rgbrown

I reckon you should be able to get it pretty accurate. Assuming friction and environmental conditions (like gravity) don't change, the only variables that could influence the running of the clock are 1) Different initial conditions (starting the pendulum higher, etc) 2) Change in torque exerted by the falling weight (e.g. more string length == more weight, etc) It should be possible to design so that any transients introduced by different initial conditions are quickly damped out, and that 2) is so small as to be completely insignificant. The really really hard part is to design an escapement that provides *just* enough energy to the pendulum to prevent it from stopping, and doesn't waste a whole lot of energy spinning around between ticks. aww, where's the fun in that?

I doubt it - the problems are more serious than just unbalanced actuators. Though I agree that a diff is a nice way to make parallel actuators self-balanced

May I just say ... good answer

Awesome! I'm glad these have come up again :) I have a whole lot of gears from Bricklink waiting for me when I get back home in September to make a clock as well. I take it you've seen all the various Lego escapements by BenVanDeWaal on YouTube? edit: yes you have, I missed your credit to him at the end of the video the first time I watched it

Not really. If the screw behaves ideally, then the friction is proportional to the normal force exerted by the screw. Increasing the number of contact points by a factor n will divide the normal force by n (assuming the load is distributed equally between the contact points), and summing them up will give (in theory) the same friction.

Potentially the defective LAs damage the motors that are driving them (if they overheat)

Awesome, thanks. So in your earlier experiment (stalling the two motors through the IR receiver), it looks like IR receiver was current limiting to 1A (unless the battery voltage dropped a whole lot)

That is correct. I had another thought - do the stresses of supporting the digging arm cause some of the framework that supports the transmission to get slightly out of alignment? This would cause more friction on the axles, and would potentially happen over a period of time ... I'm guessing this is not the problem though, as it would probably have been noticed by now ...

Rechargeable battery is the same: http://powerfunctions.lego.com/en-us/Bios/power/8878.aspx

On the power functions website it says that stalling either a single M motor or XL motor can trip the overload protection in the battery box (which it says is designed to provide 800mA), so it won't be 4A, it will be closer to 1

The obvious solution for TLG is to get you to send your instruction booklets back in exchange for the new parts and new instructions (if indeed they release new instructions) That would also make the collectability of the existing instruction books even higher!

Give them the benefit of the doubt for now Everything I've heard says that TLG customer service is very good. By the way I'm amazed at the extent to which your excavator has killed your M-motors. That's pretty impressive.

I've emailed them to ask about this ... I have an unused one in my suitcase (with no box -- I'm travelling). I'll let you know when I hear back. I don't want a refund, as I live in New Zealand, and I do want the excavator, and it will be much more expensive there than I paid for it in the USA

AAA alkalines are not going to be very happy running the excavator ... too much current draw will mean they will discharge very quickly

Thanks, hadn't seen that before. The schematic confused me for a moment though, until I realised he'd drawn his battery back to front! Is that really how it works? That schematic implies that if you put the batteries in backwards, and switch it on it will just short circuit and trip the fuse ... is that what actually happens?

What about these ones? (81.6x38 -- 8295, 8265, etc) http://www.peeron.com/inv/parts/45982

You should be seeing more than 950mA (I think) when both motors are stalled, so it looks like something is limiting current. It won't be the batteries, they'll get hot (rule of thumb for AA alkalines is < 700mA is OK, heatwise), but they'll still deliver maybe 1.2-1.3V up to around 2A. I know the battery box is not supposed to operate above 800mA, and has overload protection (I'm not sure if this is thermal, or simply tripped by high curent). But that didn't happen, and I don't know whether it also has a current-limiting feature. Maybe the IR receiver limits output current to 1A - can you compare the input voltage to the output voltage from the receiver when both motors are stalled? Maybe your schematic will tell us what's going on :)

Nice! Impossible to calculate, but possible to measure. Imagine this experimental setup: Set up the slope so that you can get the different kinds of behaviour we saw earlier. The idea is, you change the wheelspeed (by adjusting gear ratios, or using PWM speed control), and then add weights to either the top or bottom tray so that the car is stationary. You can then find the optimum wheelspeed for those conditions, by finding the speed that requires the most mass on the bottom tray to keep it still. You could sketch out the u vs f(u) profile curve from my previous post this way too ... Anyone game to humour me and try this? :)

This page http://www.icver.com/html/pro/13/25395.htm claims to sell both, and the image appears to match the one on your board (same font, similar batch number, etc ...). But, frustratingly, the datasheets for both are for LB1836M. (I downloaded lb1836m.pdf and lb1836.pdf). I'm now pretty convinced that they're the same thing ... I think it would be more obvious if they were not ... happy to be proven wrong though. The one sanity check left to do is to probe the voltage on the VCC pin (pin 1) of the motor controller

A couple of questions. That plane of track on the top side of the board, is that connected to the power functions 9V line? If so then it appears that the VCC and VS pins on the motor controller follow a rectifier diode of some sort (1N4002 perhaps?), but I can't quite tell. That would mean that the output voltage to the motors at 100% duty cycle should be about 0.5-0.7V lower than the supply, which would also be the same as the VCC pin on the the motor controller. Which would also mean that applying a voltage higher than about 11V would be outside the recommended operation range of the motor controller. By the way, I think the LB1836M and LB1836 are the same thing -- if you do an image search for LB1836M, you never find a picture of one with the M printed on it. So I think that's probably our answer. If you're applying more than 11V or more, you are probably working outside the operating limits of the motor controller, and are asking for trouble ... It will probably work, but you may be shortening the life of the controller. Forgot to say: Thanks! You are awesome for dismantling your IR receiver Edit: OK, watched the video properly. There definitely is no voltage drop. VS and VCC are definitely connected together on the PCB, so the output voltage is the voltage that you are applying to the LB1836. Even more reason not to stick higher voltages on it.

If any TLG employees are watching, we'd love to know the design limits of PF ... :)

OK, I'll take your word for it on the terminology. All I'm positing is some tangential force that varies with the slip speed when the wheels are slipping (limiting ourselves to the dynamic friction case) Because I'll keep forgetting, I'll continue referring to this as friction. But no, we don't need more variables! We have a minimal model here that explains the observations. Sure we'd need to do more if we wanted to actually estimate the exact shape of the curve. And yes, ignoring temperature changes is certainly a simplification that we could take into account if we really wanted to. Does friction increase or decrease with a rise in temperature? Sure, static friction is higher than dynamic, however in the scenario that Zblj posted, the wheels are slipping, no matter what you do. Only dynamic friction applies here. So the whole point is that the forces on the car are NOT exactly the same no matter what you do. Otherwise the car would slip down the hill no matter what you do. And I don't think it's to do with the irregularity of the tyre or the surface here, as the surface is smooth. Therefore, the dynamic friction/adhesion forces must vary with slip speed. This is a different question, when we're dealing with irregular surfaces, and I agree that it's probably right that it's more to do with the shape of the tire than any pure friction argument. This argument doesn't apply to Zblj's example though. Doesn't mean that the friction is irrelevant though -- again, as soon as you're spinning your tyres you're into the realms of dynamic friction. The interesting thing about the above model in this case is that it implies that there is an optimal spinning speed that maximises the amount of traction you get from dynamic friction. If you must spin your tyres, it can't hurt to spin at that speed. This is a fun little problem, by the way. @Zblj, I'm not a physics teacher, I'm an applied mathematician. Basically, that means if reality is too hard I have the right to make it up. I would guess if you try smooth tyres you'll see the same thing, but the slip speed will be different, and you may need to go steeper to make it slip. I would guess that smooth tyres will work slightly better (more surface area in contact at any one time)

I'd be interested to hear the outcome of this. I can't check it until I get home (in September), but judging from all the comments about how the 8043 works less well on rechargeable batteries, I'd expect that the voltage is passed straight through ...

OK, let me refine my argument. Let's not give up, we can include what you say too ! First, we'll assume we're not dealing with static friction here, all of the cases are dynamic (the wheel is rubbing the slope somehow or other). Remember, u = v - w where u is the relative wheelspeed, v is the observed speed of the car up the slope, and w is the wheelspeed (linear speed at the outside of the wheel). First, note that in the model I showed, if v is higher than the equilibrium speed (the car is moving faster than expected), then the relative speed u is lower, hence the friction is lower, and so the gravitational pull acts to slow the car down to equilibrium. And if v is lower than equilibrium speed (car is moving slower than expected), friction is higher, and the car speeds up towards its equilibrium. Therefore the equilibrium predicted by this model is dynamically stable. You can imagine arrows on the blue curve pointing towards the equilibrium, which would be how the operating conditions change with time, as follows: Suppose now that friction was decreasing with increasing relative wheelspeed, so we have a graph like this: In this case, if the car was moving faster than its equilibrium speed, u would be lower than equilibrium, friction would be higher, and the car would grip even better, making it go even faster up the slope. If the car was moving slower than its equilibrium speed, u would be higher than equilibrium, friction therefore be lower, and the car would start to slip even faster down the slope. If this type of equilibrium existed, it would be dynamically unstable (you couldn't observe it). You can imagine arrows on the curve pointing away from the equilibrium. Since zblj has observed a kind of equilibrium of the first type, it's fair to conclude that the assumptions I made in the operating region he's working in are essentially correct (increasing friction with increasing speed). However, we've all observed that if wheels spin too fast, you don't get enough grip, so there must be a region of the curve for high u where the friction is lower than the gravitational threshold. A continuity argument means that the shape of the friction vs relative speed curve must look like this: I've added arrows this time. Now's where it gets more interesting (I think). The second equilibrium on this graph is of the unstable type, so if you put the car down with a really high u (to the right of u2 (fast wheels, no "push-start"), it will start sliding downwards, u will increase even further, and you eventually crash at the bottom of the hill. However, if you set it going fast enough (give it a really good shove), then u will be low enough to be on the left-hand side of the unstable equilibrium u2, and you will effectively traverse the curve towards the stable equilibrium, speeding up even more, until the stable equilibrium is reached (lowish u). Cool huh? Conclusion. Even at really high wheel speed, it is possible for the car to have upward traction on the hill if it was going fast enough to start with. So speed is everything. We can explain it all! the power of mathematics (or something)

Your physics prof was fobbing you off, it's easy to come up with a simple model that explains this. Here goes: Let w be the wheelspeed (linear speed at the outside of the wheel), and let v be the speed of the car, with positive v defined to be up the slope. Likewise, positive w means the wheels are rotating forwards. Then the relative wheel speed u, i.e. the speed at which the wheel surface passes the slope is u = w - v If there's full traction, then u = 0. Now consider the force balance in the following picture N is just the normal force that counteracts the gravitational pull on the car. f(u) is the friction force from the wheels passing over the slope, and increases with increasing u (this is standard physics). If the car is moving at constant velocity, or stationary, then all of the forces in the figure must balance. In particular mg sinθ = f(u) The speed u* that makes this equation balance is the relative wheelspeed that the car will stabilise at, as shown in this figure: From our first equation, the speed of the car is then given by v = w - u* So if w > u*, the car goes forwards at speed w - u*. If w < u*, the car goes backwards at speed u* - w, and if w = u*, the car stays still. Finally, if w = 0 (i.e. the wheels are locked), then v = -u*, so you can measure u* as the speed at which it slides down the slope. So, yes, zblj, your theory is entirely reasonable