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R179 Discussion Thread


East New York

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2 hours ago, EphraimB said:

How many R179 trains are in service on the (J)? Wikipedia says 2 trains (16 cars). Wikipedia also says that there's 3 trains (24 cars in service).

IIRC there are 3.

Wikipedia is not a reliable source for railfan info btw. 

Edited by kosciusko
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8 hours ago, kosciusko said:

IIRC there are 3.

Wikipedia is not a reliable source for railfan info btw. 

Actually people on this forum update and maintain the wiki. However, information is usually spread here before it reaches the wiki. 

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With 3086-3089+3094-3097 in service, we now have the following trains in service:

3058-3061+3062-3065
3066-3069+3090-3093
3086-3089+3094-3097

So three trains are in service.  The following sets have been delivered but are still waiting to enter service:

3078-3081, 3082-3085, 3098-3101

 

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3 hours ago, Calvin said:

This is the cycle if I'm not mistaken:

New R179 train at ENY

Then, moves 8-car R32 to 207 St

R32 to 207 St bumps an R46 to the C

8-car R160A train to ENY

Then, another R32 (either 8 or 10) to 207 St. 

Is the (A) going to be all R32s eventually?

Also, how do they get 10-car R32s on the (A) if there were only 8-car R32s before the swap?

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11 minutes ago, EphraimB said:

Is the (A) going to be all R32s eventually?

Also, how do they get 10-car R32s on the (A) if there were only 8-car R32s before the swap?

I'm not sure if the (A) will be all R32s, there may be a few R46 sets left if I'm not mistaken, and when the 5-car R179 sets enter service, those will also be on the line, so you'll end up having a mix of R32s, R46s, and R179s.

As for your second question, all they do is just add two extra cars (The R32s and R42s travel as "married pairs", meaning that two cars will stick together as a set and can be configured with other "married pairs", similar to how 5-car sets of R160s can be matched around with other 5-car sets of R160s). That's how the (M) shuttle is running with only 6 R42 cars per train, they just took 3 8-car sets from the (J) / (Z), took off a "married pair" from each of the 3 sets and then combined those sets to form 4 6-car sets.

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8 hours ago, LegoBrickBreaker101 said:

As for your second question, all they do is just add two extra cars (The R32s and R42s travel as "married pairs", meaning that two cars will stick together as a set and can be configured with other "married pairs", similar to how 5-car sets of R160s can be matched around with other 5-car sets of R160s). That's how the (M) shuttle is running with only 6 R42 cars per train, they just took 3 8-car sets from the (J) / (Z), took off a "married pair" from each of the 3 sets and then combined those sets to form 4 6-car sets.

Where do they get the extra 2 cars from?

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11 minutes ago, EphraimB said:

Where do they get the extra 2 cars from?

The married pairs that were removed from the 8-car set were combined together.

So if each married pair is represented by a single letter, originally it was setup like this:

ABCD

EFGH

IJKL

and it became this:

ABC

EFG

IJK

DHL

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4 minutes ago, EphraimB said:

Where do they get the extra 2 cars from?

They not extras....They are married paired meaning that 2 cars gonna always be together...All you have to do is just uncoulple one married paired that will equal 2 less cars...Example a normal J train has 8 cars talking of the 32s and 42s...Remove one married pair leave's the set with 6 cars and add  the married pair you back to 8 cars....That's one of the reasons the A getting some....Besides the C becoming full length that allows the A to be ten cars... 

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15 minutes ago, P3F said:

The married pairs that were removed from the 8-car set were combined together.

So if each married pair is represented by a single letter, originally it was setup like this:

ABCD

EFGH

IJKL

and it became this:

ABC

EFG

IJK

DHL

What are they going to do with the R32s on the (C)?

Edited by EphraimB
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11 hours ago, LegoBrickBreaker101 said:

I'm not sure if the (A) will be all R32s, there may be a few R46 sets left if I'm not mistaken, and when the 5-car R179 sets enter service, those will also be on the line, so you'll end up having a mix of R32s, R46s, and R179s.

As for your second question, all they do is just add two extra cars (The R32s and R42s travel as "married pairs", meaning that two cars will stick together as a set and can be configured with other "married pairs", similar to how 5-car sets of R160s can be matched around with other 5-car sets of R160s). That's how the (M) shuttle is running with only 6 R42 cars per train, they just took 3 8-car sets from the (J) / (Z), took off a "married pair" from each of the 3 sets and then combined those sets to form 4 6-car sets.

There also aren't enough R32s around to cover the (A) entirely.  Something else that should be mentioned is that the only way this is possible (since technically the number of cars needed is increased) is to decrease the spare factor of the R32s.  Before the swap, about 160 cars were out during rush hour out of 222 (a spare factor of 28%, which is pretty high for the MTA right now).

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16 hours ago, Calvin said:

This is the cycle if I'm not mistaken:

New R179 train at ENY

Then, moves 8-car R32 to 207 St

R32 to 207 St bumps an R46 to the C

8-car R160A train to ENY

Then, another R32 (either 8 or 10) to 207 St. 

The 8-car R160’s haven’t been transferred yet. The (C) still has 96 R160’s. I think those  8-car sets are going to Coney Island...

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12 hours ago, LegoBrickBreaker101 said:

I'm not sure if the (A) will be all R32s, there may be a few R46 sets left if I'm not mistaken, and when the 5-car R179 sets enter service, those will also be on the line, so you'll end up having a mix of R32s, R46s, and R179s.

As for your second question, all they do is just add two extra cars (The R32s and R42s travel as "married pairs", meaning that two cars will stick together as a set and can be configured with other "married pairs", similar to how 5-car sets of R160s can be matched around with other 5-car sets of R160s). That's how the (M) shuttle is running with only 6 R42 cars per train, they just took 3 8-car sets from the (J) / (Z), took off a "married pair" from each of the 3 sets and then combined those sets to form 4 6-car sets.

I hear that the (A) will have very few sets of R46’s from the (L) shutdown til retirement, meaning the (A) will have mostly 60ft equipment in service. 

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2 hours ago, VIP said:

I hear that the (A) will have very few sets of R46’s from the (L) shutdown til retirement, meaning the (A) will have mostly 60ft equipment in service. 

In other words, the A will get all 5 car r179's.

Possible car distribution

A: 5 car r179's, r32's, r46's

C: r46's

G: r32's, 4 car r179's and/or 4 car r160's

J/Z: r143's, 4 car r179's and/or 4 car r160's

Edited by subwaycommuter1983
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