- Thread starter TaseMeBro
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That would work if there is track circuit in the area, which is not guaranteed in the brave new world of GPS driven PTC, and axle counters to determine section occupancy.

Even leaving a lit flare on the tracks a mile down the line in both directions could do wonders for reducing severity.

1. If crossing call number on blue sign if possible then 911 if possible.

2. If not at crossing call the RR if you know which one or make an educated guess calling probable RR.

3.. Render first aid if needed.

4. Remove anyone if possible from tracks perpendicular away from tracks.

5. Place flares if available either side of incident.

If you hear train walk toward sound on with flare and / or shirt.

6. If not needed at scene walk toward any mile marker (MM) and call RR with the mile number and if RR ID is on marker. Remember any block signal box will have identification of location. Other structures may have ID as well.

7. Otherwise walk, if possible, toward closest curve with operating flare or white shirt to flag down a train. Go around curve for some distance noting any MM. Place flare or red or white shirt on stick in middle of track.

8. Return to scene and walk in other direction of track repeating above.

Is my math right or am I totally off?

Don't know about your math. From the Metrolink Timetable, the "Maximum Authorized Speed" (passenger) is 60 in the vicinity of the Sylmar/San Fernando Station (north of the crash site). Beginning just south of the station the speed is 79 which includes the vicinity of the crash site.

Is my math right or am I totally off?

With more than a mile long coal and grain trains and the modular and mix bag freights I have often it heard -

A train of that length weight and kinetic movement takes a mile to stop without derailing (jack-knifing) cars -

So after an accident maybe 40-50 - more - cars will run over the debris scene of the accident -

something to consider if in the vicinity of that scene - stay away and stay safe especially with hazardous cargo

With this passenger train - that airplane was demolished to light weight airborne pieces -

There was no loading platform evident in the vicinity - so the train was at speed no slowing to stop and

the train set did not stop until w-a-y pass the accident scene (no reports of how far - no reports of passenger

injuries on the train making an emergency stop)

The accident reports and findings will be months away -

Will there likely be a repeat of this accident - NO -

But will the authorities be keen on how to timely stop a train from compounding an accident scene ? ? ?

There has to be a solution beyond the basic PTC elements - - -

The protocol that would work is the Metrolink dispatcher confirms a Mayday distress and that makes future communications and protocols very important.

Stopping distance is not linear with speed. The rate of deceleration isn't necessarily linear, but depending on the situation, it could be close enough.

Is my math right or am I totally off?

If we assume the

I don't have specific data available, but one quick Google search has suggested some trains have an emergency braking deceleration of 1.4 m/s2, which translates, very roughly, to 3 mph/second. In other words, to go from 90 mph to 60 mph would take 10 seconds, 60 to 30 mph would take 10 seconds, and 30 to 0 would take 10 seconds.

The total braking time would be 30 seconds (plus a bit of a factor as the brakes build up to maximum effectiveness). However, the distance traveled when you are going from 90 mph to 60 mph in 10 seconds is considerably farther than the distance traveled when going from 30 mph to 0 in 10 seconds. Remember, after the first 10 seconds of braking, you're still traveling twice as fast as you are on a train going 30 mph, meaning that you've probably covered roughly 2.5-ish (rough approximation) times the distance of a train that was going a constant 30 mph (let alone one that was slowing down from 30 to 0).

In the next 10 seconds, you've gone from 60 to 30, you've covered yet again another 1.5 (ish) times the distance of a train traveling at a constant 30 mph. Going from 30 to 0 covers 0.5 (ish) the distance of a constant 30 mph. I say "ish" because the actual distance covered is more of a curve than linear. In reality, you're covering a higher distance because of the time you spend at a higher speed while you slow down. Even using these rough numbers, a train slowing down from 90 mph would require 9 times the distance of a train slowing down from 30 mph.

Apply that to a train making a controlled stop on a 600-foot platform, and the engineer needs to start braking at least 5400 feet (i.e. over a mile) in advance of the stop. Certainly not 2100-2700 feet.

As for the delay in stopping the train I would venture that the officers on scene radioed their dispatch and then it took a minute or so to locate the phone number for the railroad and then a minute or so for the railroad to verify exact location before contacting any trains in the vicinity. As for the blue emergency signs at crossing I would say the officers were probably more focused on getting the pilot out and medical attention for him than looking around for a phone number. Not to mention was the blue sign even posted at the crossing?

The train speed also seems off as the track in dead straight for 3.24 miles prior to the crash scene. Was the train operator not paying attention (not the first time MetroLink drivers weren't)? He/she had a perfect line of sight for a long distance to throw the train into emergency. Or at least slow it down.

As someone stated we hopefully will get all the details in the NTSB report.

Call it speculating if you desire just as others have been doing in the thread. As for the blue sign yes in that snapshot (April 2021) it is posted as well from the other direction. Just if it was there at the time of the crash?Indeed. So maybe hold off on speculating that the pilot or engineer were negligent.

A simple look at Google Maps shows that the sign you question is present:

View attachment 26725

As Trogdor noted, "Stopping distance is not linear with speed." I wonder if deceleration is constant, because braking is changing kinetic energy to some other form of energy (often heat energy), and kinetic energy varies with the square of speed. Something going 90 mph has nine times the kinetic energy of something going 30 mph. But a varying deceleration makes computation more difficult, so let's assume a constant deceleration.[...] So if they take around 700 feet to slow down 30mph, it'd take about 2100 feet to slow from 90? And brakes need to kick in so.... 2500-2700?

Is my math right or am I totally off?

I have no better number, so let's use 3 mph/second as deceleration.[...] some trains have an emergency braking deceleration of 1.4 m/s2, which translates, very roughly, to 3 mph/second. [...]

The time dependent location of an object (train) moving on a straight line (track, with no curve) in the direction of travel at constant acceleration is the starting point (milepost, assumed to be zero) plus the change due to the initial speed plus the change due to the acceleration. In other words,

s(t) = s0 + v0*t + 0.5*a0*t^2 where s0 is the initial location, v0 is the initial speed, and a0 is the initial acceleration (assumed to be constant).

Consistency in units is important. Let's use miles and hours. As I said, I assume the initial location is at zero. I'll assume the initial speed is 60 mph, although it was probably higher, based on

The location of interest is where the speed is zero. If the initial speed is 60 mph and it drops by 10,800 mph every hour, then it will be zero after 60/10800 hours, or 0.00556 hours. (Three digits should provide sufficient accuracy, but feel free to use more digits if you like.) Plugging that value into the formula,

s(0.00556) = 0 + 60*0.00556 - 0.5*10800*0.00556*0.00556

Evaluating the expression, we find the train comes to a stop after 0.1667 miles, one sixth of a mile.

Whenever conducting a thought experiment like this, it is useful to ask, "Is that answer reasonable?" With an initial speed of 60 mph and a deceleration of 3 mph/second, it would take 20 seconds to slow to zero. Happily, 0.00556 hours is 20.016 seconds; the difference is a rounding error because I didn't carry more digits. A train going 60 mph will cover 1/3 of a mile in 20 seconds, so certainly the answer should be less than 1/3 of a mile. Since the assumed deceleration is constant, it seem reasonable that the distance covered would be half the distance covered with no deceleration. Happily, 0.1667 miles is half of 1/3 of a mile. Whatever mistakes I made, I seem to have made them consistently. Corrections and clarifications are welcome.

I'm gratified to see that a 90 mph train takes 9 times longer to stop than a 30 mph train, exactly in line with the change in kinetic energy.[...] I say "ish" because the actual distance covered is more of a curve than linear. In reality, you're covering a higher distance because of the time you spend at a higher speed while you slow down. Even using these rough numbers, a train slowing down from 90 mph would require 9 times the distance of a train slowing down from 30 mph. [...]

Because the deceleration is constant, you needn't say "ish". If you plot the speed versus time, you get a triangle. The distance covered is the area under the triangle. You are correct that "the distance covered is more of a curve" but the change in the rate of covering the distance is constant, as indicated by assuming a deceleration of 3 mph/second.

However, I think you slipped up somewhere in your calculations. If the train is going 90 mph, then it travels 1.5 miles every minute or 0.75 miles every 30 seconds. If it takes the train 30 seconds to slow to a stop, it cannot go more than 0.75 miles as comes to that stop.

the tower may not have had a clue as to were plane crashed , if it was on radar they would only know were it disappeared from screen .

for crossings there is no general emergency number only the number on ENS sign and ID posted is relevant. even a road name is not relevant as road name a railroad uses is not always same as official name and sometimes a road crosses tracks multiple times so which crossing is it .

for crossings there is no general emergency number only the number on ENS sign and ID posted is relevant. even a road name is not relevant as road name a railroad uses is not always same as official name and sometimes a road crosses tracks multiple times so which crossing is it .

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Ah I knew I had to be off. Thanks for the quick lesson.Stopping distance is not linear with speed. The rate of deceleration isn't necessarily linear, but depending on the situation, it could be close enough.

If we assume thetimeit takes to go from 30 to 0 is the same as the time it takes to go from 90 to 60 or 60 to 30, the distance covered will certainly not be the same.

I don't have specific data available, but one quick Google search has suggested some trains have an emergency braking deceleration of 1.4 m/s2, which translates, very roughly, to 3 mph/second. In other words, to go from 90 mph to 60 mph would take 10 seconds, 60 to 30 mph would take 10 seconds, and 30 to 0 would take 10 seconds.

The total braking time would be 30 seconds (plus a bit of a factor as the brakes build up to maximum effectiveness). However, the distance traveled when you are going from 90 mph to 60 mph in 10 seconds is considerably farther than the distance traveled when going from 30 mph to 0 in 10 seconds. Remember, after the first 10 seconds of braking, you're still traveling twice as fast as you are on a train going 30 mph, meaning that you've probably covered roughly 2.5-ish (rough approximation) times the distance of a train that was going a constant 30 mph (let alone one that was slowing down from 30 to 0).

In the next 10 seconds, you've gone from 60 to 30, you've covered yet again another 1.5 (ish) times the distance of a train traveling at a constant 30 mph. Going from 30 to 0 covers 0.5 (ish) the distance of a constant 30 mph. I say "ish" because the actual distance covered is more of a curve than linear. In reality, you're covering a higher distance because of the time you spend at a higher speed while you slow down. Even using these rough numbers, a train slowing down from 90 mph would require 9 times the distance of a train slowing down from 30 mph.

Apply that to a train making a controlled stop on a 600-foot platform, and the engineer needs to start braking at least 5400 feet (i.e. over a mile) in advance of the stop. Certainly not 2100-2700 feet.

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The pilot declared an emergency as he was taking off, too late to stop before running out of runway (after V1, if my memory of all my "Air Disaster" pandemic binge watching is correct.) He was talking to the tower.the tower may not have had a clue as to were plain crashed , if it was on radar they would only know were it disappeared from screen .

for crossings there is no general emergency number only the number on ENS sign and ID posted is relevant. even a road name is not relevant as road name a railroad uses is not always same as official name and sometimes a road crosses tracks multiple times so which crossing is it .

The grade crossing where he crashed was at the 1st perpendicular street after the end of the airport's only runway. There are no buildings or trees or other obstructions except the airport perimeter fence between the runway and the street.

The railroad tracks are parallel and about 200 feet away from the runway. It was (or should have been) in easy sight of the tower, which is located between the tracks and the runway, about 1500' to 2000' from the intersection. There are no trees, buildings or hills between the airport and the tracks, street or intersection. Maybe just a low fence and a lot of flat dirt.

No radar involved, it was all VFR.

I should clarify a couple of things. First, the deceleration rate was obtained by a quick google search, and not really validated (shame on me). Digging further, it seems that rate was for light rail trains, and a train such as Metrolink or Amtrak could have a deceleration rate 1/2 to 1/3 of that value.Oh, thank you. I needed an exercise in applied math.

As Trogdor noted, "Stopping distance is not linear with speed." I wonder if deceleration is constant, because braking is changing kinetic energy to some other form of energy (often heat energy), and kinetic energy varies with the square of speed. Something going 90 mph has nine times the kinetic energy of something going 30 mph. But a varying deceleration makes computation more difficult, so let's assume a constant deceleration.

I have no better number, so let's use 3 mph/second as deceleration.

The time dependent location of an object (train) moving on a straight line (track, with no curve) in the direction of travel at constant acceleration is the starting point (milepost, assumed to be zero) plus the change due to the initial speed plus the change due to the acceleration. In other words,

s(t) = s0 + v0*t + 0.5*a0*t^2 where s0 is the initial location, v0 is the initial speed, and a0 is the initial acceleration (assumed to be constant).

Consistency in units is important. Let's use miles and hours. As I said, I assume the initial location is at zero. I'll assume the initial speed is 60 mph, although it was probably higher, based onTinCan782's comment. (Plug in a different number if you want to assume a different initial speed.) I'll assume constant acceleration of -3 mph/second, or -(3*3600) mph/hour, based onTrogdor's comment.

The location of interest is where the speed is zero. If the initial speed is 60 mph and it drops by 10,800 mph every hour, then it will be zero after 60/10800 hours, or 0.00556 hours. (Three digits should provide sufficient accuracy, but feel free to use more digits if you like.) Plugging that value into the formula,

s(0.00556) = 0 + 60*0.00556 - 0.5*10800*0.00556*0.00556

Evaluating the expression, we find the train comes to a stop after 0.1667 miles, one sixth of a mile.

Whenever conducting a thought experiment like this, it is useful to ask, "Is that answer reasonable?" With an initial speed of 60 mph and a deceleration of 3 mph/second, it would take 20 seconds to slow to zero. Happily, 0.00556 hours is 20.016 seconds; the difference is a rounding error because I didn't carry more digits. A train going 60 mph will cover 1/3 of a mile in 20 seconds, so certainly the answer should be less than 1/3 of a mile. Since the assumed deceleration is constant, it seem reasonable that the distance covered would be half the distance covered with no deceleration. Happily, 0.1667 miles is half of 1/3 of a mile. Whatever mistakes I made, I seem to have made them consistently. Corrections and clarifications are welcome.

I'm gratified to see that a 90 mph train takes 9 times longer to stop than a 30 mph train, exactly in line with the change in kinetic energy.

Because the deceleration is constant, you needn't say "ish". If you plot the speed versus time, you get a triangle. The distance covered is the area under the triangle. You are correct that "the distance covered is more of a curve" but the change in the rate of covering the distance is constant, as indicated by assuming a deceleration of 3 mph/second.

However, I think you slipped up somewhere in your calculations. If the train is going 90 mph, then it travels 1.5 miles every minute or 0.75 miles every 30 seconds. If it takes the train 30 seconds to slow to a stop, it cannot go more than 0.75 miles as comes to that stop.

Second, that was just a general example to show the math. The 5400-foot distance was taking Cal’s note of trains going from 30 to 0 in a distance of 600 feet (which I have also not specifically validated) and extrapolating that. But in that case, that refers to a train not in emergency but rather a train doing a normal service stop, and that deceleration rate will be much lower. The basic point was to show how a train going 3x as fast doesn’t stop in 3x the distance.

No shame there.[...] First, the deceleration rate was obtained by a quick google search, and not really validated (shame on me). [...]

Ah, understood.[...] The 5400-foot distance was taking Cal’s note of trains going from 30 to 0 in a distance of 600 feet [...]

Maybe sort of like having a "RED PHONE" to call to avoid a nuclear global warming melt-down.

You’re joking right?

Five police officers came very close to losing their lives AS WELL AS a retired USAF pilot. It’s been 95 years since MayDay MayDay MayDay was established as a universal EMERGENCY call across the entire aviation world... where English is the common language.

As a result all RAPID ACTION efforts are made to respond to the aircraft emergency including shutting the airport down if needed.

What universal distress call is associated with the railroads?

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