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 NeededI’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 chassisWe 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 (79.63 KiB) Viewed 8687 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 chassisWhen 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 (100.96 KiB) Viewed 8687 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 (39.55 KiB) Viewed 8617 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.

QEDIt 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 (137.98 KiB) Viewed 8687 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.