tim,

OK, thanks. I see where I erred, in scaling down only 1 dimension of a 3 dimensional object.

Paul Hillman

----- Original Message -----
From: Anthony Thompson<mailto:thompson@signaturepress.com>
To: STMFC@yahoogroups.com<mailto:STMFC@yahoogroups.com>
Sent: Thursday, August 18, 2005 10:57 AM
Subject: Re: [STMFC] Tony Thompson & Scale Weights


Paul Hillman wrote:
> Say, in HO scale, 87.1:1, we physically scale-down the prototype
> dimensions by that ratio. What if the car were built of the exact same
> materials as the prototype, would the weight scale-down by 87.1 also?

No, it would scale by the volume, which goes as the cube root
of the scale. That would make it roughly 1.5 to 2 ounces.

Tony Thompson Editor, Signature Press, Berkeley, CA
2906 Forest Ave., Berkeley, CA 94705 www.signaturepress.com<http://www.signaturepress.com/>
(510) 540-6538; fax, (510) 540-1937; e-mail, thompson@signaturepress.com<mailto:thompson@signaturepress.com>
Publishers of books on railroad history




Yahoo! Groups Links


Paul Hillman
 

Tony,

HUM!!!??? That just brought up another interesting thought to me. Wouldn't it be more true to scale-weight then, to know the prototype-car's empty weight, divide that by the cube of 87.1, (or some other scale ratio) and weight the car accordingly, rather than per the NMRA tables, eventhough they're a pretty good, "ball-park" figure?

Paul Hillman

----- Original Message -----
From: Anthony Thompson<mailto:thompson@signaturepress.com>
To: STMFC@yahoogroups.com<mailto:STMFC@yahoogroups.com>
Sent: Thursday, August 18, 2005 10:57 AM
Subject: Re: [STMFC] Tony Thompson & Scale Weights


Paul Hillman wrote:
> Say, in HO scale, 87.1:1, we physically scale-down the prototype
> dimensions by that ratio. What if the car were built of the exact same
> materials as the prototype, would the weight scale-down by 87.1 also?

No, it would scale by the volume, which goes as the cube root
of the scale. That would make it roughly 1.5 to 2 ounces.

Tony Thompson Editor, Signature Press, Berkeley, CA
2906 Forest Ave., Berkeley, CA 94705 www.signaturepress.com<http://www.signaturepress.com/>
(510) 540-6538; fax, (510) 540-1937; e-mail, thompson@signaturepress.com<mailto:thompson@signaturepress.com>
Publishers of books on railroad history




Yahoo! Groups Links


Marcelo Lordeiro <mrcustom@...>
 

Don't forget that you will have to think on the locomotive drawbar pulling power also.
Precision scale and Intermountain makes wheel sets with ball bearings and don't forget to replace the coil springs on the steam locomotives bearings .
The correct springs allows you to raise one driving wheel without lifting the locomotive.
Doing that you garantee that all drivers have the same adesion tp the track and you will pull twice as much cars.
Marcelo Lordeiro

----- Original Message -----
From: Paul Hillman
To: STMFC@yahoogroups.com
Sent: Thursday, August 18, 2005 1:47 PM
Subject: Re: [STMFC] Tony Thompson & Scale Weights


Tony,

HUM!!!??? That just brought up another interesting thought to me. Wouldn't it be more true to scale-weight then, to know the prototype-car's empty weight, divide that by the cube of 87.1, (or some other scale ratio) and weight the car accordingly, rather than per the NMRA tables, eventhough they're a pretty good, "ball-park" figure?

Paul Hillman
----- Original Message -----
From: Anthony Thompson<mailto:thompson@signaturepress.com>
To: STMFC@yahoogroups.com<mailto:STMFC@yahoogroups.com>
Sent: Thursday, August 18, 2005 10:57 AM
Subject: Re: [STMFC] Tony Thompson & Scale Weights


Paul Hillman wrote:
> Say, in HO scale, 87.1:1, we physically scale-down the prototype
> dimensions by that ratio. What if the car were built of the exact same
> materials as the prototype, would the weight scale-down by 87.1 also?

No, it would scale by the volume, which goes as the cube root
of the scale. That would make it roughly 1.5 to 2 ounces.

Tony Thompson Editor, Signature Press, Berkeley, CA
2906 Forest Ave., Berkeley, CA 94705 www.signaturepress.com<http://www.signaturepress.com/>
(510) 540-6538; fax, (510) 540-1937; e-mail, thompson@signaturepress.com<mailto:thompson@signaturepress.com>
Publishers of books on railroad history




Yahoo! Groups Links













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Bruce Smith <smithbf@...>
 

On Thu, August 18, 2005 11:47 am, Paul Hillman wrote:
Tony,

HUM!!!??? That just brought up another interesting thought to me. Wouldn't
it be more true to scale-weight then, to know the prototype-car's empty
weight, divide that by the cube of 87.1, (or some other scale ratio) and
weight the car accordingly, rather than per the NMRA tables, eventhough
they're a pretty good, "ball-park" figure?
Only if you want to model empty cars! Then, as pointed out in another
post, mixing widely differing car weights causes problems on the real RR,
and with our exagerated curves and grades, these are multiplied. There
are modelers who compromise somewhere less than NMRA RP weights, often
50-75% of these.

Regards
Bruce
Bruce Smith
Auburn, AL


Mike Brock <brockm@...>
 

Marcelo Lordeiro writes:

The correct springs allows you to raise one driving wheel without lifting the locomotive.
Doing that you garantee that all drivers have the same adesion tp the track and you will pull twice as much cars.
I kinda hate to do this but I think I'll respond here on the STMFC. Actually, while the subject has a closer association to steam era frt cars than several recent posts, this subject should be discussed on the Steam Loco group...

steamloco@yahoogroups.com

Anyhow, many people missunderstand the principle of adhesion so I think I'll comment on it. A locomotive driver...even works for...yuk...diesels...has two primary forces acting on it to move the locomotive. The first is the force from the rod attached and the second is the "normal" force [ weight ] acting downward. The Factor of Adhesion [ FA ] is defined as weight on drivers/tractive force. Hence, a UP Big Boy had 540,000 lbs of weight acting on the 8 drivers and about 135,000 lbs of tractive effort [ at start ]. The ratio is 540,000/135,000 = 4.0. This has long been established as the weight/tractive force relationship required to given traction without slipping when steel wheel encounters steel rail...dry & clean. Guess what? It is...surprise...based on the coefficient of static friction for steel on steel [ the actual figure is given as a range of 0.15 to 0.3 equating to FA's of 3.3 to 6.6 but 4.0 has been determined in practice to be the required number ]. To see the effects of these forces on a single driver, simply divide the whole mess by 16. Thus, 540,000/16 and 135,000/16 gives 33.75/8.4375 = 4.0. Hmmm. Lets convert our 4-8-8-4 to a 4-6-6-4, leaving the weight and tractive force the same. OK...we now have 540,000/135,000 still equal to 4. Hmmm. Now lets make our engine a 4-2-2-4. We still have 540,000/135,000= 4.0. IOW, adhesion has nothing to do with the number of drivers making contact with the rail. We DO have a different problem of course. Axle loading went from 67750 lbs to 270,000 lbs and the wheels broke under the weight in addition to snapping the rails. The point of all this is that a driver could lift off the rail...as my Big Boy does on a superelevated curve...but adhesion is not effected at all. The weight simply is supported by those wheels still on the rails and force remains the same. Now...on a real locomotive, springing is VERY important for a number of reasons. First is axle loading. Second is dynamic balance. All kinds of other problems occur if axle loading exceeds that designed and an out of balance engine could wreck havoc with itself.

Mike Brock


Anthony Thompson <thompson@...>
 

Paul Hillman wrote:
HUM!!!??? That just brought up another interesting thought to me. Wouldn't it be more true to scale-weight then, to know the prototype-car's empty weight, divide that by the cube of 87.1, (or some other scale ratio) and weight the car accordingly, rather than per the NMRA tables, eventhough they're a pretty good, "ball-park" figure?
The NMRA weights, as are well documented, were chosen for good operating behavior, and had nothing explicitly to do with scale weights.

Tony Thompson Editor, Signature Press, Berkeley, CA
2906 Forest Ave., Berkeley, CA 94705 www.signaturepress.com
(510) 540-6538; fax, (510) 540-1937; e-mail, thompson@signaturepress.com
Publishers of books on railroad history