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The load is not purely inductive, because the wiring inside the motor is not super-conducting. Any current that runs through any wire that is not super-conducting will produce a certain amount of heat (the square of the current, times the resistance). This is true regardless of whether the current is AC or DC, if the wire is in an inductor or not, etc. The motor has about 7 Ohms of resistance. So the heat (in Watts) that this produces is the square of the current, times 7 Ohm. If you double the current, the number of Watts dissipated as heat in this way goes up a factor 4. This is why, when the current is applied unevenly (but with the average staying the same) then the total number of Watts dissipated as heat in the motor will have to be more than when the current is applied evenly. If the PF system has some components to smooth out the current through the motor, then that should increase efficiency. I do not know to what extend the PF system does that, I have no oscilloscope so I can't see (the train motor itself contains of course a small capacitor, that's of course already helpful. If you put a bigger one in there, you should get less of that high-pitch sound). One way to check the PF efficiency is to see if the batteries last longer on low speed settings than on high speed settings, and if so, by how much.

I have only one BNSF engine and one SF engine, but my Maersk train is headed by 2 Maersk engines, so that makes it a bit of an unfair comparison. In any case, if I put these trains next to each other, Maersk wins. I made a few minor changes to the Maersk engine. The line between black and grey, near the top of the engine, right before the two side plates this line goes up by 2 plates. That little bit of gray should be black because then the lines are smoother. Esthetically that was the only change that I made. This very simple change makes the train look noticeably better I think. The second Maersk engine in my train has no motors in it. I felt it was necessary to apply some lubrication, otherwise the 6-length axles had way too much friction for the train to run smoothly. Following the advice from someone else on this forum, I also replaced the wheels on the second engine by wheels that have much less traction (this makes it go more easily through the curves). In any case, this train runs very well now, and it's an awesome sight to see all the air vents of the two engines line up, and the side railings, this is a very nice looking train.

For my trains, I found it sufficient to add just 1 extra stud between 12V straight tracks. That's a minor modification of the electric level crossing (only need a few extra 1x8 gray tiles). When one track is curving and the other is straight, then I had to add 2 extra studs in between to avoid collisions. The track has enough flexibility to shift it over by 1 stud without making any real changes in the track layout. So I can design the track with the usual track setup software, and then build it 1 or 2 studs shifted.

Do others here see it the same way?

I have two resistors, 10 Ohm each. One of them I subject to 4.5 Volt continuously. The other one gets 0V for 0.01 seconds, then 9V for 0.01 seconds, then 0V, 9V, 0V, 9V, .. etc. So the first one is "on all the time" at 4.5 V. The second one is "on half of the time" but when it is on, it gets 9V. Question: which one will get hotter? (note: the internal resistance of the lego train motor is roughly 10 Ohm if I remember correctly, or perhaps a little bit less. I can measure it again if you want a more precise number).

You just drove up the price! (cheapest one now is $59.99) (aargh!).

I think it is fair to say that I've spent too many messages on this topic already. Since I already went way overboard on this topic, why not do some more. I was trying to decide whether I should run my cargo train with 9V or with PF (it is a 10 feet train, with 2 Maersk engines followed by 8 light-weight cars (some Maersk with 1 container, some MOC). This train has two 9V motors, and when running at a constant speed, it consumes 0.42 amp (i.e. 0.21 amp per motor). This number is pretty much the same for any speed setting that I've tried. Regardless what speed it runs, as long as you keep the speed constant, then it uses 0.42 amp. This test was done with the 9V system. I want to run it at speed setting #2, which is just above 4 volts. Now I wanted to know how long the batteries would last if I turned this into a PF train. So I computed it, in two ways: Method 1 counts Amps: I'm using 0.42 amp, or 420 mA. If I use 1100 mAh batteries, then it is a simple computation: 1100 mAh divided by 420 mA, that's 2.62 hours. So that's how long the train should run before the batteries run out. Method 2 counts energy: I'm using 0.42 amp at 4 volts, so that's 0.42 * 4 = 1.68 Watt. The battery pack contains 9V * 1100 mAh of energy, which is 9.9 Watt-hour, so 9.9/1.68 is 5.9 hours. Here is the puzzle: I computed the same thing in two ways, and get a different number. If I use amps to compute the battery life, then I get 2.62 hours. But if I use Watts, and the energy content of the batteries, then I get 5.9 hours of computed running time (assuming 100% efficiency for the PF system). I know that at least one of these computations must be wrong. But it's not so easy to understand why this is so. The one thing I know for sure is that every second, you have to get 0.42 Coulomb through the two motors. If you do not, the train won't have enough torque to overcome friction. This matches not only observations, but also makes perfect sense if you view the motor through the laws of motion and electromagnetism. If the PF system were 100% energy efficient then this train really has to run for 5.9 hours. Of course you never get 100%, but if the train ran for say 5 hours, then that would mean that the system has a good efficiency. On the other hand, if the efficiency was no better than simply wasting away the unwanted voltage in a resistor, then we should get 2.62 hours of running time. Initially, that's what I thought would happen. But there is a clear consensus that the efficiency is better than that. That means that method 1 is wrong. But method 1 is a simple computation, and it's not so easy to understand why it is wrong. The people that told me that I'm wrong are probably right, but have not given a consistent explanation (a few self-contradictory explanations have been given though). Of course, no explanation trumps observation. The only way that would really settle this is to just put in a fully charged battery pack and just let the train run until it runs out.

To run a typical lego train, you need about 200 mA, or 0.2 A, or 0.2 Coulomb per second. So for each Coulomb, it can run about 5 seconds. At the top speed setting, the motor is basically connected straight to the batteries, so for every Coulomb the motor uses, one Coulomb is taken out of the battery pack. If the batteries are 1000 mAh, that's 1000 mC/s * h = 1 C/s * h = 3600 C. So the train should run for about 5 * 3600 seconds, i.e. 5 hours. I think that for a light-weight train, this is pretty accurate. If you run the train now at a low speed, say speed setting #2, then it will still use 0.2 A so you get 5 seconds of running for each Coulomb. When the motor used one Coulomb at a slow speed setting, say equivalent to 4 volts, then it consumed 4 Joule's of energy, during 5 seconds of time. The battery pack is at 9 volts, so 4 Joule's of energy corresponds to 4/9 Coulombs of electricity at 9V. This 4/9 Coulomb is roughly 0.5 C, minus some losses. Now during those 5 seconds, how many Coulombs were taken out of the battery pack? (A) If it is 1 Coulomb, then that's 9 Joules, meaning that 5 Joules must have been wasted somewhere. (B) If it is 0.5 Coulomb, then (except for minor losses) no energy was wasted. Intially I thought option (A). But the consensus here is pretty solidly against option (A). I'll have to assume then that option (A) is wrong (if option (A) were true, then the motor should get much hotter than what we saw in the 9V system. But we have not observed that). The only other option is that, at a low speed setting, for every Coulomb that the motor consumes, less than one Coulomb is taken out of the battery pack. I don't really see a way around that. Whether the speed is high or low, either way you use 1 Coulomb for every 5 seconds. There are only 3600 C in the battery pack. The only way you could avoid wasting lots of energy at a low speed setting, is to take less than 1 C out of the battery pack for every 1 C that flows through the motor. Conversion AC --> AC is much simpler to understand. With AC -> AC it is clear that every Coulomb at 4.5V costs only 1/2 Coulomb at 9V (with some minor losses). With DC -> DC this is much more complicated to understand, but if little or no energy is wasted (which is the clear consensus) then something similar must be going on. Look, I'm just trying to understand what's happening. I value your explanations and those of others, regardless of qualifications.

On the 6 length axle, you have wheel, technic brick, 2 spots open, then a technic brick, and again a wheel. On the open spots, you put technic bushes. Use the 1/2 width technic bush (item 4265) because they hold tighter than the 1-width bush (item 3713). Fill up all the room you have left on that axle (an axle with no gear will have room for 4 of them. That will definitely prevent the axle from sliding one way or the other!).

I think I wrote the same thing but the other way around. In any case, current and torque are almost linearly related to each other, and to produce the torque needed to move the train, you have to run at least a certain amount of current through it. The capacitor and inductor must have a major impact on the energy efficiency. If they weren't there, the high-pitch humming sound that the PF train motor makes would be much louder, and you can't make sound without spending energy. I've been drawing diagrams as to how this might work, and I think that you need a diode too. Both an inductor, and a capacitor, can store a small amount of energy. I think that's the answer to the "where does the excess energy go" puzzle. During on "on" cycle, when 9V is coming from the batteries even though we want to run the motor at 4.5V, some of this excess energy is moved into the capacitor and the inductor. Then, during the "off" cycle, the current to the motor does not drop to 0. Instead, the motor still receives current (and hence, it still produces some torque). This current comes out of the capacitor and inductor. Because they are small, the capacitor/inductor quickly run out of energy, so to prevent the current from dropping too far, the on/off switching needs to be done quickly (hence the high pitch).

That's correct. The main difference between the RC motor and the PF motor is that the PF motor contains a DC motor of much better quality. So it is stronger, and more energy efficient at the same time. It will probably last much longer too. By putting in the cheapest motor they could find into the RC train, and putting a better motor into the PF train, they can now correctly claim to have improved the motor. Some time ago I had two 12V trains running on the same loop, and I remember seeing something interesting: Every time when one train goes through a curve, then *both* trains slow down. What happens is that when one train goes through a curve, then it starts to draw more mA. This extra draw of current causes the voltage on the track to decrease. This, in turn, causes both trains to slow down (the effect is quite significant with the 12V system. It is less obvious with the 9V system because the 9V controller has a regulated output voltage). In essence, the reason that the train slows down in the curve is not simply "it has to do more work, so that's why it slows down". When a train goes through a curve, the motor starts pulling more amps, that causes the voltage on the motor to decrease, and that decrease in voltage explains most of the decrease in speed. With the 12V system you can see this very clearly (like I said, when one train goes through the curve, then both trains slow down, including the one that's on the straight track). If you want to observe this in the 9V system, then make a very long oval (say: 40 straight, then 8 curves, then 40 straight, and 8 curves again). Now connect the power supply in one of the two curves. You'll see that the train slows down in each of the two curves, but it slows down way more on the curve that is far away from the power supply. This is because a long track has resistance (count about 0.08 Ohm per track, so the other side of the track is about 40 * 0.08 / 2 = 1.6 Ohm away). The higher the resistance is, the more of a voltage-drop you'll get with an increase in amps. So the higher the resistance, the more of a speed-drop you get in the curve. An extra 1.6 Ohm is something you'd certainly notice (especially in the lower speed settings).

A motor behaves very differently than for example an incandescent lamp. With a motor, the current depends mostly on the torque that the motor is producing. With an incandescent lamp, the current depends on voltage, so that's a completely different behavior. I agree with that. But does the lego PF receiver have all those things? I've read that the only thing it does is rapidly switching the power on/off/on/off... But if you want to be efficient from an energy point of view, then you have to do more than just rapidly switching on/off, you also need the other components you mentioned. To turn 9V DC into 4.5V DC without wasting half of the energy, that's a complicated thing to do. Because it means that you have to turn for example 9V with 100mA into 4.5V with almost 200mA. It is possible to do that, but it is complicated, and I question if the lego PF receiver actually does that (turn 9V, 100mA into 4.5V with more than 100mA).

If you are powering a light, then yes, but if you're powering a DC motor, then I do not believe that that is true. The quick on/off switching is definitely the most practical thing to do, because that way you can make a PF receiver that doesn't generate heat (in other words: a PF receiver that can be made small enough to be practical for lego). The reason for building it this way is because: it works, it is cheap, practical, and small. But if the battery pack produces 9V and you want to run a train at 4.5V DC, then I do not think that this system is any more energy efficient than simply dissipated half of the voltage away in a resistor (note: this applies only to the motor. If you're powering a light, then it is more efficient).

If the PWM does not dissipate the excess energy, then it'll have to be dissipated somewhere else. The whining/buzzing noise explains at least partially where that excess energy goes; obviously, you can't generate noise without spending at least some energy. Here's one way to think about the various scenarios: (1) Turn on 4.5 volts, constant voltage (as in the old 9V system). (2) Rapidly switch 0 and 9V, with each on 50% of the time (as in PF). Clearly, option (2) will produce a whining noise, and option (1) will not. There is another scenario to consider. Suppose you did this: (3) Rapidly switch between -4.5 volts, and +4.5 volts, with each 50% of the time. In option (3), the train would stand still, but it would make a whining noise. And clearly the motor would dissipate some energy as heat because the coils inside have resistance, so they must warm up when current runs through. Notice now that option (2) is really the same thing as the sum of options (1) and (3). 0 and 9 is the same as 4.5 added to -4.5 and +4.5. So the way that the PF train runs is as though you are doing (1) and (3) at the same time. The energy for (1) is spent to move the train, while the energy for (3) is dissipated as heat and a whining sound. The conclusion is that if I have two 9V trains, of equal length/weight/speed/etc, and if one is powered by PF and the other by the old 9V train controller, then the one powered by PF has to get warmer (i.e. dissipate more energy) than the one powered by the old 9V controller. I know this seems counter intuitive, after all, both trains are equally heavy and go equally fast. But still, the whining noise produced by the motor that is powered by the PF system is a clear indication that energy is indeed being dissipated in the motor. If no energy was lost, then why would it whine? Although I have not yet tested it, I'm almost sure that in PF, the batteries will not last longer in low speed than they would in a high speed setting. At speed setting #2, my cargo train moves gently over the track. At a high speed setting, it thunders over the track so fast that I fear it might derail in the curves and fall of the table. Seeing that, it seems impossible to believe the amount of current is pretty much the same in both cases. But that's exactly what my multi-meter is telling me.

This is a bit time consuming, it takes 3 hours for my Emerald Night to deplete the 8878, and I have to do that twice, once at high speed, and once at low speed. I was hoping that someone here might have already done this test. It takes less time to simply measure the current. I've done that at high speed and at low speed, and the result is the same. I have a 10-feet long train with 2 Maersk engines, and a number of light-weight cars behind it. At the moment, it has two 9V motors. I was trying to decide whether to use the 9V system or the PF system to run this train, so I measured its electricity usage. It uses 0.42 amps (so 0.21 amps per motor) and this doesn't really change with speed (at low speed it takes 0.42 amps and at high speed it takes 0.43 to 0.44 amps). If I use rechargeable AAA batteries with a capacity of 1,000 mAh (which I do not have, I'd have to buy them) and I need 420 mA to run the train, then it should run 1000/420 = 2.4 hours before the battery is empty. I don't see how one can get around that. Slow or fast, either way the train pulls 420 mA according to my multi-meter. There are only so many mAh in the battery, so it seems inevitable that the batteries will run empty after 2.4 hours, regardless of the speed setting. Of course, this still needs to be tested, I may be overlooking something in this reasoning. There is one other test that one can do, one that is not as time consuming as running the battery empty twice: we can check if (as I suspect) that a PF train motor running at low speed becomes warmer than one running at a high speed. Incidentally, does anyone here know how many mAh there are in a fully charged 8878?

I measured this, both for 9V trains and for 12V trains. The current that a train draws is (within just a few percent) practically independent of the speed of the train. As long as it is moving, it amount of current it draws depends only on things like the weight of the train, and whether it is in a curve or on straight track. But the amount of current is practically independent of the speed. So the batteries will not last longer if you run the train slowly. I think that that's exactly what happens, that the battery won't last longer at a slow speed, and my Emerald Night will take 3 hours to empty the 8878 rechargeable battery, regardless of the speed setting.

It's better to use rechargeable batteries (the deliver fewer volts, but more amps, and that's what's needed here).

Only one hour??? My train should use less batteries because there are fewer cars, and they weigh less (I'm only using one Maersk container per car). Still, I don't want to change batteries that often. The show is about 7 hours, it's a bit too much if I have to replace batteries so many times. That means I'll have to use the 9V motors, but as you noticed, it slows down on the far side of the track. To prevent that, I'll have to supply power to all corners of the track. The only way I can see how to reach these corners is to make the 9V-track connectors longer (one of them will need to be about 15 feet long, so I'll have to cut the wires, and insert additional wire in them. Perhaps that's not a "pure lego" solution, but I don't see any other way to do it because the 9V extension wires are way too short).

By the way, this is good advice too. I did encounter another problem with my train. This train is 10 feet long, 2 Maersk engines, 4 Maersk cars with 1 Maersk container each, and another 4 light-weight MOC cars. It is supposed to run a whole day. With 2 motors in the first engine, if I use PF then I'll have to change the batteries during the show. So I was thinking it'd be easier to use two 9V motors in the front engine. The trouble is, with speed setting #2 (which is about 4 volts) the train runs the right speed when it is on the part of the track that is close to the power supply. But when it gets to the other side, the voltage there drops by 2 volts. So the train stops there unless I use a higher speed setting (but then it'll go faster than I want it to go when it gets back to the side that has the power supply). So to get a constant speed along the track, I'll either have to use PF instead of 9V, or, add long wiring underneath the tables to make sure that both sides are equally supplied with electricity (which adds to setup time) (this loop has 134 track pieces, so the other side of the track is some distance away). With 2 PF motors, how long would the batteries last? Does it last longer at low speed, or is it the same?

quote of entire topic removed by moderator -TheBrickster With no load, a 12v motor draws around 0.20 amp. In my 12V trains, the motors use 0.25 amp, which indicates that they are pulling a modest load (still, these are 5 feet long trains, but I've taken steps to reduce rolling friction, e.g., I'm using 9V train wheels instead of 12V train wheels). If you try to pull a very long train with a single motor, consumption can go up to 0.50 amp and probably higher but I did not test it higher than 0.50 amp. The 9V and PF motors use less. There is a useful website with information that has all the information for them (just google: LEGO 9V Technic Motors compared characteristics). There is no overload protection, so there is no well defined peak consumption. I do not know what a safe amperage is for these motors. But I would not want to subject my motors to say 0.50 amp for an extended period of time. If they don't get warm, then you know you're OK.

I've used WD-40 for a few other things, more than a year ago, and those still run perfectly smoothly. Using microscopic quantities is best (I think it's enough if you have a layer that's just a few molecules thick), and, by using tiny amounts you also don't get it on parts where you don't want it (the track!). So I think that the lubricated Maersk engine will run smoothly for a very long time, and won't have to be re-lubricated for years. Even if the WD-40 would affect the lego, I'm sure that the net effect is still less damage (less wear) than if it was not lubricated. Every time that you run an unlubricated Maersk engine, you are constantly rubbing hard on the axle and the hole in the technic brick. Lubricated it runs 3 times smoother, so there's much less rubbing, and hence less wear on the lego's. I'm sure that that more than compensates for whatever damage the WD-40 might do to the lego ABS. I've read elsewhere on eurobricks that there are lubricants that are better for lego (I didn't think it was worth it to drive to the hobby store because I have spare technic bricks/axles, and in any case I don't really believe there'll be any damage). Two years ago I had a burned out 9V motor, and since then I do everything I can to prevent unnecessary wear on the motors. A tiny amount of lubricant on a few cheap technic bricks will drastically reduce the wear on the motor(s) and hence prolong their lifespan. I think that's especially important for 9V and 12V motors, because they don't make those anymore. PS. Is there a consensus about which lubricant is best? The value of the technic bricks is much less than the gasoline spent on a trip to the hobby store, but if I'm there anyway for something, I'd like to pick up the correct lubricant (if I know which one to pick).

I now built 2 Maersk engines, and compared them side by side. One is standard, and the other one is lubricated with WD-40 (I've read that WD-40 might not be good for plastic, so I used only a tiny amount. I put a tiny amount of WD-40 on a piece of napkin, and inserted that into the hole of the technic brick. The amount of WD-40 that is put into that hole is so small that you can't even see it.). If you give both engines the same push and let them roll out, then the lubricated one rolls 3 times further (tried this on straight track). In any case, both in straights and in curves, it takes much less effort to pull the lubricated engine compared to the standard one, the difference is immediately obvious when you push or pull them by hand. On straight or curved track, two lubricated engines will definitely be easier to move than one standard one. A train built with 2 sets of 10219 can be pulled comfortably with 1 motor, but only if you use lubrication. Without lubrication, you'll need 2 motors. (obviously, the cargo cars do not need lubrication, they roll very well as they are, and don't add much to the load for the motor).

Without major modifications this won't work, because the studs on the standard train wheel holders will bump into the front of the train base plate (so it won't go through curves). I've done a few comparisons (very inaccurate, by hand) regarding the friction. On straight track, pulling 1 un-motored Maersk engine takes much more effort than pulling 6 Maersk cargo cars (with 1 container each). You can lower the friction quite drastically by putting some lubrication into the holes of the technic brick where the axle goes into (I'm not sure what the best lubricant is though). I tried some WD-40 (note: I've read somewhere that this might be bad for lego bricks, so I can not recommend that you do this too) and I estimate that this reduces friction by about half. Big improvement, but even with that it's still a lot more friction than what you would have had with standard train wheels. Perhaps a different lubricant, one more suitable to plastic, would have given a better result, although I suspect that even with the best lubrication, a plastic axle would still have more rotational friction than a metal axle. My advice, if you want to pull a second Maersk engine, is go to a hobby store, and ask for some plastic-to-plastic lubricant. Without that, pulling that 1 engine is a MUCH higher load than pulling those cargo cars. Could you check if you get the same results? How many of your cargo wagons do you have to put behind each other to create a load that takes an equal effort to pull as 1 Maersk engine?

20 by 20 feet of space! You lucky ... My layout is 18 by 16 feet, and I get to see it only once per year (at our local train show). At home there is no room for it.

I've done the same thing with my first train MOC (sorry about the flash light, I take terrible pictures). It doesn't look like much, but it is remote-controlled, and this was made before lego came out with remote controlled trains. The tender contains the battery box and the electronics that I took out of a remote-controlled toy car that was no longer fully functioning (of the 4 functions, left/right/forward/backward, only 3 were still functioning. The train can go forward, backward, and the other button can turn the front light on/off).