Signalling the terminal exercise

[onestrangeday]

Hi Signalling Professional:

This time I have tried another exercise for signalling the terminal. Please check my work and calculation, I'd like to see whether I am on the right course or not.


thanks

[PJW]

The layout looks ok except for the depiction of route indicators on signals (which is not really important for the role for this exercise but is worthy of comment).

Signals 21, 23 & 29 can apparently read into any of the 4 platforms and therefore need enough route indicators to tell the driver exactly to whre routed. In such a scenario it would be normal to use a Standard Alpha-numeric Route Indicator rather than the Position Light Junction Indicators which you have depicted. These are a big box which illuminate to show a large single character (number or letter)- the oldest type used a matrix of individual lamps, more modern use fibre optics to distribute the light from a single quartz halogen lamp to individual points making up the character and the most modern uses multiple LEDs. The same box can display one of several (up to 6) characters and therefore in this case each would be capable of 4.

Use of the PLJI is not actually wrong but signal 29 would need pos 1,4 &5 (there would not be one for the "straight route" to 15 signal; ditto signal 23. Signal 21 however would have positions 1,2 &3 as its "straight route" is to signal 11 and tall the other routes duiverge from that to the left.

You have correctly shown signal 22 with positions 4&5 for Main and Branch respectively; however signal would have been better at platform end instead of 12.
Signal 14 has correctly got positions 1 and 4 as routes diverge one eother side of the straight route to Down Main.
Signal 16 has no obvious straight route, so giving it a route indicator for all troutes is sensible, so again you are right to give positions 1,2 & 4. You should have done the same for signal 18.

Beginning to look at your calculations. I liked your diagram and agree that the train leaving the bottom terminal platform must have got its rear end clear of the UM - DM crossover at 0.800 befor being able to route signal 29 for a second train. However having difficulty relating to the numbers calculated.

Bullet 1/2: Driver needs to regulate train speed to 25km/h prior to arriving at the beginning of the trailing points whose tips at 1.300; does seem sensiblefor simplicity to say that restriction starts at 29 signal. Hence approaching trains would be braking to that speed and prior to this braking would be travelling at 85km/h.
Think this is what you meant.
Also correct that train couldn't reaccelerate until REAR has passed over switch tip at 1.300. In reality I think better assumption is that the whole area to platforms would have a blanket 25km/h restriction. As a side benefit it also makes calculation easier!

Bullet 3: Have 150m platform,, 120m train; makes sense that train stops 15m short of 17 signal so when driver changes end the other cab is then 15m from signal 18. Hence would go 135m into the platform.

Bullet 4. Agree would accelerate to 25km/h and then continue at that speed until reat has cleared 1.400 and probably the associated curve so say 1.600

I agree your calccs section a) and at start of b)
However I lose it hen looking at T1- you seem to be saying that train will be ACCELERATING for this length???

Similarly whereas I'd have said T2 was constant speed (but was wondering whether you would have included a small time for accelerating oncerear has got off the branch, then immediately braking to brake to get back down to 25km/h again before front reaches facing points av 0.600), you claim it is braking to a stop in the platform and that would have it overspeeding at the facing points.

If we just do a sense check of your result; you say T2 = 5 seconds. Looking at your diagram, T2 is the time taken to cover about 600m, which is 120m/s which I think of as approximately 60 miles per hour. I know that 80km/h is almost exactly 50mph so that is quicker than the maximum speed and the train certainly won't be going that fast for any of that length, let alone all of it!

So although you are bradly on the right lines, alarm bells should have been sounding that the answers you were getting must be erroneous. Never switch off your mind when turning on the calculator !

So have another look and do a bit of rework. Don't forget that we need to avoid the oncoming train commencing to brake because of a red on signal 29, so work out where a train would be to be able to respect the 25km/h over the trailing points at 1.300. Factor into your calculation that 29 must have cleared early enough that it won't force the driver to slow down more than they would have done otherwise because of the speed restriction.

Then calculate T2 at a constant 25km/h.
Calculate where the driver will apply brakes to stop 15m short of left end of platform and how long the braking takes from the initial 25km/h, then calculate the rest of the section T3 at steady speed of 25km/h (in a similar manner as you did when calculating T4 for the corresponding acceleration and then steady speed).

Also factor time for the signaller / ARS to recognise that 29 can be set and for the relevant points to throw and achieve detection before signal 29 can clear again, by which time train 2 should not have commenced braking earlier than needed purely to pass over the trailing points safely.

Hope those hints are helpful,
PJW

Hi Signalling Professional:

This time I have tried another exercise for signalling the terminal. Please check my work and calculation, I'd like to see whether I am on the right course or not.


thanks

[onestrangeday]

Hi PJW:

Thank you for taking time to review and check my work, yes it is helpful.

I have modified my signalling layout and calculations please check my work.

I agree that it is better to assume the train keeps at the speed of 25km/h (7m/s) after passing the first points at 1.3km, since it is only few hundred metres away from the station (it also keeps the calculation simpler).

The layout looks ok except for the depiction of route indicators on signals (which is not really important for the role for this exercise but is worthy of comment).

Signals 21, 23 & 29 can apparently read into any of the 4 platforms and therefore need enough route indicators to tell the driver exactly to whre routed. In such a scenario it would be normal to use a Standard Alpha-numeric Route Indicator rather than the Position Light Junction Indicators which you have depicted. These are a big box which illuminate to show a large single character (number or letter)- the oldest type used a matrix of individual lamps, more modern use fibre optics to distribute the light from a single quartz halogen lamp to individual points making up the character and the most modern uses multiple LEDs. The same box can display one of several (up to 6) characters and therefore in this case each would be capable of 4.

Use of the PLJI is not actually wrong but signal 29 would need pos 1,4 &5 (there would not be one for the "straight route" to 15 signal; ditto signal 23. Signal 21 however would have positions 1,2 &3 as its "straight route" is to signal 11 and tall the other routes duiverge from that to the left.

You have correctly shown signal 22 with positions 4&5 for Main and Branch respectively; however signal would have been better at platform end instead of 12.
Signal 14 has correctly got positions 1 and 4 as routes diverge one eother side of the straight route to Down Main.
Signal 16 has no obvious straight route, so giving it a route indicator for all troutes is sensible, so again you are right to give positions 1,2 & 4. You should have done the same for signal 18.

Beginning to look at your calculations. I liked your diagram and agree that the train leaving the bottom terminal platform must have got its rear end clear of the UM - DM crossover at 0.800 befor being able to route signal 29 for a second train. However having difficulty relating to the numbers calculated.

Bullet 1/2: Driver needs to regulate train speed to 25km/h prior to arriving at the beginning of the trailing points whose tips at 1.300; does seem sensiblefor simplicity to say that restriction starts at 29 signal. Hence approaching trains would be braking to that speed and prior to this braking would be travelling at 85km/h.
Think this is what you meant.
Also correct that train couldn't reaccelerate until REAR has passed over switch tip at 1.300. In reality I think better assumption is that the whole area to platforms would have a blanket 25km/h restriction. As a side benefit it also makes calculation easier!

Bullet 3: Have 150m platform,, 120m train; makes sense that train stops 15m short of 17 signal so when driver changes end the other cab is then 15m from signal 18. Hence would go 135m into the platform.

Bullet 4. Agree would accelerate to 25km/h and then continue at that speed until reat has cleared 1.400 and probably the associated curve so say 1.600

I agree your calccs section a) and at start of b)
However I lose it hen looking at T1- you seem to be saying that train will be ACCELERATING for this length???

Similarly whereas I'd have said T2 was constant speed (but was wondering whether you would have included a small time for accelerating oncerear has got off the branch, then immediately braking to brake to get back down to 25km/h again before front reaches facing points av 0.600), you claim it is braking to a stop in the platform and that would have it overspeeding at the facing points.

If we just do a sense check of your result; you say T2 = 5 seconds. Looking at your diagram, T2 is the time taken to cover about 600m, which is 120m/s which I think of as approximately 60 miles per hour. I know that 80km/h is almost exactly 50mph so that is quicker than the maximum speed and the train certainly won't be going that fast for any of that length, let alone all of it!

So although you are bradly on the right lines, alarm bells should have been sounding that the answers you were getting must be erroneous. Never switch off your mind when turning on the calculator !

So have another look and do a bit of rework. Don't forget that we need to avoid the oncoming train commencing to brake because of a red on signal 29, so work out where a train would be to be able to respect the 25km/h over the trailing points at 1.300. Factor into your calculation that 29 must have cleared early enough that it won't force the driver to slow down more than they would have done otherwise because of the speed restriction.

Then calculate T2 at a constant 25km/h.
Calculate where the driver will apply brakes to stop 15m short of left end of platform and how long the braking takes from the initial 25km/h, then calculate the rest of the section T3 at steady speed of 25km/h (in a similar manner as you did when calculating T4 for the corresponding acceleration and then steady speed).

Also factor time for the signaller / ARS to recognise that 29 can be set and for the relevant points to throw and achieve detection before signal 29 can clear again, by which time train 2 should not have commenced braking earlier than needed purely to pass over the trailing points safely.

Hope those hints are helpful,
PJW

Hi Signalling Professional:

This time I have tried another exercise for signalling the terminal. Please check my work and calculation, I'd like to see whether I am on the right course or not.


thanks

[PJW]

I am busy for next couple of days; I shall look at the weekend

Hi PJW:

Thank you for taking time to review and check my work, yes it is helpful.

I have modified my signalling layout and calculations please check my work.

I agree that it is better to assume the train keeps at the speed of 25km/h (7m/s) after passing the first points at 1.3km, since it is only few hundred metres away from the station (it also keeps the calculation simpler).

[onestrangeday]

Hi PJW: Sure, many thanks.

I am busy for next couple of days; I shall look at the weekend

Hi PJW:

Thank you for taking time to review and check my work, yes it is helpful.

I have modified my signalling layout and calculations please check my work.

I agree that it is better to assume the train keeps at the speed of 25km/h (7m/s) after passing the first points at 1.3km, since it is only few hundred metres away from the station (it also keeps the calculation simpler).

[PJW]

Sorry for delay; yes looks much better to me. Sensible degree of rounding etc.
A few small points
1. I don't know whether it was luck or skill that placed signal 29 at 300m from the point and this happened to turn out to be the distance needed to brake to turnout speed! Actually the speed restriction through the points applies to the curved path through the turnout and its approach (as not possible to cant it appropriately for the radius of curvature because of the need to join smoothly with the straight line) and thus in reality would apply at latest say 50m prior to the switch tips, but that is just a quibble.
2. The calculations are not exactly consistent with the demarcation line drawn separating T2/T3; it was a good presentation to show what you were going to calculate, but for the benefit of anyone else looking to tie the numbers with the diagram it is worth stating that this showed your initial assumption which was actually later refined when the numbers were produced, since the braking to a stand would actually not be necessary until well into the platform.
3. As I see it, the datum for the commencement of the platform is at 400m and you have decided to stop the 120m train centrally within it and thus have a 15m allowance at each end. Surely then you have the RH end of the train at 385m and the LH at 265m. When calculating T4 then the end of the train initially at 265m needs to get up to 800m in order that the Up/Down crossover can be normalised behind it.
Given you have calculated that it takes 50m to accelerate up to 25km/h, then I think the length to be travelled at that constant speed would be 800 - 265 - 50 = 485m. Hence I basically agree with you but it wasn't obvious to me where you were getting the 440 value from, so a little further explanation would have been good.
Possibly should have stated some more assumptions (the main one being that there is no traffic from the Up Main nor use of the adjacent platform; I really don't understand why the question gave the station layout yet told you to consider only one element of the apparent traffic), but otherwise your answer demonstrates what I think the question required, showing as it does an understanding of the signalling operational logic as well as the Newton's Laws calculations.

Hi PJW: Sure, many thanks.

[onestrangeday]

Hi PJW:

no worries, thanks for your comment and suggestion. I will move the signal is bit more back instead of 300m before the points. For the demarcation line shown on the layout, I will adjust it in line with my calculation.

"Given you have calculated that it takes 50m to accelerate up to 25km/h, then I think the length to be travelled at that constant speed would be 800 - 265 - 50 = 485m. Hence I basically agree with you but it wasn't obvious to me where you were getting the 440 value from, so a little further explanation would have been good."

I think I am wrong for this, I don't know why I put 440m on my calculation??. The final answer is 485m for sure.


thanks for reviewing my work

Sorry for delay; yes looks much better to me. Sensible degree of rounding etc.
A few small points
1. I don't know whether it was luck or skill that placed signal 29 at 300m from the point and this happened to turn out to be the distance needed to brake to turnout speed! Actually the speed restriction through the points applies to the curved path through the turnout and its approach (as not possible to cant it appropriately for the radius of curvature because of the need to join smoothly with the straight line) and thus in reality would apply at latest say 50m prior to the switch tips, but that is just a quibble.
2. The calculations are not exactly consistent with the demarcation line drawn separating T2/T3; it was a good presentation to show what you were going to calculate, but for the benefit of anyone else looking to tie the numbers with the diagram it is worth stating that this showed your initial assumption which was actually later refined when the numbers were produced, since the braking to a stand would actually not be necessary until well into the platform.
3. As I see it, the datum for the commencement of the platform is at 400m and you have decided to stop the 120m train centrally within it and thus have a 15m allowance at each end. Surely then you have the RH end of the train at 385m and the LH at 265m. When calculating T4 then the end of the train initially at 265m needs to get up to 800m in order that the Up/Down crossover can be normalised behind it.
Given you have calculated that it takes 50m to accelerate up to 25km/h, then I think the length to be travelled at that constant speed would be 800 - 265 - 50 = 485m. Hence I basically agree with you but it wasn't obvious to me where you were getting the 440 value from, so a little further explanation would have been good.
Possibly should have stated some more assumptions (the main one being that there is no traffic from the Up Main nor use of the adjacent platform; I really don't understand why the question gave the station layout yet told you to consider only one element of the apparent traffic), but otherwise your answer demonstrates what I think the question required, showing as it does an understanding of the signalling operational logic as well as the Newton's Laws calculations.

Hi PJW: Sure, many thanks.

[PJW]

Happy to do so, but obviously my response time depends upon my other committments.
Quite intrigued re the source of these exercises, as they are not IRSE material as far as I know. However they are very pertinent and discussion of them useful to many I am sure.

Hi PJW:

no worries, thanks for your comment and suggestion. I will move the signal is bit more back instead of 300m before the points. For the demarcation line shown on the layout, I will adjust it in line with my calculation.

"Given you have calculated that it takes 50m to accelerate up to 25km/h, then I think the length to be travelled at that constant speed would be 800 - 265 - 50 = 485m. Hence I basically agree with you but it wasn't obvious to me where you were getting the 440 value from, so a little further explanation would have been good."

I think I am wrong for this, I don't know why I put 440m on my calculation??. The final answer is 485m for sure.


thanks for reviewing my work