Track Grade

[tree68]

CopCarSS

When you talk about adhesion locomotives, are you talking about stuff like Shays, Heislers and Climaxes?

I believe he would - rod locomotives as well.   Anything that depends on the friction between the driving wheels and the rail.

[Paul_D_North_Jr]

Chris -

Not to put any words in RWM's mouth - but he may also have been referring to some logging railroads in the western part of Michigan 's Upper Peninsula.  They had grades in that range that were operated by rod locomotives.  If they had to, the engines would "rock" back and forth to build up enough momentum to get out of the valleys.  See:

World's steepest adhesion railroad?
Trains, June 1969 page 42
15- to 20-percent grades
( GRADE, "JONES, CLINTON, JR.", LOGGING, MICHIGAN, TRN )

Also, the grades at Cass, W. Va. are in the 10 -11 % range, where the geared lokies you listed all operate, if I recall correctly.

- Paul North.

[Deggesty]

tree68

The Appalachians are simply older and more experienced.    By and large the western mountains are just whippersnappers (geologically speaking).

Larry, I like those statements. All my really serious mountain road driving has been in Appalachia. Last year, on our way to Bristol, I drove on a section of US 58 , near Whitetop Mountain, that I had not been on since 1954--and I had forgotten how it is. Ricki did not mind the left hand curves, but the right hand ones bothered her. If we remember that if West Virginia were ironed out, it would be bigger than Texas, we can realize that there are serious mountains and hollows there--Webster Springs, for example, is nestled right down in a ring of mountains.

tree68

A two percent hill is still a two percent hill, whether it's in the Rockies or the gentle hills of central Illinois...

Yes, even though the force of gravity varies with latitude and altitude. The greater the altitude, the less the force will be, and the higher the latitude, the greater the force will be.

Incidentally, have you now learned to be careful about drinking tea (or anything else) when working at your computer? Never look at the screen with liquid in your mouth.

Johnny

[erikem]

Deggesty

erikem

P.S. Someone actually thinks the Appalachians are mountains????

Yes, consider what the Western North Carolina did to get up to Ridgecrest from Old Fort, what the Spartanburg and Asheville did to get up to Tryon from Spartanburg, and what the Carolina, Clincfield and Ohio did to get down to Marion, N. C., from Erwin, Tenn.

Johnny

 

I presume you did note the smiley????

IIRC, the steepest grade on any Class 1 was the Madison, IN branch on the PRR coming in at over 5%, and while Indiana isn't as flat as Florida, it is still a pretty flat state.

One reason for the jibe at the Appalachians is that they are about the same height as the coastal range in San Diego County (tallest being somewhat over 6,000'), which is dwarfed by the mountains a few miles north in Riverside and San Bernardino counties, which in turn are dwarfed by the Sierras. Note that only one RR was built through that range and that passes through some spectacular (albeit dry) scenery. Having spent many a business trip in Magna, the chunks of granite around your neck of the woods aren't particularly known for their lack of stature.

- Erik 

[zardoz]

Deggesty

Yes, even though the force of gravity varies with latitude and altitude. The greater the altitude, the less the force will be, and the higher the latitude, the greater the force will be.

While I am aware that there is a measureable difference between the gravitational attraction to an object at the surface of the Earth compared with the attraction to an object that is above the surface (further away from the center of gravitational attraction), I believe the difference is so small as to be irrelevant pertaining to the case at hand (the effect on trains).

However, I am unaware of any studies indicating a difference of gravitational attraction due to latitude (unless you are refering to the value of the Coriolis parameter, which does vary with latitude, and that dependence is due to the Earth's shape).

[timz]

zardoz

I am unaware of any studies indicating a difference of gravitational attraction due to latitude

It depends what you mean by gravity. If you put an object on a scale at the equator, the scale will read about 0.5% less than it would at the pole.

[Deggesty]

zardoz, timz, yes, there are studies and even tables that show the differences in acceleration that come from different locations. PDN continually uses the figure 32.2 feet per second per second as the acceleration due to gravity. I thought back to what I remembered from the constant I used when I was taking the basic college course in physics (fifty-three years ago), and I thought that it was merely 32 feet per second per second. So, I pulled my 1954-1955 Chemical Rubber Handbook of Chemistry and Physics out, and found several tables. Of course, most of them are expressed in centimeters per second per second. One shows the acceleration (g) observed at many places in our country, Canada, and many other places around the world. The latitude, longitude, and elevation (in meters) is given for each location. Another one gives the correction for g as altitude increases, indicating how much the acceleration is reduced as you rise.

timz

zardoz

I am unaware of any studies indicating a difference of gravitational attraction due to latitude

It depends what you mean by gravity. If you put an object on a scale at the equator, the scale will read about 0.5% less than it would at the pole.

Yes, you are closer to the earth's core at a pole than you are at the equator, so, even though your mass has not changed, you weigh more at a pole. Advice to those concerned about too much weight: move to the equator?

There is also a table showing the length of a simple pendulum with a period of two seconds. It shows that at 0 degrees latitude, your pendulum will be 39.0141 inches long and at 90 degrees latitude, your pendulum will be 39.2207 inches long--if your keep your altitude at sea level. This is a difference of 0.527% from the pole to the equator (zardoz, do I have too many sig figs?).

Johnny

[Modelcar]

Deggesty

Last year, on our way to Bristol, I drove on a section of US 58 , near Whitetop Mountain, that I had not been on since 1954--and I had forgotten how it is

 

....Sounds like that route would be comparable to route 50 across WV.

[blue streak 1]

what about the spinning earth's centrifigual force at the equator? None at the poles. Most rockets at lauched to the east to help speed them up. Wonder if the spinning force adds a force vector outwards? Gravity would be same whether spinning or no spin. That's one reason that most sci-fi has people on small asteriods that do not spin.

[wjstix]

Dakguy201

Railway Man

It's a true percentage -- it compares one number as a percent of another number, in this case the rise (in any unit of measurement) to the run.  For example, if a given track has 1' rise in 100' of run, it would be a 1.0% grade (1 = 1% of 100), a 2' rise in 100' = a 2% grade, and so on.  A 100% grade -- a track that rises 100 feet in 100 feet of run, would be at a 45 degree from the horizontal.

Sorry, I think I may be confused.  By "100 feet of run" I take you to mean 100 feet of track.   Is "run" the distance that the track would cover if it were level ground?  

Yes you're measuring 100 feet horizontally, just as if the track were on flat ground. How ever many feet up the track rises over the 100 feet would be the grade in pct.

Dakguy201

However, if that is indeed what is intended, then the angle is not 45 degrees. 

Well, then what would the pct. of grade be?? Seems to me if you went up 100 feet for every 100 linear feet, it would have to a 45 degree angle. Zero degrees would be flat, 90 degrees would be straight up. Halfway between would be 45??

[timz]

Deggesty

you are closer to the earth's core at a pole than you are at the equator, so, even though your mass has not changed, you weigh more at a pole.

What if the earth's ellipticity were a hundred times greater than it is? Would your weight at the pole (the center of the pancake-shaped earth) be more than at the equator (the rim of the pancake)?

I think we'll do better with

blue streak 1

what about the spinning earth's centrifigual force at the equator? None at the poles.

[Deggesty]

timz

What if the earth's ellipticity were a hundred times greater than it is? Would your weight at the pole (the center of the pancake-shaped earth) be more than at the equator (the rim of the pancake)?

Oh, yes!

Johnny

[tree68]

wjstix

Well, then what would the pct. of grade be?? Seems to me if you went up 100 feet for every 100 linear feet, it would have to a 45 degree angle. Zero degrees would be flat, 90 degrees would be straight up. Halfway between would be 45??

Exactly. 

[Paul_D_North_Jr]

wjstix

Dakguy201

Railway Man

It's a true percentage -- it compares one number as a percent of another number, in this case the rise (in any unit of measurement) to the run.  For example, if a given track has 1' rise in 100' of run, it would be a 1.0% grade (1 = 1% of 100), a 2' rise in 100' = a 2% grade, and so on.  A 100% grade -- a track that rises 100 feet in 100 feet of run, would be at a 45 degree from the horizontal.

Sorry, I think I may be confused.  By "100 feet of run" I take you to mean 100 feet of track.   Is "run" the distance that the track would cover if it were level ground?  

Yes you're measuring 100 feet horizontally, just as if the track were on flat ground. How ever many feet up the track rises over the 100 feet would be the grade in pct.

Dakguy201

However, if that is indeed what is intended, then the angle is not 45 degrees. 

Well, then what would the pct. of grade be?? Seems to me if you went up 100 feet for every 100 linear feet, it would have to a 45 degree angle. Zero degrees would be flat, 90 degrees would be straight up. Halfway between would be 45??  

Stix -

Here's the original question as posed by Dakguy201 on Page 1 of 3 of this thread on 02-24-2009 at 8:31 AM: 

"Sorry, I think I may be confused.  By "100 feet of run" I take you to mean 100 feet of track.  However, if that is indeed what is intended, then the angle is not 45 degrees.  Is "run" the distance that the track would cover if it were level ground?"

As you can see, somehow that became "parsed" into the 2 separate segments that you quote above, and that's probably what is leading to the confusion here.  Your understanding as set forth in your last question above is indeed the correct understanding of how percent of grade is defined and calculated.

Dakguy201 was intending to ask the usual question about how the 100 ft. of "run" is measured - either as the "run" on the level, or else as a distance on the slope/ grade, only he called that "on the slope/ grade" measurement "100 feet of track" instead.  That made sense from his perspective then, because the track would have been angled up on the actual slope or grade, but that wording of the question may be confusing to those who are not accustomed to the nuances of this.  But anyway, that's not how it's done - the 100 feet is measured as "run" on the level, as we all know now. 

So, his next question that you also quote above -  "However, if that is indeed what is intended" - is still based on that mistaken premise, that the 100 ft. is a distance that is measured on the slope.  Well, that's not how it is intended.  Also, note that unstated in his question is that he is also contemplating a 100 ft. rise - that's the only way we get anywhere near to a 45-degree angle discussion.  (But if it were intended that way, then Dakguy201's unknowingly mistaken mental image there is sound - the 100 ft. distance on the slope would be the hypotenuse of the right-angle [90-degree] triangle.)  Unfortunately, though, that then leads us to the same place that this discussion is going - no place, or undefined*, because:  The only triangle-type shape that has a 100 ft. rise with a 100 ft. distance on the slope or hypotenuse is a vertical line with an angle of 90-degrees - those lines are the same length, and hence have to be coincident = on top of each other.  This can be confirmed from a trigonometric perspective by realizing that the arc-sine or inverse sine function - for the angle where the hypotenuse is 100 and the opposite side (rise) is also 100 - is the arc-sine of 100/100 = arcsine of 1.00 = 90 degrees.

* For a 90-degree angle, the "run" is 0, hence the "rise over run" calculation devolves to dividing the rise by 0, which is usually understood to result in an "undefined number".

I hope this answers your question, and that everyone else is thoroughly confused now. 

- Paul North.

[henry6]

Paul_D_North_Jr

wjstix

Dakguy201

Railway Man

It's a true percentage -- it compares one number as a percent of another number, in this case the rise (in any unit of measurement) to the run.  For example, if a given track has 1' rise in 100' of run, it would be a 1.0% grade (1 = 1% of 100), a 2' rise in 100' = a 2% grade, and so on.  A 100% grade -- a track that rises 100 feet in 100 feet of run, would be at a 45 degree from the horizontal.

Sorry, I think I may be confused.  By "100 feet of run" I take you to mean 100 feet of track.   Is "run" the distance that the track would cover if it were level ground?  

Yes you're measuring 100 feet horizontally, just as if the track were on flat ground. How ever many feet up the track rises over the 100 feet would be the grade in pct.

Dakguy201

However, if that is indeed what is intended, then the angle is not 45 degrees. 

Well, then what would the pct. of grade be?? Seems to me if you went up 100 feet for every 100 linear feet, it would have to a 45 degree angle. Zero degrees would be flat, 90 degrees would be straight up. Halfway between would be 45??  

Stix -

Here's the original question as posed by Dakguy201 on Page 1 of 3 of this thread on 02-24-2009 at 8:31 AM: 

"Sorry, I think I may be confused.  By "100 feet of run" I take you to mean 100 feet of track.  However, if that is indeed what is intended, then the angle is not 45 degrees.  Is "run" the distance that the track would cover if it were level ground?"

As you can see, somehow that became "parsed" into the 2 separate segments that you quote above, and that's probably what is leading to the confusion here.  Your understanding as set forth in your last question above is indeed the correct understanding of how percent of grade is defined and calculated.

Dakguy201 was intending to ask the usual question about how the 100 ft. of "run" is measured - either as the "run" on the level, or else as a distance on the slope/ grade, only he called that "on the slope/ grade" measurement "100 feet of track" instead.  That made sense from his perspective then, because the track would have been angled up on the actual slope or grade, but that wording of the question may be confusing to those who are not accustomed to the nuances of this.  But anyway, that's not how it's done - the 100 feet is measured as "run" on the level, as we all know now. 

So, his next question that you also quote above -  "However, if that is indeed what is intended" - is still based on that mistaken premise, that the 100 ft. is a distance that is measured on the slope.  Well, that's not how it is intended.  Also, note that unstated in his question is that he is also contemplating a 100 ft. rise - that's the only way we get anywhere near to a 45-degree angle discussion.  (But if it were intended that way, then Dakguy201's unknowingly mistaken mental image there is sound - the 100 ft. distance on the slope would be the hypotenuse of the right-angle [90-degree] triangle.)  Unfortunately, though, that then leads us to the same place that this discussion is going - no place, or undefined*, because:  The only triangle-type shape that has a 100 ft. rise with a 100 ft. distance on the slope or hypotenuse is a vertical line with an angle of 90-degrees - those lines are the same length, and hence have to be coincident = on top of each other.  This can be confirmed from a trigonometric perspective by realizing that the arc-sine or inverse sine function - for the angle where the hypotenuse is 100 and the opposite side (rise) is also 100 - is the arc-sine of 100/100 = arcsine of 1.00 = 90 degrees.

* For a 90-degree angle, the "run" is 0, hence the "rise over run" calculation devolves to dividing the rise by 0, which is usually understood to result in an "undefined number".

I hope this answers your question, and that everyone else is thoroughly confused now. 

- Paul North.

 

So how much fuel is consumed by an SD70MAC on the Brazilian Ry vs. CP in Winnepeg or ARR in Denali?  And does this mean retainers have to be set differently on Stephen's Pass?  Really, I am overwhelmed buy the information and don't see how 0.5% weight difference from the Equator to the North Pole has any major impact on rail loadings and operations.  Or am I just that stupid?  (Go ahead, I'd deserve it if there is a real answer.)

[Paul_D_North_Jr]

erikem

I don't remember the exact percentage for aero drag as far as the total train resistance of the TGV, but do remember it was huge ( >90%). I would strongly suspect that the TGV would have to slow down a bit when hitting a 3% grade, but bear in mind that anything under a 1,000' change in elevation is pretty much a momentum grade. I would also guess that even with the slowdown, the loss in time going up a 3% grade is less than the slowdown due to curves and circuitry with a lesser grade.

- Erik  [snip of P.S. question about the Appalachians being mountains]

From another post above:

Seriously, any grade with a rise or fall of 100 ft. or more (= 2 miles at 1%, or 1 mile at 2%, etc.) is significant.  That is enough of a rise to essentially use up most or all of the momentum ("velocity head") of a typically powered train at speeds of up to 60 MPH, and that amount of fall has the potential to be a runaway hazard for an uncontrolled train (no brakes) to reach the same speed range.  The formula is pretty simple (below), and I hope to post some examples later today.

Velocity Head (ft.) = [Speed (MPH) x 1.47* ft./sec. per MPH)], squared / [64.4** ft. / second, squared]

*1.47 ft. sec. = (5,280 ft. / mile) / (3,600 secs. per hour = 60 mins. x 60 secs.)

**64.4 ft. / second, squared = 2 x G = Gravitational Constant = 32.2 ft. per second, squared.

For a TGV at 150 MPH, from the above formula the "Velocity Head" = [ 150 x 1.47 ] squared / 64.4 = 755 feet.  What this means is that if the engineer shut off the throttle just as a TGV started up a grade at 150 MPH - and there was no resistance from rolling or aero drag - it would coast on up the hill, all the while trading speed for height, until it slowly came to rest at a point 755 feet higher than where it started.  Since a 3 % grade is about 150 feet per mile, that would mean coasting about 5 miles up such a grade.  Of course, in the real world there are those resistances that were ignored above - and the engineer would also not shut off the throttle, but instead would leave it wide open so as to keep the speed up as much as possible.  So as Erik points out, the TGV would gradually slow down some until it reaches a new equilibrium or "balancing speed", where the power consumed by the grade + rolling resistance + aero drag has risen to where it just equals the power output of the train's traction motors.

For 180 MPH, the comparable Velocity Head figure is 1,087 feet, which supports and confirms Erik's "anything under a 1,000' change in elevation is pretty much a momentum grade" point above.

For some other common speeds:

10 MPH - Velocity Head = 3.4 ft.  Note:  This - and the next one below - is why it doesn't take too much of a rise to keep a car from rolling out of a "bowl track" in a hump yard or other locations, as long as it isn't being pushed along by wind or have some other source of momentum, such as being "kicked in" from one direction or the other.

20 MPH - Velocity Head = 13.4 ft.

30 MPH - Velocity Head = 30 ft.

40 MPH - Velocity Head = 54 ft.

50 MPH - Velocity Head = 85 ft.

60 MPH - Velocity Head = 121 ft.

70 MPH - Velocity Head = 164 ft.

80 MPH - Velocity Head = 215 ft.

90 MPH - Velocity Head = 272 ft.

100 MPH - Velocity Head = 336 ft.

One more point for those of you familiar with the Tri-Met's "MAX" LRV system in the Portland, Oregon metro area:  When I was last out there a couple years ago, at the "Gateway" / NE 99th Ave. station* on the "Blue Line" route to Gresham, there sure seemed to be a pretty good grade on the "Red Line" Airport track that does a high U-turn around from the west into the station itself to join up with the Blue Line there - I'm guessing at about an 8 % grade, though maybe the contra grade on the roadway below made the track grade look steeper than it really is.  I wasn't able to get a photo illustrating that at the time, though, and haven't been able to find anything on-line that addresses that.  Is my recollection anywhere near correct ?  Does anyone know of an on-line photo of that ?  I'd appreciate any info on this.  Thanks !

* - On the northeasterly side of the area, a couple miles southeast of the Airport, where I-205 - Veterans Memorial Highway connects with I-84/ U.S. 30 - Banfield Expressway, specifically where the MAX Airport line crosses over the ramps from NB I-205 and NE Glisan St. to WB I-84/ U.S. 30, etc.

- Paul North.

[blue streak 1]

Paul_D_North_Jr

One more point for those of you familiar with the Tri-Met's "MAX" LRV system in the Portland, Oregon metro area:  When I was last out there a couple years ago, at the "Gateway" / NE 99th Ave. station* on the "Blue Line" route to Gresham, there sure seemed to be a pretty good grade on the "Red Line" Airport track that does a high U-turn around from the west into the station itself to join up with the Blue Line there - I'm guessing at about an 8 % grade, though maybe the contra grade on the roadway below made the track grade look steeper than it really is

Paul: Also the mid - grade station (forgot name) confused me the last time I rode it. Was surprized that the dynamic/regenerative braking was so effective.