Abstruse CSB Theory

Alan Turner
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Re: Loco Suspension, fitting CSBs

Postby Alan Turner » Sat Feb 12, 2011 3:10 pm

Russ Elliott wrote:Just started reading your pdf, Alan:

The basic underlying assumption is that the wheel loads can be calculated before the calculation of
the deflections and that the wheel loads remain unaffected by the deflection.

No. The assumption is (like any spring) that wheel load will change according to the deflection.


But as the wheel load is an input into the spreadsheet how do you calculate it?

Alan

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grovenor-2685
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Re: Loco Suspension, fitting CSBs

Postby grovenor-2685 » Sat Feb 12, 2011 3:52 pm

The naive assumption would be that each wheel carries 1/3 of
W. That cannot be correct. A more so realistic assumption would be that the outer wheels carry 1⁄4W
and the middle wheel carries 1/2W

Alan,
I have a problem with this, either of these results could be produced by appropriate positioning of the fulcrums, and we definitely want a result closer to the former than the latter, and preferably with the middle wheels carrying a bit less than 1/3. But the wheel load distribution is not something to be assumed but rather something that the calculations must account for.
The current programme although upside down, does allow the user to define the axleload distribution they want. What is missing is a means of relating these axleloads to the CofG position.
Changing from a definition of axleloads to one of loads carried by each fulcrum just makes this more difficult as there are more fulcrums to share the load and their positions are not available at the input to the problem but are actually the desired output.
The basic underlying assumption is that the wheel loads can be calculated before the calculation of
the deflections and that the wheel loads remain unaffected by the deflection.

Not necessarily calculated, but certainly specified, as the purpose of the calculation is to achieve the specified wheelload. That load will be the load at rest on level track, none of this addresses the dynamic situation.
The basic underlying assumption is that the wheel loads can be calculated before the calculation of
the deflections and that the wheel loads remain unaffected by the deflection.

How will you calculate the loads on the fulcrums in your version, seems to me to be the same problem?
Regards
Keith

Alan Turner
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Re: Loco Suspension, fitting CSBs

Postby Alan Turner » Sat Feb 12, 2011 4:15 pm

Keith,

the underlying complexity with what we are trying to do is the fact that the load applied by each of the load points (fulcrum) is not independant of each other. This is because they are attached to a ridged beam (the chassis). In fact what is realy happening is that a fixed displacement is applied to the CSB by the fulcrum points. Now this is a very complex problem to solve (although any structural analysis programme will do it - if you have the odd £1k to spare) and it cannot be solved by naive assumptions about load distributions. My approach suffers from the same problem. the difference with mine is that PROVIDED you ensure equal displacements of the fulcrum points then I am suggesting that the relation to load distribution and displacement are a reasonable approximation of the actual displacement and load.

You can do it your way as well but you cannot dictate the wheel loads and again you have to ensure equal displacement (or lying on a straight line) of the fulcrum points.

Alan

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grovenor-2685
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Re: Loco Suspension, fitting CSBs

Postby grovenor-2685 » Sat Feb 12, 2011 7:51 pm

You can do it your way as well but you cannot dictate the wheel loads and again you have to ensure equal displacement (or lying on a straight line) of the fulcrum points.

Alan,
I don't have a way ;) I don't own the spreadsheet and had nothing to do with authoring it. But it does appear to give results that work when applied.

All I was trying to do with your paper was to try and understand it.

Surely the purpose of the spreadsheet is to assist with the location of the fulcrums, because of the complexity it does not try to calculate the positions for you but allows you by trial end error to see the effect of various locations. As I understand it the deflection is calculated from the wheelload you input, so you need to understand that the actual deflections will be equalised when the loco is on the track and consequently the wheelloads will not be as input. Super accuracy is not important, what is important is to be able to see when the wheelload on the centre axle is a little less than that on the ends so that the running will be stable. The spreadsheet illustrates this by the relative deflections but in the inverse sense so you aim for a larger deflection on the centre axle.
Perhaps it would be possible to turn this calculation around to use a fixed deflection and calculate the wheelload instead. However this alternative approach would need input of the position of the CofG would it not? Which is allowed for in the existing spreadsheet by the wheelload input.
Regards
Keith

Alan Turner
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Re: Loco Suspension, fitting CSBs

Postby Alan Turner » Sun Feb 13, 2011 9:36 am

Keith,

use of the collective you - bad habit of mine.

The basic problem comes down to this:

loads.png
loads.png (24.9 KiB) Viewed 8335 times


They are two differnt problems.

Alan

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grovenor-2685
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Re: Loco Suspension, fitting CSBs

Postby grovenor-2685 » Sun Feb 13, 2011 11:01 am

They are two differnt problems.

Different, yes but related, are you saying that the answers given using the spreadsheet derived from the first problem will be significantly different from those that might be derived using the second? If so, by how much and does it matter?
Regards
Keith

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Will L
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Re: Loco Suspension, fitting CSBs

Postby Will L » Sun Feb 13, 2011 1:10 pm

Alan Turner wrote:I offered to set out some thoughts on CSBs. I do so here in the attached pdf file.

Alan


Sorry I haven't been ignoring you, my PC died and I've only just recovered it.

I'll have a look at you PDF tonight Alan but I have an area group meeting this afternoon.

A though for Keith/Rob. Given that this sort of technical discussion is enjoyed by some of us, but other find it very off putting, I'm wondering if we could split the thread down into the blow by blow posts, plus all the directly related posts, and a second "CSB Clinic" thread which could carry all the technical discussion and be clearly welcoming for people to ask their own questions

Alan Turner
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Re: Loco Suspension, fitting CSBs

Postby Alan Turner » Tue Feb 15, 2011 8:38 am

grovenor-2685 wrote:
If so, by how much and does it matter?


Does it matter? Within our context probably not because I think that any reasonably intuitive modeller could set up a CSB without recourse to the spreadsheet.

However if we are to mathematically model the CSB system then I think we should attempt to simulate it as best we can.

The problem I have with the present spreadsheet approach is its underlying assumption that the wheel loads are independent of each other and that they are an input. They are not independent of each other as they are connected by a ridged beam (the chassis) and they are an outcome not an input.

Having given this problem some thought over the week-end I believe a more appropriate model is this:

Beam dia.PNG
Beam dia.PNG (22.03 KiB) Viewed 8242 times


Here we represent the CSBs as springs acting on a ridged beam.

In order for the chassis to sit level then the centre of gravity of the weight must be coincident with the centre of action of the springs.

That is a relatively easy calculation.

The spring constants of the CSBs can be calculated by use of the standard equn's for beams as so:

Beam dia 2.PNG
Beam dia 2.PNG (4.32 KiB) Viewed 8242 times


and

Beam dia 3.PNG
Beam dia 3.PNG (4.53 KiB) Viewed 8242 times


The spring constants of both are easily calculated. I consider the encastra condition justified as the slope of the CSB will be horizontal at the fulcrum (due to the desire to have near equal defections at each wheel).

Hence once the springs are calculated it is a relatively easy matter to calculate the necessary spring rates to ensure coincident of the centre of action of the spring with the centre of action of the weight.

I will prepare a paper with the necessary equations and publish it later. I might even have a chance to write a spreadsheet that does the calculations.

Alan

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Russ Elliott
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Re: Abstruse CSB Theory

Postby Russ Elliott » Tue Feb 15, 2011 4:43 pm

Alan Turner wrote:The problem I have with the present spreadsheet approach is its underlying assumption that the wheel loads are independent of each other and that they are an input. They are not independent of each other as they are connected by a ridged beam (the chassis) and they are an outcome not an input.

The wheelbase and wheel load inputs to the current spreadsheet are merely the means of establishing the longitudinal positions of the frame fulcrum points. They are not independent of each other nor are they assumed to be 'independent' of each other. The actual value(s) of the wheel loads are not particularly important, e.g. the spreadsheet will output the same frame fulcrum positions whether the wheel loads are input as 35g each or 40g each or 20g each. Wheel load inputs can be varied if desired, e.g. 35g (front wheel), 32g (middle wheel), 35g (rear wheel). Or alternatively, the frame fulcrum positions, after inputting an initial set of equal wheel load values, can be changed to effect a 32g middle axle, for example - the methodology is essentially the same.

What inputs would you envisage?

I consider the encastra condition justified as the slope of the CSB will be horizontal at the fulcrum (due to the desire to have near equal defections at each wheel).


No, no, no! Such a beam would be an entirely different type of CSB, where the end moments are fixed. We are not dealing with such a beast: it would be (beam diameter for diameter) about twice as long as what we have got currently! All one can assume, in the context of the type of CSB we are dealing with (i.e. M is zero at the outermost fulcrum points), is that the deflection of the beam will be zero at the frame fulcrum points. The slopes of the beam (the first differential of the curve) might not be exactly horizontal at the intermediate frame fulcrum positions, and they certainly won't be anywhere near horizontal at the outermost frame fulcrum positions.

It is true that the CSB can be replicated as a rigid thing ('the chassis') sitting on top of three springs (for the 3-axle case). The determination of the position of the CofG to implement chassis horizontality is easy, as you say. What I think is impossible to determine is the new set of spring forces if that CofG is intentionally moved, because the overall situation, albeit still in equilibrium, then becomes statically indeterminate, i.e. there isn't a unique solution to the resulting set of moment equations. An iterative (trig-based) reversal method could derive this, but I'm not sure of the mathematical methodology needed to address the general (non-trig) case. I'm currently consulting mathematicians on this matter, btw, but at the moment they seem to be a bit stumped!

It begs the question though as to what we are trying to achieve here. Are we simply saying that someone who does not attempt to balance a loco (within reasonable limits of course), or intentionally shifts the balance of a loco, will not achieve what was intended. Isn't that self-evident?

Does it matter? Within our context probably not because I think that any reasonably intuitive modeller could set up a CSB without recourse to the spreadsheet.

Been there 10 years ago, got the t-shirt, and had a bunch of rows about the disastrous 'guessing-game' approach. I felt strongly about properly reflecting the physics, Roger (likewise) did the spreadsheet gawd bless 'im, and the penny eventually dropped. But hey, if someone insists on having a CSB but also insists on not using a spreadsheet or the plots readily available, or insists on guessing everything because they don't know how it works or they aren't prepared to know how it works, that's their problem. You can hardly expect me to be sympathetic to that approach, can you?

Please note however I'm not saying there aren't different approaches (compared with the method used in the spreadsheet) to the solving of CSBs.

Alan Turner
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Re: Abstruse CSB Theory

Postby Alan Turner » Wed Feb 16, 2011 10:56 am

Russ Elliott wrote:No, no, no! Such a beam would be an entirely different type of CSB, where the end moments are fixed. We are not dealing with such a beast: it would be (beam diameter for diameter) about twice as long as what we have got currently! All one can assume, in the context of the type of CSB we are dealing with (i.e. M is zero at the outermost fulcrum points), is that the deflection of the beam will be zero at the frame fulcrum points. The slopes of the beam (the first differential of the curve) might not be exactly horizontal at the intermediate frame fulcrum positions, and they certainly won't be anywhere near horizontal at the outermost frame fulcrum positions.


The outer most fulcrum is a pin and therfore the moment is zero but if the slopes at intermediate fulcrums are horizontal then that is a condition of fixity. However, I agree it would be better to allow the moment to be free - it just brings an added complexity but I can see how it can be handled now after playing about with the equations. I am at a loss however to understand why it would result in a CSB twice as long. If anything I would have thought it should bring about a shorter or thiner wire.

However I recognise that my approach of having encastra ends was a simplification and it would be better to calculate the moments, as per the spreadsheet (by moment distribution) or directly by Clapeyron's equtions. However that cannot be done directly as the wheel loads are dependant of the spring stiffness supporting each wheel. There is therefore an iteration between spring stiffness (ie the fulcrum positions), fulcrum moments and wheel loads.

It may be easier to solve this problem by developing the stiffness matrix for the system and go that way. I shall however pursue my original approach a little longer before going down that road.

Nevertheless the problem I have with the present spreadsheet approach still remains, the wheel loads must be an outcome not an input.

The model applies its weight through the fulcrum points. If the centre of weight is coincident with the centre of stiffness of the springs bearing on the wheels then the fulcrum points will undergo an equal deflection and hence the CSB will bend up an equal amount above each wheel (assuming the the wheels are on straight track and the chassis does not itself bend - reasonable given the disparity in stiffness of the CSB and the chassis).

The load on each wheel will therefore be dependant on the stiffness of the spring above it. As that stiffness is a function of the geometry of the fulcrum points (and yes the EI of the wire) the wheel loads cannot be predetermined.

Alan

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Will L
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Re: Abstruse CSB Theory

Postby Will L » Thu Feb 17, 2011 12:36 am

Personally I have enjoyed reading this immensely. I do have to say that quite a bit of it has gone over my head too, but I think, after a bit of a false start, we do seem to be getting somewhere.

Alan Turner wrote:...Nevertheless the problem I have with the present spreadsheet approach still remains, the wheel loads must be an outcome not an input.


While I have to agree with you, Alan, that one would instinctively like that to be true, I think it was the difficulty in achieving that result which gave rise to the upside down approach, which does allow weight as an input. What the upside down approach does do is to allow the interaction between adjacent sections of the CSB to be modelled and give a result as a deflection. As you yourself said:-

The load on each wheel will therefore be dependant on the stiffness of the spring above it. As that stiffness is a function of the geometry of the fulcrum points (and yes the EI of the wire) the wheel loads cannot be predetermined.


But the deflection can be, and if we calculates the deflection upside down with equal weights, the deflection calculated will be proportional to the weight carried when the right way up. Given the assumption of a CofG central over the wheelbase, and just for the moment that the centre axle is in the middle, it surely must be true that an axle deflected by the same amount on the spread sheet will carry the same weight. So if you can get the spreadsheet to show equal deflection on all wheels then the weight carried by each wheel will be equal. So not perhaps such a naive assumption.

Let us not forget, the function of the spread sheet is to allow us to find fulcrum solutions that produce the desirable equal deflection result. And this it does no matter what you may think of its structural analysis credentials, and I'm sorry but I don't think the fulcrum points it suggest are anything like as intuitive as you may think. The fact that there are many such solutions and that they all involve a minimum of 4 variables (the fulcrum points) suggests that numerical analysis may not provide any simple solutions. As it happens I do have a non spread sheet way of calculating results, but this involves yet more assumptions and a not very constant value somewhere between .56 and .6 which I have derived empirically. See this post.

So having got a perfectly good solution to a rather assumption bound case (both CofG and centre axle central), we need to consider real worlds systems where neither the CofG or the centre axle turn up in the middle.

As it turns out the off centre axle is never far enough off centre to cause too much trouble, particulay if you keep the centre axle slightly softer than the outer two. The slight variation in deflections at either end implies the chassis won't sit perfectly flat but the gradient front to back can be kept acceptably invisible in practice.

Which leaves us with the off centre CogG. I have practical experience which says that letting the CofG stray too far from the centre can produce unexpected results. Hence it would be nice to be able to find a way of dealing with this eventuality, but it will still need to be based on finding a fulcrum point solution which gives a level chassis, not in predicting the weight distribution once the fulcrum points are known.

Will

Alan Turner
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Re: Abstruse CSB Theory

Postby Alan Turner » Thu Feb 17, 2011 8:49 am

Will L wrote:
But the deflection can be, and if we calculates the deflection upside down with equal weights, the deflection calculated will be proportional to the weight carried when the right way up.


Ah but you can't do that because all wheel springs (i.e. the CSB above each wheel) is forced to deflect an equal amount (assuming coincidence of weight with centre of spring stiffness and rigidity of the chassis). The load carried by each wheel is therefore the product of this deflection and the stiffness of its spring.

In fact if you were to increase the deflection of say the centre wheel by putting a matchstick underneath it would attract more load and relive the outer wheels. The CSB is of course setup assuming the wheels are resting on level track.


Alan

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Will L
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Re: Abstruse CSB Theory

Postby Will L » Thu Feb 17, 2011 8:10 pm

Alan Turner wrote:Ah but you can't do that because all wheel springs (i.e. the CSB above each wheel) is forced to deflect an equal amount (assuming coincidence of weight with centre of spring stiffness and rigidity of the chassis). The load carried by each wheel is therefore the product of this deflection and the stiffness of its spring.


Ah but I have an ah but too. While I'm sure your right for any old set of fulcrum points with significantly differing deflections, what we are actually dealing with here would be a specific fulcrum set, devised so that the outer pair of wheels have a deflection under the same weight as close as identical as we can manage, and the inner wheel within a few percent the same. In these limited circumstances, which happen to be the ones we are operating in, I think you can say that the weight carried will be close enough to directly proportional to the calculated deflection as to make no practical odds.

Yes if you run over a match stick, the weight born by the various wheels will change significantly, meaning the one on the match stick will get more weight and depress more, while the ones on the track will receive less weight and depress less. But so long as the match stick is less than 1mm thick, they do all stay in contact with the track and isn't that just the sort of behaviour that we wanted?

The proportionality of deflection to weight carried is the nub of the argument I was trying my best to make last night. If you come to the conclusion I might have a point, you might want to re-consider if where I took it is valid as well.

Remember what we need here is a practical process which works well enough. Taking advantage of special cases is fine so long as we understand when the special case no longer applies. Which I think we have, so long as we can keep the CofG neatly in the middle of the chassis. Which I think is where I came in.

Will

Alan Turner
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Re: Abstruse CSB Theory

Postby Alan Turner » Thu Feb 17, 2011 9:33 pm

Will L wrote: Which I think we have, so long as we can keep the CofG neatly in the middle of the chassis. Which I think is where I came in.

Will


but that is not necessary if the wheel loads are calculated as you can arrange for the resultant of the wheel loads to be coincident with the CG of the loco weight.

The calculation of the CSB is actually quite simple it is the determination of the loads being applied by the fulcrum points that is the difficulty. That is because they are not independent of each other but interdependent due to them all being connected by a ridged beam (the chassis).

Once the point loads are determined then the CSB calculations are:


Equation.jpg


You will note that for constant stiffness of CSB (EI) the wheel loads (and moments) are independant of CSB stiffness. Stiffness only affects resultant deflection.

If however you follow what I set out in my original "thoughts on CSBs" I propose a restriction which I believe overcomes the difficulty with determining the loads, namely you ensure that all fulcrum points have the same deflection. At that point the wheel load distribution is correct for level track.

what you then need to do is arrange things so that the centre of action of the wheel loads is coincident with the CG of the loco weight.

I am working on a spreadsheet, I will let you know how I get on.

Alan

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Will L
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Re: Abstruse CSB Theory

Postby Will L » Fri Feb 18, 2011 11:42 am

Alan Turner wrote:but that is not necessary if the wheel loads are calculated as you can arrange for the resultant of the wheel loads to be coincident with the CG of the loco weight.


Yes I think that's what were after

... I propose a restriction which I believe overcomes the difficulty with determining the loads, namely you ensure that all fulcrum points have the same deflection. At that point the wheel load distribution is correct for level track.


Ok, but we have to be a little careful, to equalise the deflection exactly, fulcrum points need to be placed with considerable precision, while practical construction demands no greater accuracy than 0.5mm.

what you then need to do is arrange things so that the centre of action of the wheel loads is coincident with the CG of the loco weight.

I am working on a spreadsheet, I will let you know how I get on.


I look forward to it. I won't be able to comment further for a while, but I will be back.

Will

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Russ Elliott
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Re: Abstruse CSB Theory

Postby Russ Elliott » Sat Feb 19, 2011 3:35 pm

Alan - on your diagram (and equations), I think you've got the direction of the MB moment the wrong way up.

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Will L
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Re: Abstruse CSB Theory -In defence of the CSB Spread Sheet

Postby Will L » Wed Jun 29, 2011 10:34 pm

Previously in this thread, the validity of designing CSB chassis using the spread sheet available from the CLAG web site, or else where on this forum from me, has been questioned. Better methods have been suggested but have not yet fully materialized.

In practice our use of the spread sheet has implied making assumptions, that is
1. assuming all wheels carry equal weight .
2. assuming that the loco Centre of Gravity (CofG) should be placed centrally over the CSBed wheelbase.

It has to be said that from a theoretical point of view these assumptions are not comfortable, particularly as the spread sheet comes at the problem from a counter intuitive direction. While practical experience suggests that these assumptions work, in as far as we get good working results, they do seem to artificially restrict ones options. By adopting the central CofG design approach are we failing to consider other, possibly better, solutions. After all it is perfectly practical to specify a CSB chassis which does have differently weighted wheels, yet sits level and runs happily. So what are you supposed to do with a loco where the CofG doesn’t naturally fall at the CSB midpoint? Then there is the whole thorny topic of load bearing carrying wheels.

In the CSBs and the Single Bogie thread, I have been thinking about how one would deal with chassis for which particularly the CofG location assumption does not appear to be helpful, e.g. chassis with load bearing carrying wheels. The result of all this thinking is that I feel I can now explain why the way we have been using the spreadsheet is fully justified.

So the purpose of this post is to show that an equal weight distribution on all axles is a perfectly reasonable aim for a CSB chassis, that to achieve it placing the CodG central on the chassis is the only way to go, and just why the spreadsheet is a practical tool for doing this. I can now also explain exactly what we are gaining from careful control of the CofG location and I have a method for dealing with chassis with more than just driving wheels, but as this post is already big enough, this additional stuff will come in a different post on another day.

Revision Needed

I’ve put this post on the Abstruse CSB Theory thread because I know this sort of discussion is only a minority interest, and forewarned by the thread title, most people who decide to read it will already have a fair understanding of CSB theory. If you haven’t and your still reading, it may we worth while doing a little revision by reading some previous posts. Start from here if its all new , or if your just a bit rusty on what the spread sheet was about you can start a bit further down. At least you will then know what spread sheet I am talking about. Or if you prefer Russ Elliott’s literary style much the same ground is covered on the CLAG web site here.

Also at this point I need to introduce you to, or remind you of, the “principle of moments”, which has a lot to say about they way a loco's weight is distributed across its wheels. I can do no better than point you at Scalefour Digest 41.0 “The principles of model locomotive suspension”, and particularly annex 2 on determining the location of a locos CofG. This is more of Russ’s work on the CLAG web site. It shows you how you can establish where a loco's CofG is, if you know the weights carried by each wheel.

From this you would have thought that, given a known position for the CofG, the same mathematics would allow you to establish how much weight is carried by each wheel/axle. Well, you can for a compensated chassis, see large tracts of digest 41.0, and you will note that it is not just CSB suspension that can appear mired down in complex mathematics! However springing muddies the maths rather, because the amount of weight carried by each wheel/axle now may also depend on the strength of the spring betwixt wheel and chassis (the spring rate). So for most CSB chassis you can’t calculate the weight carried by each wheel simply from knowing the CofG location.

Justifying the equal weight on each axle assumption – 2 axle CSB chassis

We will start with a 4 wheel/2axle chassis. This is actually the CSB trivial case. Assigning the fulcrum points does not require any fancy calculations. The centre fulcrum goes half way between the wheels and the outer two should be placed symmetrically (i.e. the same distance) either side. They shouldn’t be too close to the axle but otherwise you can put them more or less where you like. I do have a 2 axle version of that spread sheet, but the only useful thing you can do with it is to calculate what size of CSB wire you will need, given the weight carried and the position of the fulcrum points. You can probably manage very well without this, though it has its uses which I will be coming back to in due course, but not in this post.

This diagram was original used in the CSBs and the Single Bogie thread.
CSB C12 draw 3.jpg
CSB C12 draw 3.jpg (79.63 KiB) Viewed 7842 times


In these simple circumstances the presence of springing makes no difference. if the CofG is central between the wheels, then the wheels will be equally loaded. You can use the principle of moments calculations to verify this. Conversely, if the CofG isn’t central then the wheels wont be equally loaded.

If the CofG is set centrally, and the fulcrums were not set symmetrically, i.e. making the spring rate different on each side, The chassis will not sit level but the wheels will be equally loaded (unless it sits so crookedly it materially affects the position of the CofG in which case…. Oh lets not go there, we want our chassis to sit level don't we?).

Thinking ahead, of course a 4 wheel chassis may well be a bogie, in which case the CofG is replaced by a load transmitted through the bogie pivot, but the principles don’t change.

Justifying the equal weight on each axle assumption – 3 axle CSB chassis

When we move on to a 6 wheel/ 3 axle chassis, things are definitely more complicated and the springs do matter. In these circumstances there are a range of possible weights that can be carried by each wheel that will satisfy the principle of moments calculations for any given CofG location. This gives an infinite number possible solutions to the question, how much weight is born by each wheel with the CofG just here. As in any circumstance where there is an infinity of possible answers, the next step has to be to limit some of the variables. So lets look at a symmetrical chassis and keep that CofG central.
CSB gravity 1.jpg
CSB gravity 1.jpg (100.96 KiB) Viewed 7842 times

As the outer two CSB spans, and their relationship to the centre axle, are identical they will have the same spring rate. From that it should be clear that the outer two axles will carry equal weight. If this is not clear to you, consider what would happen if the spring rate over the central axle is reduced to zero. Now the centre axle would carry none of the weight and whole set up becomes equivalent to a 4 wheel/2 axle solution. In more general terms, how the weight is shared between the two outer axle and the inner one will depend on the relationship between their spring rates. A whole range of possible answers is available. Weights ranging from nothing to the total weight can be placed on w2. Any weight not placed on w2 will be shared between w1 and w3. All of these will give results that will satisfy the principle of moments calculation.

Now we are getting somewhere, because that the desired “equal weight on each wheel” result is in there among all the other possibilities, and the principle of moment calculation to demonstrate it is true is not hard to do. This calculation will also remind you that there is only one place the CofG can be for this to be true. That is dead central.

What the spreadsheet really does

At this stage we need to return to the spreadsheet. You will remember that this works by placing the chassis on its back, placing weights on each wheel and calculating the resulting deflection of the spring for each wheel. To help you remember, the following diagram shows the spreadsheet inputs and outputs, but if that doesn’t help then perhaps you might want to read the revision suggested earlier.
csb draw 07.jpg
csb draw 07.jpg (39.55 KiB) Viewed 7772 times

The calculation the spread sheet performs is to determine the spring defection given the weights applied and a set of fixed fulcrum points. Most importantly it takes into account the interaction between adjacent wheels across the fixed fulcrums which are implicit in how CSBs work. Clearly when the calculated deflections vary considerably from wheel to wheel, they bear very little relationship to any real world situation when the chassis is right way up and standing on a (presumably) flat peace of rail. However these deflection results do have some validity in as far as they are directly proportional to the effective spring rate of the section of the CSB to which they apply, so equal defections imply that the effective spring rate is equal on each wheel.

If we have followed our assumption of applying equal weight to each wheel, and we have equal defections on each wheel, then the wheel rims will be as level as if they were the other way up and standing on a piece of track. Therefore, we have a chassis in the same state as if it was the right way up standing level on a flat peace of track, with the CofG dead central and a equal load on each wheel.

QED

It may be only one case out of many possibilities, but it is the one we want, its useful, and we don’t need any of the others.
You can get the same answers for 4 or more axles.

Deviations from perfection.

Yes I did assume that the 6 wheel chassis was symmetrical, and it is true that very few actually are. So what are the implications for asymmetrical chassis?

If we go back to the principle of moments, It is certainly possible to calculate a location for the CofG for three equally loaded axles which are not symmetrical, but perhaps not surprisingly this isn’t quite dead centre of the wheel base. Fortunately the maths works out relatively easily and gives a generalised result for both 3 and 4 axle cases (and probably 5, and 6 axle cases too but I haven’t chosen to go there). These results aren’t easy to explain in words but the following diagram should show how to calculate the answers clearly enough.
CSB gravity 2.jpg
CSB gravity 2.jpg (137.98 KiB) Viewed 7842 times

If your inclined to doubt these results, it is easy enough to verify. Take a chassis and work out the CofG location assuming equal weight on each axle à la Digest 41.0. Then use the results above, you should get the same answer.

If you decide you want to use these results, I plan to reissue my version of the spread sheet shortly and that version will do the necessary sums for you. However I’m quite happy with the view that this may be overcomplicating things for the average CSB users, and going for the mid point of the chassis will be close enough in most cases.
Last edited by Will L on Thu Jun 30, 2011 4:44 pm, edited 1 time in total.

Alan Turner
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Re: Abstruse CSB Theory

Postby Alan Turner » Thu Jun 30, 2011 10:40 am

I really must get on and complete my spreadsheet as previously promised.

However I can't agree with your approach to the "equal weight" aspect of your explanation.

The fundamental problem is to think of weights as an input. They are an output because you are not in fact applying loads; you are applying deflections and the weight (or force more properly) that gets applied to the wheel axle is a result of that deflection i.e. F=dxK where K is the spring stiffness. In the case of flat track the deflections are equal.

Taking your diagram
csb%20draw%2007.jpg
csb%20draw%2007.jpg (55.88 KiB) Viewed 7809 times


Your weights need to be replaced by equal deflections (assuming flat track for the purposes of this argument). The weight then carried is calculated as above. The spring stiffness’s being derived from your spreadsheet or by taking a stiffness matrix approach (which is what I am trying to develop).

If you examine the diagram the spring stiffness of the outer two axles is by observation much less than the middle axle (if only because the outer two have pin ends). By applying equal deflection, which is what the flat track will do, the load carried by the middle axel will be much greater than the outer two. I would hazard a guess at a ratio of 1:2:1.

Regards

Alan

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Will L
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Re: Abstruse CSB Theory

Postby Will L » Thu Jun 30, 2011 5:11 pm

Alan Turner wrote:I really must get on and complete my spreadsheet as previously promised.


Yes Please

Alan Turner wrote:However I can't agree with your approach to the "equal weight" aspect of your explanation.

The fundamental problem is to think of weights as an input. They are an output because you are not in fact applying loads; you are applying deflections and the weight (or force more properly) that gets applied to the wheel axle is a result of that deflection i.e. F=dxK where K is the spring stiffness. In the case of flat track the deflections are equal.


I'm not pretending its a perfect model of any configuration of chassis Alan, a good enough proxy will do and this I think I have. Remember what we actually need from the spreadsheet is not the weights in or out, but the fixed fulcrum positions that will give us close to equal weight distribution. They aren't an output either. More than that, there isn't just one right answer, so the maths can't be expected to give us one. See this post a bit further back, and the diagram in the one that follows it. Using the weights as input and inspection to discover an acceptable set of fulcrum points that give us equal deflections as an output, we then have a recipe for a chassis design that performs as we require. What more would a more rigorous analysis give us ?

Alan Turner wrote:...If you examine the diagram the spring stiffness of the outer two axles is by observation much less than the middle axle (if only because the outer two have pin ends). By applying equal deflection, which is what the flat track will do, the load carried by the middle axel will be much greater than the outer two. I would hazard a guess at a ratio of 1:2:1.


Woops sorry, I used an old diagram which wasn't drawn with a equal weight outcome in mind or to scale for that matter. It just shows where the spreadsheet inputs and outputs fit against a generic chassis design . Your right, as drawn weight distribution would not accompany equal deflections. I'll correct the relative dimensions, but beyond being a bit misleading in the circumstance, I don't think this changes the underlying argument.

Anybody interested in spotting the difference should note that the original diagram has now changed, but the one in Alan's response has not.

Will

Alan Turner
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Re: Abstruse CSB Theory

Postby Alan Turner » Sat Jul 23, 2011 11:32 am

Well having threatened to write my version I have finally done it.

First of all let me acknowledge the assistance given to me by Will Litchfield who acted as both a guinea pig for the development versions and also in giving invaluable advice about the user interface.

What is attached is a beta development version and it is time limited (so if you are using it for real print off the results). With feed back I receive from now to Scaleforum I will endeavour to take it on board and then issue a version for use after that.

For this version macros have to be enabled.

This version (apart from details of the model) only needs to know the deflection that you want the CSB to have and what wire you are using.

You adjust the fulcrum points to achieve the axle load (note NOT wheel load) distribution you want and to ensure the Centre of Gravity of the model is coincident with the Centre of Action of the axle loads. This latter point is important; the results are not valid otherwise.

Regards

Alan
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CSB caculator - beta test v1.0.xls
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Chris Mitton
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Re: Abstruse CSB Theory

Postby Chris Mitton » Sat Jul 23, 2011 6:43 pm

Hi Alan, thanks for this.....look forward to trying out the numbers for my engines.

But.....re the weight of the model, I assume that (a) you have to omit the weight of the wheels, axles, axleboxes and coupling rods since the springs don't carry them, and (b) you forgot to weigh them separately before you assembled the loco!

On the other hand, does this matter? What you're really interested in is the weight distribution, not the absolute weight.

Regards
Chris

Alan Turner
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Re: Abstruse CSB Theory

Postby Alan Turner » Sat Jul 23, 2011 9:15 pm

Chris Mitton wrote:Hi Alan, thanks for this.....look forward to trying out the numbers for my engines.

But.....re the weight of the model, I assume that (a) you have to omit the weight of the wheels, axles, axleboxes and coupling rods since the springs don't carry them, and (b) you forgot to weigh them separately before you assembled the loco!

On the other hand, does this matter? What you're really interested in is the weight distribution, not the absolute weight.

Regards
Chris


You are strictly correct and the weight required is the sprung weight of the model.

However as you probably have some wheels, axles and hornblocks lying around why not weigh these and deduct their weight.

Regards

Alan

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Russ Elliott
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Re: Abstruse CSB Theory

Postby Russ Elliott » Thu Sep 15, 2011 8:29 pm

Alan - I'm couldn't get your Excel file to fit into my modest Excel window screen. Is it possible for you to bring in the side margins a bit, particularly on the right-hand side? The big problem I have though is the way I can't get to the bottom of the file (in order to vertically resize the window).

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Russ Elliott
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Re: Abstruse CSB Theory -In defence of the CSB Spread Sheet

Postby Russ Elliott » Thu Sep 15, 2011 9:01 pm

Btw, on Will's point about our spreadsheet assumption that "all wheels carry equal weight", this isn't necessarily true. The wheel/axle weight inputs can be made different on the spreadsheet if desired. Our 'slightly slackening off' the middle axle technique by means of allowing it some extra deflection could equally be achieved by giving it a lesser input weight - the resulting plot in either case will be identical. In explanation to Alan's initial theoretical objection, these 'input' values are what we are seeking as output values (forces) on the wheel. I liken it to hitting a certain key on a keyboard - it's both an 'input', and with a lot a technical wizardy inbetween, the desired character appears as an output.

P.S. Apologies for being incommunicado for so long. Effectively, I've been without a computer (at least, no Excel, no drawing software, no html editor, no reliable internet etc) for about 6 months.

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Will L
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Re: Abstruse CSB Theory -In defence of the CSB Spread Sheet

Postby Will L » Thu Sep 15, 2011 11:54 pm

Russ Elliott wrote:Btw, on Will's point about our spreadsheet assumption that "all wheels carry equal weight", this isn't necessarily true. The wheel/axle weight inputs can be made different on the spreadsheet if desired. Our 'slightly slackening off' the middle axle technique by means of allowing it some extra deflection could equally be achieved by giving it a lesser input weight - the resulting plot in either case will be identical. In explanation to Alan's initial theoretical objection, these 'input' values are what we are seeking as output values (forces) on the wheel. I liken it to hitting a certain key on a keyboard - it's both an 'input', and with a lot a technical wizardy inbetween, the desired character appears as an output.

P.S. Apologies for being incommunicado for so long. Effectively, I've been without a computer (at least, no Excel, no drawing software, no html editor, no reliable internet etc) for about 6 months.


Welcome back Russ I've been wondering where you were.

While it is indeed true that the original speed sheet does allow us to vary the weights, which could be used as you describe, trying to do anything more radical wasn't really on as it gave you no clue if the entered weight distribution was actuality reproducible for any possible CofG location, or what that CofG location would be.

The joy of Alan's spreadsheet is that it will allow you to specify a chassis with a CofG in any reasonable location, get it to sit level and and know what the weight distribution would be.

You'll be pleased to hear that both old and new spreadsheets give consistent answers for CofG's placed centrally to produce near equal axle loadings, and I did checked out a what I hope was a representative sample of the plots posted on the CLAG website to be sure this was true. Now, with Alan's spread sheet, if anybody badly wants a a loco with the CofG well away from the chassis centre, we can now do a plot for that too.

I also think that by hiding the works Alan has produced a tool which should be less intimidating for the non technical.

Will


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