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General Track circuit calculations
#1
PLEASE NOTE THERE IS A MORE UP TO DATE CALCUALTOR BELOW

There is often a lot of difficulty with the tack circuit calculation question. This is I suspect because people tend to over complicate it (it is after all not much more than a battery and a few resistors), although when you start to talk about "equivalent" circuits, students take on that look of "I should know something about those, shouldn't I".


.xls   TC.xls (Size: 19.5 KB / Downloads: 707)

The attached spreadsheet is not intended as a definitive calculator, it is there as an indication of how you might lay out the TC question (in this case to calculate the reed resistor) and break down the (relatively few) sums which are needed to arrive at the answer.

Fiddle with the numbers and see the sort of effect that it has on the other parameters (given that this is calculating the feed resistor, if you want to have a look at the effect of the Drop shunt of altering other things, change those values and fiddle the number for the drop shunt to get back to a feed resisance of 9ohms).

If you don't understand any of the sheet, please ask!
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#2
Peter, should the formula in H4 for calculating 'BR' be H2*(H3/1000) as opposed to H2/(H3/1000)?
mel
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#3
Ah ha, the "ballast resistance calculation" dilemma. Probably one of the most common things that people make a mistake on, and the bit that we did right at the end of tonight's study group.

Remember that the ballast resistance is effectively made up of loads of resistors in parallel, so if the TC is longer, the total will be smaller. The value for ballast resistance is quoted in the units of "ohm km" and here as a value of 2.5 ohm km. This means "the length of the tc in km times the resultant equivalent value of resistance must be 2.5", hence the formula given is correct but appears odd at first sight.

Thanks for asking the question, it shows someone is thinking about things.

You are never wrong, but there may be some things that you have not quite had the chance to learn properly yet.

P.S. For resistance read "impedance" to be fully correct and not lazy like me.
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#4
Yes- this is a very common error.

Think of it like this: although we call it ballast resistance in actual fact a lot of the current can run on the surface / through the sleepers (especially if thinking wet timber sleepers into which are screwed the chairs holding bullhead rail in place). Think of each of these as being say a 1kohm resistor; these are spaced approx every metre along the length of the track circuit and thus it will become blindingly obvious that the greater the length of track circuit the LOWER the reistance because there are more parallel paths available for the sneak current. Keep this picture in your mind and you should never make the mistake again!

Peter's point about impedance is very relevant. At dc (and to a very fair approximation at 50Hz ac) the rail resistance can be ignored, thus there is no volt drop along the length of the TC hence a effect of the lost current can be approximated by a theoretical equivalent ballast resistance across the rails. At higher frequency (even Reed frequencies of a few hundred Hertz) this is not true and at Aster / TI21 frequencies (circra 2000Hz) the voltage drops off very markedly along a track circuit (not linearly either- falls sharply when leaving the transmitter but then the curve flattens out). Whole concept of ballast resistance is far too simplistic to be useful; one useful by product though is that it can be easier to find the short circuit on a failed track circuit by plotting the rate of decline in voltage when walking every 10m along the track.

PJW

Peter Wrote:Ah ha, the "ballast resistance calculation" dilemma. Probably one of the most common things that people make a mistake on, and the bit that we did right at the end of tonight's study group.

Remember that the ballast resistance is effectively made up of loads of resistors in parallel, so if the TC is longer, the total will be smaller. The value for ballast resistance is quoted in the units of "ohm km" and here as a value of 2.5 ohm km. This means "the length of the tc in km times the resultant equivalent value of resistance must be 2.5", hence the formula given is correct but appears odd at first sight.

Thanks for asking the question, it shows someone is thinking about things.

You are never wrong, but there may be some things that you have not quite had the chance to learn properly yet.

P.S. For resistance read "impedance" to be fully correct and not lazy like me.
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#5
A couple of small refinements to the earlier spreadsheet which hopefully might make some things clearer - with each step written out in full.
There are two versions, one
.xls   Feed Resistor.xls (Size: 21.5 KB / Downloads: 353) working out the feed resistor for a given set of parameters eg question 1 in 1997 (although that was for an ac track, but most of the principles are the same - maybe someone could post an answer for the ac track); and the other {attachment removed because it is wrong!, see post below} calculating the theoretical maximum length of the track circuit as in question 6 in 2007.


Attached Files
.xls   TC length.xls (Size: 21.5 KB / Downloads: 307)
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#6
Peter Wrote:and the other calculating the theoretical maximum length of the track circuit as in question 6 in 2007.

Hmmm, looking at it, not quite!

The calculations were right, but a bit meaningless - certainly not what was wanted for the question quoted - I had left it calculating the length of the TC at which the relay would drop with a given drop shunt which does not really mean much. Sorry for any confusion there.

Here,
.xls   TC length for relaible operation.xls (Size: 20.5 KB / Downloads: 491) , instead is a modified sheet which gives you the steps for calculating the length of a reliable track circuit given the parameters quoted, except the question was for a TC with an additional, identical feed end relay. The model only has one relay so the parameters entered are those of both relays combined in parallel (ie 10ohm vice 20ohm and PU current of .23A vice .115A)
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#7
A very minor point, but something that confused me for a second - in "T C length.xls", I believe cell A11 should say DA/PU rather than PU/DA - unless it's referring to the ratio of resistance rather than current. Similarly for "Feed Resistor.xls".
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#8
BedfordBoy Wrote:A very minor point, but something that confused me for a second - in "T C length.xls", I believe cell A11 should say DA/PU rather than PU/DA - unless it's referring to the ratio of resistance rather than current. Similarly for "Feed Resistor.xls".

You are indeed correct in your assertion since it is generally the current ratio that is quoted in the question. Sorry for the confusion, but well done on figuring it out.
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#9
We are beginning to get attempted answers for the mod5 Exam Workshop later on this month and got a bit of a surprise; we got some "bonus extra" track circuit calculations from other past papers. Hence posting them here.


Attached Files
.jpg   TC calc 1995 Q1.jpg (Size: 260.7 KB / Downloads: 358)
.jpg   TC calc 2003 Q3.jpg (Size: 302.91 KB / Downloads: 258)
.jpg   TC calc 2005 Q3a.jpg (Size: 250.79 KB / Downloads: 226)
.jpg   TC calc 2005 Q3b.jpg (Size: 276.13 KB / Downloads: 212)
.jpg   TC calc 2005 Q3c.jpg (Size: 249.41 KB / Downloads: 266)
PJW
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#10
(01-07-2009, 06:46 PM)PJW Wrote: We are beginning to get attempted answers for the mod5 Exam Workshop later on this month and got a bit of a surprise; we got some "bonus extra" track circuit calculations from other past papers. Hence posting them here.

OK

Taking them in the order PJW listed them:

1995 (actually I think this is 1999, but that does not really matter as you have written the question out - incidentally something that is not necessary in the exam and only wastes time). This one was a pretty sparkling attempt. Your diagrams are correct and your logic is sound and the answer you get is spot on what I had previously calculated in a post for a similar question. The only comment I would make by the way of improvement would be for there to be a few more words in there about what you are calculating eg some words about the fact that you are looking to find the value for the fault which causes the rail voltage to fall to the point at which the relays will drop away.

2003 - A good start. You correctly got to the point where you worked out that for the relay to pick up, the value of the ballast resistance was 2.14ohms, although from what you did in the last part and what you did in the 2005 question, I suspect that is more my luck than actually realising what you needed to calculate. I would start off my saying what you meant by "reliable operation" - something like the point at which the relay will re-pick after a train, maybe noting that this was the limit of reliable operation, not a realistic situation in which to leave the track.
Having worked out the critical point to be where the ballast resistance is 2.14ohms you fluffed converting this to a length which surprised me because in the other questions where you had to go the other way (ie were given a figure and a length you correctly came up with the Rb figure). You had a value in ohms and needed to translate that to a figure in km and the linking constant is measured in ohm km. Considering the dimensional units of each we must have

ohm km / ohm = km

ie you have to divide the given value of 3 ohm km by the ohmic value you calculated to give the corresponding length. The fact that you got a 6km DC track circuit should have rung some alarm bells with you. It should be more like 1.4km.
You then tried to calculate the drop shunt but made the same error here as you did in the 2005 question. You used the figure given as the pick up current around which to calculate the drop away. Deliberately they have not told you that for a typical track relay, the drop away current is about 68% of the pickup hence you need to calculate the value of shunt that will reduce the rail voltage to a point much less than you did. Again, an alarm bell here in that you stated that the drop shunt was an open circuit.

In 2005 your maths (with one exception, see later) and layout are spot on apart from the fact that you were trying to calculate the drop shunt using the rail voltage value for the pickup. The 3 ohm drop shunt you calculated is not unsafe, but is higher than a typical value. Carrying on using this erroneously high rail voltage, you try to calculate the drop shunt required and come up with a value of 38 ohms (which is to be expected as the TC is easy to shunt and likely to right side fail), however, look at the line in the middle of page 3. You have:

15Rs = 15.57Rs + 21.9 (correct for the numbers that you have)
But you then say
0.57Rs = 21.9, hence Rs = 38.43.

The flaw is that you should have written
- 0.57Rs = 21.9

This, of course gives you a negative value for the drop shunt which is meaningless, but in fact what you are describing is a situation where, under steady state, the TC will already have dropped and maths being what it is, is helpfully telling you that to get to the point at which the relay would drop (if it were not already down!) you have to apply a negative resistance. This is of course the point at which you say "this cannot be right", look back in your working and realise there is a mistake higher up.

I think you also missed the part where you were asked to comment on the reliability and how it can be improved (oh and note that you showed a variable feed resistor in your diagram when the text specifically states this is fixed, a significant fact if you are then going to comment about how to improve the performance of the wet track).

In summary, they are three pretty good attempts in terms of circuit analysis, but you do need to gen up on the characteristics and performance of track relays (pick and drop points), get comfortable working out ballast resistance and watch out for arithmetic slips.

If there is anything that does not make sense, please let me know and I'll try to explain it better.

Peter
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