Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

How can an vehicle move faster than the wind that is powering it?

  1. Oct 20, 2008 #1
    As I'm sure we all know, that when you go sailing your fastest speed isn't directly downwind but when you are traveling at an angle to it. This allows sailboats and iceboats to go several times faster than the wind speed that is powering them.

    But how is it that a vehicle such as THIS, that is powered by the wind, can go DIRECTLY downwind faster than the wind that is powering it? Apparently, this is true of iceboats as well. Could someone please explain this so me.
  2. jcsd
  3. Oct 20, 2008 #2


    User Avatar
    Science Advisor

    It's probably a hoax. I suppose the principle on which it is meant to operate is that the blades are at an angle to the wind, and this allows them to go many times faster than the wind (like a sailboat or iceboat going cross-wind). But of course, once the entire contraption gets going at the same speed as the wind, it will experience no wind at all. What powers it then? I couldn't guess, but whatever it is, it proves that this vehicle can travel when there is no wind at all.
  4. Oct 20, 2008 #3
    It's a hoax all right. But the idea can be: there's a flywheel which can store energy when the wind is strong, then when there's weak or no wind, the flywheel makes the cart move.
  5. Oct 20, 2008 #4


    User Avatar
    Science Advisor

    I stopped reading when they put the car on a treadmill. I wonder if it took off?
  6. Oct 20, 2008 #5
    Pixel, consider this as an ideal situation not an realistic one. This "contraption" should still be able to constantly move faster than the wind assuming the wind has a constant velocity. Also consider the ground to be perfectly level and the cart's motion is only being influenced by the wind.

    Also, does this also hold true for an ice boat? Consider the same phenomenon if the iceboat has a zig-zag type path and its magnitude velocity is greater than that of the powering wind. Could the ice boat's velocity component in the direction of the wind ever exceed the speed of this wind?

    This isn't a trick question or a riddle, but a simple application of physics. There are a lot of people that believe this is not a hoax, but I consider most people on this forum to be experts in the area of physics so I would like your expert opinions.
  7. Oct 20, 2008 #6
    Looks like a hoax with an RC car pulling it with some fishing line.
  8. Oct 20, 2008 #7


    User Avatar
    Science Advisor
    Homework Helper

    Ironically it could go faster than the wind if the wind was from the side.
    Then the power only depends on the wind speed and the size of the turbine - which you can make arbitrarily large.
  9. Oct 20, 2008 #8
    Yes, I agree. And is why sailboats typically do not move directly down wind in order to achieve fast speeds.

    I found this topic on anther forum, and this is one of the forum members explanation/logic about how this thing might work.

    Is their something wrong with this persons logic? Does it disobey any physical laws?
  10. Oct 20, 2008 #9
    You have to let go of the idea that it's the wind that is powering it by making the propeller turn. If that is the case it can never work.

    Check out this:


  11. Oct 20, 2008 #10
    Topher refuses to believe that ice-boats can maintain a downwind tack (at an angle to the wind) with speed sufficient to give them a downwind velocity component faster than the wind. I know it seems non-intuitive to some; but it's just a normal day out for the ice-boating community.

    Here's just one document that talks a bit about it:

    http://www.nalsa.org/Articles/Cetus/Iceboat Sailing Performance-Cetus.pdf

    And a couple of diagrams from the article:



    I'm well aware this doesn't relate *directly* to going STRAIGHT downwind faster than the wind, but once you realize your downwind component can be much faster than the wind, the steps to get to that nifty cart in the video are somewhat easier.

    Attached Files:

    Last edited by a moderator: Oct 20, 2008
  12. Oct 20, 2008 #11


    User Avatar
    Homework Helper

    and he would be correct. The diagrams appear to be wrong. In the second diagram, the boat is probably going 8 degrees from perpendicular to the wind, not 8 degrees from parallel to the wind.

    Assuming Vb of 70 mph, and an angle of 8 degrees downwind of perpendicular, the boat would be going 9.74mph downwind (a bit over 1/2 the downwind speed), and 69.32mph cross wind. Relative to the air, the cross wind component remains the same at 69.32mph, the "up wind" component (boat is moving downwind slower than the air) is 18 mph - 9.74 mph = 8.26 mph. The boats speed relative to the air or it's "apparent wind" is sqrt(69.32^2 + 8.26^) = 69.81mph.
    Last edited: Oct 20, 2008
  13. Oct 20, 2008 #12
    And you base this on what exactly? This will be news to the iceboat community where they do it all the time. And how would you explain the GPS data?
  14. Oct 20, 2008 #13


    User Avatar
    Gold Member

    Did you allow for "apparent wind" in determining this? The angle of the wind from the PoV of the vehicle will be quite different from the angle of the wind from an exterior PoV. (This is why sailboats seem to have a headwind when in fact, they're more like 45 degrees from head-to-wind).
  15. Oct 20, 2008 #14


    User Avatar

    Staff: Mentor

    I'm not sure about the underlying issue (I believe sailors actually do go downwind at an angle for the purpose of increasing their speed to their destination), but the GPS data does not support the contention. I don't know what the calculation is based on (there is no explanation of where the data came from), but it does not match the GPS output.
  16. Oct 20, 2008 #15


    User Avatar
    Homework Helper

    From that article: "Beta (β) is the angle between the apparent wind and the yacht vector." If Beta is small, then the apparent wind is mostly a headwind, with a small crosswind component:

    If Va is 55mph, and Beta is 8 degrees as stated in the second diagram, then the apparent wind components, relative to the boat, break down into a head wind of cos(8) x 55 mph = 54.465 mph, and a cross wind of sin(8) x 55 mph = 7.655 mph. Still this doesn't give an overall picture of what's going on.

    Wings divert the air, they don't accelerate air backwards in the opposite direction of travel. If the wind speed is 18mph, then the downwind component of the ice boat must be less than 18mph. If the speed of the boat is 70mph, the angle between wind direction and boat direction must be greater than 75.1 degrees (arccos(18/70)).

    I'll research this and reply later.
    Last edited: Oct 20, 2008
  17. Oct 20, 2008 #16
    Topher, so much of debate is about definitions. I'd like to ask you a question that will help me (and perhaps others) determine exactly what your question and position is.

    Many folks wonder ... “Can a ice boat (or other sailing craft) travel downwind faster than the wind itself?”. Most of us will concede that the ‘land speed’ of these craft greatly exceeds that of the wind, but there is much debate as to the downwind component of the path or the velocity made good (VMG) of the craft.

    My scenario setup and question to you Topher, is this:

    On a frozen lake, with a steady 15mph wind from north to south a start point is defined. Directly south, one mile away a finish point is defined. The ice boat takes a ~45degree track (say SW) and comes across the start point at full speed. At the moment the ice boat passes the start point, a neutral buoyancy balloon is released, floating to the south in the wind. At the ~halfway point, the ice boat tacks to the SW and takes aim at the finish point.

    Can the ice boat win the race to the finish point, or will the balloon always win?

    Thanks Topher (and anyone else who cares to answer).

  18. Oct 20, 2008 #17
    Thinairdesign, you are right, it is always good to make sure you have the problem or question clearly defined. The scenario you mentioned is a good one and is an example where me and spork with many others from RR disagree.

    Assuming the balloon always has the same velocity as the wind and the ice boat is only being powered by the wind, then yes the balloon will always win. The best thing that can happen is a tie but the ice boat will never beat the balloon.

    If anyone doesn't clearly understand the problem, let me know and I can draw a simple diagram.
  19. Oct 20, 2008 #18


    User Avatar
    Homework Helper

    Bad analogy, since the boat could build up a lot of speed perpendicular to the wind before crossing the start line and then covert this speed into downwind speed, essentially coasting against a relative headwind.
  20. Oct 20, 2008 #19
    No question about it. That's exactly what sailors do, because their downwind VMG (velocity made good) is significantly better on a downwind tack than on a direct downwind run.

    I can't imagine how you come to that conclusion. The true wind is 10 mph, the ice-boat speed is 40 mph, and the angle between them is 35 degrees. That gives the ice-boat a downwind velocity component of 28 mph. That's nearly 3 times the wind speed.

    In the vector diagram they show an ice-boat going 70 mph at 25 deg from direct downwind in an 18 mph wind. This gives the ice-boat a downwind velocity component of more than 63 mph (in other words 3.5 times the wind speed).

    Guys, there's no debate on this part of the question. Ice boats do this all the time. They chuckle at the notion that there are people that think it can't be done. Additionally, it's very easy to show *how* it can be done given the most basic performance parameters and a few vectors.

    The performance I'm describing is steady state.
  21. Oct 20, 2008 #20
    Ok, excellent ... it's always good to be on the same page with the questions.

    Your answer is very common -- there's something about this "outrunning" our power source that just screams "over-unity", but in the end it turns out that most folks gut reactions are just plain wrong.

    Given a reasonable range of wind (say 5mph to 35mph), the balloon just gets it's *** kicked three ways to Sunday.

    I've been through this discussion many, many times. I've showed the charts, gps plots, vectors, etc, etc. Finally just very recently I got motivated and joined the Yahoo groups for Land Sailors just to get confirmation from the folks that go out on Sunday and actually do this.

    I posted the exact question I posted for you. Here are a few of the responses and the link to the discussion: http://sports.groups.yahoo.com/group/2nalsa/

    First, I got a response via email from the President of the NALSA (North American Land Sailing Association). It follows:
    Following is a sampling of the responses from the thread itself:

    Perfomance in these craft is all about L/D after all and it turns out it doesn't even take a high performance craft to beat the wind -- even the most basic can do it. Of course the higher performance machines leave the balloon cryin' for it's mommy.

    I urge you to go to the forum yourself and quiz these guys if you are a doubter. The math supports it, the vectors support it, the gps plots support it, and the land sailors will get a good chuckle out of someone telling them they are violating the laws of physics as they rip across the lake bed with VMGs of 2-3x of the wind speed.


    PS It's not just land sailing craft that can do this ... the very highest performing racing sailboats also can achieve VMGs of greater than the winds, though just barely. A personal friend of mine (Google Stan Honey) is a world renoun racing navigator (and engineer) with countless world records to his name -- he tells me he's raced numerous sailboats with the performance to beat the balloon. He also chuckles at the notion it can't be done.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: How can an vehicle move faster than the wind that is powering it?