Merged DDWFTTW in a “Real” Wind

[A.T.]

Are propeller blades generally shaped so that they could be screwed into a block with deep threads without getting stuck?

If you mean a rigid block with deep threads, then no, there is no guarantee it won't get stuck. It depends on the shape of the threads. Every propeller is optimized for a certain propeller advance ratio (not to be confused with the vehicle advance ratio of the vehicles discused here, which also includes the gear ratio between prop-shaft and ground). And there are other considerations mentioned by spork.

However, you put can a propeller in jello and apply a torque on the shaft, while allowing the prop to move freely along the shaft axis. The propeller will create its own threads, based on least resistance, and move a certain distance per revolution. That gives you the effective pitch.

You can do this regardless of how the blades are shaped or if there are any blades at all. You can actually do that with any object, you just define an axis and constrain the rotation to it.

That effective pitch in jello might be different than in air. But it explains the principle. If the propeller was very light you could do the same in an block of air.

 

[profpangloss]

I don't follow your reasoning. I can see that the power available to the cart might decline as some function of the angle of deviation of the wind, and RMS is a form of averaging which negates the effect of positive and negative deviations tending to cancel each other out, but from the look of the graphs I've seen so far I don't think the deviations were both positive and negative values - the offset angle was always to the same side. So if that's not why RMS is required then perhaps you think there's some specific reason to take the quadratic mean of the deviations to model the available power rather than, say, the mean of the absolute values of offset, but so far I haven't seen anyone describe why that should be.

(The graphs I have seen do indeed show that the wind variation was both positive and negative with respect to this propeller cart.)

As to why RMS averaging should be used:

The power available from the wind is a function of both wind direction and velocity. It is beyond dispute that if any standard sailing craft were to be moving directly down wind at wind speed, there would be no wind power available. Take a look at this summary for the Greenbird:

Greenbird (Jenkins' land/ice boat) obviously has the fastest speed sailing ... But even at 45-degrees to wind direction he's able to get 3X wind speed, and 2X wind speed at 20-some degrees, and 1X at 6-some degrees. ...

You see that the craft’s speed increases the further off of DDW it gets. There isn’t even any mention of it’s speed when DDW as it is understood it must be less than wind speed. It must be a minimum of 6 degree off of DDW to achieve 1x wind speed. But the Greenbird was not trying to prove anything regarding fastest speed closest to DDW, it was intended to set a land yacht speed regardless of wind direction and even wind speed.

Once you start to claim fastest speed DDW, then obviously the wind direction relative to the cart direction matters a great deal, and it matters a great deal how you measure that! If the wind is changing direction so as to swing positive and negative, and you do a simple arithmetic average, it may well average out to zero. Do you think it would be correct in that case to claim the craft went DDW because the wind direction averaged out to zero? I certainly hope you would agree that such an average is meaningless.

When the variant is positive and negative, the only proper way to take an average value is to use the RMS method of taking the square root of the mean of the squares. That will give you a fair value for how far off of DDW the craft was for the entire run. It is also the proper way to calculate the wind power that was available to the craft.

 

[profpangloss]

Yes, the torque on the wheels from the propeller tends to slow the cart, via the wheels' interaction with the ground. But the torque on the propeller from the wheels tends to speed it up, via the propeller's interaction with the air. (The propeller is blowing air backward, propelling the cart forward.) How can we tell which effect is greater?

Great question! Is the propeller thrust greater than the reverse torque produced by the tendency of the propeller to be turbined by the wind? I don't see how there can be any meaningful thrust below wind speed. If you saw the "popcorn video" that I Ratant posted, you saw that when the popcorn was released it blew past the struggling propeller cart very quickly! The wind was blowing through the propeller from back to front! It is possible some popcorn blew right through the propeller although I don't claim to be able to see that in the video. I have to admit I found that particular video both very funnny and very damaging to the claim of DDWFTTW. How much thrust can the propeller be producing with a tailwind blowing right through it?

 

[Jack by the hedge]

Forgive my ignorance about rms and its relevance in this debate:
The humber angle, if I get it, is that all variations from ddw add to the available power, and that the ones to the right don't cancel out the ones to the left, as an averaging might tend to do?

Some ways back, I thought I heard mentioned that the cart actually does better ddw than with the variations. Is that correct?

It's a point of contention whether the cart would perform better or worse off-axis to the wind. (Well, when I say contention, I mean there seems to be one person who says it'll do better and everyone else expects it to do worse.) However, I think everyone can agree that the power supplied to the cart varies in some way as a function of its angle to the wind.

In that case, we'd want to know how large the deviation from the wind direction was and, as the cart is symmetrical, it shouldn't matter which way the wind deviates, only the amount of the deviation. If the wind was +10° off axis for half the time and -10° off axis for half the time then the average deviation would be 10°, not zero.

RMS is one way of eliminating the + and - directions (by squaring the values) but you could equally well just take the positive absolute value of all the data. Any agreement on which, if either, is more valid may have to wait for some agreement on how the power varies with angle to the wind.

 

[Jack by the hedge]

If the wind is changing direction so as to swing positive and negative, and you do a simple arithmetic average, it may well average out to zero. Do you think it would be correct in that case to claim the craft went DDW because the wind direction averaged out to zero? I certainly hope you would agree that such an average is meaningless.

Yes indeed. I absolutely agree. As I said above, if the wind was, say, +10° off axis for half the time and -10° off axis for half the time then the average deviation would be 10°, not zero.

When the variant is positive and negative, the only proper way to take an average value is to use the RMS method of taking the square root of the mean of the squares. That will give you a fair value for how far off of DDW the craft was for the entire run. It is also the proper way to calculate the wind power that was available to the craft.

I don't agree that it's the only proper way. You could for example take a simple arithmetical mean of the absolute values of the data. Without a model of how the power varies with angle to the wind, I don't see how you can say what method is valid. We even have one contributer who insists power passes through a local minimum at DDW.

 

[A.T.]

It's a point of contention whether the cart would perform better or worse off-axis to the wind. (Well, when I say contention, I mean there seems to be one person who says it'll do better and everyone else expects it to do worse.)

And that one person posted this vector diagram to explain how the vehicle will exploit the crosswind for "propulsion":

 

[Jack by the hedge]

Great question! Is the propeller thrust greater than the reverse torque produced by the tendency of the propeller to be turbined by the wind? I don't see how there can be any meaningful thrust below wind speed. If you saw the "popcorn video" that I Ratant posted, you saw that when the popcorn was released it blew past the struggling propeller cart very quickly! The wind was blowing through the propeller from back to front! It is possible some popcorn blew right through the propeller although I don't claim to be able to see that in the video. I have to admit I found that particular video both very funnny and very damaging to the claim of DDWFTTW. How much thrust can the propeller be producing with a tailwind blowing right through it?

Moving off from a standstill, the wind acts on the propeller like a simple sail. (In fact it continues to do so right past windspeed.)

 

[profpangloss]

Moving off from a standstill, the wind acts on the propeller like a simple sail. (In fact it continues to do so right past windspeed.)

If it acts like a simple sail, yet has a reverse torque on wheels due to turbine force, how can you expect it to ever reach wind speed? I think a simple sail cart would have a higher final velocity, and it too does not ever reach wind speed. (That is DDW in both cases).

 

[Jack by the hedge]

If it acts like a simple sail, yet has a reverse torque on wheels due to turbine force, how can you expect it to ever reach wind speed? I think a simple sail cart would have a higher final velocity, and it too does not ever reach wind speed. (That is DDW in both cases).

The reverse torque means that it will not accelerate as quickly as a simple cart with an equivalent propeller fixed in place, but the pitch of the turning propeller means that it presents a constantly retreating surface to the air in contact with the back of the blades. That means that it continues to accelerate (slowly) past windspeed, since the local condition at the blade surface effectively means that it has not yet reached windspeed.

 

[profpangloss]

The reverse torque means that it will not accelerate as quickly as a simple cart with an equivalent propeller fixed in place, but the pitch of the turning propeller means that it presents a constantly retreating surface to the air in contact with the back of the blades. That means that it continues to accelerate (slowly) past windspeed, since the local condition at the blade surface effectively means that it has not yet reached windspeed.

Well, it is one thing to assert something another thing to prove it. My opinion is the reverse torque, plus the fact the propeller is acting as a sail, and the fact that no sail carts can achieve DDWFTTW, means this cart also cannot achieve DDWFTTW. The NALSA trial data doesn't indicate to me that DDW was achieved. So I remain unconvinced.

 

[3point14]

Well, it is one thing to assert something another thing to prove it. My opinion is the reverse torque, plus the fact the propeller is acting as a sail, and the fact that no sail carts can achieve DDWFTTW, means this cart also cannot achieve DDWFTTW. The NALSA trial data doesn't indicate to me that DDW was achieved. So I remain unconvinced.

Are you saying that it's not possible to go DDWFTTW or that this cart doesn't achieve it?

 

[profpangloss]

I don't agree that it's the only proper way. You could for example take a simple arithmetical mean of the absolute values of the data. Without a model of how the power varies with angle to the wind, I don't see how you can say what method is valid. We even have one contributer who insists power passes through a local minimum at DDW.

I also believe that the cart is least effective DDW because of the turbine force on the propeller in that situation.

I should not have said the only correct way! Probably the best method would be to use an numerical integration technique such as Simpson’s or the Trapezoidal rule or even go whole hog and do a Rayleigh distribution. But I think RMS is the acceptable standard method for this application.

 

[profpangloss]

Are you saying that it's not possible to go DDWFTTW or that this cart doesn't achieve it?

I would never say that something is absolutely impossible. Certainly using stored energy a vehicle can be built to achieve DDWFTTW trivially. But I don't see how it can be done in real time, steady state with no stored energy, and I certainly don't see how this propeller cart can even reach wind speed, let alone exceed it.

 

[Jack by the hedge]

I also believe that the cart is least effective DDW because of the turbine force on the propeller in that situation.

I suppose that would depend on whether it's the propulsive force or the counteracting turbine force which declines faster as the wind moves off axis.

 

[Jack by the hedge]

... I certainly don't see how this propeller cart can even reach wind speed, let alone exceed it.

Do remember that from the point of view of the propeller blades, it never actually reaches windspeed.

 

[Subduction Zone]

profpangloss why are you running away from my simple series of yes/no questions? If you don't want to understand how the treadmill is an excellent example of DDWFTTW of the small cart I can understand that. It indicates that you would prefer to remain ignorant and spout nonsense. But why someone would want to spout nonsense, that part I do not understand.

 

[spork]

I suppose that would depend on whether it's the propulsive force or the counteracting turbine force which declines faster as the wind moves off axis.

Once we're going faster than wind speed the wind does not produce a counteracting turbine force. At this point a chain break would leave the propeller free-wheeling in the correct direction.

 

[Jack by the hedge]

Once we're going faster than wind speed the wind does not produce a counteracting turbine force. At this point a chain break would leave the propeller free-wheeling in the correct direction.

Wouldn't that freewheeling speed necessarily be slower than the running speed, though? So long as the prop blade maintains a positive angle of attack it's turning faster than it would freewheel, so it ought to be experiencing some degree of opposing force.

 

[spork]

Wouldn't that freewheeling speed necessarily be slower than the running speed, though?

Yes.

So long as the prop blade maintains a positive angle of attack it's turning faster than it would freewheel, so it ought to be experiencing some degree of opposing force.

That "opposing force" at this point is drag - no different than the drag that has to be overcome to turn the prop on a Cessna as it flies along on a no-wind day. At this point the wind is not trying to turn it in the opposite direction as a turbine. That only happens below wind speed.

 

[Dr. Keith]

I'm not sure that is true, and I'm not convinced the wind speed was only 10 mph at the propeller height, and especially when it was changing direction so rapidly. In your experience as a sailor, doesn't a rapid change in wind direction accompany a gust?

When it does it disrupts the sail because the sail is no longer at the best angle to the wind. Rapidly changing wind direction is a bitch when sailing because you don't have any time to adjust angles to exploit the wind. Claims that BB somehow takes advantage of rapidly changing wind direction with fixed pylons push this further into the realm of way cool.

I think he means it would be a drag element IF the cart was doing what it was supposed to be doing, going DDW. If it is moving at an angle to the wind, the prop would make a reasonable sail.

Doesn't that strike you as a bit absurd.

Do you have any aerodynamic analysis that would even suggest a spinning prop would make a reasonable sail at an angle to the wind? Even IRatants tests showed "best performance" for his prop sail was DDW.

Besides, Humber has been consistent that the prop was a drag element at anything over WS, so how does a drag element becom an efficient sail?

None of this sounds really plausible to me! In my opinion, the wind was gusting and changing direction so effectively the cart was tacking a wind that was much higher velocity that 10 mph. It could have been twice that. Now how amazing does 28 mph look?

But there is no evidence that the wind was higher than 10mph. And even if it was double that at the top of the prop, that means that at least half of the prop was well below WS and acting as a drag element according to Humber. Assuming anything like a natural gradient across the swept area of the prop with 10 at the center and 20 at the tip and then take the average windspeed over the swept area and it will be lessk than 10 just based on the shape of wind gradient curves. That is trivial, as Humber would say.

I understand Humber's arguments. This entire thing, from treadmills to changing reference frames and claiming ground power, really strike me as a bunch of voodoo physics and I will be happy to see it all go through the shredder. Time will tell.

I don't think you or Humber have ever applied all of his arguments to the NALSA trial. He has made several objections and if you take each of them at face value the BB could not have even approached the average WS across the swept area of the prop. The only source of faster than WS propulsion that he has allowed is the pair of symmetrical pylons that are completely below the measured WS and would be at an average WS of less than 5mph if you assume the huge gradients you are quoting. Think about that. Even if they were asymmetrical, as would be most advantageous in a sailing environment, they could hardly push the drag element of the prop to anything approaching the WS measured at their tips. Even at 60 degrees off DDW, much less at the 20-40 degrees off DDW that Humber is citing. It is simply too much drag.

Unless, at least one of numbers arguments is wrong.