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Steam and pistons

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Posted by Anonymous on Friday, January 20, 2006 10:43 PM
Aside from the technical aspect of question asked one way they got that 80 plus train moving, if not forward as intended, was to apply power in reverse to remove slack ,and possibly free up any brakes that might be sticking, then back to forward with all that slack coming into play forward and with a another rail sanding and their off to the races..
Sorry about getting off of the main subject guys but retired after 30 yrs. on Santa Fe I had to " jump in" with my second post.
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Posted by Virginian on Friday, January 20, 2006 5:51 AM
Take for example a 30" diameter cylinder. With 300 psig steam, you have a maximum force of slightly more than 212,000 pounds available. Yes you have to translate this to the wheels, and it's slightly less on the return stroke because of the rod, and all kinds of other things, but this is a pretty darned healthy number to start with ! Throw in articulateds and double-headers, etc., and you can move many tons over many mountains.
What could have happened.... did.
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Posted by marknewton on Friday, January 20, 2006 4:24 AM
QUOTE: Originally posted by Virginian

Further consideration -
the pistons in an automotive engine work the same basic way


There's one important difference. Steam loco pistons and cylinders are double acting - steam bears on both sides of the piston during the cycle, unlike an automotive engine, which is single acting.

Cheers,

Mark.
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Posted by marknewton on Friday, January 20, 2006 3:41 AM
QUOTE: Originally posted by selector


The bumpers do not meet until the couplers are essentially compressed, otherwise the cars could not be shunted in tight turns in yards. If they could be at significant angles to each other, the bumpers would get in the way. So, there must be a gap, and that gap affords the sequential starting described earlier.


No, that's not correct. Buffers - not bumpers - are sprung, and they do make contact most of the time, not just on sharp curves. The heads are dished to minimise buffer locking in this situation. Industrial locos and stock used on very sharp curves often had grossly oversize buffer heads to deal with this problem.

A gap between buffers is not required for sequential starting as you stated. Passenger stock had screw couplings, which were tensioned to eliminate free slack between vehicles. Many goods vehicles were close coupled for the same reason. The slack necessary for starting the train is built into the drawgear itself, just like North American AAR couplers.

(Apart from me, there is at least one other contributor to this thread with practical experience of shunting and running trains with hook drawgear - G'day, Roger T.!)

All the best,

Mark.
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Posted by edkowal on Thursday, January 19, 2006 5:22 PM
The reason that those buffers are wide and dish shaped is so that they do not interfere with each other on curves.

Historically, European couplers were similar to the link and pin couplers used on American railroads before the Janney and AAR coupler came into use. They used links and hooks, rather than links and pins. And, if memory serves, there were different techniques of coupling, for different purposes. On freight trains, the cars were loose coupled, but on passenger runs, the couplers were tightened up between cars until the buffers were in contact.

More modern equipment is probably different, and I'll leave further elaboration to the folks that live there, and have much more knowledge.

For a good photo of what I'm talking about, visit here: http://www.modelbaneteknik.dk/model/index-e.htm Note the turnbuckle type link in the middle of the chain, to tighten the spacing between cars.

-Ed

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Posted by selector on Thursday, January 19, 2006 3:55 PM
The bumpers do not meet until the couplers are essentially compressed, otherwise the cars could not be shunted in tight turns in yards. If they could be at significant angles to each other, the bumpers would get in the way. So, there must be a gap, and that gap affords the sequential starting described earlier.
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Posted by NeO6874 on Thursday, January 19, 2006 2:11 PM
theres still the offset between the couplers - so the loco pulls car 1, which then pulls car 2, which then pulls car 3, etc.

-Dan

Builder of Bowser steam! Railimages Site

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Posted by Anonymous on Thursday, January 19, 2006 1:49 PM
QUOTE: Originally posted by Roger Traviss

Keep in mind that NO locomotive ever moves the whole train at one time. When a locomotive starts, it picks up one car at a time and that car adds its inertia to the movement which adds a small amount to assist moving the next car and so on


how does this relate to those european trains with their buffers to hold the cars apart?
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Posted by Anonymous on Thursday, January 19, 2006 10:32 AM
Keep in mind that NO locomotive ever moves the whole train at one time. When a locomotive starts, it picks up one car at a time and that car adds its inertia to the movement which adds a small amount to assist moving the next car and so on and so on.

On a 50 car train, there's a lot of inertia from 50 cars moving at 20mph. That's why it takes a lot of brake power to stop them.
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Posted by Pruitt on Thursday, January 19, 2006 8:30 AM
Ignore my earlier post. Sometimes I miss the painfully obvious.

The crankpin offset is 1/2 the piston stroke by definition. So the offset is in the formula.
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Posted by NeO6874 on Thursday, January 19, 2006 8:02 AM
QUOTE: Originally posted by edkowal

The rolling resistance of a modern freight car on roller bearing trucks is only about 1% of the dead weight of the car.

It takes a lot less force than you would think to keep it moving on level track. If you follow that first statement, it means that the force required to keep a 100 car freight train moving is only equal to the weight of ONE of its cars!!

-Ed


i remember being in like 5th grade and going to my uncle's rail yard (he was supervisor to the yard of some steel company.. well still is.. but havent been there in ages). went up to a caboose that was sitting on one of the spurs about 4 lengths from the end of the spur.

he released the brakes and i pushed it (i wanted to see it move...), after getting over the inertia of the car it rolled on its own about halfway down the track before he reapplied the brakes (granted there might have been some grade to the ground.. but still)

-Dan

Builder of Bowser steam! Railimages Site

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Posted by Fluesheet on Thursday, January 19, 2006 7:54 AM
Re: Tractive Effort, I built a spreadsheet about a year ago to calculate tractive effort - mainly to compare a heavy LS&I consolidation to a Chinese 2-10-2. I ended up putting on the web so I could get to it from work if needed - it's documentation is meant for me, so feel free to ask if anything is not clear.

Find it at http://www.calvars.com/unilat/Tractive%20Effort.xls

If for any reason that link doesn't work, try http://www.calvars.com/unilat/steam.htm and follow the tractive effort link.

An N&W class J is used as an example since I knew it's "advertised" tractive effort.

Fluesheet.
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Posted by Virginian on Thursday, January 19, 2006 7:46 AM
Further consideration -
the pistons in an automotive engine work the same basic way. And in a Grand Prix engine you have 8, 10, or 12 pistons as small as salt shakers screaming at up to 20,000 + RPMs, rocketing the F1 cars to over 200 MPH in a flash. If there was ever an example of proven tachnology, this is it.
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Posted by marknewton on Thursday, January 19, 2006 5:55 AM
QUOTE: Originally posted by Brunton

Either the formula or the definition of the terms is inaccuate. One of the things directly affecting Tractive Effort is the ratio of the (offset of the Side Rods from the wheel axles) to the (wheel diameter). Maybe D in the equation above takes this into account, but then D wouldn't be the actual wheel diameter.


I presume you're referring to crankpin throw? Except in some very special cases, it's always the same as the piston stroke. Why would this matter?

Cheers,

Mark.
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Posted by Pruitt on Thursday, January 19, 2006 5:49 AM
QUOTE: Originally posted by edkowal

The formula:

Tractive Effort = ( c P d d s ) / D

where c = constant mentioned above
P = boiler pressure
d = piston diameter
s = piston stroke
D = wheel diameter

Either the formula or the definition of the terms is inaccuate. One of the things directly affecting Tractive Effort is the ratio of the (offset of the Side Rods from the wheel axles) to the (wheel diameter). Maybe D in the equation above takes this into account, but then D wouldn't be the actual wheel diameter.
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Posted by marknewton on Thursday, January 19, 2006 5:26 AM
QUOTE: Originally posted by GBeylick

I've wondered these many decades just how steam generated by a locamotive has the power to push against a piston of a steam engine to which a hundred or more cars are attached.Huge steam pressure has to build inside the cylinder...So why then doean't the ends of the cylinders blow off?


The cylinder covers are a very tight fit, and secured with many large diameter studs. On many modern US steam locos with cast engine beds, the back cylinder covers were an integral part of the casting. Steam locos are very strongly built.

Cheers,

Mark.
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Posted by Virginian on Tuesday, January 17, 2006 12:57 PM
It's all physics and thermodynamics. And believe me, it does work.
BTW, so is a hurricane.
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Posted by Anonymous on Tuesday, January 17, 2006 10:34 AM
Thank you all for helping clear up some of my wonders about steam locomotives and just what it is that gets them to pull such a load.

GGB
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Posted by edkowal on Monday, January 16, 2006 11:06 PM
Did a short Internet search and found a formula for calculating tractive effort for steam engines. Tractive effort found in this way would be a theoretical number, actual tractive effort developed by a built locomotive would by measured by a dynamometer car at the tender's rear coupler.

The equation uses a constant "c" in it, which represents the inefficiencies in translating the steam's power to actual pull. The value for "c" was usually set equal to 0.85 in North America, while a more conservative value of 0.6 was used in Europe.

The formula:

Tractive Effort = ( c P d d s ) / D

where c = constant mentioned above
P = boiler pressure
d = piston diameter
s = piston stroke
D = wheel diameter

If you follow through the calculation using appropriate units, you'll see that the value obtained for Tractive Effort will be in pounds, a unit of force.

Some examples of measured tractive effort values are as follows

for the UP Big Boy 135,375 lbs
N&W Y6a 152,200 lbs
modern passenger steam engine 70,000 to 80,000 lbs

Tractive Effort was really a measure of the ability to start a train. If the boiler was not well designed, it might not be able to generate enough steam to keep the train moving at speed, even though it could get the train started. For example, the Virginian's 2-8-8-8-4 Triplex, although it had a tractive effort of 195,560 lbs in simplex, starting mode could not continue to exert this force above about 5 mph, due to an undersized boiler. (Values given as examples here are from Wikipedia's entry on the topic.)

Horsepower was a better measure of the ability to move a train at speed.

The tradeoff between tractive effort and horsepower was a continuing source of vexation for locomotive designers throughout the history of steam locomotive design.

-Ed

Five out of four people have trouble with fractions. -Anonymous
Three may keep a secret, if two of them are dead. -Benjamin Franklin
"You don't have to be Jeeves to love butlers, but it helps." (Followers of Levi's Real Jewish Rye will get this one) -Ed K
 "A potted watch never boils." -Ed Kowal
If it's not fun, why do it ? -Ben & Jerry

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Posted by edkowal on Monday, January 16, 2006 10:30 PM
The rolling resistance of a modern freight car on roller bearing trucks is only about 1% of the dead weight of the car.

It takes a lot less force than you would think to keep it moving on level track. If you follow that first statement, it means that the force required to keep a 100 car freight train moving is only equal to the weight of ONE of its cars!!

-Ed

Five out of four people have trouble with fractions. -Anonymous
Three may keep a secret, if two of them are dead. -Benjamin Franklin
"You don't have to be Jeeves to love butlers, but it helps." (Followers of Levi's Real Jewish Rye will get this one) -Ed K
 "A potted watch never boils." -Ed Kowal
If it's not fun, why do it ? -Ben & Jerry

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Posted by Texas Zepher on Monday, January 16, 2006 3:13 PM
QUOTE: Originally posted by GBeylick
[Huge steam pressure has to build inside the cylinder with the piston fighting against the weight of the locomotive and its rolling stockl. So why then doean't the ends of the cylinders blow off? Just what gives the locomotive a boost to get its wheels moving ,and its strength to pull mile-long rolling stock?

1. The power in a bit of water vapor under pressure has been greatly under-estimated. Have you ever watched Myth Busters? If so, do you notice how they get very concerned with anything over 5 psi. They explode things with only 7 psi. Locos regularly ran 200-300 psi.
2. I think the strength of the iron/steel of the boiler and cylinders has been underestimated. Just look up some pictures of the accidental boiler explosions to see how much energy they contain.
3. Seems like the difficulty of rolling a loco/train down the track has been seriously overestimated. I mean the whole idea of having the tracks is because it is so much easier to roll things on. It isn't like pushing your car on rubber tires on a rough asphalt road. I believe it was Timkin who took a new locomotive equiped with their roller bearings out on a demonstration tour. At each demo they tied a rope to the front of the loco and would choose a woman from the spectators to come and pull on the rope. Sure enough she could easily pull the locomotive.
4. The wheels do have some slip. Steam engines often spun their wheels trying to get a train moving.
5. The train couplers have slack in them so it isn't necessarily staring the whole train at once but one car at a time. That is why it was the worst thing to stall on a hill. This advantage went away.
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Posted by egmurphy on Monday, January 16, 2006 3:00 PM
There's also a trick to getting started with a real heavy load where the engineer backs the locomotive up to bunch the slack in the couplers of the cars. Then when he starts forward he's only pulling the first car (for an instant), then two cars (for another instant), etc. It allows the locomotive to build up some momentum before actually having to be pulling the total weight of the train.

The locomotive would slip its wheels before blowing the cylinders apart.

Regards

Ed
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Posted by jimrice4449 on Monday, January 16, 2006 2:14 PM
The steam is at high pressure, on modern engines as mucn as 300 PSI. Figure the area of the piston and multiply by 2 (4 if you're talking about a simple articulated)and then factor in the mechanical advantage of the main rod connection to the driver (think of a lever w/ the axle as the fixed fulcrum and the crank pin as the force and the distance between the two as the lever)
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Steam and pistons
Posted by Anonymous on Monday, January 16, 2006 12:27 PM
This is not a model railroad question. I'm sure there are many on this forum with an answer to my query.

I've wondered these many decades just how steam generated by a locamotive has the power to push against a piston of a steam engine to which a hundred or more cars are attached. The weight and resistance to move the piston is enormous, I should think. And on many locamotives, the cylinders aren't very large. Just how is a locamotive able to move an inch?

Huge steam pressure has to build inside the cylinder with the piston fighting against the weight of the locomotive and its rolling stockl. So why then doean't the ends of the cylinders blow off? Just what gives the locomotive a boost to get its wheels moving ,and its strength to pull mile-long rolling stock?

I was asked this question several weeks ago and couldn't come up with an answer that satisfied some children because I, myself, never knew the answer.

GGB


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