Dot Physics

Why would I think that?

So suppose you saw something that looked like this:

This is a ball shot out of a shooter device. Well, it is a vypthon animation of a ball. What would you do if you came to see this video? If I had not made it, I would say it is an unrealistic video. It does not agree with my basic model of how things move after being thrown or shot or whatever. Interestingly (but unrelated) there was a set of physics questions that showed different possible paths of a thrown ball. The path representing the motion above was a common choice.

Like I said, I made that animation. Here is another one. In this second animation, there are two balls. The red ball is exactly the same motion as before. But, in this case there is also a blue ball that is shot with the same initial velocity but accelerates as a normal projectile motion object would.

So, if I see something like the motion of the red ball, here are the possible things that could be going on:

  • It is wrong. Either the simulation is wrong or the observation is wrong. Really, this isn’t too uncommon. If I drop two balls of different masses in front of a class from the same height at the same time, quite a few students will say they saw the more massive ball hit the ground first.
  • The basic physics is wrong. For this case, the basic physics says that a ball shot from a launcher will have a constant acceleration in the downward direction. This acceleration would make the ball move in a parabolic path. If there is air resistance, the path would not be parabolic but it would also not move in a straight line like that. This is a tough option. If we say the basic physics is wrong, then we have to go look at all the other cases where this worked.
  • There is something else going on here. For this case, maybe there are some other forces that are not obvious. With these non-obvious forces (like maybe an electric force or something) the basic physics ideas are still valid.

Now, what about another case? Here we come back to this “Directly Down Wind Faster Than the Wind” business. I originally posted about this quite some time ago (like 3 servers ago). This stuff is the same as the red ball above. To a physicist, it looks like energy would not be conserved. Oh, stop right there. I know what you are going to say because I have heard it before: but you get energy from the wind. I still don’t get it.

Does this kind of machine violate the conservation of energy? Does this machine show that energy is not conserved? Or, maybe something different is going on here. Hopefully you can see why some physicists would say “errr?”

Ok – so I keep getting emails and comments saying this down wind faster than the wind machine works. That would make my claim from earlier that it did not work wrong. So, which of the above options would this situation fall into? I would say that there must be something else going on here. Abandoning conservation of energy would be dangerous (because it works so well).

Note: I will not be discussing the physics of this machine. Why? Is it not interesting? No, there is some interesting stuff going on here. I will not discuss the physics because it doesn’t really match with my agenda for this blog. My main goal is to look at cool stuff and use it to try to explain some physics. I tried that before with this machine and failed. If I were to try again, I do not think it would be too productive. Let me just say that the down wind machine builders are some pretty pumped up people. Being pumped up is a good thing. I am pumped up about VPython. But when they are way more pumped up than I am, some feelings get hurt.

Two more quick comments. Why is there no wikipedia article for down wind faster than the wind? Why are there down winders so anxious for me to say it works? If it works, does it matter what I say?

Comments: This is the kind of post that has the ability to generate pumped uped comments. To be a good comment hoster, I am going to start with the likely comments. If you want to write one of these comments, you can simply type the number corresponding to the comment you would like to make. You know, to save time.

  1. I, for one, welcome our new down wind faster than the wind powered cars.
  2. You are in idiot. You don’t understand physics. You should be fired.
  3. Why can’t we attack you like you attack Sport Science?
    • Reply: good point. Maybe I should stop attacking Sport Science. Help me, I can’t stop.
  4. I see what you are up to. You didn’t ACTUALLY say you were wrong. You are looking for some logical loop hole so you won’t have to actually say you were wrong. Why can’t you just be a man and say you were wrong.
  5. You say you don’t understand down wind faster than the wind machines. It is relatively simple. You see, there is this wind and you get energy from the wind. How hard is that? Sail boats go faster than the wind.
  6. I, for one, do not believe in down wind faster than the wind. Will you please look over my calculations that show this can not work?
  7. I am a true believer. Please support my plans to convert all car and air planes over to down wind faster than the wind machines. It will lead to world peace just like cold fusion.
  8. When Zerg returns, it will be in a directly down wind power vehicle.
  9. I have these plans for a machine that will make energy. I want to submit it to a physics journal, but no one will accept it because I am not in an academic position. Would you be the co-author and submit it for me?

Comments

  1. #1 Thomas
    May 31, 2010

    5

  2. #2 MartinM
    May 31, 2010

    5 is actually pretty close to a complete explanation of how the cart works. Think Rama.

  3. #3 spork
    May 31, 2010

    I like how you kind of ridicule things you don’t understand – even when you admit you don’t understand them. But I really like how you call us hoaxers, and then refuse to admit you even MIGHT be wrong. You dedicated an entire blog to to topic of how great it is for someone to be able to admit they are wrong, and you can’t even admit the possibility. That leads me to #2 (sorry – you asked).

    >> 8.When Zerg returns, it will be in a directly down wind power vehicle.

    >>
    9.I have these plans for a machine that will make energy. I want to submit it to a physics journal, but no one will accept it because I am not in an academic position. Would you be the co-author and submit it for me?
    <<

    Shouldn’t you at least TRY not to be an idiot?

  4. #4 spork
    May 31, 2010

    >>
    Here we come back to this “Directly Down Wind Faster Than the Wind” business… To a physicist, it looks like energy would not be conserved
    < <

    Not to a physicist that understood physics and cared to give it a moment's thought.

    >> Why are there down winders so anxious for me to say it works?

    Actually, it’s a much better story to have a physics professor claiming it doesn’t work. You’re not alone. There are other professors of physics, aero, M.E., etc. that think it doesn’t work.

    I think the videos on the Ivanpah lakebed pretty much speak for themselves. I’ve tried to give you a friendly nudge in the right direction more than once – off line. All I got was a condescending response or no response at all.

  5. #5 rob
    June 1, 2010

    at band camp, i heard that if a ddwwtf delorean hits 88 mph it goes back in time and runs over a butterfly. then Zerg will never arrive!

  6. #6 Maxx
    June 1, 2010

    You nailed it right there when you said “Why are there down winders so anxious for me to say it works? If it works, does it matter what I say?” That is the whole story, right there. They cannot prove this woks, so they must have credible physicists say that it does. They cannot tolerate anyone who says it does not work! Do you need any more proof that they are cranksters? The entire argument is based on a false premise, and that premise is that you can ever know that you are going directly down wind! The fact is, you cannot ever know this, let alone prove this, so the argument is false. There is no doubt that you can go down wind faster than the wind, and this has been well established for many decades by sailing craft, including iceboats and water craft. Going down wind faster than the wind is academic and I can make a plastic bag do this! The key word is “Directly” down wind, and this is impossible to prove unless the craft is in a wind tunnel, something these cranksters will not allow to be tested! So they run the carts at an angle to the wind and go faster than the wind and claim they have gone faster than the wind directly down wind. It is a fool’s game! If they put a small cart in a wind tunnel, it will be shown to never reach wind speed. Yes, they are nothing but cranksters and not really worth the time it takes to reply to them. Just ignore them and they will go away!

  7. #7 MikeB
    June 1, 2010

    By first hand experience I can say you are incorrect, Maxx. It’s already been proven to work but it is not intuitive and I also didn’t initially believe it would work. The team is “anxious” for the Prof. to understand how it works because the concept has tripped up a number of smart people and there is some entertainment in addressing their incorrect arguments about how the thing works.

    Small versions of the ddwfttw cart have been tested on treadmills in still air, which is the exact equivalent of directly down wind. I own one of these carts and it performs exactly as shown on several YouTube videos. I can also assure you that if a suitable wind tunnel were available the team would welcome the chance to use it.

    The fact is that this particular design is optimized for directly downwind and any angle of travel off of that reduces its performance, not improves it as you suggest.

    Pray tell how you think you can make a plastic bag travel downwind faster than the wind, powered only by the wind. This is the claim that shows you to be a crank.

  8. #8 cha cha
    June 1, 2010

    I am a physicist too. And when I heard of that DDWFTTW thing I too so absolutely convinced that it was a hoax that I even didn’t take a look at the video. Every day there are hundreds of cranks on the net that claim to have constructed a working perpetuum mobile.

    But after an hour or so I got it. It’s completely in accordance with all known laws of nature, and by no means it’s a perpetuum mobile. Indeed the physics is quite simple, and EVERY physicist should at least admit that the theory works. Here is the true physical explanation:

    As long as the wind has a relative speed over the ground, everybody is able to draw energy from that system. Since there exists no frame of reference where the wind has no speed relative to the ground, it is possible to draw energy from that system even from frames of reference that are travelling faster than the wind in the direction of the wind.

    The car works very simple: it accelerates the wind backward, thus decreasing the relative speed of the wind over the ground. So it draws energy from the wind even when it is faster than the wind.

    If Rhett Allain is a phycist, than he knows all this since long and makes a fool of all who join his published opinion that this car can’t do what is claimed.

  9. #9 Maxx
    June 1, 2010

    You are a physicist too? Maybe a cha cha physicist? If wind is moving with respect to ground, there is always energy to be had with respect to ground. But you are saying everybody can draw from this energy, no matter what frame they are in. That is not correct! If you are in frame of wind you can draw energy from the wind? If so, you can draw energy from still air while you are still with respect to air! No, you are not physicist, you are crankster and you should go away now.

  10. #10 spork
    June 1, 2010

    >>I will not discuss the physics [of the DDWFTTW cart] because it doesn’t really match with my agenda for this blog. My main goal is to look at cool stuff and use it to try to explain some physics.

    If you understood how this works, you’d find this to be extremely fitting to your proposed agenda.

  11. #11 cha cha
    June 1, 2010

    Maxx, there is no need to get insulting. I am a physicist, but since Rhett obviously is a physicist too that doen’t mean anything.

    Please, don’t change frames. Just take a safe place on the side of the street and remain inside that frame for the time of this experiment:

    1. Can you feel the wind blowing? Yes, you can even measure it. It’s blowing directly down the street. So we can tell that the wind has some kinetic energy, right?

    2. Ok, now see this car coming down the street – huuuush, there it goes! Quickly take a look at your wind speed measurement device! Can you believe that! The car DECREASED the speed of the wind over the ground. And all that has been measured inside your very same frame!

    Conclusion: the car took energy from the wind! Or did you do that?

    And now a question especially for you: Where did that energy go?

  12. #12 spork
    June 1, 2010

    So Maxx, I see you like to lie about what we would or would not do. And I see that you also like “proof by absolute assertion”. Both are bad form. I have to wonder what your qualifications are.

    As to putting the cart in a wind-tunnel – wind tunnels allow us to observe vehicles that move INTO the wind from a stationary vantage point. The treadmill allows us to observe a vehicle that’s going DOWNWIND at windspeed from a stationary vantage point. But if you have access to a wind tunnel we’ll make sure it gets testd there – if only to prove you to be a liar about what we will and will not allow.

  13. #13 Mark
    June 1, 2010

    I don’t care about this ddwfttw (an abbreviation that has MORE syllables than the actual words!!!) stuff… I want a peek at the VPython code so I can see what you did to that red ball.

    This whole idea of creating a VPython model and letting students talk about what aspects MIGHT be unphysical is a great idea for an open-ended topic loaded with potential ambiguity. Cool!

  14. #14 rob
    June 1, 2010

    cha cha i don’t think you sound like a physicist either.

    your explanation doesn’t have the physicsy feel about it. it comes off more as a lay persons regurgitation of something they don’t understand but are trying to pretend they do.

    your explanation shares a lot in common with other explanations i have read about this topic.

  15. #15 spork
    June 1, 2010

    >> your explanation doesn’t have the physicsy feel about it.

    His explanation is fine. But we understand that some people are only inclined to accept an explanation they can’t begin to follow – because it has that “physicsy” feel. Let me know if you want one of those. I’ll be happy to oblige.

  16. #16 spork
    June 1, 2010

    Incidentally, the reason cha cha’s explanation doesn’t have the “physicsy” feel is because he’s able to do what Rhett tries so hard to do – and that is to explain sometimes difficult concepts in a more intuitive manner. It’s not actually necessary to be less rigorous and innaccurate when doing so – but that does seem to be Rhett’s trademark.

  17. #17 cha cha
    June 1, 2010

    Rob, maybe my explanation doesn’t have the “physicsy” feel because since some years I am teaching to physics to absolute ignorant students in a school for carpenters :-)

    I will give you an even easier explanation for what is happening with the DDWFTTW-thing: imagine a highway going down a long hill, and a lot of cars rolling engine off downhill. And among those cars is a bicycle driver, rolling down at the very same speed as the cars. And now the bicycle driver does something funny: he pushes against a car, thus accelerating and moving faster that the car. And then he pushes himself again from the car ahead of the first one and so on. Can you see it? The bicycle is driving faster than the cars, BECAUSE it accelerated the cars backward.

    This is essentially the very same principle that is used by the DDWFTTW thing. The only difference is that the DDWFTTW thing uses air molecules instead of cars to push itself forward.

  18. #18 cha cha
    June 1, 2010

    Here is a really good graphical explanation:

  19. #19 cha cha
    June 1, 2010

    sorry, image-tag doesn’t work here: http://i46.tinypic.com/2ujo9jn.gif

  20. #20 spork
    June 1, 2010

    >> The only difference is that the DDWFTTW thing uses air molecules instead of cars to push itself forward.

    —-

    There is one other critical difference… your biker is using his muscles to push himself along the row of cars, while the cart has a sort of clever way of using the wind energy to do that. The diagram you linked actually illustrates the difference nicely. Rather than push off each car, the biker could throw a lever down in front of each car such that one end of the lever sticks into the ground, the middle of the lever is pushed by the car, and the other end of the lever whacks the biker in the butt and accelerates him.

  21. #21 Robert
    June 1, 2010

    The sociology and psychology of this puzzle is reminiscent of the Monty Hall problem. The expert’s intuition just turns out to be wrong. And as with Monty Hall, most people only get it when work through the details by themselves. (I didn’t understand Monty Hall until I started writing a program to simulate it – halfway through, it suddenly hit me why switching was the better strategy. I then developed a simple yet elegant (IMO) demonstration, which I tried out on several people. None of them found it convincing – they, too, only got it by developing their own models, analogies, etc.)

    With that caveat in mind, here’s the simplest argument I have seen for why dwfttw is theoretically possible. It’s not original, but I don’t recall who came up with it – I encountered it in a comment on one of Mark Chu-Carroll’s threads.

    We agree, I hope, that sailboats and especially iceboats can travel faster than the wind. And, crucially, they can travel with a velocity component in the wind direction that is faster than the wind. (Lots of demonstrations of the mathematics on the web – the ultimate speed is determined by drag.)

    Now let’s connect two iceboats with a long rope. Send each of them out at an angle of 45 degrees to the wind, one going northeast and one going northwest (the wind is from the south.) Just before the rope is fully stretched, have the sailors reverse their tillers so that they are going towards each other. And again, before they collide. The system as a whole is moving downwind faster than the wind, and it can do so because its individual components are travelling at an angle to the wind.

  22. #22 TT
    June 1, 2010

    >>I will not discuss the physics [of the DDWFTTW cart] because it doesn’t really match with my agenda for this blog. My main goal is to look at cool stuff and use it to try to explain some physics.

    Too late — you HAVE attempted to discuss the physics of this vehicle several times on your blog now and have gotten it 100% wrong every single time. It’s a shame that we taxpayers get such a raw deal with guys like you claiming to instruct our youth.

    #2 btw.

  23. #23 spork
    June 1, 2010

    >>The sociology and psychology of this puzzle is reminiscent of the Monty Hall problem… it suddenly hit me why switching was the better strategy.

    The Monty Hall problem actually has an interesting twist beyond that normally discussed. Most people don’t pose the problem clearly enough so that switching is truly always the best answer. If not worded VERY precisely, the answer gets more comlicated.

    >>here’s the simplest argument I have seen for why dwfttw

    We’ve used a very similar approach on many people. We use a telescoping pole rather than a rope. Surprisingly, even some scholars accept it right up to the point where we say “so this is a vehicle that goes DDWFTTW”. Others say “sure – but you could never make such a pole – so it proves nothing”.

  24. #24 Seebs
    June 1, 2010

    Conservation of energy just says that the energy has to come from somewhere. If the cart is causing air to be slowed down with respect to the earth, it is perfectly reasonable for this to result in the cart being accelerated in the other direction, with no conservation of energy problems.

    The problem here is that you’re attacking a claim that no one made. No one made a claim that they could get “free energy” from this, or that this should be used as a primary mode of transportation, or anything like that. It’s a cool trick that’s a bit counterintuitive. It’s not free energy, it’s not a violation of conservation of energy, and it’s just plain not that big a deal.

    That said! I would love to talk about this with someone who actually knows some physics and can walk me through the math, because while I’ve convinced myself it ought to be possible, and I consider the multiple GPS-verified runs to be pretty convincing, it’s certainly possible that I’ve made a mistake. Sadly, thus far, all the people I’ve run into who think DDWFTW is impossible have been unwilling to actually have a conversation about their objections. Feel free to be the first, and to prove me wrong. I assure you, if you can provide convincing arguments, I’ll happily go argue with spork for a few months.

  25. #25 spork
    June 1, 2010

    >>I would love to talk about this with someone who actually knows some physics and can walk me through the math, because while I’ve convinced myself it ought to be possible, and I consider the multiple GPS-verified runs to be pretty convincing, it’s certainly possible that I’ve made a mistake.

    I’d be happy to go through the math with you. There are several approaches, but it’s really pretty straightforward.

  26. #26 Robert
    June 1, 2010

    spork:

    “We’ve used a very similar approach on many people. We use a telescoping pole rather than a rope. Surprisingly, even some scholars accept it right up to the point where we say “so this is a vehicle that goes DDWFTTW”. Others say “sure – but you could never make such a pole – so it proves nothing”

    Yes, that’s the form of the argument that I saw on Mark’s blog, so credit goes to you. I replaced the pole with a rope precisely because it seemed more practical. Of course the rope/pole plays no essential role, it is just a crutch designed to get people thinking about how the system as a whole moves.

  27. #27 Hank Antler
    June 1, 2010

    This video shows how the prop vehicle is using exact same principle as a sailboat in broad reach. There is no perpetual motion in teh cart anymore than in a sailboat that has downwind speed component faster than the wind.

    If one accepts that a boat off wind can do it then it should be obvious to anyone with physics mind that the dilemma is in the mechanics not whether its possible.

    and the vid:
    http://www.youtube.com/watch?v=UGRFb8yNtBo

  28. #28 Fran
    June 1, 2010

    I haven’t actually seen any of this DDWFTTW stuff, but I thought of a way to think about it that may or may not be helpful.

    Imagine being a passenger in a thing moving at the SAME speed as the wind. You could go faster by pushing against something. You could push against the ground, or you can push against the air molecules (which are still, in your frame of reference).

    It sure seems to me to be more efficient to push against the ground than to push against the air molecules. That’s like trying to swim through the air. And once you are going faster than the wind, there will be air drag, and you will be trying to push backwards either against the big massive earth or against tiny molecules now moving backwards relative to you. How much energy can you get from tiny molecules if their relative speed is small compared to you, and you are not some giant low friction wind turbine?

  29. #29 spork
    June 1, 2010

    >> I haven’t actually seen any of this DDWFTTW stuff, but I thought of a way to think about it that may or may not be helpful.

    Unfortunately, your analysis isn’t quite right. Pushing against tiny air molecules can be pretty effective actually. That’s what the propellers and fans do on those giant aircraft afterall. And if you’re going downwind faster than the wind you have the added advantage of pushing on something that you’re moving slow relative to (the air) rather than the road that’s screaming by underneath. What do you think would take more work, running on an inclined treadmill that’s going 10 mph or one that’s going 15 mph. That’s the same issue as pushing on the air that you’re moving past slowly vs. the ground that you’re moving past much more rapidly.

  30. #30 Maxx
    June 2, 2010

    >>>There is one other critical difference… your biker is using his muscles to push himself along the row of cars, while the cart has a sort of clever way of using the wind energy to do that. The diagram you linked actually illustrates the difference nicely. Rather than push off each car, the biker could throw a lever down in front of each car such that one end of the lever sticks into the ground, the middle of the lever is pushed by the car, and the other end of the lever whacks the biker in the butt and accelerates him.

    Posted by: spork | June 1, 2010 6:24 PM<<<<

    Very nice fairy tale spock! The problem is that air molecules are not equipped with levers and even if they were how do you plan to get them to jump through the hoop for you? How can you make them plant their levers in the ground and kick your bike rider in the butt and do it in a continuous organized manner so he accelerates? You must give the lever to the bike rider and he must provide the energy to employ it. In other words, you are right back to the same problem of the bike rider using his own muscles and energy to push off the air molecules. The wind cannot provide the energy to lever the wind as that is a clear violation of conservation of energy, you cannot use 5 Joules of energy to get 10 joules back you will always get back less than what you use! This is all fairy tales and has no basis in fact or in physics. As for the videos, you can never prove the cart is going directly down the wind, so they prove nothing. Only way is to put a cart in a wind tunnel and then you see it can not reach the wind speed. Discussing this is a very big waste of time and is best left to the cranksters only.

  31. #31 spork
    June 2, 2010

    Sorry Maxx – you’re simply wrong on too many levels to address individually. If you actually wanted to discuss this in a responsive manner, I’m reasonably confident we could reach agreement. But it seems you have no interest in that. Prove me wrong – please!

  32. #32 cha cha
    June 2, 2010

    @ spork
    : There is one other critical difference… your biker
    : is using his muscles to push himself along the row
    : of cars, while the cart has a sort of clever way
    : of using the wind energy to do that.

    Yes, you are halfway right. But the bicyclist does not only use his muscles, but also uses the kinetic energy from the cars. The car is not a massive wall, so when the bicyclist pushes himself away from the car, the car will slow down a little bit, thus lossing kinetic energy that is used by the bicyclist.

  33. #33 spork
    June 2, 2010

    >> you are halfway right. But the bicyclist does not only use his muscles, but also uses the kinetic energy from the cars. The car is not a massive wall, so when the bicyclist pushes himself away from the car, the car will slow down a little bit, thus lossing kinetic energy that is used by the bicyclist.

    —-

    Hmmm… I’m still not entirely comfortable with the analogy because it’s no different than throwing his water bottle aft except for two details:

    1) the car presumably weighs more than his water bottle
    2) he had to get his water bottle up to speed, whereas the car was already at speed when he pushed off of it.

    This is why I like the analogy of using levers between road and car a bit better. I think it more accurately reflects how the cart manages to go DDWFTTW without having to provide any of its own energy (basically leveraging the wind against the road). Make sense?

  34. #34 cha cha
    June 2, 2010

    @ Maxx
    : The wind cannot provide the energy to lever the
    : wind as that is a clear violation of conservation
    : of energy, you cannot use 5 Joules of energy to
    : get 10 joules back you will always get back less
    : than what you use!

    You are the one who is telling fairy tales. Nobody here claims to get 10 J of energy back after input of 5 J.

    The amount of energy which can be drawn from the wind depends only on the propeller area and the difference in pressure that the propeller can create between the both sides of that area.

    The amount of energy that is needed to accelerate the car above wind speed and to keep this high speed, depends only on mass, acceleration, driving distance and friction.

    If you don’t get enough energy to accelerate the vehicle above wind speed, then simply increase your propeller. The only limit for getting more energy is the size of the wind system and the material strength of your construction.

    : Discussing this is a very big waste of time and
    : is best left to the cranksters only.

    So why are you trying to discuss this all your time?

    Just to make you a little bit more upset I will introduce another vehicle design that also could drive ddwfttw:

    It’s simpler than the original DDWFTTW design, since it doesn’t need a fixed mount between wheels and a propeller. You can use a simple soapbox design with 4 freely turning wheels. Attach dynamo systems at the wheels to get electric energy as soon as the wheels are turning. Take heat guns, remove the heat elements from them, mount them to the vehicle so that they draw the fresh air from the front of the car and blow the air backward. Wire the heat guns to the dynamos.

    To start the vehicle, simply rig a sail. If the dynamos cause too much friction, you can switch them off or increase yor sail area. When your speed comes close to wind speed, your sail is of no use anymore, so pull it down.

    At least in theory this design can move faster than the wind. The only difference to the DDWFTTW car is that it uses heat guns and not a propeller to decrease the speed of the wind over the ground. When running at wind speed, the dynamos create an amount of electric energy according to the speed of the car over ground, while the heat guns only need much less energy to accelerate the air backward, because they are already running at wind speed, making the apparent wind zero.

  35. #35 cha cha
    June 2, 2010

    @ spork
    : This is why I like the analogy of using levers
    : between road and car a bit better. I think it
    : more accurately reflects how the cart manages
    : to go DDWFTTW without having to provide any
    : of its own energy (basically leveraging the
    : wind against the road). Make sense?

    Yes, of course. From my experience there is no single analogy that is suited to explain the system equally well to all laymen. Some people will understand it this way, others will it understand better another way.

  36. #36 cha cha
    June 2, 2010

    @ Maxx
    : The wind cannot provide the energy to lever the
    : wind as that is a clear violation of conservation
    : of energy

    BTW, Maxx, every sail does exactly that: it levers the energy of the wind and uses this energy to move the boat.

    So you are telling that sailing a boat downwind is physically impossible?

  37. #37 rob
    June 2, 2010

    cha cha: didn’t mean to offend. it is just that i have seen the arguments various people have made and have been underwhelmed. i was hoping someone would post the math or link to the math that uses the conservation laws, some correct free body diagrams and momentum theory or something to derive an equation that shows that it is possible. i imagine the derivation will show the max speed of the cart depends on the windspeed, similar to how you can derive the maximum efficiency of a windfarm air turbine generator. i haven’t see that yet.

    Rhett: check out the site http://www.fasterthanthewind.org it is a blog showing the steps a group took to make a downwind prop vehicle. one of the groups is developing air wind turbines on a tether, so they appear to have the know-how to build the cart.

    they have videos of the cart on a dry lakebed. it *does* appear to be going faster than the wind. they apparently have gps and windspeed indicator that can calculate the true wind speed and apparent wind speed to show the car is going faster than the wind. i think you will like the sample plot they have of gps and windspeed data. plus, they seem to have some sort of deal with the Discovery channel to showcase the cart.

  38. #38 cha cha
    June 2, 2010

    @ Rob
    : i was hoping someone would post the math or link
    : to the math that uses the conservation laws, some
    : correct free body diagrams and momentum theory or
    : something to derive an equation that shows that
    : it is possible.

    Rob, the math is very simple, but IMO this can be expressed simpler in plain english. Math is just a language that is used by scientists to describe what can be seen. You can do that in any other language just as well.

    OK, let’s do some work:
    the power that is needed by the propeller to create a pressure difference ahead and behind itself is determined by it’s area and the apparent wind speed:
    P = dp * A * v_a
    (dp = pressure difference; A = area;
    v_a = apparent wind speed = v_t – v; v = vehicle speed;
    v_t = true wind speed (over ground))

    the power that drives the vehicle: P = dp * A * v

    excess power: P_e = dp * A * (v_a – v) = dp * A * v_t

    You can see, that the excess power that is produced by the propeller at any (!) vehicle speed always only depends on the propeller geometry and the true wind speed over ground. In plain english: as long as the wind has a relative speed over the ground, you can draw energy from the wind-ground system from ANY frame of reference, since there exists no frame where the relative speed of wind over ground is zero.

    Since we have a lot of excess power, we are able to accelerate the vehicle up to a speed where the power created by friction equals the excess power. If this is below the true wind speed, we simply need to increase the geometry of our propeller. The only limit is material strength and the size of the wind system.

  39. #39 spork
    June 2, 2010

    >>they seem to have some sort of deal with the Discovery channel to showcase the cart.

    No deal with Discovery. They simply did a story on our project. This morning Wired published this:

    http://www.wired.com/autopia/2010/06/downwind-faster-than-the-wind

    Definitely some errors in there, but that’s the reality of press.

  40. #40 J-J
    June 2, 2010

    To simplify things, suppose I have a tricycle, and I’m going to have a truck pull it. I’ve got a long rope, hundreds of meters long, and I’m going to tie one end to the rear bumper of the truck. I’m going to attach the other end to my tricycle. Can I do this in such a way that the tricycle will move faster than the truck (until it overtakes the truck, of course), or would that violate conservation of energy? If you like, we can even do this up a moderate hill.

  41. #41 hank antler
    June 2, 2010

    P=FV (power=force times velocity)

    10 m/s wind, cart going 9 m/s ddw. Apparent wind at prop is 1 m/s, apparent speed for wheels in relation to the ground is 9m/s.

    Lets see what would it look like if we got 1000N force from the prop to push the cart forward. At 1m/s (relative speed to air) P=100N*1m/s = 1000W

    How much do we have to break the wheels to get 1000W?
    F=P/V

    F=1000W/9m/s (the ground to wheel speed) =111N

    You can repeat this at different speeds. It should be pretty obvious that it really is about leverage. You can gear down the faster ground to wheel speed into slower prop to air speed and have power to spare when getting force/thrust gain.

    and mr docphysics might have realized this too if he hadn’t stuck to flat earthers’ mindset.

    And when it comes to the red dots from a gun animation – is this a lame way of twisting you being blatantly wrong not being embarrassing? Man up. Don’t pretend you were “morally right” when you were wrong. Puzzles are non-intuitive by nature – that is the whole point,

  42. #42 rob
    June 2, 2010

    cha cha, your analysis above is too simple and appears to be linear in the true wind speed, which is not the case.

    the pressure difference term (dP) in your equation is dependent on the relative speed of the air going into the propeller and the air going out of the propeller. if you draw a streamline going in and out of a propeller then keep track of the pressure and speed of a parcel of air along the streamline while conserving the momentum and energy of the parcel, you will find that the pressure difference depends on the speeds and what you call excess power P_e is at least a quadratic polynomial dependent on the actual windspeed and the vehicle speed. thus, your simple mathematical explanation is wrong and you cannot draw any conclusions about the feasibility of ddwfttw travel from it.

    the fluid dynamics and kinematics of the downwind vehicle is more complicated than any of the simple hand wavy or mathematical explanations i have seen anywhere on any ddwfttw forums. and none of them are very convincing as to the feasibilty of going faster than the wind.

    however, the videos of the vehicle on the dry lake bed certainly do seem to show experimentally that faster than wind travel is possible. experiment trumps theory.

    but i would still like to see a good theoretical solution to the problem.

  43. #43 cha cha
    June 2, 2010

    @ Rob
    : the pressure difference term (dP) in your equation is
    : dependent on the relative speed of the air going into
    : the propeller and the air going out of the propeller.

    No, Rob, the basic principle of the energy flow indeed is almost as simple as I wrote it. The excess power depends on the true wind speed, no matter which speed the vehicle is going. This is the only important thing to understand. And I think you forgot that the propeller only works against the apparent wind speed, which is (v_t – v), while the propeller itself is always driven in dependency to v. The difference between (v_t – v) and v is always v_t, qed.

    BTW: The term dp (a small p for pressure, not the big P for Power) depends on the geometry of the propeller and of course on the propeller’s rpm, which again depends on vehicle speed v.

  44. #44 rob
    June 2, 2010

    there is another subtlety to the ddwfttw problem that i never see addressed either.

    everyone always talks about the kinetic energy available in the wind. it turns out it is not possible to extract all the wind kinetic energy to power the cart, yet most everyone does their analysis with this assumption.

    the reason is because in addition to energy being conserved, momentum has to be conserved. when you couple the two conservation principles together when analyzing how much wind energy is available, you will find that there is a maximum kinetic energy available to extract to convert into the kinetic energy of the cart.

    here is a related example that illustrates this.

    as the wind blows through a wind turbine generator, the generator cannot extract all the kinetic energy available in the wind. there is a limit, called Betz’s limit. you can derive it and show that a turbine cannot extract more than 59% of the wind power.

    this means is that if someone has a solution showing how ddwfttw carts are possible. if they haven’t considered momentum conservation, then the solution is wrong.

  45. #45 rob
    June 2, 2010

    cha cha:

    check out http://www.grc.nasa.gov/WWW/K-12/airplane/propth.html.

    that site shows the thrust of a propeller that is powered by an engine. for the ddwfttw cart just replace the engine with a linkage to the carts wheels where power is used and not produced. (thus, it is really a negative sign difference)

    you will see that the dP term in the thrust equation is a quadratic function of the velocities. there is an analgous derivation for the dwfttw carts with the same sort of velocity dependence.

    your P_e equation above does not take this into account, and therefore is incorrect.

  46. #46 cha cha
    June 2, 2010

    @ Rob
    Thrust is a ***force***, it’s measured in Newton (kgm/s^2). I calculated the ***power*** that is necessary to create a pressure difference. Power is measured in Watt (kgm^2/s^3). Indeed it is completely irrelevant how this pressure difference is created, it has nothing to do with a propeller or any other technical device.

    My calculation is only to show that there is plenty of excess power, which can be used to accelerate the car above wind speed, and that this excess power is a linear function of the true wind speed, and that it is completely independend from the speed of the vehicle.

    If you want to calculate the real thing, you need to kow many, many parameters, and among those you need to know the efficiency of the real propeller, which again depends on wind speed over the propeller blades (which is a function of apparent wind speed, rpm, and geometry). We don’t have any knowledge of these values. For your purpose it should be sufficient when you know that a propeller can achieve as much as 90% efficiency.

  47. #47 spork
    June 2, 2010

    Rob, you seem to confuse analyses you can’t follow with “hand wavy” or wrong. It’s simply not the case. You’re applying incorrect principles (such as Betz limit) to a propeller (we don’t HAVE a turbine), and then concluding that our explanations are incorrect. I assure you I’ve posted straightforward and perfectly accurate analyses of this thing many times.

    If you actually want to understand what’s going on here – ask.

  48. #48 rob
    June 2, 2010

    @cha cha:

    power can be expressed as force times velocity. if you write that out:

    Power=(thrust of propeller)*(speed of cart)

    if compare it to your equation for power, you can see that the dp*A in your equation *is* the thrust. so it really is okay to consider the thrust, since they are related.

    and yes, it *is* relevent how the pressure difference is created, since that pressure difference applied over the area A of the propeller is what actually makes the car move. if you don’t understand the pressure difference that the propeller generates you cannot understand how the cart can even move, let alone outpace the wind.

    to understand how the car moves you need to derive the pressure difference using momentum theory. then you will see that the pressure difference is directly related to the incoming air velocity squared and the air exit velocity squared, and not just the true wind speed like you claim.

    your calculation does not show that there is plenty of excess power. for instance, what happens if the two squared velocity terms in dp are equal? the power is zero.

    also, your caculation assumes extracting *all* the kinetic energy available in the wind. i discussed above that you also need to consider momentum conservation cause you *cannot* get all the kinetic energy out of the wind. (to do so would acutually make it a perpetual motion machine)

  49. #49 cha cha
    June 2, 2010

    Rob,

    I tried to explain to you that there is no quadratic function of velocities. I understand quite well what my equations are telling, indeed I made them up for myself.

    And no: it is NOT important, how the pressure difference is created, when you want to know how much power you need to achieve that. The power is simply pressure difference times area times speed. And this is ALWAYS the same, completely independend from the machine you are using. The only difference between different machines is their efficiency, and the efficieny of a propeller is very high.

    And yes, Rob, I know quite well that pressure is force by area, and that pressure times area is force, and that this force is driving the car. But you don’t seem to understand that I don’t need to know how much pressure difference the propeller makes. I only need to know the ***relation*** between the power that is created by the wheels and the power to run the propeller at that speed. The simple answer is: I get much more energy than I need. And so I have plenty energy over to accelerate the car over the given speed.

    : your calculation does not show that there is
    : plenty of excess power. for instance, what
    : happens if the two squared velocity terms
    : in dp are equal? the power is zero.

    There is no squared velocity term, since I don’t calculate dp. I am calculating the necessary power at any given dp. I don’t care how you create the pressure difference, because for my calculation this is completely irrelevant. I need less power to drive the vehicle with the propeller than I get from the wheels. That’s all I need to know.

  50. #50 spork
    June 2, 2010

    >> I need less power to drive the vehicle with the propeller than I get from the wheels. That’s all I need to know.
    <<

    Yes, it’s just about that simple. If you consider the typical efficiencies involved (transmission, prop, rolling resistance, aero drag) and exactly what they mean, you can do a very simple energy analysis and find that you’ll be accelerating handily when at wind speed.

    There’s no more need to go into momentum theory here than there is to do so on a Cessna. Prop performance is pretty well understood.

  51. #51 J-J
    June 2, 2010

    It’s certainly not necessary to extract all of the energy from the wind. For starters, how much power does it take to have the cart moving along at wind speed, steady state? Pretty close to zero. DDWFTTW requires extracting only a modest amount of power from the wind in order to have it move a bit faster than that. 3x wind speed DDW takes some good engineering.

  52. #52 cha cha
    June 3, 2010

    @ spork

    : >> I need less power to drive the vehicle with
    : >> the propeller than I get from the wheels.
    : >> That’s all I need to know.
    :
    : Yes, it’s just about that simple.

    … at least for the physicist. Then comes the engineering stuff … :-)

  53. #53 rob
    June 3, 2010

    cha cha:

    “There is no squared velocity term, since I don’t calculate dp. I am calculating the necessary power at any given dp.”

    question for you: is dp constant?

  54. #54 cha cha
    June 3, 2010

    Rob, you don’t seem to read what I wrote: my equations are valid for ANY given dp.

    dp is the cause for the force that drives the car. The car will always adjust it’s own speed over the ground in a way, so that the given dp creates exactly the force, that is lost by friction and other efficiency losses. Always! No matter if the speed over the ground is zero, if it is wind speed (so that the apparent wind speed on the car is zero), or if it is quite well above wind speed.

    At ANY speed of the car we have excess power which is a function of the true wind speed over the ground. If the excess power is not sufficient to accelerate the car above wind speed, then simpy increase the size or improve the geometry of your propeller. It’s as simple as that!

  55. #55 jr
    June 3, 2010

    @cha cha #38

    Assuming I follow you correctly, I think your maths might be broken.

    So, I think your P_e is calculated as your prop power minus your power needed to drive the vehicle at v. so you end up with P_e = dp*A(v_a-v) and then you say that this is equal to dp*A*v_t. But you define v_a as (v_t-v) so shouldn’t your P_e actually be equal to dp*A(v_t-2v)?

  56. #56 rob
    June 3, 2010

    cha cha:

    all i wanted to know is if dp in your equation is constant.

    yes or no.

  57. #57 MikeB
    June 3, 2010

    Rob, you wrote “power can be expressed as force times velocity. if you write that out:

    Power=(thrust of propeller)*(speed of cart)”

    But you appear to have left out a critical detail…wind speed over the ground.

    Power available taken from the wheels of the cart = (drag force at wheels)*(speed of cart over ground).

    At the propeller, power=(thrust of propeller)*(speed of cart with respect to the air).

    Allowing for some frictional losses and propeller inefficiency, it still only requires a moderate wind speed for the downwind system to work in several machines we have seen.

    Consider the cart moving downwind at wind speed. Power delivered to the propeller will then be used to produce “static thrust”. You can assume some cart ground speed and wheel drag force, calculate power, apply that to a reasonable prop size, calculate static thrust or obtain it from a chart of propeller performance and see that thrust will exceed drag and therefore push the cart beyond wind speed.

  58. #58 cha cha
    June 3, 2010

    @ rob

    : all i wanted to know is if dp in your equation is
    : constant.

    I know that you wanted to know that, since your question was expressed in understandable words. And I told you and explained to you that this is completely irrelevant. Didn’t you read that? Since my equation isn’t a differential equation, it can’t be used to calculate the dynamics. It’s just good for calculating the excess power at and given speed. A certain speed v always corresponds with an exactly defined dp, since dp only depends on v and v_t.

    It’s absolutely trivial that dp or the propeller thrust depends on the prop’s angular frequency, and it’s also obvious that also the apparent wind speed at the propeller plays an important role. Since the prop’s angular frequency is a linear function of the vehicle speed v, and since the vehicle speed is not constant, since the vehicle accelerates from 0 m/s to v_max, it is an absolutely trivial observation that dp is NOT CONSTANT as long as the vehicle continues it’s acceleration. When the car reaches it’s final speed, where friction and air drag equal dp*A, dp remains constant.

    Are you happy with that?

  59. #59 cha cha
    June 3, 2010

    @ JR

    : But you define v_a as (v_t-v) so shouldn’t
    : your P_e actually be equal to dp*A(v_t-2v)?

    Uups … I probably did put something into the wrong order. I am used to sail my boat upwind and calculate the apparent wind speed in absolute values :-). I’ll take a close look at the mistake later. Thank you.

  60. #60 rob
    June 3, 2010

    @Rhett:

    it’s like talking to Nigel Tufnel. the only answer you get is “These go to eleven.”

    i shoulda saved time with your option 5.

  61. #61 hank antler
    June 3, 2010

    @Rob – don’t runaway like that. You have been presented with very reasonable explanation – not “it goes to 11″ faith based crap.
    think this: FV (on prop) = FV (on wheels) (-losses of course)

    It is a really simple device that works exactly on same principle as a sail on a faster than the wind boat.

    Lets take a boat that travels SouthWest on wind from North. It reaches a speed at which its speed component directly South matches the wind speed. What is the apparent wind like on the boat at this condition? Directly from West is the answer. The sail will have airflow over it and will act as a wing and help the boat gain speed past and beyond wind speeds. Where does the energy come from? Well it is affecting whole lot air and slowing it down.

    same way the cart has an airfoil (prop blade) that is restricted to traveling off angle to the wind (as it spins) and because of that spinning it will experience apparent wind past the wind speed.

    Its the naysayers who seem to ave decided that its non-physical or impossible and refuse to read explanations carefully.

  62. #62 rob
    June 3, 2010

    cha cha:

    “When the car reaches it’s final speed, where friction and air drag equal dp*A, dp remains constant.”

    yes. at it’s the cart’s final speed dp is constant. the net force on the cart is zero and it will travel at constant speed. but your analysis is incorrect and you cannot conclude from it that the final speed is greater than the wind speed.

    your analysis shows the power a function of windspeed, which you rightly conclude means as long as there is wind the car has enough power available to go faster than the wind.

    however, as i pointed out, the dp term in the power is not a constant until final speed is reached. until then it is a function of v. that means you actually do have to consider how dp changes as the carts velocity increases.

    since the dp term varies as the square of the cart’s velocity, the power is a 3rd degree polynomial in v. since a 3rd degree polynomial can have a relative max and min, it could be that the power goes to zero or hits a minimum as cart speed goes up. in that case, when the wind power input and friction power loses balance, the cart speed may very well be less than wind speed. until the mathematical details are worked out, you cannot say for sure what the case may be.

    i ran into some snags when i was working out the math, and put the problem aside. however, now that i saw the video it really does look like the cart can go faster than the wind, which is cool. now i want to re-explore the problem to see if the upper speed depends only on dissipative forces or if it is some multiple of the wind speed. that is why i have looked into momentum theory as it is applied to wind turbines and propellers. the analysis of the cart is going to be very similar to the analysis of turbines and airplane propellers.

  63. #63 cha cha
    June 3, 2010

    @ Rob
    : however, as i pointed out, the dp term in the
    : power is not a constant until final speed is
    : reached.

    As I said: that’s trivial. When the car stops or runs at very low speed, there is almost no dp – in this case the propeller blades act as walls. This makes it very difficult for the car to start just by wind energy.

    But all this is completely irrelevant, as I pointed out several times already. When you don’t understand this approach to the problem, another approach may be much more helpful. Here is another one:

    You seem to forget that the wind blows all the time and never stops. The car is driven by the true wind AND the propeller. The car’s speed is the source for the energy that drives the propeller. But the propeller doesn’t need this amount of energy to run the car, because it get’s support from the true wind. The propeller works only against the apparent wind speed. So it’s obvious that there always is excessive power as a function of the true wind speed.

    Or another approach: Consider the propeller working like a rocket: it accelerates air backward, and because of impulse conservation the rocket is accelerated forward. Again the true wind helps, so the energy collected from the wheels is bigger than is needed by the rocket engine. Again you get excessive power as a function of true wind speed.

    IMO the last approach is the easiest to understand that the real source of energy for the car is the true wind speed. The propeller allway throws air backward, thus slowing down the real wind speed. After the car drove through it the whole wind system has less energy than before.

  64. #64 MikeB
    June 3, 2010

    It’s become pretty obvious that the speed of the cart is limited by propeller efficiency and various frictional losses but that it can certainly accelerate past wind speed. It’s been pointed out several times that, assuming some wind speed, the power available from drag at the wheels when applied to the propeller will produce static thrust in excess of the drag. I don’t know if you would consider the loss of thrust, at constant propeller horsepower and increasing airspeed, to be part of what is termed “efficiency”. But it’s interesting to note that, at a constant wheel drag the power available goes up directly with speed over the ground.

  65. #65 spork
    June 3, 2010

    >>it really does look like the cart can go faster than the wind

    It really can.

    >>now i want to re-explore the problem to see if the upper speed depends only on dissipative forces or if it is some multiple of the wind speed.

    It is both. For a fixed pitch and gearing there is a “design speed” that is a given multiple of wind speed. If I can vary the gearing and prop pitch, then the top speed as a multiple of wind speed is limited only by the internal losses. In other words – it could theoretically go any multiple of wind speed (given low enough losses).

    >> that is why i have looked into momentum theory as it is applied to wind turbines and propellers.

    I think you’re over-thinking it. Prop performance numbers are readily available.

  66. #66 jr
    June 4, 2010

    @cha cha

    OK. Well if my interpretation of your model is correct, then your model does not do a very good job of explaining ddwfttw.

    For a given value of dp it means that as the vehicle speed increases, the additional power generated (P_e) decreases. This continues, so that by the time the vehicle is travelling at half the windspeed, P_e = 0. As the vehicle speed increases further, then P_e becomes negative and continues to be negative as the vehicle speed exceeds the wind speed.

    I don’t see how that produces ddwfttw. It actually mirrors pretty much the intuitive explanation for why ddwfttw probably shouldn’t work.

  67. #67 rob
    June 4, 2010

    @jr:

    i agree. i focused on the problem of ignoring the pressure difference across the prop, but i did notice what you did. i think it is a problem with defining +/- for the directions of the vehicle velocity, wind velocity and apparent velocity.

  68. #68 cha cha
    June 6, 2010

    @ jr

    : As the vehicle speed increases further,
    : then P_e becomes negative and continues
    : to be negative as the vehicle speed exceeds
    : the wind speed.

    My mistake is a rather simple one: v is not the vehicle speed, but is the same value but negative.
    So the power that drives the vehicle at speed v is:
    P = dp * A * v_d
    with v_d = -v.

    excess power: P_e = dp * A * (v_a – v_d) =
    = dp * A * (v_a + v) = dp * A * v_t

    It’s really quite obvious that there is always excess power that depends on v_t, since the wind never stops blowing. It’s like a plane with wind from the back …

  69. #69 Cloxxki
    June 16, 2010

    1.

    Wow, that’s a lot of responses.

    “Props” for finally formally admitting to have been wrong.
    Calling people wrong on the internet is the largest global sport, and as you present yourself as a man of scientific education, you’re a popular target to get this wrong.

    It’s indeed not so hard. How moch energy does it cost to push an aerodynamic cart, on wheels, forward at 3x wind speed? And how much drag does a prop have, when it turns like the DDWFTTW cart’s does? In total, the amount of wind available through the prop, wil come pretty darn near that figure.
    The hard to understand part, is probably how the whole thing goes faster than the wind then. If you figure in that the prop covers most of the frontal surface area, and that it’s actually turning fast enough to be thrusting backwards rather than acting as a turning garage door for drag…. You see that the large prop (think for ease of understanding of a 17 foot turbine) is extracting energy sufficient to power a funny looking cart with tall useless bridge on it, up to near 3x wind speed. Not strange at all, compare the surfaces. The funky gearing allows to take the “turbine” on board, but it will need to be used as prop rather than as turbine, or it won’t work as well :-)

  70. #70 Cloxxki
    June 16, 2010

    I have another bicycle analogy to offer.

    Imagine not a car rolling alongside the bike, but a tank, on caterpilar style tank tracks. Rolls just fine otherwise.

    Now the cyclist has a messenger back, tied around his torso. But there is another strap, for regular carrying flying around, which manages to get caught up in the upper tank track alongside him.

    This tank track is laying STILL on the ground, and moving twice the vehicle’s ground speed, forward, on the upper side. So, the cyclist is nicely slung forward, and he may just hope that the strap unties itself before the track track goes for a short rest on the road surface.

  71. #71 Commenter
    June 17, 2010

    Rhett: “(…) but you get energy from the wind. I still don’t get it.”

    Rhett: “I will not be discussing the physics of this machine. (…)I tried that before with this machine and failed.”

    Maybe MIT professor Mark Drela can help you:
    http://web.mit.edu/aeroastro/news/magazine/aeroastro-no3/2006drela.html
    His analysis:
    http://www.boatdesign.net/forums/attachments/propulsion/28167d1231128492-ddwfttw-directly-downwind-faster-than-wind-ddw2.pdf
    His conclusion:
    “This confirms that the DDWFTTW condition V/W > 1 is achievable with a wheeled vehicle without too much difficulty”

    Seriously Rhett: How do you want to teach physics, if you don’t understand it yourself, and even worse: are not willing to learn?

  72. #72 Maxx
    June 19, 2010

    Posted by: hank antler: < <<< 10 m/s wind, cart going 9 m/s ddw. Apparent wind at prop is 1 m/s, apparent speed for wheels in relation to the ground is 9m/s.
    Lets see what would it look like if we got 1000N force from the prop to push the cart forward. At 1m/s (relative speed to air) P=100N*1m/s = 1000W
    How much do we have to break the wheels to get 1000W?
    F=P/V
    F=1000W/9m/s (the ground to wheel speed) =111N
    You can repeat this at different speeds. It should be pretty obvious that it really is about leverage. You can gear down the faster ground to wheel speed into slower prop to air speed and have power to spare when getting force/thrust gain.
    and mr docphysics might have realized this too if he hadn't stuck to flat earthers' mindset.
    And when it comes to the red dots from a gun animation - is this a lame way of twisting you being blatantly wrong not being embarrassing? Man up. Don't pretend you were "morally right" when you were wrong. Puzzles are non-intuitive by nature - that is the whole point,>>>>

    This is the type of vacuous argument that firmly convinces me this ddwfttw cart is a farce and a hoax! The claim being made here is of a 1000 Newton force to push the cart forward while only developing a111 Newton drag force at the wheels! Clearly whoever wrote this nonsense does not even understand a simple lever, but unfortunately he is not alone, as Mark Drela has made exactly the same mistake in his analysis!

    If there is a 100 Newton force on the air over a distance through the air of one meter, the Work done on the air is 1000 N/M (1000 Joules). This is also the Work done on the cart, as the cart moves a distance of 9 meters over the ground. The force on the cart is not 1000 Newtons, but is calculated as 1000 Joules / 9 meters = 111 Newtons. That is exactly the same as the drag felt at the wheels!

    You cannot claim the propeller is acting as a lever because of unequal distanes travelled and still claim to have equal force on both the cart and the air! You cannot have it both ways!!! If the distance travelled through the air and on the ground is unequal, then so is the force felt on the air and on the cart.

    With the propulsive force equal to the drag force, there is only the resistive force left to work on the cart and it will slow down until it is below the wind speed. It will slow down until an equilibrium is reached between between all forces acting and this will be considerably less than the speed of the wind.

    The more I read about this nonsense, the more I feel it is time for Mark Drela to step forward and take some responsibility for this internet farce. It is time for physicists to stand up and denounce this as the work of crackpots. I respect Rhett Allain because he has not fallen for this farce!

  73. #73 Commenter
    June 21, 2010

    @ Maxx

    “The claim being made here is of a 1000 Newton force to push the cart forward while only developing a 111 Newton drag force at the wheels!”

    So what? There is no law of force conservation.

    “…Mark Drela has made exactly the same mistake in his analysis!”

    It is not a mistake. The MIT professor just understands physics better than Rhett and you. Any decent physicist/engineer comes to the same conclusion as Drela. Here a lecture from the Technical University of Denmark:

    http://www.youtube.com/watch?v=4ZjX_DIosM8&feature=PlayList&p=6C046C08399626CA&playnext_from=PL&index=0&playnext=1

    (DDWFTTW starts in the middle of part 4)

  74. #74 Maxx
    June 21, 2010

    Cowardly Commenter wrote: < < @ Maxx

    "The claim being made here is of a 1000 Newton force to push the cart forward while only developing a 111 Newton drag force at the wheels!"

    So what? There is no law of force conservation.

    "...Mark Drela has made exactly the same mistake in his analysis!"

    It is not a mistake. The MIT professor just understands physics better than Rhett and you. Any decent physicist/engineer comes to the same conclusion as Drela. Here a lecture from the Technical University of Denmark:

    http://www.youtube.com/watch?v=4ZjX_DIosM8&feature=PlayList&p=6C046C08399626CA&playnext_from=PL&index=0&playnext=1

    (DDWFTTW starts in the middle of part 4)>>

    Yes there is no law of conservation of force, that is exactly my point! The simple law of the lever, which you obviously do not understand, is the work is the same on both ends, as distance is exchanged for force.

    Once again, the claim is that the propeller travels a lesser distance thru the air than the cart travels over the ground. That is OK so far. However, the claim is also that the force exerted on the air is the same force that propels the cart forwrd! That is obviously nonsense. If the force on the air is 1000 Newtons over a distance of 1 meter, that is 1000 Newton/meters of Work done on the air and that is also the work done on the cart! If the cart moves 9 meters, the force on the cart will be 1000 N/M divived by 9 meters, or 111 Newtons of propulsive force, as I stated. With no losses, that is the same as the wheel drag. In reality, the wheel drag will be greater than 111 Newtons. In any case, frictional forces will work against the cart to slow it down.

    Please do not argue from authority, as Mark Drela said this or that! Drela is wrong!!

    Let him come on this forum to defend himself, if he has the courage. Meanwhile, you should learn some physics yourself and stop believing everything you see on youtube.

    Stop believing in crackpots and fairy tales and argue this from physics, if you can!

  75. #75 Commenter
    June 21, 2010

    @ Maxx

    “However, the claim is also that the force exerted on the air is the same force that propels the cart forward! That is obviously nonsense.”

    Newtons 3rd Law:

    The force exerted on the cart by the air (propeller thrust), is of the same magnitude as the force exerted on the air by the cart. The same applies to the interaction with the ground, but the force magnitudes can be lower there, due to higher relative speed (gearing). Therefore propeller thrust is greater than breaking force.

    “Drela is wrong!!”

    I don’t see any errors in professor Drela’s analysis. Nor in Gaunaa’s university lecture.

    “Stop believing in crackpots and fairy tales and argue this from physics, if you can!”

    That is what Rhett should finally do. But he apparently gave up on understanding physics. And you just are spewing insults and gibberish, making Rhett look even worse, with one crazy guy as his last supporter.

  76. #76 Maxx
    June 21, 2010

    @ Commenter

    Please! Don’t tell me about Newton’s Third Law of equal and opposite forces while at the same time claiming leverage! Learn what it means to have leverage and then you see how silly your claim is! How can you claim the distance the propeller is travelling is less than the cart is travelling and then go on to claim equal and opposite forces? Do you realize how stupid that is? Once again, and try to understand this time please, if the distance the propeller is travelling thru the air is less than the distance the cart is travelling over the ground, the leverage is at the propeller/cart interface. The force on the air is greater than the force on the cart because the work is the same.

    Drela is wrong and once again let him speak for himself if he has the conviction.

    Stop claiming authority and argue the physics but first you need to understand simple leverage.

  77. #77 Commenter
    June 21, 2010

    @ Maxx

    “Don’t tell me about Newton’s Third Law of equal and opposite forces while at the same time claiming leverage!”

    Why not? Because you are not able to keep more than two force apart? I’m using only magnitudes here:

    Leverage :
    cart_force_on_air > ground_force_on_cart

    Newton 3rd:
    cart_force_on_ground = ground_force_on_cart
    cart_force_on_air = air_force_on_cart

    Taken together that means:
    air_force_on_cart > ground_force_on_cart
    in words:
    Propeller thrust is greater than breaking force

  78. #78 eyytee
    June 27, 2010

    Rhett,

    I’m not sure what you are trying to tell us with your ball animations. But here is an animation that explains DDWFTTW using balls (which represent air molecules):

    http://www.youtube.com/watch?v=vVMqa7Mft0k

  79. #79 Commenter
    July 29, 2010

    From the North American Land Sailing Association site
    http://www.nalsa.org/

    “Get out your slide rules and physics text books…On July 2, 2010 on El Mirage Dry Lake, Blackbird sailed directly down wind at a speed of 27.7 mph in a 10 mph wind to set a first record for the ratio of Boat Speed to true wind speed of 2.8. BlackBird was designed and built by the Thin Air Designs team (Rick Cavallaro and John Borton) and sailed by Rick. Links to follow soon.”

  80. #80 S.K.Graham
    August 30, 2010

    Hi Rhett,

    As a physicist (in larval stage, or is it pupae?) I think I can explain this in terms that will eliminate your intuitive objections.

    Ask yourself: can ball moving with velocity Vi1=[1,1] collide with a ball moving with Vi2=[.5,0] in such a way that the velocity of the first ball, after collision is Vf1=[vf1x,vf1y] where vf1x>1?

    It is easy to set up initial conditions such that a collision occurs with this result. Of course ball #2 must start out ahead of ball #1 in the x direction. Start the balls at [0,0] and [.41,1] respectively, and give them a radius of .05 each. Give them any relative mass you like. Make it an elastic collision. Final x component of velocity vf1x will be >1.

    This is an example of a ball (ball #2) with a slower x velocity imparting additional x velocity to a ball (ball #1) that already has greater x velocity than ball #2.

    Step 2. Now suppose the first ball is *constrained* to move along the line through [0,0],[1,1] (the 45o line through origin) — suppose it is a bead on a rigid wire. Taking constraints into account, can we set up the collision such that vf1x>1 and vf1y>1?

    Again it is easy to set up initial conditions that result in such a collision. The conditions I gave for the first case are an example. The constraining forces between bead and wire allow exchange of momentum between bead & wire and the wire will exchange momentum with whatever it is connected to (the earth, ultimately), so conservation of momentum is not an issue here.

    I’m sure you can create some VPython scripts that illustrate both cases above quite nicely. Completely physical.

    Now, ball #1 is like a propeller blade of the proposed DWFTTW device. It is not moving straight downwind (x direction), it is moving at an acute angle to the x direction. Ball #2 is like a molecule of air. There is wind ahead of the cart (or boat) so there are always molecules of air that the cart (or boat) can “catch up to” and collide with such that the molecule of air is making contact on the “upwind” side of the propeller blade (or sail), and you get a result such that vf1x of the sail can increase. (but what about collisions on the propeller blade from the “downwind” side that could push it backward? see below)

    For a sailboat moving in a straight line acute angle from the x-direction (direction of wind is positive x direction) the keel provides the constraint (approximately but sufficiently) to the linear path. For the wind-cart, the fact that the propeller and wheels are connected by a transmission provides an equivalent constraint — the propeller blades are rigidly constrained to move in a specific spiral path (path has shape of coil spring).

    Now. A sailboat on an acute angle tack, or a propeller blade on its spiral path, with x-component greater than wind, is “catching up” to the air molecules that are ahead of it in the x direction. Why is it not also colliding with air molecules on to sail’s (or blade’s) upwind face and getting “knocked backwards”. The answer to this is easier to understand with a tacking boat. Say the boat is moving at 45o. The plain of the sail (for simplicity assume the sail is a rigid board, which would work just fine for this purpose) must be at an angle greater than that 45o but small enough such that the X-intercept of the plane of the sale increases slower than the speed of the wind (equivalentlly, the plane of the sail must have an angle less than the angle of the apparent wind in the sailboat or propeller blade’s frame of reference). It is easy to see there is always such an angle, because the rate of increase of X-intercept of a sail at same angle as boat (sail parallel to keel) will not change — so we can have an X-intercept as small at we like. In other words, the plane of the sail falls behind the molecules of air that are directly ahead, so it does not collide with them. It catches up to and gets in front of molecules of air that are displaced laterally (relative to wind direction) in the Y direction.

    This is definitely not intuitive, but neither conservation of energy, nor conservation of momentum, are violated. There is no “free energy” or “perpetual motion” here. The DWFTTW car (or boat) is not useful for much other than racing downwind.

    See my blog for an intuitive cartoon-vector diagram showing the basics of a sailboat moving at roughly 45o to the wind, with a windward component of velocity greater than the wind, but still getting a net forward thrust (2nd diagram): http://rightnice.blogspot.com/2010/08/racing-wind.html

    The youtube video linked a couple comments up is an attempt to illustrate that same points I’ve made here, but I think it is a bit too visually confusing — that video would not convince me until after I had worked out the physics myself. One nice thing about that video is it illustrate my point about the plane of the propeller blade moving slower in the x-direction than the wind (bear in mind that the video is in the cart’s frame of reference).

    I think people are interested in changing your mind on this subject because you are a physicist with a higher-than-normal public profile (which, granted, is not saying much) due to your blog. There is also, no doubt, a desire to see you eat crow, due to your “Mythbusters” comment. :)

    (for what it is worth, only a week ago, I was entirely unaware of this question of DWFTTW, and I have no vested interests, other than that it is an interesting, counter-intuitive, bit of physics)

  81. #81 S.K.Graham
    August 30, 2010

    I should add the the *constraints* in the above are critical. Without the constraint of the ball (bead on wire) any collision that gave it additional x-velocity would also reduce its y-velocity, and if its y-velocity is zero, *then* it can never again get a collision-from-behind by a slower moving ball (which all are moving in X direction only).

    Without a constraint on the motion of a tacking sailboat provided by the keel, the lateral forces would push it sideways, reducing it’s lateral component of velocity. If it only has windward velocity, then the sail will be slack if speed >= windspeed and the boat would slow (or rather, it could never reach such speeds in the first place unless towed or using a motor).

    Without a constraint on the motion of the prop blades, forces on the blades would quickly cause it to stop rotating. The blades then would only have an X-component of motion and no longer benefit from collisions-from-behind from air molecules too slow to catch up. (or rather, again, it could never reach speed >= windspeed without a motor, tow, or other propulsion)

    The keel, and transmission+wheels friction with ground all provide constraints critical to making the DWFTTW device work.

  82. #82 S.K.Graham
    August 30, 2010

    wait wait… I want to change my comment to #8

    Zerg will conquer all!

  83. #83 S.K.Graham
    August 31, 2010

    Minor correction in my numbers… initial position for ball#2 is [.39,1], not [.41,1]. Just in case anyone actually works this out.

    Note that in the case of the ball, there are many possible collision which do not accelerate ball#1 in the x or y direction, but rather the opposite.

    However, the sail or prop blades are approximately planar and oriented at angle such that all collisions on the planar surface provide thrust in positive direction along the constrained path, neglecting collisions at the edge of the sail or blade which are negligible compared to the collisions along the surface.

    This explanation also assumes frictionless collisions, but additional drag can be made low enough in practical applications such that nothing intrinsically changes.

  84. #84 Michael C
    September 14, 2010

    Rhett, you write “Why is there no wikipedia article for down wind faster than the wind?”. There is one now:

    http://en.wikipedia.org/wiki/Sailing_faster_than_the_wind#Sailing_dead_downwind_faster_than_the_wind

  85. #85 Matt Ryan
    November 8, 2010

    I have a question: say we modified the DDWFTTW cart so that its propeller were enclosed in a vacuum-filled box. The propeller of this vehicle would thus be unable to provide drive, however some of the Slower-Than-The-Wind drive experienced by the cart would still go into spinning up the prop. Would this cart’s ground speed still approach the wind speed?