Merged DDWFTTW in a “Real” Wind

[profpangloss]

I'm trying to make sense of humbers objections, as best as I can decipher them. So far as I can tell there are no other land or ice yachts that can travel within 40 degrees of DDW at 2.8x WS where the WS is on the order of 10mph..

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?

Further, humber has more than once "shown" that the propeller is nothing more than a drag element when over WS.

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.

Therefore, if we take the sum of humber's arguments this cart is traveling at higher speeds so close to DDW than any comparable cart can achieve while being burdened with a 4-5meter diameter prop dragging in the wind.

Does that really sound plausible?.

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?

The record will stand or fall on its own. I hear there is a crack analytical team working on the numbers as we speak. I'm trying to understand why humber thinks his arguments have "debunked" the BB rather than catapulting it from cool to amazing
.

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.

 

[humber]

But they did not use RMS averaging. I can show you right here why that is important: Suppose the wind alternated between + 40 deg and - 30 deg during that 10 seconds. A simple average would be 10 degrees and does not represent the true wind power that was available to the cart An Rms average would be 35.35 degrees and that is a true indication of how much wind power was available to the cart the entire 10 seconds. Do you now see why RMS is important? It is not only used in electronics! If you have ever done any mechanical alignments, such as shaft runout, you will find it is the RMS average that matters. That is just one example. There are countless others.

It have made the same point many times, but Carteers have a new way of knowing.

One other thing they find "amusing", is that my claim that the push to 11mph adds directly to the final speed. The acceleration is linear, the cart does not slow in between the push and when the wind changes, and never reaches a constant velocity.
The average speed over the whole trial is then about 8.5mph.

 

[profpangloss]

I don't follow. The propeller of the Blackbird doesn't face the wind; it has its "back" to the wind. Isn't that just what you're suggesting it should do?

Turning a propeller around on the shaft does not change its basic operating mode. (It will have some effect because leading egde is now trailing edge, but the pitch is the same and it will turn the same way). To get it to work as a turbine you need to change the direction of rotation. That means changing the gearing to the wheels so the turbine rotates in the direction the wind wants to turn it, as the cart goes down wind in the direction the sail force wants to push it. The turbine will be working to assist the sail force, not working against it.
The propeller cart is not working in that way. The sail force pushes the cart down wind but the propeller wants to be turbined the opposite way, putting a torque on the drive shaft and on the wheels that opposes their motion down wind.

It is like Larry and Mo are pushing a piano from one end and Curly is pushing from the other end, opposing them. If they are stronger, the piano moves the way they are pushing it, but not very efficiently. Get Curly over on the same side and things get better. Thats what happens when you use the spinny thing as a turbine and not a propeller!

As long as the turbining of the prop is working against the cart, it would be best to try and weaken that as much as possible. Turbines are most effective when facing directly into the wind stream. If you want to weaken the turbine, get it off of DDW! You will still have a decent sail force, but now you have weakened the opposing turbine force. In short, the propeller cart should work best off of DDW.

Sorry for the long tortorous post, hope you get my drift.

 

[humber]

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?

That is definitely an error and there will be speed changes with the direction changes. It is also possible that the wind measurements came from the fixed station. I have asked for that to be confirmed. On a desert floor, on a hot day, the wind will tend to shear over the warmer surface air.
All measurement were taken at hub height. NIST standard is 10m

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.

Yes, IF.
I think the gearing forces a regulation mechanism on the cart, but it is a sail cart, and a tall one.

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?

If averaged over the course the average is 32 degrees RMS
Excluding the extremes, ( which would lower the top speed in reality) 20 degrees RMS.

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.

Voodoo it is.

 

[69dodge]

The propeller cart is not working in that way. The sail force pushes the cart down wind but the propeller wants to be turbined the opposite way, putting a torque on the drive shaft and on the wheels that opposes their motion down wind.

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?

 

[humber]

b. the difference between the rms of wind direction and the rms of cart direction was only 7 degrees. They were biased in the same direction.

Then you don't understand RMS, but it's at least 20degrees.

c. the vast majority of the "other side" peaks were caused by wind vane oscillation, not changes in wind direction. And everyone knows that. So by using either rms or abs mean, you would be measuring the magnitude of the measurement error, rather than actual wind direction.

Those errors suppress the actual reading.

 

[humber]

The record will stand or fall on its own. I hear there is a crack analytical team working on the numbers as we speak. I'm trying to understand why humber thinks his arguments have "debunked" the BB rather than catapulting it from cool to amazing.

That's because you have rose colored glasses. It is not amazing from any perspective.

 

[Llyricist]

Then you don't understand RMS, but it's at least 20degrees.

So you've changed your story? Before when you added instead of subtracted you got 15 degrees for the test period. I'm using the very same numbers that you calculated months ago, what changed?

Those errors suppress the actual reading.

LMAO, no they don't, in fact it is trivial to take the arithmetic mean to determine the actual direction about which the vane is oscillating.

 

[humber]

So you've changed your story? Before when you added instead of subtracted you got 15 degrees for the test period. I'm using the very same numbers that you calculated months ago, what changed?

Then you STILL don't understand RMS.

LMAO, no they don't, in fact it is trivial to take the arithmetic mean to determine the actual direction about which the vane is oscillating.

Sure it is. I am not letting you know anything about that.

 

[Myriad]

That's because you have turned lead to gold. It is not amazing from any perspective.

Man, this dude is hard to please. Turned lead to gold? Waste of time. Desktop cold fusion? Boring. Cured the common cold? Yawn. Set a direct downwind wind powered speed record well above wind speed? Trivial and impossible (both, simultaneously), but let's discuss it for 10,000 pages anyhow.

Respectfully,
Myriad

 

[humber]

Man, this dude is hard to please. Turned lead to gold? Waste of time. Desktop cold fusion? Boring. Cured the common cold? Yawn. Set a direct downwind wind powered speed record well above wind speed? Trivial and impossible (both, simultaneously), but let's discuss it for 10,000 pages anyhow.

Respectfully,
Myriad

Magic carpet treadmills.

 

[3point14]

Turning a propeller around on the shaft does not change its basic operating mode. (It will have some effect because leading egde is now trailing edge, but the pitch is the same and it will turn the same way). To get it to work as a turbine you need to change the direction of rotation. That means changing the gearing to the wheels so the turbine rotates in the direction the wind wants to turn it, as the cart goes down wind in the direction the sail force wants to push it. The turbine will be working to assist the sail force, not working against it.
The propeller cart is not working in that way. The sail force pushes the cart down wind but the propeller wants to be turbined the opposite way, putting a torque on the drive shaft and on the wheels that opposes their motion down wind.

It is like Larry and Mo are pushing a piano from one end and Curly is pushing from the other end, opposing them. If they are stronger, the piano moves the way they are pushing it, but not very efficiently. Get Curly over on the same side and things get better. Thats what happens when you use the spinny thing as a turbine and not a propeller!

As long as the turbining of the prop is working against the cart, it would be best to try and weaken that as much as possible. Turbines are most effective when facing directly into the wind stream. If you want to weaken the turbine, get it off of DDW! You will still have a decent sail force, but now you have weakened the opposing turbine force. In short, the propeller cart should work best off of DDW.

Sorry for the long tortorous post, hope you get my drift.

Are all the posts you've made in this forum in this thread?

 

[Llyricist]

Then you STILL don't understand RMS.

It is the square root of the average of the squares. It is used when there is a quadratic relationship involved.

the RMS of wind direction during the test period was about 11, the RMS of the cart direction was around 4, those are values you calculated, and I calculated too, the difference is 7.

Of course if you take the RMS of deviation instead of the deviation of the RMS it works out to 8.8 degrees.

All still within the required 10 degrees and well beyond the capabilities of any other wind powered cart.

Sure it is. I am not letting you know anything about that.

Of course not.

 

[69dodge]

You assume some coefficients for hull aero drag, rolling drag, airfoil lift/drag and the areas (airfoil, frontal hull). That gives you the net force as function of velocity. The maximal velocity is the one where that net force is zero.

Thanks. That makes sense.

I said I understood how you got the theoretical maximum speed, but maybe I'll briefly outline how I analyzed it, just to make sure.

I imagined that the propeller, instead of moving through fluid air, rotates inside a horizontal cylinder that's threaded on the inside, and the tips of the propeller blades mate with the threads. So, if the propeller rotates, it will also move forward relative to the cylinder, similar to the way a propeller spinning in air moves forward relative to the air. (The cylinder has a slot cut into the bottom of it, lengthwise, to allow the propeller support to pass. So, its cross-section is like the letter C, but with the opening at the bottom.) The propeller is geared to the cart's wheels, which roll on the ground, as usual. The question then becomes, if the cylinder moves at a certain speed relative to the ground, how fast will the cart move relative to the ground?

We can choose units of length so that, as the wheels rotate once through 360 degrees, they move forward along the ground a distance of 1. Suppose the propeller is geared to the wheels so that it rotates x times for every rotation of the wheels. And suppose that the propeller blade pitch, and the matching cylinder thread pitch, is such that a single rotation of the propeller moves it forward a distance of y relative to the cylinder. Then, for every rotation of the wheels, the cart moves forward a distance of 1 relative to the ground (because of the wheels), and it also move forward a distance of xy relative to the cylinder (because of the propeller). In other words, the cart's speed relative to the cylinder is xy times its speed relative to the ground.

Let v be the speed of the cart relative to the ground, and let w be the speed of the cylinder relative to the ground. Then, the speed of the cart relative to the cylinder is v - w, and we have

v - w = xyv
v(1 - xy) = w
v = w/(1 - xy)​

I gather that xy is what you called the "vehicle advance ratio"?

 

[69dodge]

the RMS of wind direction during the test period was about 11, the RMS of the cart direction was around 4, those are values you calculated, and I calculated too, the difference is 7.

Of course if you take the RMS of deviation instead of the deviation of the RMS it works out to 8.8 degrees.

I don't think it's correct to calculate the RMS of the wind direction alone, and of the cart direction alone, and then subtract the two. I think, that way, you'd get different answers depending on what reference direction you were measuring their directions against, which doesn't make sense. You should take the RMS of the difference between the wind direction and the cart direction. (Assuming RMS is the right thing to do here in the first place, which I don't know.)

 

[Llyricist]

I don't think it's correct to calculate the RMS of the wind direction alone, and of the cart direction alone, and then subtract the two. I think, that way, you'd get different answers depending on what reference direction you were measuring their directions against, which doesn't make sense.

well that could be handled by determining the bias of each value, and subtracting or adding depending on whether the biases were on the same side or not. Both biases are on the same side in this case. So subtraction is the way to go.

You should take the RMS of the difference between the wind direction and the cart direction. (Assuming RMS is the right thing to do here in the first place, which I don't know.)

Yes, which was kind of the point, doing it the way you suggest is the way I would do it, and comes up with the higher number, the 8.8 above.

And without a demonstrable connection to a relevant quadratic relationship, there is of course no reason to prefer RMS other than the higher number it yields.

Problem is, the rules actually say the difference of the averages rather than the average of the differences, which doesn't make any difference with arithmetic mean, but does make a difference with RMS.

 

[Jack by the hedge]

But they did not use RMS averaging. I can show you right here why that is important: Suppose the wind alternated between + 40 deg and - 30 deg during that 10 seconds. A simple average would be 10 degrees and does not represent the true wind power that was available to the cart An Rms average would be 35.35 degrees and that is a true indication of how much wind power was available to the cart the entire 10 seconds. Do you now see why RMS is important?

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.

 

[humber]

It is the square root of the average of the squares. It is used when there is a quadratic relationship involved.

the RMS of wind direction during the test period was about 11, the RMS of the cart direction was around 4, those are values you calculated, and I calculated too, the difference is 7.

Of course if you take the RMS of deviation instead of the deviation of the RMS it works out to 8.8 degrees.

All still within the required 10 degrees and well beyond the capabilities of any other wind powered cart.



Of course not.

Noooo.....

 

[humber]

I don't think it's correct to calculate the RMS of the wind direction alone, and of the cart direction alone, and then subtract the two. I think, that way, you'd get different answers depending on what reference direction you were measuring their directions against, which doesn't make sense. You should take the RMS of the difference between the wind direction and the cart direction. (Assuming RMS is the right thing to do here in the first place, which I don't know.)

It is about available power. That is all the is needed to know.

 

[humber]

I don't think it's correct to calculate the RMS of the wind direction alone, and of the cart direction alone, and then subtract the two. I think, that way, you'd get different answers depending on what reference direction you were measuring their directions against, which doesn't make sense. You should take the RMS of the difference between the wind direction and the cart direction. (Assuming RMS is the right thing to do here in the first place, which I don't know.)

But there is no reference direction. The wind change is all the is needed. If does not matter to the cart if it were + 20 then -10 or the other way around
or any other combination. All that matters is the total available power.
You all seem to think the cart is a wonder one way or the other, but the fact is, there is a lot of energy in that wind, and the cart makes poor use of it.