• Pennsy S2 Turbine Drivers (wheels)

  • Discussion of steam locomotives from all manufacturers and railroads
Discussion of steam locomotives from all manufacturers and railroads

Moderators: slide rules, Typewriters

  by Allen Hazen
 
Mtu Andrew--
In your second-to-last post you say
"There is a finite (if small) amount of play in this system. The bearings have a greater-than-zero clearance, and even the steel can compress ever-so-slightly.
However, even if the S2 had connecting rods exactly 180 degrees opposite, with no play or compressibility in the drive train whatsoever, it wouldn't matter in this case."

The only thing I'd disagree with is the "even". With no play or compressibility, the rod would have to transmit vertical forces just as efficiently as horizontal ones. With the play that a real system has, I think the system will be a bit less efficient at transmitting vertical forces: what it would have to do if the pins on opposite sides were 180 degrees apart and the locomotive was starting from a dead stop with the rods in mid position (at same height as axle centers). But, as I think I have already said, I don't know how significant this would be.
  by jgallaway81
 
With the concern of the rods being in the "non-productive" position of being in line with the axles, I think, ironically, that would in fact make the 180* position a more favorable one over quartering.

In a quartered system, the turbine would cause axles 2 & 3 to turn, forcing axles one and four via the rods, one in a vertical motion, the other side in a horizontal motion. This would actually put an uneven torque on the axle, in a vertical axis at the centerline of the engine (parallel to the smokestack, at the centerline of the axle). This would cause premature & uneven wear on the driveboxes, liners, pedestals etc.

In a 180* alignment, even if both rods were perfectly inline with the axle, the fact that axles 2 & 3 are geared direct would cause the rods to place an up & down force on the axle, providing a weaker, but more uniform torque to the axle. I would expect to see a force diagram in the shape of a sine wave, but with the entire wave form in the positive section of the chart, demonstrating a given amount of torque, which would weaken as the force left a perpendicular direction and then strengthened again as the force returned to parallel to the rail.

It would be interesting to see if there was any indication of the engine having unusually high repairs required to the drive boxes & other related parts.
  by timz
 
I woulda thought the light bulb would have snapped on over somebody's head by now.

Look at one side of the engine, stopped with the rods on that side at 12 o'clock. The turbine applies torque to the #2 and #3 driving axles, but (for whatever reason) not enough to move the engine. Say the torque is enough to produce 10,000 pounds of tension in the side rod between #3 and #4 axles.

Now say the rods are at 2 o'clock instead of 12 o'clock, 60 degrees away from their original position. The turbine is applying the same torque as it was before to axles 2 and 3, and the same torque as before is being transmitted to axle 4. The tension in the side rod has to be 20,000 lb.

You can see where this is going. If the rods are 89 degrees right of the 12 o'clock position, still with the turbine applying unchanged torque to axles 2 and 3, the tension in the side rod will have to be 573,000 lb to transmit the unchanged torque to axle 4.
mtuandrew wrote:Even if that force is perpendicular to the connecting rod's axis [as it is when the rods are level with the axles], the rod is still being driven in a particular direction at any given instant - in this case, straight up or down - and it transfers that force to any additional coupled wheels.
The end of the rod is being driven sideways. The other end of the rod isn't being driven anywhere, unless the rod can cope with infinite tension/compression.
  by Eliphaz
 
the force vector on the #4 pin is always tangent to the rim of the wheel, not parallel to the rod.
at 12 o'clock, the force in the rod is pure tension. At 90 degrees from top dead center the force on the pin is entirely downward, perpendicular to the rod, producing a bending moment at the #3 pin, but tension in the rod at that position is nil.
  by mtuandrew
 
timz wrote:I woulda thought the light bulb would have snapped on over somebody's head by now.
mtuandrew wrote:Even if that force is perpendicular to the connecting rod's axis [as it is when the rods are level with the axles], the rod is still being driven in a particular direction at any given instant - in this case, straight up or down - and it transfers that force to any additional coupled wheels.
The end of the rod is being driven sideways. The other end of the rod isn't being driven anywhere, unless the rod can cope with infinite tension/compression.
Ok, I think I see where you're going with your argument - a piston rod with the pin at the 12 o'clock position will continue to build tension exponentially. However, you're doing your math wrong - the force exerted along the connecting rod's axis decreases as it rotates from 12 o'clock. The right to left force reaches zero (minus frictional losses) at 9 o'clock, since the force is being directed straight downward. Eliphaz explains this better than I do:
Eliphaz wrote:the force vector on the #4 pin is always tangent to the rim of the wheel, not parallel to the rod.
at 12 o'clock, the force in the rod is pure tension. At 90 degrees from top dead center the force on the pin is entirely downward, perpendicular to the rod, producing a bending moment at the #3 pin, but tension in the rod at that position is nil.
timz, I think your flaw is that you assume that one wheel can begin to rotate while the others stay stationary, thereby stretching the connecting rod. The wheels don't work like that, since they're rigidly coupled together just as if they were chained or geared into one unit (like the bike example mentioned above.) Besides, the maximum amount of tension that can be placed on the connecting rod is equal to the maximum power created at one axle, times the length of the lever arm (the distance from the driven axle to the rod pin.)
  by timz
 
Eliphaz wrote:At 90 degrees from top dead center the force on the pin is entirely downward, perpendicular to the rod, producing a bending moment at the #3 pin, but tension in the rod at that position is nil.
mtuandrew wrote:the force exerted along the connecting rod's axis decreases as it rotates from 12 o'clock. The right to left force reaches zero (minus frictional losses) at 9 o'clock, since the force is being directed straight downward.
A rod on a frictionless pin can't transmit a force perpendicular to its length, of course, but probably no use trying to explain that. But if you understand what torque is (force times arm length) you ought to realize the tension in the rod increases (not decreases) as the rods move from 12 o'clock toward 3 o'clock.

Say the engine is stationary, trying but failing to start, with the rods 89.9999 degrees away from 12 o'clock. You guys figure the tension in the rod is an ounce or two. We tie a length of thread between the two pins. Once the thread has taken up the one-ounce tension can we remove the rod, and the engine will continue to pull as hard as ever?
  by Eliphaz
 
A rod on a frictionless pin can't transmit a force perpendicular to its length
care to site a refferance to support that statement?

the thread analogy doesnt apply.
at 90 degrees, the movement of the rod is downward, perpendicular to its length, in fact. it is being impelled in that direction by the pins on the wheels 2 and 3.
In that position the rod is a cantilever, not a string.
  by Desertdweller
 
It sounds to me like the direct-drive turbine suffers from the same problem as rod-driven steam locomotives: lack of starting TE. The turbine needs a controlled-slip coupling of some sort to enable it to spool up. Otherwise, the engine is dependent on track speed to get into its power curve. Once it gets a train into motion, it can run with it at speed like any rod-driven steam locomotive.

For those old enough to remember, this was the same problem encountered by the Chrysler Turbine Cars tested in the mid-60's. Poor low-end torque, and massive fuel consumption at low speeds.

Marine steam turbine applications (those involving geared turbines) solved the problem by using multiple turbines. Two or three turbines for forward motion, using dedicated turbines for low, cruising, and high speeds. An additional turbine for reverse.

For locomotives, I would think a turbo-electric drive would be much better. Some ships used this to reduce mechanical complications over geared turbines. One turbine sufficed for all speed ranges and reverse.

With a generator driving axle-hung traction motors, the turbine could operate in its optimum speed range. If there was still a problem with starting effort, the drive could be augmented like in an ES-series locomotive by using a battery bank that could give a boost to the motors. This would also allow dynamic braking.

If a 21st-Century Jawn Henry were built, using these features, I suspect it could be operationally successful. The biggest hurdle would probably be the use of non-standard technology.

Les
  by timz
 
Eliphaz wrote: at 90 degrees, the movement of the rod is downward, perpendicular to its length, in fact. it is being impelled in that direction by the pins on the wheels 2 and 3.
In that position the rod is a cantilever, not a string.
How about at 60 degrees?

Say the rods are back at 12 o'clock and the engine is stopped. Remove the middle section of the rod and replace it with a fat cable-- fat enough to handle the tension needed, but flexible, so you can't compress it and you can't cantilever it. The pin connections are still frictionless, and total length of the rod-cable combination is the same as the rod it replaced. We all agree that as long as the rods are at 12 o'clock the cable will work fine?

Now the engine is stationary and trying to start with the rods at 2 o'clock, with the cable still in place instead of the rod. What will happen?

(Answer: no change-- the cable works fine. And the cable would continue to work fine with the rods at 89.999 degrees, if it could handle the enormous tension that you don't think exists. No need for any cantilever.)
  by mtuandrew
 
timz wrote:
Eliphaz wrote: at 90 degrees, the movement of the rod is downward, perpendicular to its length, in fact. it is being impelled in that direction by the pins on the wheels 2 and 3.
In that position the rod is a cantilever, not a string.
How about at 60 degrees?

Say the rods are back at 12 o'clock and the engine is stopped. Remove the middle section of the rod and replace it with a fat cable-- fat enough to handle the tension needed, but flexible, so you can't compress it and you can't cantilever it. The pin connections are still frictionless, and total length of the rod-cable combination is the same as the rod it replaced. We all agree that as long as the rods are at 12 o'clock the cable will work fine?

Now the engine is stationary and trying to start with the rods at 2 o'clock, with the cable still in place instead of the rod. What will happen?

(Answer: no change-- the cable works fine. And the cable would continue to work fine with the rods at 89.999 degrees, if it could handle the enormous tension that you don't think exists. No need for any cantilever.)
You're right that at 89.999 degrees the tension or compression is not zero, which is what would allow a cable to act as a connecting rod for half of a wheelset's rotation. However, the tensile force approaches zero as the rod pin (on the driving wheel) rotates toward 90 degrees, reaching zero at 90 degrees and then steadily gaining in compression as the rod pin moves towards 180 degrees.

timz, I highly encourage you to take a statics class, or talk to an engineer about moment couples (the exertion of force around a center pivot) and their impact on a connected rod constrained in its range of motion. I don't say that to be mean, but I've taken a few college classes on statics and mechanics of motion, and feel that Mr. Hazen, Eliphaz and I have a better grasp (if imperfect) of the situation. I feel as though you might be able to better understand the concept if a qualified person was walking you through this explanation, rather than just posting it on a message board. Good luck.
  by timz
 
mtuandrew wrote:I've taken a few college classes on statics and mechanics of motion
Great-- then you should be able to understand the situation once you think about it a bit. You figure you do know what torque is? Force times the length of the lever arm-- right?

So say the S2's rods are succeeding in transmitting constant torque from driver #3 to driver #4, and when the rods are at 12 o'clock the tension in the rod is 10000 lb. When the rods are at 2 o'clock, still transmitting the same torque, the arm is half as long as it was at 12 o'clock so the force along the rod has to be double what it was.

Once you realize what "the lever arm" is, the scales will drop from your eyes. If you don't know what it is then you don't know what torque is, and we'll have to try a different approach.
  by Eliphaz
 
The "lever arm" is the segment of the wheel including the axle and drive pin.
lets consider a force triangle (google it) for this system:
the lever is marked in red, the force vector, which I have tried to draw of a constant lenght, is the thick blue line, perpendicular to the lever arm.
the "tension" component of the transmitted force is the thin horizontal line parallel to the rod.
the downward component of the transmitted force is the vertical thin blue line.
as the wheel rotates the force transitions from horizonal to verticle, and the relative magnetudes of the two componants are represented by the sides of the triangles.
attached, my crappy PAINT sketch

when you see the lever at 89 degrees, keep thinking of the little kid standing up on the peddle. no wire being pulled, rod pushing straight down on the end of the lever arm.
You do not have the required permissions to view the files attached to this post.
  by timz
 
I assume when you drew the siderods as one continuous rigid rod from driver #1 to driver #4, you just weren't taking the time to draw it correctly. We all agree the rods are only rigid from each driver to the next-- right? In our discussion let's just consider one pair of drivers and the one rod between them.

Your problem is you still imagine the rods can be cantilevers-- you still think they can exert a force perpendicular to their length, which they can't. What's your answer to the question I asked before? What will happen when the rods are at 2:00-- will the cable work or not?

"How about at 60 degrees?

"Say the rods are back at 12 o'clock and the engine is stopped. Remove the middle section of the rod and replace it with a fat cable-- fat enough to handle the tension needed, but flexible, so you can't compress it and you can't cantilever it. The pin connections are still frictionless, and total length of the rod-cable combination is the same as the rod it replaced. We all agree that as long as the rods are at 12 o'clock the cable will work fine?

"Now the engine is stationary and trying to start with the rods at 2 o'clock, with the cable still in place instead of the rod. What will happen?

"(Answer: no change-- the cable works fine. And the cable would continue to work fine with the rods at 89.999 degrees, if it could handle the enormous tension that you don't think exists. No need for any cantilever.)"
  by Eliphaz
 
OOPS !Image no, youre right Tim, I have been describing a single solid bar extending across all four wheels from the very beginning. It takes a picture some times.
in my experience with stationary equipment a machine like a rotary screen would be set up just like that with bearings precisely located and rigidly mounted, and a single bar driving the whole train, but on a locomotive each connector must be a separate link to allow the wheels to conform to the road. the four axles are connected by three independant links, there is no cantilver, and no downward force component. however the tensile force in each rod does drop to zero at 90 degrees. the applied force is a sine wave and passes through zero at 90 degrees:
the two sides must be quartered because there are dead centers at 90 and 270 degrees on each side.

Good on ya Tim, for sticking to your guns. you caught me here:
A rod on a frictionless pin can't transmit a force perpendicular to its length, of course, but probably no use trying to explain that.
but then didnt succeed in explaining un til I drew a picture.
and the whole time I had been deliberately ignoring a two-axle configuration, because the single rod four-axle had caught my fancy.

I come here to learn, as well as, I hope, to contribute once in a while. this has been a very amusing thread.
  by timz
 
Next question: what's the objection to having a continuous rigid rod across four drivers? We all know conventional steam locomotives didn't do that, but truth be told I don't know for certain the S2 didn't do that. Anybody think of a reason such a plan could make sense on a turbine engine when it doesn't on a reciprocating? Or is it out of the question for both?