I've heard 3 times faster, but not 6 times, and a little more than 2 times faster, but not 5 times faster VMG.

The secret lies in the 45º angle the iceboat is traveling to the tailwind. The pilot will adjust the angle of the sail to just the point of turning over to optimize the constraint of the skates, pushing the boat forward. And at 45º and 3 times the speed of the wind, the wind is still "felt" by the sail. (That's why the iceboat is trying to tip over)

The cart is going directly downwind and will run out of wind if it tries to outrun the wind. The only thing the tailwind can be pushing on is the "air cushion" coming from the back of the prop.

I'll go back to my electric powered prop example. If we took a cart with the same weight & drag ratios as the cart in question, and placed only an electric engine powering a prop and the prop pushed air backwards at 10 MPH, and we placed this cart in a 10 MPH tailwind, how fast would it go?

I'm guessing 15 MPH, as 5 MPH is lost to friction. This is the same as Jack's cart, which also has a prop pushing air backwards at about 10 MPH. Why wouldn't the reason behind why both go 15 MPH be the same?

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...at 45º and 3 times the speed of the wind, the wind is still "felt" by the sail.

Not from behind it ain't!

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The cart is going directly downwind and will run out of wind if it tries to outrun the wind.

WHO CARES what the cart is doing? What matters is what the wings are doing. And in this case they are tracing EXACTLY the same path as the ice-boat sail.... 1 foot downwind and 1 foot cross-wind.

Incidentally, if we're going to talk about the technical realities of the aerodynamics involved, it would be best not to say things like "run out of wind". I don't know what that means (and I'm pretty sure no one does).

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The only thing the tailwind can be pushing on is the "air cushion" coming from the back of the prop.

The air is "pushing" on the blades of the prop that are swinging through it in exactly the same way the air pushes on the sail. The vector diagram I posted fits both situations perfectly.

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Why wouldn't the reason behind why both go 15 MPH be the same?

Because this is based entirely upon guesswork - not aerodynamics.

Hudman, have a look at this site:

http://www.idniyra.org/articles/IceboatSailingPerformance.html

Here's the specific excerpt:

"The apparent wind/yacht angle (β) is where iceboats, and particularly Skeeters, are King. The apparent wind angle (β) is surprisingly low for very efficient boats like Skeeters (6 to 7 degrees). This is equivalent to sailing at 8 to 10 times the wind speed and they are, in fact, capable of this feat in light winds on good ice. In DN's and fast dirt boats β is more like 10 to 12 degrees."

The wind that the sail feels and uses is coming from the front - how could it be otherwise with that much ground speed? The air flows from the front of the iceboat to the rear, and the sail is at a very shallow angle to the direction of travel.

The air that influences the sail is coming from in front of the sail. This is most important for you to understand.

Spork, I just read the physics forum thread. I find it amusing but at the same time a bit unsettling that the thread was locked. Do physicists normally ignore data that conflict with their world view?

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Not from behind it ain't!

Indeed, not from behind. But from a 45º angle, yes. And if not, why is it trying to tip over?

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"run out of wind". I don't know what that means (and I'm pretty sure no one does).

Fair enough. And for those that can't figure it out. It means that the cart has outrun the wind and the wind is not hitting the cart (or prop) anymore from behind, hence it has "run out of wind."

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The air is "pushing" on the blades of the prop that are swinging through it in exactly the same way the air pushes on the sail.

How you figure? The tailwind is 10 MPH, the cart is 15 MPH. How can that 10 MPH tailwind catch the 15 MPH cart? The prop would be swinging through a relative 5 MPH headwind instead.

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Because this is based entirely upon guesswork - not aerodynamics.

By that statement, are you saying my electric powered prop cart, pushing 10 MPH aganst a 10 MPH tailwind wouldn't be traveling at about 15 MPH?

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The wind that the sail feels and uses is coming from the front

So which way is it trying to tip then?

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Indeed, not from behind. But from a 45º angle, yes. And if not, why is it trying to tip over?

The wind is coming from slightly more than 45 degrees off of straight ahead - in both cases - since both cases are identical.

Why is it trying to tip over? Because force times moment-arm creates a torque or moment that will tend to tip things over.

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Fair enough. And for those that can't figure it out. It means that the cart has outrun the wind and the wind is not hitting the cart (or prop) anymore from behind, hence it has "run out of wind."

It's not that I can't figure it out - it's that the phrase has no meaning. When you say it has "outrun the wind" that implies that it has gone so far that it's now in a region where there is no wind - but I know this is not what you mean. You say it's no longer hitting the cart from behind - and that's true. You say it's not hitting the prop from behind - and that's true as well. It's now hitting the prop from an angle just off of 45 degrees from straight ahead - exactly like the ice-boat sail - since the two are identical.

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The tailwind is 10 MPH, the cart is 15 MPH. How can that 10 MPH tailwind catch the 15 MPH cart? The prop would be swinging through a relative 5 MPH headwind instead.

No, the prop tip is moving to the side at 15 mph as the cart moves forward at 15 mph. That makes the prop go at a 45 degree downwind angle - exactly like the sail on the ice-boat - since the two are identical. Draw the vectors (or look at mine). The wind is 10 mph from behind, while the prop tip traces a 45 degree downwind path at 21.2 mph. You'll easily see exactly how the wind hits the back of the prop.

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By that statement, are you saying my electric powered prop cart, pushing 10 MPH aganst a 10 MPH tailwind wouldn't be traveling at about 15 MPH?

Nope. I'm saying exactly what you said - that you're guessing. You're electric powered cart will be moving at a speed between 0 and 20 mph depending on it's efficiency and friction.

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So which way is it trying to tip then?

Let's say the boat is traveling due east in a wind that's coming from due south. It wants to tip it's mast to the north (tipping toward port-side). Now let this boat bear off the wind until it's on a 45 degree downwind course and going twice the wind speed (it's a really high performance boat). The tipping moment will still be toward it's port-side, and the sail will be filled from the same side at all times. At no time does the sail luff or move through the eye of the relative wind.

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You say it's not hitting the prop from behind - and that's true as well. It's now hitting the prop from an angle just off of 45 degrees from straight ahead

Since the wind is coming from directly behind the cart, and is not hitting the prop, then it's not hitting the prop. The wind, to the cart is not 45 degrees, it's directly in line with the cart. The prop blades are at 45 degrees, but again, if the wind is no longer reaching the prop, it's no longer reaching the prop. The prop can't tack from a wind that is no longer there.

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No, the prop tip is moving to the side at 15 mph as the cart moves forward at 15 mph. That makes the prop go at a 45 degree downwind angle - exactly like the sail on the ice-boat - since the two are identical. Draw the vectors (or look at mine). The wind is 10 mph from behind, while the prop tip traces a 45 degree downwind path at 21.2 mph. You'll easily see exactly how the wind hits the back of the prop.

I follow what you're saying, but, again, if the tailwind is not reaching the prop anymore, I can't see how the wind hits the back of the prop.

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Nope. I'm saying exactly what you said - that you're guessing. You're electric powered cart will be moving at a speed between 0 and 20 mph depending on it's efficiency and friction.

The efficiency, friction, drag, and weight were to be equal to Jack's cart. And it is only a guess, but I believe it would be faster than 10 MPH, but not quite 20 MPH. The average of those is 15 MPH.

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and the sail will be filled from the same side at all times.

The sail will be filled from the starboard side, yes, and remain so. And I agree, even though it is going twice the wind speed, the sail still "feels" the wind and remains filled. The cart is not doing the same thing.

No, but the propeller is. Ignore the cart!

Yes, I agree that the iceboat can't go faster than the wind directly downwind (to answer your next post - I must be psychic!).

oldguy1, sorry I missed your reply in there. I read your link & did a Google search, as I have never ice-boated before. (Although I'd love to do it!) And I stand corrected. While some sites agreed with the information I found earlier, most sites agreed with your link.

Now, what would the speed of the iceboat be if it were sailing directly downwind with a 10 MPH tailwind? Regardless of the angle of the sail, I believe you'd agree with me that it cannot outrun the wind this way. Taking away anomolies like gusts & crosswinds that is.

Why can't it outrun the wind while sailing directly downwind? Because if the sail no longer "feels" the wind, since it has outrun it, it's power supply is gone. I believe this is the same as the cart, regardless of the position/speed of the prop. If the prop cannot "feel" the tailwind, like the iceboat in my example, it cannot get any more power from the tailwind.

Back to my "air cushion" example, if we have a 10 MPH tailwind & 10 MPH pushing against it from the prop on the cart, for a 20 MPH collision, we can sail faster than the wind in a direct downwind configuration. How much faster is dependent upon drag and efficiency, and according to Jack, his cart is 1.5 X the wind speed.

The icesail or landsail operate as long as the apparent wind has a component perpendicular to the true direction of travel large enough for the sail to divert and generate thrust. When the true downwind component is greater than the wind, the sail diverts the apparent wind in the direction of true upwind. If the true wind is 10mph, and the landsail's downwind component is 25mph, then the sail upwind component of diverted apparent wind has to be greater than 15mph, otherwise there would be a net drag in the direction of the wind.

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propeller at 45 degrees

The angle of attack on a propeller varies with radius, higher near the axis, and lower at the tips, so that the pitch (how far the prop would move through jello per revolution) is near constant.

As I've stated before, induced wash reduces the effective angle of attack and apparent wind speed on a propeller. I don't know if the losses are enough to prevent DWFTTW.

If the propeller driven cart is moving DWFTTW, then the propeller has to generate thrust in the direction of the wind. If the wind is 10mph and the cart is moving downwind at 15mph, the propeller has to divert it's apparent wind upwind at a speed greater than 5mph, otherwise there would be a net drag.

I'm still waiting for a conclusive video to be made. I think this would be worthy of a MythBusters segement. It wouldn't cost that much to test, and I'd be happy to see anything significantly over 1.0 on a DWFTTW treadmill test. Even if the cart didn't work, the segment could include videos of landsails going DWFTTW.

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As I've stated before, induced wash reduces the effective angle of attack and apparent wind speed on a propeller. I don't know if the losses are enough to prevent DWFTTW.

Well, go back to spork's cylinder example.

Do you agree that an iceboat can have its downwind component of motion in excess of the wind speed when tacking at 45°?

If the world were a cylinder and the wind was blowing down the length of the cylinder over the entire surface, could such an iceboat not spiral around the cylinder, continuing its 45° tack, and its motion in the direction of the wind, down the length of the cylinder would exceed the wind speed?

Can we not envision two of these iceboats diametrically opposed, spiralling down the length of the cylinder in unison?

So, where's the difference between the two spiralling iceboats, moving forwards faster than the wind that's "propelling" them, and the two prop blades doing the exact same thing, spiralling their way through the air?

Any of the objections based on effective angle of attack and apparent wind speed, should be equally applicable to the iceboat scenario. And yet they are known to have the downwind component 3 times or more of the actual wind speed. They don't seem to mind that the relative wind gets reduced a little when tacking.

Seems to me that a lot of confusion is coming from speed of the wind vs speed of the craft/sail and so on. Same thing that was happening over on the physics site. But over there someone asked about the forces involved, which I thought was a great question and much more appropriate.

If I remember correctly, BB and spork have occasionally discussed L/D ratios and have stressed the fact that the cart pulls its motive force from the speed difference between the ground and the air. Jack Goodman listed his cart as having a 1.75:1 L/D ratio. At one point I think oso9282 also posted the amount of energy that could be harvested from that speed difference; he felt that was important because of his wind turbine background but was offering it as proof that the cart didn't work. It helped me see that the cart was indeed possible instead, and what exactly was happening during the treadmill test!

My point? I think this has a lot more to do with the amount of force that is captured at any given wind speed and how that pushes the cart forward to the speed that the total drag in the system balances that force. The higher the wind speed over the ground, the more force can be captured and the higher the potential speed for the same drag. The lower the total drag of the vehicle, the higher the top speed, whether upwind or downwind in either case.

Maybe now isn't the ideal time to bring this up, but maybe it'll help someone else.

Here's the links to the little cart I built. I still have not tested it yet, as I can never find a steady wind blowing in a constant direction. Plus it has not remote control steering. I would try it on a treadmill if I had one.

http://i80.photobucket.com/albums/j169/HUDMAN_2006/WindMachine1.jpg

http://i80.photobucket.com/albums/j169/HUDMAN_2006/WindMachine2.jpg

http://i80.photobucket.com/albums/j169/HUDMAN_2006/WindMachine3.jpg

As you can see, I went with a turbo-prop design. And my friction losses will be high, with no bearings and all that plastic-on-plastic.

Let's ponder the treadmill idea.

We have a treadmill in a room with zero wind. We place the cart on the treadmill, power it up, (we'll have to have a backboard to keep it in place while the prop spins up) and turn it loose.

First, it will go backwards against the backboard, then as the prop spins up, it will propel itself off the front of the treadmill. The prop will push air out the back of the craft, and that will propel the cart forward.

In this case, there is no sailing involved, only thrust. That seems to support my idea of why it works. All I need is a treadmill & time to test it, and I'll see if my little craft will drive off the front of the treadmill.

And for those confused by the cylindrical Earth mentioned earlier, I have a diagram.

http://i80.photobucket.com/albums/j169/HUDMAN_2006/IceBoatonCylinder.jpg

Be sure to include some way of measuring the speed of the air coming off the back of the prop at several different distances. I'm sure you'll be surprised!

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Now, what would the speed of the iceboat be if it were sailing directly downwind with a 10 MPH tailwind? Regardless of the angle of the sail, I believe you'd agree with me that it cannot outrun the wind this way.

Why can't it outrun the wind while sailing directly downwind? Because if the sail no longer "feels" the wind, since it has outrun it, it's power supply is gone. I believe this is the same as the cart...

But this most definitely is NOT the same as the cart. If you want to make a *direct* downwind ice-boat the same as the cart you can do it. You just have to put the sail on a long side-to-side slider. Now, as the ice-boat goes *directly* downwind, you have to make the sail slide from one side to the other at the same rate that the hull goes downwind. Bingo, you've now got an ice-boat with a more or less traditional sail, that can now go *directly* DWFTTW.

But since it's a bit awkward to make your sail translate rapidly from side to side (so it can tack even if the boat doesn't), it might be easier to put two of these sails on a shaft and let them rotate about that shaft. This way they can always have the transverse motion needed.

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If the prop cannot "feel" the tailwind, like the iceboat in my example, it cannot get any more power from the tailwind.

The prop will "feel" the relative wind in exactly the same way that an ice-boat's sail "feels" the relative wind - and from the same direction - and on the same surface. Props, sails, wings, etc. can only ever "feel" the relative wind; never the true wind.

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You just have to put the sail on a long side-to-side slider. Now, as the ice-boat goes *directly* downwind, you have to make the sail slide from one side to the other at the same rate that the hull goes downwind.

What's the difference between a fixed sail that cannot feel the tailwind and a moving sail that cannot feel the tailwind?

I think the moving sail would act like a propeller more so than a moving propeller would act like a sail.

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What's the difference between a fixed sail that cannot feel the tailwind and a moving sail that cannot feel the tailwind?

I give up - what is the difference? I'm assuming that's a trick question since the sail, wing, prop, keel, blades, etc. never feel the true wind. They always feel the relative wind. And in the case of the prop blades on the cart, and the sail on the ice-boat tacking 45 degrees downwind, and the sail that's translating side-to-side on the ice-boat that's going directly downwind, the relative wind is always a quartering headwind.

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I think the moving sail would act like a propeller more so than a moving propeller would act like a sail.

Three identical cases:

1) spinning prop going downwind faster than the wind
2) fixed sail on an ice-boat going faster than the wind on a 45 degree downwind course.
3) A sail translating side-to-side on an ice-boat going directly downwind faster than the wind.


Tell me this... if you were a fly on the tip of the prop as it's going over the top, while the cart is going directly downwind faster than the wind - on a foggy day - how could you tell whether you were on the prop tip or the tip of the ice-boat sail as it tacks downwind on a 45 degree downwind course? You couldn't tell by how the relative wind hits you - because that will be identical in both cases.

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Originally posted by roofingguy:

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As I've stated before, induced wash reduces the effective angle of attack and apparent wind speed on a propeller. I don't know if the losses are enough to prevent DWFTTW.

Well, go back to spork's cylinder example. Do you agree that an iceboat can have its downwind component of motion in excess of the wind speed when tacking at 45°?

yes, if speed/drag and lift/drag, factors are high enough.

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So, where's the difference between the two spiralling iceboats, moving forwards faster than the wind that's "propelling" them, and the two prop blades doing the exact same thing, spiralling their way through the air?

Convetional propellers do not have the high lift to drag ratios of a sail or wing. I've already mentioned that operating in their own induced wash and energy wasted in rotating air as factors that reduce the lift to drag ratio of a propeller. Combine these losses in a propeller based system with the losses in the drive train and the result maybe that the cart simply isn't efficient enough to go DWFTTW.

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feel the tailwind

In order for there not to be a net drag in the downwind direction, the sail has to divert the apparent wind upwind a bit faster than the downwind speed minus the wind speed. For example, if the downwind speed is 25mph, and the wind speed is 10mph, the sail has to divert the air flow so the upwind component off the sail is greater than 15mph.