{"id":712,"date":"2008-12-07T14:54:45","date_gmt":"2008-12-07T14:54:45","guid":{"rendered":"http:\/\/scientopia.org\/blogs\/goodmath\/2008\/12\/07\/the-real-bozo-attempts-to-atone-why-the-ddwftw-car-works\/"},"modified":"2008-12-07T14:54:45","modified_gmt":"2008-12-07T14:54:45","slug":"the-real-bozo-attempts-to-atone-why-the-ddwftw-car-works","status":"publish","type":"post","link":"http:\/\/www.goodmath.org\/blog\/2008\/12\/07\/the-real-bozo-attempts-to-atone-why-the-ddwftw-car-works\/","title":{"rendered":"The Real Bozo Attempts to Atone: Why the DDWFTW Car Works"},"content":{"rendered":"

Technorati Tags: ddftw<\/a>, bozos<\/a>,
\nmarkcc-screwups<\/a><\/span><\/p>\n

So, as I said in the edit to my previous post about the wind-driven cart, I seriously blew it. The folks who pointed out the similarity of the cart to a tacking sailboat were absolutely correct. The guys who built this cart, and recorded the demo were absolutely right<\/em>, and I was stupidly wrong, in multiple ways. First of all, I thought this was a really simple system. I couldn’t possibly be more wrong about that – this is anything but<\/em> simple; in fact, it’s a remarkably interesting and elegant demonstration of how complicated and counterintuitive fluid dynamics can be. Second, I completely misunderstood the simple mechanics of the device; I originally thought that the propellor was spinning in the opposite<\/em> direction – and I completely missed it when people repeatedly tried to explain that error to me. And third, I completely screwed up my own mathematical model of how something like this works.<\/p>\n

So – to repeat: the guys who did the demo of this are clearly not<\/em> bozos; the only bozo in this situation is me, for screwing it up so badly, on so many levels. I sincerely apologize for calling them bozos and mocking them. I made a whole series of really stupid errors, and took an unreasonably long time to recognize that fact.<\/p>\n

It should be obvious<\/em> that there’s some way to go downwind faster than the wind, because as so many people pointed out, sailboats do it. It was frankly stupid of me to even argue about this – it’s really pretty boneheadedly obvious. The question never should have been “can it be done?”, but rather just “does this device do it?”<\/p>\n

And the answer to that is “Yes”. This thing does do it. It’s not magic, it’s not perpetual motion. In fact, it’s really astonishingly simple, once you realize that the behavior of things moving through air is quite different from the simple rigid system that it appears to be equivalent to.<\/p>\n

In this post, I’m going to try to do two things:<\/p>\n

    \n
  1. Explain the faulty reasoning that led me to think it couldn’t work, and why it’s wrong.<\/li>\n
  2. Explain why the thing really works.<\/li>\n<\/ol>\n

    In another post later this week, I’m going to try to explain why I think the mathematical element is so important. There’s a ton of people who’ve got devices that really look<\/em> convincing, and that have really convincing arguments for why they should work; only they don’t, because they’ve missed something. <\/p>\n

    <\/p>\n

    The explanation that got me started on really understanding this was based on a rigid geared cart, which had one wheel on the ground below, and one wheel touching a moving belt above. While this is really what twigged me to how things could work, I also realized that it’s a good demonstration of why I thought it wouldn’t work. Here’s the diagram:<\/p>\n

    \"rigid-wheels.png\"<\/p>\n

    We’ve got a rigid cart. It’s help up by free wheels on either side, to make sure that it doesn’t tip. It’s got a belt on top, which is moving at a velocity of vupper<\/sub>. The upper belt is touching the upper wheel, which makes it rotate. The middle wheel is touching a center wheel, whose radius is equal to the upper wheel. The middle wheel touches a bottom wheel, which has a different radius. The only purpose of the center wheel is to reverse the direction of rotation, so that the rotation of the top and bottom wheel are the same – so that the upper track is making the bottom wheels move the correct direction. Assume that the wheels are actually gears, and that the energy transfer between gears is lossless. The upper and middle wheels have equal gearing. The bottom wheel is geared differently – one rotation of the middle wheel causes more than<\/em> one rotation of the bottom wheel, and<\/em> the bottom wheel is slightly larger than the other two.<\/p>\n

    If you look at this with a static analysis, you can work out what appears to be an interesting result:<\/p>\n

      \n
    1. The upper belt is moving at velocity vupper<\/sub>, which means that the rim of the upper wheel at the point of contact should move at a velocity of -vupper<\/sub>, which will give it a rate of rotation of vupper<\/sub>\/2πrupper<\/sub><\/li>\n
    2. Because of equal gearing, the rotation rate of the middle wheel is the same.<\/li>\n
    3. The bottom wheel is more complicated, since its gearing is different. Let’s assume that it’s turning K times for each turn of the middle wheel. So the bottom wheel should turn at Kvupper<\/sub>\/2πrupper<\/sub>.<\/li>\n
    4. The speed of the bottom wheel at its rim touching the ground is 2πrlower<\/sub>Kvupper<\/sub>\/2πrupper<\/sub> = rlower<\/sub>Kvupper<\/sub>\/rupper<\/sub>.<\/li>\n
    5. The forward speed of the cart is the same as the speed at the rim of the lower wheel.<\/li>\n<\/ol>\n

      So, this makes it look like if rlower<\/sub>K > 1, then the maximum speed of the car should be greater than vupper<\/sub>.<\/p>\n

      Is this the end of the story? Nope. Because that’s a static analysis of something dynamic. It doesn’t actually work out quite that smoothly, because the rigidly connected moving parts that constrain each other.<\/p>\n

      What happens when Vcart<\/sub> = (1\/2)vupper<\/sub>? The rim velocity of the upper wheel relative to the upper track is reduced by 1\/2. So it’s getting less push. But not only is it getting less acceleration, it’s limiting<\/em> the speed of the bottom wheel, because they’re rigidly connected. If the top wheel slows down, the bottom wheel needs to slow down. If the cart is moving at a speed equal to vupper<\/sub>, then the upper wheel isn’t turning<\/em> at all. It’s stationary relative to the upper track. Which means that the lower wheel can’t be turning either! Because of the rigid gearing, you’re creating a drag, because the upper wheel slows down relative to the upper track as the vehicle accelerates, which slows down the bottom wheel. <\/p>\n

      So there’s an equilibrium point where the connection to the top switches from acceleration to drag; that equilibrium point is going to be determined by the ratio between the gears – and it’s going to be less than the speed of the upper belt. So the static look at the gear system can give you a sense of how the wind powered vehicle might work, but its rigidity limits its speed to less than the speed of the belt providing the force that’s pushing it.<\/p>\n

      So: I think that with the rigid gear system, it doesn’t work. However, the wind-based system isn’t rigid, and that’s the key. The propellor system is similar<\/em> (but not identical to) the rest-based analysis of the geared system.<\/p>\n

      That’s what I initially thought was happening with the wind. The wind was behaving as the upper belt, and as the cart accelerated, the amount of force that could be extracted from the wind reduced.<\/p>\n

      Another piece of my misunderstanding was the real nature of the drive mechanism – which again can be demonstrated by the geared system.<\/p>\n

      Suppose that the wind is spinning the propellor. Then the rotation of propellor can’t<\/em> also be providing a propulsive force. That would be like the geared cart working by having the upper wheel rotated by friction with upper belt pushing the heel, and also<\/em> having the cart propelled by the rotating wheel pushing against the belt. That’s a double-dipping explanation, where the same force is being counted twice, as if it’s simultaneously pushing in opposite directions.<\/p>\n

      So that, in a nutshell, covers the errors I made in believing that the thing didn’t work. Now, let’s move on to why it does<\/em> work.<\/p>\n

      Let’s look at a new diagram.<\/p>\n

      \"propellor.png\"<\/p>\n

      This is the cart. There’s a propellor the back. It’s connected by a gearing mechanism to the wheels. In a somewhat simplified model, the wind provides a force by blowing against the propellor – not by rotating it, but by treating it as a dead sail.<\/p>\n

      So, initially, you get a forward accelerating force, which is proportional to the wind velocity. As it accelerates, the rotation of the wheels starts to spin the propellor. The spinning of the propellor pushes air backwards. The force from the spinning propellor is proportional to some properties of the propellor (generally summarized as an “advance ratio”. We’ll describe the propellor in terms of its force ratio, which is proportional to the advance ratio; the force ratio of a propellor tell us how much force the propellor generates at a given rate of rotation. We’ll call it “A”) times the rotation rate of the wheels, which is proportional to the velocity of the cart. The propellor being moved by the wheels is really the key to this: what that’s doing is providing a way of exploiting the connection to the ground, allowing the cart to (as so many people pointed out) provide a way of extracting energy from the difference<\/em> between the motion of the ground and the motion of the wind.<\/p>\n

      Here’s where it starts getting tricky. What’s the total force provided by the wind? It’s basically the forward thrust of the moving wind plus<\/em> the backward thrust of wind moved by the propellor.<\/p>\n

      Suppose the cart is moving at velocity Vc<\/sub>. The wheels are rotating at vc<\/sub>\/2πr, where r is the radius of the wheel. The rotation rate of the propellor is going to be the rotation rate of the wheels, times the gear ratio. We’ll call the gear ratio G. So, the propellor is spinning at Gvc<\/sub>\/2πr, and generating a force GAvc<\/sub>\/2rπ.<\/p>\n

      Now, suppose that the cart accelerates to windspeed. Then the force from the propellor is GAvwind<\/sub>\/2πr.<\/p>\n

      At this point, the pure wind force has fallen to 0, because the cart isn’t moving relative to the wind. But the propellor is still producing a force dependent on G, A, and r.<\/p>\n

      So for the right values of G, A, and r, you’ll still be accelerating.<\/p>\n

      What stops it from being perpetual motion? As you accelerate above windspeed, you’ll start to suffer a vaccum effect – the wind is moving slower<\/em> than the vehicle, so it’ll effectively be sucking away some of the force generated by the propellor. So you’ll need to start subtracting<\/em> a force proportional to the relative windspeed from the amount of force you can generate from the propellor. Eventually, that vaccum force will equal GAvcart<\/sub>\/2πr. As you get closer and closer, the amount of force that you’ll be generating by spinning the propellor will be completely consumed – and you’ll be at the maximum attainable speed in that amount of wind.<\/p>\n

      What this all means is that you’ve essentially gotten energy from the friction of the wheels against the ground, and turned it into propulsion. The ground friction is providing a way to make the propellor “tack” against the wind. The nature of the mechanism means that the initial acceleration of the cart is slower<\/em> than a pure sail-driven system, because the way the propeller spins adds resistance to the wheels via the gearing. But ignoring losses to friction, if the wind is constant, it won’t stop it from accelerating – it’ll just slow the rate of acceleration.<\/p>\n

      Once it’s moving, what’s going on can be described in a bunch of different ways. They’re all equivalent, but the one that seems easiest to model to me is to think of the “sail” as being not just the propellor, but the cushion of moving air that it’s pushing behind it. So it’s very much as if the “sail” is moving backwards relative to the cart. (An equivalent view is that the angle of the propellor blades are tacking relative to the wind, and that tacking angle combined with the rigid connection to the propellor spindle translates into a forward force.) Either way, the rotation of the propellor provides a way of getting an accelerating force from the wind even when the forward velocity of the cart matches the velocity of the wind.<\/p>\n

      So the folks who keep comparing to tacking in a sailboat are right. This is, essentially, the same effect.<\/p>\n","protected":false},"excerpt":{"rendered":"

      Technorati Tags: ddftw, bozos, markcc-screwups So, as I said in the edit to my previous post about the wind-driven cart, I seriously blew it. The folks who pointed out the similarity of the cart to a tacking sailboat were absolutely correct. The guys who built this cart, and recorded the demo were absolutely right, and […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"jetpack_post_was_ever_published":false,"_jetpack_newsletter_access":"","_jetpack_dont_email_post_to_subs":false,"_jetpack_newsletter_tier_id":0,"_jetpack_memberships_contains_paywalled_content":false,"_jetpack_memberships_contains_paid_content":false,"footnotes":"","jetpack_publicize_message":"","jetpack_publicize_feature_enabled":true,"jetpack_social_post_already_shared":false,"jetpack_social_options":{"image_generator_settings":{"template":"highway","default_image_id":0,"font":"","enabled":false},"version":2}},"categories":[35],"tags":[],"class_list":["post-712","post","type-post","status-publish","format-standard","hentry","category-markccs-errors"],"jetpack_publicize_connections":[],"jetpack_featured_media_url":"","jetpack_shortlink":"https:\/\/wp.me\/p4lzZS-bu","jetpack_sharing_enabled":true,"jetpack_likes_enabled":true,"_links":{"self":[{"href":"http:\/\/www.goodmath.org\/blog\/wp-json\/wp\/v2\/posts\/712","targetHints":{"allow":["GET"]}}],"collection":[{"href":"http:\/\/www.goodmath.org\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"http:\/\/www.goodmath.org\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"http:\/\/www.goodmath.org\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"http:\/\/www.goodmath.org\/blog\/wp-json\/wp\/v2\/comments?post=712"}],"version-history":[{"count":0,"href":"http:\/\/www.goodmath.org\/blog\/wp-json\/wp\/v2\/posts\/712\/revisions"}],"wp:attachment":[{"href":"http:\/\/www.goodmath.org\/blog\/wp-json\/wp\/v2\/media?parent=712"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"http:\/\/www.goodmath.org\/blog\/wp-json\/wp\/v2\/categories?post=712"},{"taxonomy":"post_tag","embeddable":true,"href":"http:\/\/www.goodmath.org\/blog\/wp-json\/wp\/v2\/tags?post=712"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}