# Single ball bearing wheel axle



## rtrains1 (Jun 13, 2009)

Help!

I'm at my friend Roy's train shop and they have ball bearing wheel axles, but only one wheel on the axle has the bearings.(?)

How does this work? Would you put the bearing wheel to the outside of most of your layout curves or to the inside?

If you have a two axle truck do you put one bearing wheel on onside and the second on the opposite side?

Best,
TJ
(using Roy's account to post this)


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## Totalwrecker (Feb 26, 2009)

Doesn't matter so much which side the bb goes on, this design is to address the diff in insde and outside diameter track. You'll still need free rolling bearings. 
Sets with dual address both diameters and bearings.


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## Greg Elmassian (Jan 3, 2008)

To re-iterate what TW said, you still have to have free rotation of the axle in the sideframes. Check that or you will not have accomplished anything. 

My experience with ball bearings is they help on the flat, but if you have grades or tight curves, they do not add much. 

Smart idea to put just one wheel bearing, keeps cost down and gives you the differential action. 

If you buy them, please report your findings here. 

Regards, Greg


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## tj-lee (Jan 2, 2008)

TW, Greg, 

Thanks for the quick responses. I'm a bit dense and still don't see why only on wheel on the axle has the ball bearings. TW mentioned cost savings which I understand, I just don't get why one free'er spinning wheel is good if the other wheel does not spin free'er as well. What am I missing here? 

And for two axle trucks, do I need to replace both axles with a new one wheel ball bearing axle with the ball bearing wheel on one side and a new one wheel ball bearing axle with the ball bearing wheel on the other side, or just replace one axle, or what? 

Sorry to be so thick on this. 

Best, 
TJ


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## Greg Elmassian (Jan 3, 2008)

The outer rail is "longer" than the inner rail, just like the path of the outer tire on a car on a curve. 

In a real train, the fillet (between the tread and the flange) provides a larger diameter, thus allowing the "differential" action, i.e. the surface speed of the wheel is faster, because it is a larger diameter. 

In our models, our curves are way too sharp for this to work. So, let the wheels turn independently, differential action. 

Regards, Greg


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## tj-lee (Jan 2, 2008)

Greg, 

Ah, the light comes on. Thanks for being patient with me. 

Best, 
TJ


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## W3NZL (Jan 2, 2008)

TJ, 
Some years ago I made up and tested differentialed wheels against standard wheels sets... I found that with the well designed
aftermarket wheel sets there was little, if any benefits to be had with differentialed sets... With the run-of-the-mill factory metal 
wheels, there was some slight improvement to be had with differentialed wheels, but the benefits were so small that I considered 
them not worth the effort or expense...
Paul R...


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## Torby (Jan 2, 2008)

I bought such wheelsets for my shorty coaches hoping they'd be easier for my little loco to pull on tight curves. I misfigured the cost and spent more than the cost of a bigger loco. Oops. They roll very smoothly, but they didn't make any noticable difference. Nice power pickups.


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## Pete Thornton (Jan 2, 2008)

In a real train, the fillet (between the tread and the flange) provides a larger diameter


Greg, 
You are technically correct, but the real engineering is in the taper of the tread - or was that what you meant? 

The wheel is very slightly tapered away from the flange, so that when you go round a curve and the wheel tries to ride up the outside rail, the taper provides a larger wheel diameter nearest the flange, and the inner wheel is running on a smaller diameter portion of the wheel. That makes the wheel tend to turn in the direction of the curve. Really smart, eh? Except that it only works 100% on a curve exactly the right radius!


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## Greg Elmassian (Jan 3, 2008)

I meant in the prototype, the "extreme" part of the differential action is performed by the fillet on curves. I think slight curves can be handled by the wheel taper alone. The taper on prototypes is much less than our models. 

I was trying to use the "extreme" case in the prototype, because that's clearly where we are in our layouts, even with 20' diameter curves I believe! 

On our models, nothing seems to be "enough" to give us that differential action from the tight turns. 

So, you are really more correct Pete, together the wheel taper and the fillet handle everything in the prototype I guess. 

Regards, Greg


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## tj-lee (Jan 2, 2008)

Paul R, Torby, 

Thanks for the feedback on the cost vs. benefit. I really appreciate the outstanding advice found here on MLS. 

Best, 
TJ


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## Semper Vaporo (Jan 2, 2008)

Posted By Greg Elmassian on 06/15/2009 1:14 PM
I meant in the prototype, the "extreme" part of the differential action is performed by the fillet on curves. I think slight curves can be handled by the wheel taper alone. The taper on prototypes is much less than our models. 

I was trying to use the "extreme" case in the prototype, because that's clearly where we are in our layouts, even with 20' diameter curves I believe! 

On our models, nothing seems to be "enough" to give us that differential action from the tight turns. 

So, you are really more correct Pete, together the wheel taper and the fillet handle everything in the prototype I guess. 

Regards, Greg


I, too, thought that the taper of the wheels helped the cars through curves, but I have always wanted to know what degree of curvature this differential action would accommodate. This discussion here has prompted me to go try to figure it out.

My brain hurts now.









I first found a wheel profile diagram (redrawn/cleaned up and attached below). I left off the dimensions to the centers of the curves and fillets, and their radii to make the diagram simpler.

I then made a BAD assumption that the rail contacted the wheel tread at the mid-point of the tread width and then I calculated the circumference of the wheel at several locations along the width of the wheel (given a 36-inch diameter wheel).

My BAD assumption was that the rail normally contacted the wheel at point "C", and that if the wheels/axle/car moved to one side or the other that the contact point could be at any of the other points I had selected. I then assumed that the wheel on the opposite side contact point would move to the complimentary corresponding spot. The circumference then would be the distance traveled in one revolution and the difference in the two wheel circumferences could then be used to calculate the degree of curvature that would be produced.

It was then that I realized the rail should ever actually move to point "A" because the wheel on the other end of the axle would then have its flange rolling on the top of the rail... a bad situation!

So I added the lower case lettered points "a" and "d". Point "a" is the place where the rail would contact if the opposite side was half way up the flange, and point "d" is the place where the rail would contact if the opposite side were at point "B".

THEN I discovered a diagram that shows my main assumption is WRONG, WRONG, WRONG.

The "normal" rail to wheel contact point is "somewhere" between points "C" and "D", depending on what the profile of the rail is... the upper surface of which is usually an arc with a 12- to 14-inch radius and the radius of each edge varies with the style of rail too.

Then I found a measurement that throws a monkey wrench into all of the above and means my lettered points are stupid and makes all my calculations totally USELESS. That measurement is the distance from the mid point of the flange depth (marked as the "Gauge Point" in the diagram) and the edge of the rail and it is only 13/32nds of an inch. This means the wheels can only move from the centered position 13/32nds of an inch in either direction... a total play of 13/16ths of an inch... just over 3/4 inch of side-to-side play!

If the flange fillet is not riding very far up on the rail the difference in circumference is less than 1/2 inch! i.e.: the wheel that is closest to the flange travels 1/2 inch farther than the wheel farthest from the flange.

I broke my brain at that point... what degrees of curvature would be formed by two wheels 4-ft 8.5-in apart if one is 113.2-inches in circumference and the other is 113.7-inches?


To further confuse matters, I quote Deb's Law... "Theory and the Real World are not related."

A "Real World" consideration that screws up the whole thing is that the gauge of the track is never "perfect" -- nor is the rail profile. The actual contact point is constantly changing as the wheels roll on the track. Also, please note the RED line in the diagram below. It shows the extreme case of the wheel profile due to wear that represets "time to re-profile/replace the wheel"... obviously if wheels wear anything like this in service then all this Theory of how the wheel profile helps a train traverse a curve is "hog wash" (my thoughtfully chosen words).


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## Greg Elmassian (Jan 3, 2008)

One thing is telling, the "worn" profile shows the areas of greatest wear, which supports the assumption that the "average" contact points are somewhere between C & D. 

One thing is clear, you get your "differential" action by the outer wheel being "larger" which means it rides up close to and possible on the fillet. 

The wear profile you show seems to confirm this very well. 

So, what conclusion did you come to personally? 

Regards, Greg


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## Les (Feb 11, 2008)

Semp,

Enjoyed your diagram, whether or no your assumptions were correct. I learned something important from this thread, well, two things:

1) I never, ever considered the 'differential effect' caused by curves.

2) Because, years 'n years ago as an avid-reader kid, I read someplace "... the cone shape of a wheel is to help center the car between the rails when they go around curves, which is the cause of cars 'rocking' as they travel down the track..." (I make *no* claim for/against the accuracy of this statement!)

But I certainly can see the validity of Greg's comment, and the usefulness (?) of a differential wheel/axle set. (For my little enterprise, it's not worth it.)

I got a chuckle, too: where I read 13/32nds, I blinked. I thought, "Self, 13/32nds is a whole bunch of 32nds considering my 45mm ga track. That's nigh onto 3/8." I sipped my coffee (only 2nd cup) and reread everything. Then I heard the 'thock'. I thought, "Huh, he's talking prototype." 

So, for me, this was a good learning topic. Esp the 14 deg radius. Not that I'll use it, I think lack of weight on wheel would tend to make models derail a lot.

Thanks for posting that diagram.

Les


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## Steve Stockham (Jan 2, 2008)

Fascinating discussion! I guess it comes down to "point of diminishing returns." Why use BB wheels? As I understand it, it's mainly to reduce friction (i.e. resistance) to allow your locomotives to pull more. When I first started out in large scale I was given a set with plastic wheels which I quickly decided needed to be upgraded to metal wheels! This produced immediate noticeable results and I firmly believe that the results justified the expenditure.

Ball bearing wheels are supposed to cut down on rolling resistance and this is their main justification but they are _expensive!_ How to justify the cost? I run Fn3 (1:20.3) and the cars come with solid metal wheel sets. Good lurication is a must as the weight of these cars is significant. I decided to use LGB BB wheel sets on them as I already had a bunch of sets that I was using on my old 1:22.5 rolling stock. These also have the power pick-up tabs so I don't have to use the flat wire rubbing the axle or the plunger rubbing against the backside of the wheel and that makes a _tremendous_ amount of difference!

Would I spend $20 or so to have BB wheels on a flatcar? No. With proper lubing the wheels will roll fine enough for me. The weight of the flatcar doesn't justify the cost in my opinion so I will have reached the "point of diminishing returns." It may be different for you but that's my take on the subject.


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## H-man (Jan 4, 2008)

All, 

The big benifit would come when you have a battery car or some other car with some heft to it to keep the car rolling freelyand not binding on curves and reduce rolling resistance. 
I don't think thi was addressed a benefit. 

P.S. Origionally the prototype used friction berings and when they were justified as the cars went through rebuilding they too were up-graded with berings on the axles. (with the older style journal boxes with the lids off). This to cut down on rolling resistance and failures of friction berings "Hot Box" failure and potential derailment. 

Howard


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## Semper Vaporo (Jan 2, 2008)

My own conclusion is that with brand new wheels with perfect profile, on freshly profiled rail, of perfectly gauged track, the coning of the wheels would help the train run straight and true on straight track and maybe, JUST maybe, help the car trucks get started going around a curve during the transition from straight to curve. But I don't think it helps all that much in any sort of curve that actually accomplishes getting the train to smoothly follow a change in compass direction.

With one side only traversing 1/2-inch more than the other side out of 113 inches I don't see that as much of a curve. Even when I was calculating circumferences with my bad assumption the greatest difference in distances traveled was only 2.25 inches out of 113 (points "a" and "E") and again, that just doesn't translate to much of a curve in my mind.

At present, I just cannot figure out how to determine the radius or degree of curvature that either of those values would translate to. I am sure it can be done, but I will have to go study A LOT! to figure out how to do it. I was kind'a hoping some math wizz would jump in here and GIVE me the formula that I could plug in the numbers to calculate it. If nobody does, I guess I'll have to spray some "Rust Buster" in my ears to loosen up the synapses and do it myself... (but not today, I still hurt from yesterday!)


I should also point out that the RED line in my diagram is just an example wear pattern that I got from my reference materials and I just kind of "free-handed" it in the drawing. Some of the other examples showed the flange itself worn down to half as wide as the original as well as some other almost unbelievable shapes. 

The various RR regulating bodies (FRA and others) over the years have had differing profiles as well as different limits of allowable wear before a wheel is declared unusable and would need to be replaced or be put on a wheel lathe to have a new tread profile cut (which obviously would make a 36-inch wheel into a 35-or 34-inch wheel) and there were limits of how much a wheel could be trimmed before it was declared totally unfit for reprofiling and have to be scrapped.


In MY "opinion", after a wheel has a few miles on it, and the rail has had a few trains run on it, and after the sun comes out to heat and kink the rail to screwup the accuracy of the gauge, all this theory about "coning of the wheels" has no validity anymore.

For "toy trains" I still think a coned wheel is better than one with a flat parallel tread, but only because it "looks" right to MY eye. 

For rolling ease and derailing abatement I think good track work is of paramount importance. A heavier car would be better than a light weight car. Well oiled Ball bearings are better than well oiled friction bearings. Allowing the wheels to rotate separately is the real question here, whether with ball bearings or not. Given that I don't think the coning has any useful effect I have to ASSUME allowing the wheels to roll separately is not a problem (like I USED to think it was prior to this little exercise!). I suppose that with really tight curves, allowing the wheels to rotate separately would reduce the wear on the wheels.

Hmmmm??? How often do people have to replace wheels due to tread wear? Any empirical evidence that separately rolling wheels last longer than solid axle sets? I have not run any of my cars enough to show any wear on the wheels... lots of grease and grime on them from running Live Steam exclusively, but no "wear" that I have noticed... not that I have done inspections! 

Guess I'll have to add "Wheel Knocker" to my list of occupations, along side of: 

Founder, Proprietor, President, CEO, CMO, CFO, Foreman of Motive Power, Engineer, Fireman, Head Brakeman, Rear Brakeman, Foreman of Right-of-way Construction and Track Maintenance, Hostler, Roundhouse crew, Blacksmith, Head of Public Relations, Yardmaster, Switch tender (former), Sole Stockholder, Disaster Inspector, Bridge Designer, Bridge Builder, Dispatcher and Wheel Knocker; 
CMBY RY


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## SE18 (Feb 21, 2008)

I've always been fascinated by the fact that wheels only have a 1" flange keeping them on the rails (along with the taper and rail profile). It seems so miniscule when looking at a large train. Common sense would tell you the thing would fly off the tracks going around curves. 

But common sense doesn't dictate physics.


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## Semper Vaporo (Jan 2, 2008)

Posted By SE18 on 06/16/2009 1:09 PM
I've always been fascinated by the fact that wheels only have a 1" flange keeping them on the rails (along with the taper and rail profile). It seems so miniscule when looking at a large train. Common sense would tell you the thing would fly off the tracks going around curves. 

But common sense doesn't dictate physics.


I grew up with the 1950's Lionel "O-27" gauge stuff... (with the knife edged "paper cutter" flanges on the wheels) so as a kid that became my "normal" for comparing other toy trains to.

I will never forget the first time I got a good look at the real thing (as an adult, long after the Lionel's were gone) I thought something was drastically wrong with the wheels with the short, stubby, rounded over flanges!


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## Schlosser (Jan 2, 2008)

Maybe i've missed it, but nobody has mentioned speed or velocity. If the "differential" aspect of the wheel's tapered tread is to become effective, the truck must be pushed, slammed might be a better word, against the outside rail. This will not happen if a train is pulled slowly around a curve. 

There was a Reform School just just south of the house I grew up in. It had a spur with a tight curve leading to the Alton RR and when coal was delivered, the screaming of the wheels on that tight curve was spectacular. You would think PUSHING the cars around the curve might also push the wheels to the outside rail, but not so. 

So I would say that the tapered tread's main purpose is for "centering" the truck between the rails; all else is gratuitis. 

Art


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## Les (Feb 11, 2008)

That was the idea expressed in that quote I posted, that the 'sloped' radii of the wheels kept the train centered: as it wandered one way, the point of contact with the differing radius (i.e. towards the flange) made the cicumference of the other wheel 'seem' smaller (it was, in effect) which gave the same effect as a small wheel on the same axle as a big one on a car, there was 'pull' on the small side, more drag, and the axle 'hunted' for equalization. This 'hunting' gave the whole car a rocking motion due to relieved forces in the springs.

That's clear as mud. But hey, I just got up from my nap.









Les


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## Les (Feb 11, 2008)

Semp wrote:

With one side only traversing 1/2-inch more than the other side out of 113 inches I don't see that as much of a curve. Even when I was calculating circumferences with my bad assumption the greatest difference in distances traveled was only 2.25 inches out of 113 (points "a" and "E") and again, that just doesn't translate to much of a curve in my mind.


I kinda sorta agree. There's drag, but it about has to be negligible. The thing is, on a fixed axle, how many revolutions less did the inside wheel turn than the outside one? That would mean it (outside one) was dragging the inside one to turn faster. Or the inside one was dragging back on the faster-turning outside one. If you calculated the revolutions each made, and subtracted, you'd have a number. What else to do with it from there, I haven't a clue. You'd have to know the coefficient of friction of the wheels.

And I doubt it makes a hill of beans.


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## Semper Vaporo (Jan 2, 2008)

Let me try to address a couple of the recent comments.

If the RR car is not being acted upon by outside forces, other than being propelled (either pushed or pulled) along the rail, it would want to roll in a straight line, given uniform diameter wheels. When it approaches a curve in the track the wheel on the outside of the curve will come in contact with the outside rail... essentially the rail turns in front of the wheel's straight path. The other wheel would fall off the rail that is curving away from the wheel's straight path if it weren't for the other wheel's flange hitting its rail and bumping the whole axle in the direction of the turn.

The taper of the wheels does tend to help in this action. The wheels are turning at the same rate because they are rigidly attached to the same axle... if one rotates, the other will rotate the same amount no matter what (we are talking real RR stuff here, not models). If the wheels were allowed to roll on their own (no rails to guide them) and they are the same diameter the whole unit would roll in a straight line. If one wheel is bigger than the other and they are allowed to roll freely, the whole unit will turn in the direction of the smaller wheel. If they are forced to remain moving in a straight line, then one wheel will be slipping, either "peeling out" or "skidding" if the other one is rolling in perfect contact with what it is rolling on.

With the taper of the wheels, the diameters at the contact point with the rail is slightly variable and in a complimentary relationship between the wheels. If one wheel moves to a larger diameter the other moves to a smaller diameter. Thus, as stated above, the whole unit turns toward the smaller diameter which due to the direction of the taper means the wheel at the larger diameter moves to a smaller diameter and the wheel at the smaller diameter moves toward a larger diameter. This can produce a bit of a "hunting action" as momentum will tend to cause the wheels to overshoot the quiescent point and cause the unit to turn back the other way.

Even if the wheels were to reach the perfect point where the two diameters were exactly the same, the non-uniformity of the gauge of the track would eventually cause one or the other wheel to be rolling at a different place in the taper and it would have to hunt for the quiescent point again.

When the rails start to curve to one side or the other (a turn in the track), the quiescent point is disturbed and the contact point of the wheel on the outside of the curve is moved closer to the flange which is a greater diameter and the other wheel's contact point is moved toward a smaller diameter. Thus the outside wheel tends to travel farther and the inside wheel travels less distance which means the unit will either turn or one of the wheels will lose traction and either "skid" or "peel out".

If the curve is sharp enough the outside wheel's flange will contact the rail and will squeal.

I did do some more research into the curve diameter that the 36" diameter wheel will traverse based on the differences in diameter in the 3/4 inch play of the tapered area. I am NOT real sure of my calculations... I never understood Algebra when they tried to teach it to me and the deeper they got into the subject the less I understood. But here I have two formulas with two unknowns. I seem to remember that one should be able to solve for the two unknowns but I can't remember how.

The formula is:

Length of an Arc = Angle * pi / 180 * Radius

The two unknowns are the Angle and the Radius.

Thus I have two formulas where I know the Length of the Arc (the length of the inside curve of the track and the length of the outside curve)

113 inches = A * Pi / 180 * R ---- Inside curve

113.5 inches = A * Pi / 180 * (R+56.5") ---- Outside curve

The 113 and 113.5 are the different circumferences and the 56.5 is the gauge (distance between the two wheels).

I gave up fiddling with the Algebra and programmed the two formulas into an Excel spreadsheet where I could vary the two unknowns (i.e.: make guesses!)

It didn't take too long to settle into an angle of 0.5070423 degrees and a radius of 12769 inches (1064 Feet and 1 inch).

Now to convert that to degrees of curvature in 100 feet. I think that comes out to 5.28452 degrees.

According to a page on the Trains Magazine web site:

http://www.trains.com/trn/default.aspx?c=a&id=211

"Curves of 1 or 2 degrees are the most common on mainline railroads; the sharpest curve a common four-axle diesel can take is about 20 degrees when coupled to other rolling stock, more than 40 degrees when by itself. Mountainous territory, however, generally dictates curves of 5 to 10 degrees, or even sharper. Branch lines and minor spurs may have an even greater number of sharper curves."

SO... I guess the taper of the wheels DOES take care of most common curves and the flange bumping into the rail takes care of the rest. 

I am pleasantly surprised!


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## Les (Feb 11, 2008)

Semp,

I believe we are saying the same thing. At any rate, I have several lengths of RR track from my B'smithing adventures. Every one of them has a rounded side along the top of the 'T', and a more-or-less square (with small bevel) on the other side. That tells us that yes, the flanges rode hard on the rail, and which side was on the inside of the track.

Incidentally, those make really nice small anvils for most small jobs. And gate stops. I had a shortie holding the kitchen door from blowing shut. Thought I was rather cool. Friend Wife came home, gave that iron one heavy look, and next thing I knew there was a brick with a knitted cover holding the door open. Go figure.

Les


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## RimfireJim (Mar 25, 2009)

Semper Vaporo:
You are pretty much "on track" with your results. A minor point: the radius of the curve is measured to the centerline of the track, not to the inside rail. For the large radius involved in this particular calculation, that changes the result only slightly. I get a result of 5.375 degrees of curvature.

Regarding your two equations/two unkowns: In this case, most easily solved by "substitution". If you multiply your second equation out, you get:

113.5"=A*Pi/180°*R+A*Pi/180*56.5" 

From the first equation, you know that A*Pi/180*R=113", so substitute that into the second equation and you get:

113.5"=113"+A*Pi/180°*56.5"

Subtract 113" from both sides, divide both sides by Pi/180°*56.5" and you get:

A=0.5"*180°/Pi/56.5

A=0.507°

We don't really care about A, because it is just a result of traveling some distance along the arc; what we really care about is R. Substituting 0.507° back into equation 1 and solving for R gives:

R=113"*180°/Pi/0.507°
R=12769"


(Note that the Pi/180° term could have been left out and the angle would have been in radians instead of degrees but we still would have found R=12769", only easier.)

The "R" we got is the inside radius. The mean radius (centerline) is:

Rm=R+56.5"/2
Rm=12797.25"=1066.438 feet


To convert that to degrees of curvature, the equation is:

DoC=2*arcsine(100ft/2/Rm)
DoC=5.375° 


where DoC is the angle subtended by a 100 foot chord.

The more important point, and relevant to the original poster's question, is whether or not our model train wheels have enough taper and lateral play to generate sufficient circumference differences for the radii we use on our models. A "sharp" 20° curve is 9ft radius in 1:32 scale, 14ft in 1:20.3 scale. Our model wheels have approximately the same taper as the prototype (I think. . .), but generally are over-scale on width, so more lateral play is possible. I'd waste some more time and figure it out, but I don't have the data.


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## Semper Vaporo (Jan 2, 2008)

Posted By RimfireJim on 06/19/2009 5:35 PM
Semper Vaporo:
You are pretty much "on track" with your results. A minor point: the radius of the curve is measured to the centerline of the track, not to the inside rail. For the large radius involved in this particular calculation, that changes the result only slightly. I get a result of 5.375 degrees of curvature.

Regarding your two equations/two unkowns: In this case, most easily solved by "substitution". If you multiply your second equation out, you get:

113.5"=A*Pi/180°*R+A*Pi/180*56.5" 

From the first equation, you know that A*Pi/180*R=113", so substitute that into the second equation and you get:

113.5"=113"+A*Pi/180°*56.5"

Subtract 113" from both sides, divide both sides by Pi/180°*56.5" and you get:

A=0.5"*180°/Pi/56.5

A=0.507°

We don't really care about A, because it is just a result of traveling some distance along the arc; what we really care about is R. Substituting 0.507° back into equation 1 and solving for R gives:

R=113"*180°/Pi/0.507°
R=12769"


(Note that the Pi/180° term could have been left out and the angle would have been in radians instead of degrees but we still would have found R=12769", only easier.)

The "R" we got is the inside radius. The mean radius (centerline) is:

Rm=R+56.5"/2
Rm=12797.25"=1066.438 feet


To convert that to degrees of curvature, the equation is:

DoC=2*arcsine(100ft/2/Rm)
DoC=5.375° 


where DoC is the angle subtended by a 100 foot chord.




I KNEW THERE'D BE SOMEONE AROUND HERE WITH THE SMARTS TO DO IT RIGHT!







Thank you! (Truely, thank you.







)

I knew my numbers were slightly off AFTER I got done with my silly way to find the answer... but I wasn't about to go try it again with the correct values since the original assumptions were just guesses anyway (e.g.: the circumfrences of the wheels at the actual point where they touch the rails because of the randomness of the point of contact, rail shape, wear, gauge, etc.) so I just let it go... 

The "practical answer being"

Just over 5 degrees of curvature, give or take a wee bit before the flanges start doing damage to themselves and the rail while doing their duty of keeping the train on the track in curves.

I figured the Pi/180 was to convert radians to degrees (or something like that) but it was in the formulas I found and I didn't want to risk removing the term and prove my idiocy. Leaving them in at least leaves some doubt about that.

But, now my po' lil' punkin' haid is powerful hurtin' aga'n.


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## RimfireJim (Mar 25, 2009)

Posted By Semper Vaporo on 06/19/2009 5:54 PM

The "practical answer being"

Just over 5 degrees of curvature, give or take a wee bit before the flanges start doing damage to themselves and the rail while doing their duty of keeping the train on the track in curves.



By all rights, the flanges on our models should be squealing like stuck pigs as they (the models, not the pigs) negotiate the ridiculously out-of-scale curves found on most layouts. But, if we had to listen to that all the time, we'd be crazier than we already are!


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## Totalwrecker (Feb 26, 2009)

Centrifical force? 

We don't have scale weight, we will never duplicate the force involved.


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## Semper Vaporo (Jan 2, 2008)

Old man Newton (Isaac, not Fig







) says that an object in motion will continue in a straight line unless some force acts upon it to alter its course. ("Centrifugal force" would be the name applied when in a curve... "Centrifugal force" being that which propels an object away from the center of rotation, [the curve]). 

The rail car will travel in a straight line once motion is commenced... it is THE RAILS that turn IN FRONT OF the car that makes the wheels run into (ride up on) the outside rail. If the wheels are tapered then the outside wheel will travel farther than the inside wheel and thus apply a force to turn the car ("Centripetal force" would be the name applied when in a curve... "Centripetal force" being that which propels an object toward the center of rotation). The flange does not contact the rail unless the curve is sharp enough to be beyond what the wheel taper can accomodate. 

Granted, if there is little weight to the car then there will be little wear on either the flange or the rail and as a consequence there will be very little noise (flange squeal).

Guys can remember the meaning of the two forces, "Centrifugal" and "Centripetal", by considering an object being swung around on the end of a string... the object is a "Gal" trying to get away, and the string attached will "Trip" you.


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## Schlosser (Jan 2, 2008)

A ball bearing on only one wheel; is a “differential effect” really needed, I wondered. And I also wondered how many more times the 'outside' wheel would turn compared to the 'inside' wheel on LGB R1 curves, the only kind on my sometimes indoor empire. 

So the circumference of the circle of track divided by the circumference of the wheel provides the number of turns the wheel makes. Pi (3.14159) times the diameter is the circumference. But the Pi in the numerator cancels the Pi in the denominator, so we're left with the C of the track divided by the C of the wheel. 

Or: Ct/Cw = nbr of turns. 

Now the R1 diameter is 1200mm, and an LGB passenger car has a 30mm (approx) diameter wheel. 

(1200+22.5)mm/30mm equals 40.75 turns. 

(1200-22.5)mm/30mm equals 39.25 turns. 

Hmm, is 1 ½ turns something to worry about? If the cars are hugging the outside rail, the difference is even less. 

And another item could be of interest: In a loop of track, no crossing like a figure 8, the outside (longer) rail is always longer by the gauge multiplied by 6.28 (2Pi) or 283mm - just a bit over 11 1/8 inches; never longer, never shorter. 

This is so even if the radius or diameter is in tens of feet; in a figure 8 configuration, both rails are the same length. 

In addition, the car wheels are not always binding on the outside rail on a curve. When being pulled up a hill, the force in on the inside rail, and the taper of the tread is of no value here. When going up a hill, straight-lining (string-lining) is a big problem, especially if empty flat cars are in the middle of the train. Tehechapi had a sharp curve where empties in the middle of a train were a problem. 

Art


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## Schlosser (Jan 2, 2008)

Gee whiz, why can't I find mistakes before I post them. The prior post should read "so we're left with the D of the track divided by the D of the wheel. 

And the equation should read "Or: Dt/Dw = nbr of turns. 

Seems to be a disconnect between the brain and the fingers.


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## Steve Stockham (Jan 2, 2008)

Okaaay........(I'm going to do like I did in Algebra II in High School and sneak out the back door while no one's looking!) Seriously, what's the bottom line? Is it worth it or isn't it? And please, can you give it to us in plain non-Vulcan english?


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## Schlosser (Jan 2, 2008)

Steve, I thought the answer was clear. Les wondered about how many turns the outside wheel needed to turn compared to the inside; one and a half turns, approx., was the answer. And if the cars hug the outside rail in a turn, it's even less. Small potatoesl; to me, that's like expressing the national debt to three decimal places - who gives a hoot. 

And skidding some 11 inches in a 1000 foot or so right-of-way is insignificant - to me. I've got better things to spend my time and money rather than buying and installing wheels!! 

But I did think it's strange that radius has no effect on the skidding or turns ratio; but that and 2 dollars won't even buy a cup of coffee at Starbucks! 

For me, it's not worth it. It might be worth it if you have an unlimited supply of money with very long trains that are run very fast. 
Art


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## RimfireJim (Mar 25, 2009)

Posted By Schlosser on 06/24/2009 3:00 PM

And another item could be of interest: In a loop of track, no crossing like a figure 8, the outside (longer) rail is always longer by the gauge multiplied by 6.28 (2Pi) or 283mm - just a bit over 11 1/8 inches; never longer, never shorter. 

This is so even if the radius or diameter is in tens of feet; in a figure 8 configuration, both rails are the same length. 

Art



This is like the "trick" question: Two critters travel around the earth at the equator. One is always on the surface, the other flies one foot above the surface. Assume the surface is smooth. How much farther does the flying one travel after a complete circuit? 
Answer: 2 X 1ft X Pi (from C=Pi X (Do-Di)), or approx. 6.28ft


The "trick" part comes from people thinking that, since the diameter of the earth is really big, the flying one surely must have traveled a lot more than 6ft more than the other one.


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## Greg Elmassian (Jan 3, 2008)

Art, slow speeds will lessen the effect against the rails, but it will not take much to get over to the flange... 

Yes, the taper in the tread is mostly centering, the differential action needs the larger diameter near the flange... 

But I think all of this was said already? 

Regards, Greg


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