Blakbird

At the time, these instructions were available with LEGO B.I.T. (Building Instruction Tool) which was a program like a mini web browser which would individually download and display the alternate instruction images. Sadly, I no longer have either this tool or the images of these instructions, but someone may. You'd think the LEGO company would still have these around somewhere, but I have no idea how you would go about getting them.

That piece came out in 1988. You can read a bit more about it here: http://www.ericalbrecht.com/technic/1988.html#Parts

The boat is very well done, but the real star here is the video. It is brilliant! Very well done indeed. I rarely laugh so hard at work.

I seem to recall that the version at Brickfest was built by Bob Kojima (although it was not his original design, he duplicated it). It was truly mesmerizing to watch, but alas I do not have a video. If I could remember the real title of it, we could probably find a video.

Ah, it is always nice to have friends on the inside. You are very fortunate to have been able to acquire this set so early. The rest of us may have to wait months in agony. If only I was an SeTechnic member..... Did Peek et Poke also have the other new models, or only the 8043? In case, congratulations again on your acquisition and the excellent and thorough review.

This is a great review with wonderful pictures Anio. But the most obvious question is not answered. How did you manage to get this model already?! The level of complexity and sheer number of gears and functions almost makes me weep. In fact, I think I have to stop typing now....

I really like the look of this. There is no doubt that it is a Humvee. I like your choice of using the old 20x30 wheels and tires. They may be foam, but they do have excellent traction on many types of terrain. When I was doing the "stock" version of the Rubicon (to look like from the factory rather than the rock crawler), I was having trouble finding any wheels that looked right. I ended up using the 62.4x20 which are the right style but much too small. I had somehow forgotten about the old foam tires. I am now rendering the Rubicon with the big 24x43 foam tires and it looks perfect! Thanks for jogging my memory. The only detriment to these wheels and tires is that they are quite rare these days. I have many of them, but I am unwilling to pilfer them from any of my old sets so I guess I will have to try to get some more!

It's called pony ear because some of the first LEGO horses such as those used in the 375 castle were brick-built and used this for the ears.

Possibly, but I don't think it ever rotates fast enough for balance to be a real concern. More likely the opening is simply to save weight (and ABS) compared to having it solid. The taper is a draft angle to allow removal of the part from the mold.

Actually, the middle of it hold 3 bevel gears. The pin in the middle support one of the gears (often called the spider gear). The other part is basically a stop. It keeps the spider gear from falling out and it keeps you from pushing in the axles on the other two gears too far.

It is a big coincidence that you wrote this post today because I just rebuilt this model last night after my move to a new house and was thinking how excellent it was. It is not possible to re-issue the model in exactly the same form because the wheels and tires are no longer in production. There aren't any new wheels and tires with the same (or even similar) size. All of the other parts still exist, though the studded Technic beams have become somewhat unusual. The model also uses the forbidden "pony ear" technique for the tail light which LEGO would not use in a model today. With the exception of the rear wheels, all of the parts in this set are very common so it would be easy to reproduce it from parts even if you don't own the set by looking up the instructions on Peeron.

I have built several of these models and can confirm that they are exemplary. The Prinoth Leitwolf snow groomer is one of the best models I have ever built.

If you ever design a spring from scratch, you find that the stiffness (spring rate) is a function of: The material (nearly constant 30e6 psi modulus for steels) The wire diameter The coil diameter The pitch Once you have a spring made, all of these things are fixed. If you want the spring to fit over the LEGO shock, the coil diameter is also fixed, so all you have to play with is wire diameter and pitch. You can see that the different LEGO shock springs have different pitch. If you make the wire diameter too large, then the solid height is very long and it won't compress much.

Wow, you're fast. I just posted those. The real photograph was used for reference because in one of the renders I made some changes to color, roll cage design, and tires to make the model more scale accurate (but less good as a trial truck).

Woohoo! I'm always pleasantly surprised when I am able to successfully explain something complex. Your equations are essentially correct. The slope/stiffness of the system is therefore bilinear and continuous. Again, to be very specific the first equation is actually also F=kx, but in that case k is the stiffness of the piston, not the spring. In the second equation k is the stiffness of the spring.

This is fascinating thought exercise (and incidentally I design and analyze aircraft springs and mechanical systems for a living). You're really close with your thought process. You are correct that most materials (at least ductile metals) are linear elastic at least up until their proportional limit stress at which point they exhibit plasticity and will not return to their original shape but rather exhibit hysteresis. When speaking of practical springs, we're never going to be loading them near the proportional limit, so we'll always consider them linearly elastic. In other words, stress and strain are proportional. A spring (or beam) with double the load will have double the deflection. This is in fact exactly the reason that the spring rate cannot change as a function of initial length. The slope of the stress/strain curve (Young's Modulus) is a constant in the linear elastic range and no amount of preload will change that unless you yield the spring (which ruins it). The issue here comes in the fact that you are saying it is impossible to add external force to the system and NOT produce some strain. This is a correct statement, but it does not therefore follow that the spring compresses even at an applied load below the preload threshold. The reason for this is hard to explain without a free body diagram, but I'll try. All static systems must be in equilibrium. When a preloaded shock absorber is at it's free length, the spring is under a certain amount of compression. The body of the shock must be in an equal and opposite state of tension. In this case, the yellow cylinder and the black piston are being forced apart by the same force that produces compression in the spring. What is preventing them from separating? The internal stop. The little "forks" inside the rod are reacting against the slots in the cylinder. This load path is carrying the 125g of tension in my example. The reason that you can apply an external load without compressing the spring is that the load is not being carried by the spring. The first 125g of external load goes into relieving the existing tension that is being carried by the stop. Only when this tension has been nullified can the spring again become part of the load path of external compression. Incidentally, there IS strain in the system as this external load is being applied, but it is a small tensile strain in the piston and cylinder which is being relieved. The piston and cylinder are so stiff when compared with the spring that this strain is not visually noticeable. This is why you don't see it when you just try to solve the problem while holding one in your hand. Hooke's law is not being violated. Upon further reflection, I decided to include a free body diagram. The purple force (S) represents the spring. The blue force (E) is external. The red force (P) is the amount of preload being carried by the stop. Now consider equilibrium. When the external load is zero (unloaded), the spring force is balanced by the preloaded stop. S=P When a significant load (greater than the preload) is applied externally, the spring compresses and there is no longer any load on the stop. In this case the external load is equal to the spring load. E=S However, when the external load is less than the preload, it is reacted by decreasing the preload in the stop. P=S-E The spring force stays the same but the preload goes down until the stop separates. If you really want to get super technical (which I guess we are) then the preloaded piston and the spring are both strained with external load as a function of their respective spring rates (springs in parallel). However, the spring rate of the piston is so much large than the spring rate of the spring that the proportional strain is effectively all in the piston until the preload is relieved. I've reviewed plenty of stress reports of hydraulic actuators by professional engineers who have still not managed to grasp this principle, so don't feel bad!

I disagree. Spring rate cannot be changed, it is inherent in the properties of the spring. It is true that the totally compressed force remains the same. The bushing changes both the starting length and the starting load but not the rate. The additional load per unit inch of deflection is the same with or without the bushing and must always be based on spring physics. Here is an example. Case 1: Unmodified sample shock absorber. Free Length: 3 cm Free Load: 0 g Compressed Length: 1 cm Compressed Load: 1000 g Delta Length (stroke): 2 cm Delta Load: 1000 g Spring Rate: 1000 g / 2 cm = 500 g/cm Case 2: Modified sample shock absorber (0.25cm bush). Free Length: 2.75 cm Free Load: 125 g Compressed Length: 1 cm Compressed Load: 1000 g Delta Length (stroke): 1.75 cm Delta Load: 875 g Spring Rate: 875 g / 1.75 cm = 500 g/cm In either case we have a 500 g/cm spring, but in the second case we need to apply at least 125g of load before the spring starts moving which gives the impression of more stiffness. Another way to put it would be that the spring rate (stiffness) is a function of the spring, but the spring loading is a function of installation (free and compressed length). The modifications proposed change the installation but not the spring.

Strictly speaking the "stiffness" or spring rate of the shock is the same whether you put a bushing on it or not. The spring rate (typically denoted with a "k") is a constant for most springs and is the amount of linear (or rotational) motion per unit force. In imperial units this is typically pounds/inch or Newtons/centimeter in SI. Adding a bushing increases the preload on the spring by making it start in a more compressed state and therefore with a higher initial load. The additional force to make it compress a certain amount remains the same. With all that being said, this is still a great idea! It will have the effect of providing more support for heavy models. Shock absorbers for R/C cars and trucks almost all contain a similar mechanism for changing the length. Now if only we can find a way to make a stiff spring softer.....

This thread contains lots of info on this topic.

Yes, each of the old "expert builder" sets came with an "idea book" wrapped around the instructions that had photos of other ideas, often combining the parts from multiple sets. I built one of them, a body for the chassis, by using the parts from the helicopter: One of my favorite "alternates" of this model is the mini version done by Arvo. I very much want to duplicate this, but I have no experience in micro building and have not been able to fully figure it out from the photos.

It's always interesting to see a modified model, and I'm a fan of combining Model Team and Technic. Most of the time, I find this is the most successful when the model looks like Model Team (a smooth, detailed construction) but has Technic functionality hidden inside. Some of M_longer's models are excellent examples of this (especially the Kenworth W900), and another of my favorite examples is the "US Truck" by Ingmar. In this case, the studded elements have just been put on the top of an existing Technic model. I like the result, but I don't find it as impressive as some of the other examples. There are others who have modified 8297 as well, and I think my favorite is this "pimped" version (in Dutch).

I'll finish the model and let you take a look. In a couple of cases the photos were blurry or I had some trouble seeing exactly which parts were used, but I think I've been able to figure them all out (I already asked you about the wheels and tires). I'm also adding all the instruction steps as I do the build so the MLCAD file will also be an instruction file. I made slight alterations to the order of steps for the file though to make parts movable. For example, each series of parts which are on the same axle are made into a submodel so I can rotate them as a group to get all the gears to line up and to make the pendular suspension and steering movable. So far so good!

I'm in the midst of making an MLCAD model of the whole thing. When I'm done, I'll post the parts list.

Not sure why no one has commented. I love this model and will be building one as soon as I can, although my current backlog of models to build is quite long. I'll probably also do this one in MLCAD so I can render it.

That's a really nice model, but those images need to be resized before posting them here. Those are enormous!