If Train A is traveling East at 35 MPH and Train B is traveling West at 45 MPH ...

You guys were "cracking me up" yesterday. Clearly modelers on this Forum include those who paid attention in high school math class and learned something useful for life, those who paid attention but sadly didn't get it, and the rest were busy perfecting their class clown routines :-)  

The real trains subtopic had an interesting true story about earlier speeders that made me think of the OP title. It went something like this:  A guy was on a manual track car going say, east; a scheduled train was behind him.  He stopped on the tracks to chat and when asked whether the train behind him was on schedule, noticed his watch had stopped.  It's 2:00 PM. Using only the angle of the sun and the length of shadows, how fast will he have to pump/pedal before he does a "Gomez"?  Bonus points for using scale time for your clock and pedaling effort.  And, no he can't just get off the track.  Where's the fun in that?

But seriously, bringing this topic back on =ahem= track -- another simple math routine I use is to figure out how many real miles of track my living room layout, spurs, and etc. convert to.  This always comes up when I buy some more sections. This math exercise usually ends in head scratching and wacky my calcuator on the side to ensure it's working.  How can I fill up my entire living room (OK small entire living room) with all that track that in my mind can represent a run from Pittsburgh to Philly and Harrisburg and back, and STILL have less than two real miles?!

TRRR

TRRR, let's go back to those thrilling days of yesteryear  Who are you more likely to remember, the class clown or the cat who won the math medal?

As to your second question, just wait.  As the cities you named grow, the distance between them lessens.  Some day, your space will be adequate.  Of course, by then you may have a different perspective

Reading this has given me a headache. Its to early in the morning for this........ My brain hurts.

CQ, they could make smoke even when not burning, since (with one exception) they had oil-fired boilers for train heat steam.

How much smoke does an oil fired boiler for heat make?

I've seen a fair number of GG-1's along the route from NYC towards New Haven, and don't remember seeing smoke.
I know a fair number of people who heat their homes with oil. It's rare to see smoke coming out of their chimneys.

How much smoke does an oil fired boiler for heat make?

That  depends on the type and quality of the fuel being burned and upon the level of maintenance.  Dirty burners or clogged air supply very definitely can cause smoke to be issued.

tackindy posted:

If a Pennsy GG1 is going due east at 20 mph in the shadow of a building 10 stories tall with a wind whipping around it in a clockwise direction at a speed of 40 mph, which direction is the smoke going to blow from the engine and at what speed?

The answer is there is no smoke!  GG1's are electric, they don't smoke!  Well except for that one picture of the burning one but if that one's going 20mph you better be too busy jumping off it to know how fast the smoke is!

So getting back to math, the better question might be how fast will the engineer jump once he detects smoke?

Depends on the engineer.  Casey Jones would ride it until stopped (see what It got him).  Others would set the emergency brake and jump.  Others would just jump.  Still others would have an immediate heart attack.

Math was my absolute worst subject in school. Usually got Ds or even Fs. I do find myself using math quite a bit. At work I have to read micrometers and calculate numbers to set up machines. Luckily we have formulas that are already figured out. Of course I always use a calculator.

If it is 1500 miles from Philadelphia to Wichita Falls, Texas and you take a train that is traveling at 50 MPH to get there, it would take 30 hours to get there, non-stop.  How fast would the train have to travel to make the average speed for the entire round trip, 60 MPH ?

The answer to follow.   

TomlinsonRunRR posted:

So getting back to math, the better question might be how fast will the engineer jump once he detects smoke?

.."IF"

.....There's still the chance at a guy staying with his machine till the end. 

OR:

Do your own physics homework!   Mines done & I'm watching TV ....

(creepy coincidences on the Wikipedia titles eh? )

https://en.wikipedia.org/wiki/...s_for_a_falling_body

https://en.wikipedia.org/wiki/Terminal_speed

Dan Padova posted:

If it is 1500 miles from Philadelphia to Wichita Falls, Texas and you take a train that is traveling at 50 MPH to get there, it would take 30 hours to get there, non-stop.  How fast would the train have to travel to make the average speed for the entire round trip, 60 MPH ?

The answer to follow.   

Okay, I had the figures backward in the original posting.  It should have read like this;

If a train is traveling at 30 MPH on a 1500 mile trip between Philly and Wichita, it would take 50 hours to get from Philly to Wichita, non stop.  How fast would the train have to travel to make a round trip average speed of 60 MPH ?

A train traveling at 60 MPH on a 3000 mile round trip would take 50 hours.  Since the 50 hours was used on the one way trip, that leaves no time for the return trip !  So unless the train could travel at the speed of light on the return trip, the average speed of 60 MPH is impossible.  Even if the return speed of 186000 MPS were possible, the average speed would still not be 60 MPH, but something extremely close to it.  

This was a riddle I learned of in high school.  That was over fifty years ago, so I apologize for the original error. 

I'm still confused Somethings still wrong .

   I see statements I'll just assume to be right, but I only see one question, and that almost answers itself. It is not "locked down" yet, still a variable question and answer. You might have to account for some time needed to be made up upon accelerating from a dead stop upon starting. I almost see the equation, haven't done numbers, but...,

The Question:   "How fast would the train have to travel to average 60mph" = X

The 60mph average, will also require X to be greater than 60mph, to make up lost time at the start etc.

61mph "max." for "a while" would be fast enough to average 60mph over 3000 miles...non-stop...mines electric . So....

X < 61mph (X is less than 61mph)

X > 60mph (X is greater than 60mph)

If the train has a running start and is limited to a constant speed, it has to do exactly 60mph, to average 60mph, if it goes 1 mile, or a million

AGHRMatt posted:

I use math quite a bit. Aside from calculating to scale up/down equipment from different scales, I use trig quite a bit to determine curve module sizes for various arcs/radii and appropriate curves/tangents to achieve an S-curve offset, etc.

By the way, the call of the question usually reads something irrelevant to the fact pattern like "who's working the grill at Pat's Steaks in Philadelphia?"

You electro dudes make me wonder what planet you guys are from.

Matt your second answer is " GENO " is at the grill!

Adriatic posted:

I'm still confused Somethings still wrong .

   I see statements I'll just assume to be right, but I only see one question, and that almost answers itself. It is not "locked down" yet, still a variable question and answer. You might have to account for some time needed to be made up upon accelerating from a dead stop upon starting. I almost see the equation, haven't done numbers, but...,

 

The Question:   "How fast would the train have to travel to average 60mph" = X

The 60mph average, will also require X to be greater than 60mph, to make up lost time at the start etc.

61mph "max." for "a while" would be fast enough to average 60mph over 3000 miles...non-stop...mines electric . So....

X < 61mph (X is less than 61mph)

X > 60mph (X is greater than 60mph)

If the train has a running start and is limited to a constant speed, it has to do exactly 60mph, to average 60mph, if it goes 1 mile, or a million

 

Yes, it is assumed that the train is traveling at 30 MPH from a dead start.  Many neck injuries !   There is no such thing as lost time in this riddle, the railroad runs perfectly all the time, every time, LOL.   Thirdly, the riddle is a trick question.  By posing it the way it is, it throws the person being asked for the answer, off.  

Have another look at it.

Dan, I think you still have the riddle wrong.  Nowhere in the givens is there a time limit for the entire round trip. 

If I get Dan P's riddle, the "trick" is that the riddle talks about the 60 MPH for the round trip and then tries and confuse the puzzler by comparing the completion time of the round trip with the completion time of a one-way trip at 30 MPH that took 50 hours.  It then concludes that there was no time for the round trip because the one-way train used up the "time".  Huh? That's ridiculous and hence the word trick.

The puzzle is comparing apples to oranges. Red-herring alert: The one-way average is 30 MPH; the round-trip average as Adriatic says is 60 MPH.  1500/30 = 50 hours or 2+ days and 1500/60 = 25 hours or 1+ day.  But the one-way train's time has nothing to do with the round trip's time.  They are separate trains. It should say so on their tickets :-). They literally ARE train A and train B. The wording is trying to conflate their running times to confuse us into thinking the problem can't be solved without the speed of light. Right?

Now, Google Maps tells me that I can make that same trip in 22 hours in my car with a boat load of construction sites in between the two points.  From this we can conclude two things:  1. there's something wrong with those trains  2. the all-seeing eye of Google has never seen me drive.

Tomlinson Run Rail or Road

Dan Padova posted:

Dan Padova posted:

If it is 1500 miles from Philadelphia to Wichita Falls, Texas and you take a train that is traveling at 50 MPH to get there, it would take 30 hours to get there, non-stop.  How fast would the train have to travel to make the average speed for the entire round trip, 60 MPH ?

The answer to follow.   

Okay, I had the figures backward in the original posting.  It should have read like this;

If a train has traveled at 30 MPH on a 1500 mile trip between Philly and Wichita, it has taken 50 hours to get from Philly to Wichita, non stop.  How fast would the train have to travel to make it back to Philly, for a round trip average speed of 60 MPH ?

A train traveling at 60 MPH on a 3000 mile round trip would take 50 hours.  Since the 50 hours was used on the one way trip, that leaves no time for the return trip !  So unless the train could travel at the speed of light on the return trip, the average speed of 60 MPH is impossible.  Even if the return speed of 186000 MPS were possible, the average speed would still not be 60 MPH, but something extremely close to it.  

This was a riddle I learned of in high school.  That was over fifty years ago, so I apologize for the original error. 

 

OK,..I edited it just a bit, I think  it "fits" better now. But does it still seem "rightly wrong" to you this way too?

   I kept seeing a possible comparison of 2 trains, on 2 round trips, while reading before.

    I put the 1st leg of the trip in past tense, and the 2cnd leg in the future; and clarified it's still the same train, by now including that the train is just beginning the return trip.

   I have mild Asperger's, the math usually comes easy, but "little stuff" in language, that most people can "read around", is often a huge stumbling block for me . Just trying to get it straight "for me", ya know .

(thanks Dan )

Tomlinson Run Rail or Road hit the nail on the head.  It is a trick question.  I used to know a few of these riddles.  When you first hear them, they sound legitimate, but if you dissect them and call the Riddler out on them, they are full of holes. 

Several different math problems have been stated here now.

If you want to see the train on its return trip, you need a special camera:

https://youtu.be/mfgsQX78hg8

Dan Padova posted:

Dan Padova posted:

If it is 1500 miles from Philadelphia to Wichita Falls, Texas and you take a train that is traveling at 50 MPH to get there, it would take 30 hours to get there, non-stop.  How fast would the train have to travel to make the average speed for the entire round trip, 60 MPH ?

The answer to follow.   

Okay, I had the figures backward in the original posting.  It should have read like this;

If a train is traveling at 30 MPH on a 1500 mile trip between Philly and Wichita, it would take 50 hours to get from Philly to Wichita, non stop.  How fast would the train have to travel to make a round trip average speed of 60 MPH ?

A train traveling at 60 MPH on a 3000 mile round trip would take 50 hours.  Since the 50 hours was used on the one way trip, that leaves no time for the return trip !  So unless the train could travel at the speed of light on the return trip, the average speed of 60 MPH is impossible.  Even if the return speed of 186000 MPS were possible, the average speed would still not be 60 MPH, but something extremely close to it.  

This was a riddle I learned of in high school.  That was over fifty years ago, so I apologize for the original error. 

 

RJR posted:

Dan, I think you still have the riddle wrong.  Nowhere in the givens is there a time limit for the entire round trip. 

Dan is correct. To achieve an average speed of 60mph for a 3000 mile round trip, the total elapsed time would be 50 hours, but it already took 50 hours to go 1500 miles at only 30mph. So the train can't travel fast enough to double its average speed for the entire 3000 miles.

Think of it this way: if a train continues to fall way behind schedule, eventually it will be so far behind that it can't get to its destination on time no matter how fast it goes (within reason) for the remainder of the trip.

I think this is the original math problem:

If it is 1500 miles from Philadelphia to Wichita Falls, Texas and you take a train that is traveling at 50 MPH to get there, it would take 30 hours to get there, non-stop.  How fast would the train have to travel to make the average speed for the entire round trip, 60 MPH ?

As RJR wrote, there is no time limit on the round trip. The fact that it would take 30 hours at 50 MPH to travel 1500 miles is not relevant.
The math to calculate the required speed to make the average come out to 60 MPH is as follows:

Stated another way we know that 1/2 the trip was made at 50 MPH, and that the desired average is 60 MPH so:

(1/2*50) + (1/2*X) = 60, where X is the required speed.
(25) + (1/2*X) = 60 ...... multiply 1/2 * 50
(1/2*X) = 35 ......... subtract 25 from both sides
X = 70 ......... multiply both sides by 2

So the train would have to travel 70 miles per hour on the return trip to make the average speed 60 MPH.

Yes CW, it is a solvable non-trick question if you use totally different numbers. But if the train travels a major portion of the route at too slow a speed, it cannot achieve a substantially higher average speed for the entire route at practical speeds.

Think of it this way: if a train continues to fall way behind schedule, eventually it will be so far behind that it can't get to its destination on time no matter how fast it goes (within reason) for the remainder of the trip.

I agree.  Amtrak proved it to me a few months ago.

Dan has applied a principle known to anyone who has ever commanded a ship in wartime:  keeo changing course and/or speed.

Yes CW, it is a solvable non-trick question if you use totally different numbers.

Please point out which numbers are different, and where the logic is incorrect.
Or did I capture the wrong problem (in italics)?

C W Burfle posted:

I think this is the original math problem:

If it is 1500 miles from Philadelphia to Wichita Falls, Texas and you take a train that is traveling at 50 MPH to get there, it would take 30 hours to get there, non-stop.  How fast would the train have to travel to make the average speed for the entire round trip, 60 MPH ?

As RJR wrote, there is no time limit on the round trip. The fact that it would take 30 hours at 50 MPH to travel 1500 miles is not relevant.
The math to calculate the required speed to make the average come out to 60 MPH is as follows:

Stated another way we know that 1/2 the trip was made at 50 MPH, and that the desired average is 60 MPH so:

(1/2*50) + (1/2*X) = 60, where X is the required speed.
(25) + (1/2*X) = 60 ...... multiply 1/2 * 50
(1/2*X) = 35 ......... subtract 25 from both sides
X = 70 ......... multiply both sides by 2

So the train would have to travel 70 miles per hour on the return trip to make the average speed 60 MPH.

I'm with you.

C W Burfle posted:

Yes CW, it is a solvable non-trick question if you use totally different numbers.

Please point out which numbers are different, and where the logic is incorrect.
Or did I capture the wrong problem (in italics)?

CW, you posed the question with transposed numbers which makes a difference:

If it is 1500 miles from Philadelphia to Wichita Falls, Texas and you take a train that is traveling at 50 MPH to get there, it would take 30 hours to get there, non-stop.  How fast would the train have to travel to make the average speed for the entire round trip, 60 MPH ?

Dan had the question like this:

If a train is traveling at 30 MPH on a 1500 mile trip between Philly and Wichita, it would take 50 hours to get from Philly to Wichita, non stop.  How fast would the train have to travel to make a round trip average speed of 60 MPH ?  [I am supposing they mean to complete this same trip, as CW did].

The train would need to travel the entire 3000-mile round trip in 50 hours to achieve an average speed of 60mph. But it has already taken 50 hours to go just half the total distance. To average 60mph for the entire 3000 miles it would have to travel the remaining 1500 miles in zero time, which is impossible.

C W Burfle posted:

I think this is the original math problem:

If it is 1500 miles from Philadelphia to Wichita Falls, Texas and you take a train that is traveling at 50 MPH to get there, it would take 30 hours to get there, non-stop.  How fast would the train have to travel to make the average speed for the entire round trip, 60 MPH ?

As RJR wrote, there is no time limit on the round trip. The fact that it would take 30 hours at 50 MPH to travel 1500 miles is not relevant.
The math to calculate the required speed to make the average come out to 60 MPH is as follows:

Stated another way we know that 1/2 the trip was made at 50 MPH, and that the desired average is 60 MPH so:

(1/2*50) + (1/2*X) = 60, where X is the required speed.
(25) + (1/2*X) = 60 ...... multiply 1/2 * 50
(1/2*X) = 35 ......... subtract 25 from both sides
X = 70 ......... multiply both sides by 2

So the train would have to travel 70 miles per hour on the return trip to make the average speed 60 MPH.

The above math is flawed, check it out:

Rate x Time = Distance

50mph x 30 hours = 1500 miles

70 mph x 21.43 hours = 1500 miles (hours are rounded slightly)

Distance / Time = Rate

1500 miles / (30 hours + 21.43 hours) = 58.33 mph

=============================================

Here is the correct way to calculate with these numbers:

An average speed of 60 mph over 3000 miles would take 50 hours. The first 1500 miles at 50mph took 30 hours. So the remaining 1500 miles needs to be covered in (50 hours - 30 hours) = 20 hours, to get the average for the entire trip up to 60 mph.

1500 miles / 20 hours = 75 mph

To review the whole trip:

Rate x Time = Distance

50mph x 30 hours = 1500 miles

75mph x 20 hours = 1500 miles

Distance / Time = Rate

3000 miles / 50 hours = 60 mph average

Thanks.
I copied and pasted the problem. I didn't realize it was not the original problem.
I came up with the wrong formula anyway. Thanks for showing the correct one.

Dan Padova posted:

If a train is traveling at 30 MPH on a 1500 mile trip between Philly and Wichita, it would take 50 hours to get from Philly to Wichita, non stop.  How fast would the train have to travel to make a round trip average speed of 60 MPH ?

A train traveling at 60 MPH on a 3000 mile round trip would take 50 hours.  Since the 50 hours was used on the one way trip, that leaves no time for the return trip !  So unless the train could travel at the speed of light on the return trip, the average speed of 60 MPH is impossible.  Even if the return speed of 186000 MPS were possible, the average speed would still not be 60 MPH, but something extremely close to it.

RJR posted:

Dan, I think you still have the riddle wrong.  Nowhere in the givens is there a time limit for the entire round trip. 

If the train is going to achieve an average speed of 60mph on a 3000 mile trip, that defines a 50 hour time frame for the entire trip. If the train has already taken 50 hours to go only half that distance, it has no time left to complete the journey and achieve an average speed of 60mph within that 50 hour time frame. Anything more than 50 hours time for the entire 3000 mile round trip will mean an average speed of less than 60 mph.

RJR, I had the same thought as you until I started figuring the numbers.

The way I figure it, they're not asking for times or distance to be figured, only average speed. If it traveled at 30MPH on the first leg then to average 60MPH for the entire trip it needs to travel at 90MPH on the return leg,    

(30 MPH + X MPH ) / 2 = 60 MPH

2 x( (30 MPH + X MPH ) /2)=  60 MPH x 2

30 MPH + X MPH  =  120 MPH

X = 120 MPH - 30 MPH

X = 90 MPH

Jerry

TomlinsonRunRR posted:

when will they meet?  Answer: never, they are travelling on separate tracks!   But seriously, it is amazing how often I think of this high school math problem when I hear a train whistle, as I just did for the Boston to Fitchburg MBTA commuter train at the ungodly hour of 7:05 AM on a Sunday.  Really, who needs to get to either location that early??  (A few minutes later, I heard the second train so I didn't even need to sharpen my pencil).  Being math challenged, I never appreciated that problem at the time, but as I say, I often think of it now.

Question: How many ways do you find yourselves using math as modelers, electricians, layout builders, rail fans, or engineers; and what are your favorite math problems or solutions to solve?  Which ones do you grapple and struggle with but plow through to the bitter end because you need to?  Or do you just say "what the heck? It's good enough for rock and roll" and wing it?  Measure twice and cut/snip/solder or only once?

Personally, I'd like a nice formula for converting 1:1 measurements into 1:48 when I want to model something that scales down to values smaller than a quarter of an inch.  If it isn't divisible by two, it makes my brain hurt to try and convert to something less than a foot. (For example, the colored metal commuter rail signal flags.)  if I had a nice formula, I could write a computer program to do it for me and avoid the risk and mess of my brain exploding. And then there are my civil engineer great uncle's rulers in divisions of 10s, 20s, and 50s that I cherish. I'm sure they'd come in handy if only I had a clue ...

Tomlinson Run Railroad: where we love applied mathematics, we just can't apply it.

 

If they're on the same track I'm going to go and get my brother. He's never seen a train wreck before.

Nick Chillianis posted:

If they're on the same track I'm going to go and get my brother. He's never seen a train wreck before.

Finally, a true genius among us! 

All things considered, we really don't need much math help around here.

We need an English professor  

If Train A is traveling East at 35 MPH and Train B is traveling West at 45 MPH ...

On my railroad, they'd never meet. On my layout, I operate by the standards the real railroad operated on, and that means a 20MPH speed restriction for mainlines. With the curves I have at some points, one (or both) of the trains would have derailed before they met and even if they managed to hold onto the rails, they would have been stopped by the dispatcher before they met for flagrantly violating operating rules!

baltimoretrainworks posted:

The way I figure it, they're not asking for times or distance to be figured, only average speed. If it traveled at 30MPH on the first leg then to average 60MPH for the entire trip it needs to travel at 90MPH on the return leg,    

(30 MPH + X MPH ) / 2 = 60 MPH

2 x( (30 MPH + X MPH ) /2)=  60 MPH x 2

30 MPH + X MPH  =  120 MPH

X = 120 MPH - 30 MPH

X = 90 MPH 

Jerry 

That's bad math. Double check your work with these formulas:

Rate = Distance x Time

Rate / Distance = Time

1500 miles / 30 mph = 50 hours

1500 miles / 90 mph = 16.66 hours

Average speed = total miles / total hours = 3000 miles / (50 hours + 16.66 hours) = 45  mph

... The way I figure it, they're not asking for times or distance to be figured, only average speed.

But the numbers are all related and dependent on each other !!!  It's basic math !!!

http://wx4.org/to/foam/warstories/grabbag.html

Ooops, political. Please delete if politically incorrect and blame WX4.

OK, this is sort of related ... Two Chinese trains pass at 261 miles per hour ...

http://tracks.lionel.com/high-...ains-pass-at-261mph/

There's something almost poetical about the portion of the video that shows the two trains passing.

Tomlinson Run RR (no math[s], I promise! :-)

Ace posted:

http://wx4.org/to/foam/warstories/grabbag.html

Ooops, political. Please delete if politically incorrect and blame WX4.

No.  The Congressman could not reach Union Station because he rode the Washingtom METRO!

Or the question would be sent to committee for a year!

Math story problems? Still sucking the life out of me after all these years!!

Ugh