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Split Thread The validity of classical physics (split from: DWFTTW)

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It is corrected, though you may be it may just be possible that you are wrong? One small step for jjcote, one giant step for this thread.
All indications so far are that your diagram was wrong, I have corrected it, and so far I don't seem to be wrong.

Both the real wind cases are incorrect.
This I find to be quite remarkable, because as far as I can see, the "real wind" cases are the same as in your diagram. Are you claiming that you were wrong, or does correctness in the humberverse perhaps vary with time or with who is saying it?

The difference between the flows of 10m/s and 6m/s, is 4m/s at standstill. This must remain the case at windspeed because it is a simple translation;
10ms + 10ms = 20m/s
6m/s + 10m/s =16m/s
Difference = 4m/s
First, this demonstrates a total lack of understanding. If the wind is out of the south at 10 m/s, and you move north at 10 m/s, then you feel no wind, not a 20 m/s wind. However, you seem to have considered a case that is not illustrated (wind out of the south at 10 m/s and the paddle is moved south at 10 m/s), and correctly calculated that the paddle will still indicate the same result in that case. Perhaps you're starting to get how this works.

The speed of the belt-flow is therefore ( 10m/s - 4m/s) and like the wind 4m/s slower than the fastest component, and so an effective 6m/s in the direction of the belt. Referenced to still air that is 6m/s back with the belt.
Not slower than the fastest component, but rather relative to the air. In all cases, at the height that we have picked, the boundary layer speed will be closer to the bulk of the air, and not as close to the speed of the surface. But that's not so important, compared to the direction. Pick a different height so that the boundary layer speed is 5 m/s if that simplifies things for you. The important thing is that the paddle will turn the same direction in all cases.
 
(quoting the original post in the first quote block. The rest are simple quotes. But everything is from the same post)

It is about power transfer in general, Greg, not only electrical. See Carnot Cycle, thermodynamics, load matching, load lines...

Not "in general" Humber, you specifically said "maximum power theorem". One doesn't cite an actual theorem and then say you were only talking about power "in general".

So, I may not be the best person to work with you because I generally have certain expectations of people beyond formulas and equations. A big one for me is I generally expect people (myself included) to acknowledge when they make mistakes. I screwed up earlier talking about inertial frames and ended up making a confusing glob of words with reference frames. They're not the same. Someone pointed it out, I admitted I screwed up. I screwed up with another post with the scenario where the cart used a generator and electric motor and it always generated 12 watts. Except at one point I wrote 24 watts. Someone pointed it out. I admitted I screwed up. I had a cart design using one moving part and fixed vanes. It doesn't work. I admitted I screwed up.

You cited a specific theorem that has only one meaning, specifically relating to electrical resistors. It has nothing to do with newtonian objects moving around. And I would expect someone to say "Yeah, I was wrong" when its pointed out, not "I was talking about power in general".

What it means to me is trying to teach you something means not only overcoming whatever hurdle there may be in understanding that thing, it means also overcoming whatever hurdle there is in getting you to admit a mistake so we can move on to whatever the correct thing is.

If you say 24 watts when it's really 12 watts and you refuse to actually say "Yeah, its 12 watts, I was wrong when I said 24", then I have no way of knowing if you still think its 24 or if you've realized your mistake, understood the problem, and have correctly figured out that its 12 watts.

And if I can't tell if you've corrected a mistake, then I have no way of knowing if we're ever making any progress. And I like to make progress. So, you're avoidance of coming out and admitting when you're wrong has removed one of my big incentives to engage in conversation with someone.

For me, it's not just about equations and formulas and being right, its about being willing to be wrong, learn from that, and move forward. We're not movign forward. Or if we are, I have no way of knowing. If you say "I used the wrong term", then at least I know we're making some sort of progress. If not, then you're basically asserting that "maximum power theoem" has something to do with newtonian objects moving around, and it doesn't.

Greg:
If that's it, then why did you pretty much ignore my scenario with the windmill on the groudn generating one kilowatt in a 10 mph wind, and the same windmill mounted on a giant treadmill moving at 10 mph indoors (no wind) also generating one kilowatt?

Humber:
Patently true. Why would you even bother to mention it? Did you expect that I would be surprised?

well, then we've just demonstrated that they are inertial frames.

An inertial frame is a frame where Newtons three laws hold true. They hold true for a windmill on the ground in a 10 mph wind, and they hold true for a windmill on a treadmill movign at 10 mph. In either case, the windmill produces the same amount of power. So Newton's laws apply to both. If the windmill produced more power in one scenario than another, then we'd have to accoutn for an extra force. If the treadmill was accelerating, then that would cause the windmill to generate more power, which would mean it isnt' an inertial frame. But newton's three laws hold true in both cases, so their both inertial frames.


Greg:
Does this whole thread come down to you disbelieving in inertial frames?

Humber:
No. Inertial frames are not applicable. It's about the machine that is the treadmill, and that makes false appeals to that idea. Sounds impressive at Arby's, though.

except you said "I am not interested in the cart. yo-yo's are a distraction." So, we've shown how the cart is an inertial frame using the power generated by a windmill, and then you say the cart wont' work. But then you're not interested in the cart.

Perhaps you can see why I might be having a hard time understanding what it is you wnat to talk about here.

Humber:
The power is generated by the action of the wind on the prop. That is 100% of what is available. Wind/prop and wind/ground are not two separate identities.

Ah, so, the treadmill isnt' the problem. THe cart is the problem. The power is generated by the difference between the wind and the ground. If you have a standard windmill planted in the ground and the wind isn't moving relative to the ground, if there is no difference between wind and ground, then the windmill can generate no power.

The cart cannot go faster than teh wind if there is no wind over the ground. Because there is no difference in velocity between wind and ground, so there is no power to be extracted.

The cart can only extract power if there is a difference between wind and ground. And because the cart uses a simple mechanism to extract power based on the difference, it doesn't matter how fast the cart is going, there is always a difference between wind and ground speed.

If the cart is on a treadmill indoors, and the treadmill is moving at 10 mph, then the cart has power available equal to the cart stationary in a 10 mph tailwind. Windmill on the ground in a 10 mph wind. Windmill on a treadmill moving at 10 mph. Either way it has power available because there is a difference between the air and ground.

Because of the way the prop is geared to the ground wheel, it will extract power based on the difference betwen wind and ground, no matter what the speed of the cart is.
 
Ok, no new scenario. Can you tell me this. At the level where the real wind is blowing at 8 m/s, how does its height compare with the 6 m/s and the 10 m/s, and what velocity does it have down belt w.r.t. room/still air? See, if I apply your method, and keep it 2 m/s different from the full speed, as you said the 6 m/s flow should be 4 m/s from full speed, then it's going R to L at 8 m/s.
(1) The wind boundary is fastest adjacent to the wind, but slowest at the ground.
(2) The belt-wind is fastest at the belt, and slowest adjacent to the "wind"

For equal profiles, the speed of 6m/s or 8m/s or some other value, will be at different heights w.r.t the surface. Because the belt surface is the road surface, that is the reference point not the "wind", meaning that the profile is also wrong, being inverted.
This is pedcadillo solved by making the paddle the full height of the boundary layer in each case, so the resulting force will be the same in belt and wind.


But the 8 m/s BL in real wind must be higher up, away from the belt, so we have the 8 m/s reversed BL above the 6 m/s BL. Do we apply the rule throughout, and have a 0 m/s wind on the belt and 10 m/s reverse BL (going right to left) just below the still air? Doesn't that create an awful sheer force, where the opposite flows of air pass each other?:) ETA;...I mean, where the 9.999999 m/s BL going right to left, dragged by the belt, passes the still air in the room? Like I asked before, if you just filled in the blanks on eomy earlier diagram, we'd already have solved these little problems.

Makes no difference, the paddle sees the entire flow, and the inverted boundary in itself a distinguishable difference. Either way a difference.
 
Is that really humber's paddle wheel argument? Why does he use a different relative speed of the boundary layer over the surface when the surface is a belt instead of the ground? Is this one of those "lets solve a different problem and get a different answer and claim that the problems are different" sort of things?

And why does he screw up so badly in the lower left panel? Are there to many vectors for him to keep track of?
Yes and yes. A belt is not the ground, and still air is not a real wind. The information is all there, no new scenarios please (which might demonstrate even more clearly how dumb I'm being; in fact, just drop the whole thing and let me be right; it's just a stupid toy; etc.).
 
(1) The wind boundary is fastest adjacent to the wind, but slowest at the ground.
(2) The belt-wind is fastest at the belt, and slowest adjacent to the "wind"

For equal profiles, the speed of 6m/s or 8m/s or some other value, will be at different heights w.r.t the surface. Because the belt surface is the road surface, that is the reference point not the "wind", meaning that the profile is also wrong, being inverted.
This is pedcadillo solved by making the paddle the full height of the boundary layer in each case, so the resulting force will be the same in belt and wind.




Makes no difference, the paddle sees the entire flow, and the inverted boundary in itself a distinguishable difference. Either way a difference.
Ah, more of your unintelligible BL BS. I must be onto something.

Please answer clearly:

If the 6 >>> layer on earth becomes a <<< 6 layer on the belt, where is the 8 >>> layer in vertical position: above or below (on earth) and is it then <<< 8 on the belt? That's all. It's not rocket surgery.
 
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Humber, I want you to go to page 12 of my document and look at a drawing there of a gear.

http://www.greglondon.com/tumbleweed/tumbleweed.pdf

Can you see that the wheel gear moves at some rate determined by the velocity of the "air" and "ground" gears? That the wheel gear can move faster than the "wind" gear?

Someone made a really great video that demonstrates the gearing here:

http://www.youtube.com/watch?v=9Yt4zxYuPzI

It doesn't use propellers and air you can't see, it relies on a simple cart made out of lego-blocks with wheels you can see and "air" and "ground" you can see. In the video, the guy shows the cart moving faster than the "air", faster than the ruler.

If the lego-cart simply used a sail to catch the wind, then it could only go as fast as the wind. If it simply attached to the ruler, it owuld only be able to go as fast as the ruler. But it connects the air and ground through a gear ratio, a block and tackle, a simple machine, to leverage the power so the cart can go faster than the wind, faster than the ruler.
 
Humber, I realise there's something new to distract you - or rather, quite old now - in the form of Michael's videos, but if you wouldn't mind answering the questions, please, I'm sure we'll have the misunderstanding cleared up in no time, and you will, of course, be proved right. I just want to get to the bottom of it.

Is the land-based 8 m/s to the right above or below the 6 m/s? (above/below)

Does it go at 8 m/s right to left above the treadmill, w.r.t. the still air/room/earth? (Y/N...it goes at). Thanks.
 
I want to thank the folks that are still trying to set humber straight (I particularly enjoy the requests for clear answers). Not because we have ANY hope of getting through to him, but simply because this is what fuels the machine that produces quotes like these:

You are a bunny. Try again. I have corrected jjcote's mistakes.

blah, blah, blah. Find your own. Don't get hysteresis-terical.
You have NO IDEA what you are talking about, and are trying to bluff, Captain.
I know quite a lot about rheology, so your failure is 100% guaranteed. Over.

And these are the quotes that make this whole thread worthwhile.

100 pages here we come!
 
Someone made a really great video that demonstrates the gearing here:

http://www.youtube.com/watch?v=9Yt4zxYuPzI

Thanks for the compliment Greg: I'm the guy who made the ruler video.

I should point out that Humber's problem is less with the DDWFTTW cart than with the fact that it is tested on a treadmill. He refuses to accept the idea that a treadmill moving under the still air in a room creates a wind just as real as a natural wind blowing across the ground. I made a second video to illustrate this principle of relative movement: "Under the ruler 2: the ground moves". Some people have told me that it helped them understand the treadmill idea, but there's no hope that it will help Humber. He has decided that he is right and is now devoting his time to "correcting" every other person on this thread.
 
OK, boundary layer measurement and determining whether we are in equivalent frames or not.

Go to the imaginary hardware store. Pick up some 1"x1" moulding. We want to build a tower of sorts. It will have a base that is 1 foot wide and 1 foot deep. And we want it to stick up into the air as far as we think there may be a boundary layer difference. Just to be crazy, lets make it 6 feet tall.

We need some kind of wind measuring device. A bunch of them actually. spinning anenometers, or simple hanging plates that get deflected by the wind. Or little propellers tied to little generators which connect to voltmeters. I don't care. We want them spaced evenly along the height of the tower. Put one on the tower every inch.

Now, put the tower on the ground, and have a 10 mph wind blowing on it. The velocity right at the ground will be zero. Then lets say that the velocity measured equals the height in inches above the ground. That the boundary layer effect is 1 mph per inch, with the ground always having stationary air.

height above ground = measured airspeed.
0 = 0
1 inch above ground => 1 mph measured windspeed
2 inch above groudn => 2 mph measured windspeed.
3 inch above ground => 3 mph measured windspeed.
...
9 inch above ground => 9 mph measured windspeed.
10 inch above ground -> 10 mph measured windspeed
11 inch above ground -> 10 mph measured windepseed
12 inch above ground -> 10 mph measured windepseed
13 inch above ground -> 10 mph measured windepseed
...
6 feet above ground -> 10 mph measured windspeed.

I don't think this is an accurate representation of boundary layer effects, but it's easy to work with the numbers.

OK, now, lets take our tower and place it on the treadmill. We'll put the treadmill indoors so there is no wind. And we'll set the treadmill to turn at 10 mph. What do we measure?

0 = 0
1 inch above treadmill=> 1 mph measured windspeed
2 inch above treadmill=> 2 mph measured windspeed.
3 inch above treadmill=> 3 mph measured windspeed.
...
9 inch above treadmill=> 9 mph measured windspeed.
10 inch above treadmill-> 10 mph measured windspeed
11 inch above treadmill-> 10 mph measured windepseed
12 inch above treadmill-> 10 mph measured windepseed
13 inch above treadmill-> 10 mph measured windepseed
...
6 feet above treadmill-> 10 mph measured windspeed.

And then the tower crashes off the end of the treadmill and falls to teh ground.

Everyone agree with this so far?

OK, now, go to the imaginary store. Go into the toy section, and pick up a nice remote controlled truck. You want something big and powerful enough to carry the tower on top of it and have enough juice to move at 10 mph, and be able to do it long enough to make some measurements.

So, mount the tower on top of our RC truck. Put the truck on the treadmill. Turn the treadmill up to 10 mph and start driving the truck so that the truck speedometer reads 10 mph. So that it doesnt' move relative to the ground.

What does our windtower read for measurements?

Well, if our boundary layer causes 1 mph per inch above teh ground, then the treadmill is going to grab the stationary air in the room and start dragging it along with the treadmill.

Right at the treadmill, the wind will be moving at 10 mph, the same speed as the treadmill.

1 inch above the treadmill, the tower will measure 9 mph wind
2 inch above the treadmill, the tower will measure 8 mph wind
3 inch above the treadmill, the tower will measure 7 mph wind

8 inch above the treadmill, th tower will measure 2 mph wind
9 inch above the treadmill, teh tower will measure 1 mph wind
10 inch above the treadmill, teh tower will measure a 0 mph wind
11 inch above the treadmill, the tower will measure a 0 mph wind
12 inch above the treadmill, the tower will measure a 0 mph wind
13 inch above the treadmill, the tower will measure a 0 mph wind
...
6 feet above the treadmill, the tower will measure a 0 mph wind

The air in the room isn't moving relative to the room. So, if we get far enough above the treadmill, the tower will say the air is stationary. Close to the treadmill, the treadmill drags some air along with it, and we get some boundary layer windspeed that the tower measures.

Since we said 1 inch per mph, we have to get 10 inches above the treadmill before we measure 0 mph wind, before we see that the air is no longer moving.

Everyone with me to this point?

OK, now, those numbers don't match, do they? Well, no. But the reason they don't match is because we didn't have our tower moving at 10 mph over the ground with a 10 mph tailwind.

So, lets go back outside. Call up the imaginary weather program. Set the windspeed to 10 mph from the south. Put the truck with the tower on a flat parking lot facing north. Start the truck moving and get it up to moving so that the speedometer says its movign at 10 mph. It is now moving at 10 mph in a 10 mph tailwind. What do the wind speed indicators on the tower say?


1 inch above the road, the tower will measure 9 mph wind
2 inch above the road, the tower will measure 8 mph wind
3 inch above the road, the tower will measure 7 mph wind

8 inch above the road, th tower will measure 2 mph wind
9 inch above the road, teh tower will measure 1 mph wind
10 inch above the road, teh tower will measure a 0 mph wind
11 inch above the road, the tower will measure a 0 mph wind
12 inch above the road, the tower will measure a 0 mph wind
13 inch above the road, the tower will measure a 0 mph wind
...
6 feet above the road, the tower will measure a 0 mph wind


If the boundary layer is 1 mph per inch above the surface, then the air near the ground will be stationary to the ground, which means our tower on the truck moving at 10 mph groudn speed will measure a 10 mph wind. 1 inch above the groudn, our tower on the truck moving at 10 mph will measure a 9 mph wind speed.

A foot above the ground, the boundary layer no longer has any effect, and the tower on the truck moving at 10 mph over the ground with a 10 mph tailwind measures zero wind. 2 feet up, 3 feet up, 6 feet up, the tower measuring points detect no wind.

Note that the wind speed measurements of the tower on the truck on the treadmill moving at 10 mph indoors with no wind is exactly the same as the windspeed measurements of the tower on the truck on the road movign at 10 mph with a 10 mph tailwind.

Everyone with me so far?

So, if the measurements, including issues with boundary layers, are exactly the same between these to scenarios, then they are exactly the same for purposes of demonstrating the operation of a propeller/wheel cart.

Humber, if your only concern is that the treadmill and the ground are not the same, these measurements should show you that they are identical.
 
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Humber, if your only concern is that the treadmill and the ground are not the same, these measurements should show you that they are identical.


Ya gotta love the boundless optimism! :D

But frankly I think you guys are going about it all wrong. humber won't be fooled with such a simple approach (i.e. facts and common sense). The best we can possibly hope for is to lead him the long way around, and get him to directly contradict himself. Of course he will still explain how the direct contradiction just proves his point, but that's half the fun of this perpetual motion machine.
 
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(quoting the original post in the first quote block. The rest are simple quotes. But everything is from the same post)
Not "in general" Humber, you specifically said "maximum power theorem". One doesn't cite an actual theorem and then say you were only talking about power "in general".
(1) Do your own work, Greg, or accept mine
(2) Electrical maximum power theorem is one manifestation of that theorem.
Conversion if necessary, can be made using in the standard list of analogues.
http://www.eas.asu.edu/~holbert/analogy.html

So, I may not be the best person to work with you because I generally have certain expectations of people beyond formulas and equations. A big one for me is I generally expect people (myself included) to acknowledge when they make mistakes.
I do not want to hear any more of this crap. Admitting mistakes is just that.
Why make a fuss? Is it that I should have the expectation that you are infallible? Those who prevaricate or deliberately try to misunderstand, are taken for short ride down humber lane...or runway.

I screwed up earlier talking about inertial frames and ended up making a confusing glob of words with reference frames. They're not the same. Someone pointed it out, I admitted I screwed up. I screwed up with another post with the scenario where the cart used a generator and electric motor and it always generated 12 watts. Except at one point I wrote 24 watts. Someone pointed it out. I admitted I screwed up. I had a cart design using one moving part and fixed vanes. It doesn't work. I admitted I screwed up.
What will you do should you make a big mistake, Greg? Spontaneously combust?
Perhaps the pain of admission would be more tolerable, if you did not so often simply assume you are right.
(I have not looked at that, model yet, but I bet there are more mistakes...)

You cited a specific theorem that has only one meaning, specifically relating to electrical resistors. It has nothing to do with newtonian objects moving around. And I would expect someone to say "Yeah, I was wrong" when its pointed out, not "I was talking about power in general".
Except that it is not wrong. This is typical. Narrowing the definition to try and shift the blame. Considering that you were, and still are, entirely ignorant of this theorem, perhaps you are wrong again.
Let me help you out. I was wrong, power transfer is organized and controlled by elves. Happy now?

You are being illogical; why ask a liar to tell the truth? If I were to be such a liar, do you expect that I would wilt under the glare of your piercing interrogation?
Either way it's all about you and hubris! None of you see that.

What it means to me is trying to teach you something means not only overcoming whatever hurdle there may be in understanding that thing, it means also overcoming whatever hurdle there is in getting you to admit a mistake so we can move on to whatever the correct thing is.
I gave you more than enough information, Greg. How do you think I learned it? Do your homework. Easy science has long gone.

If you say 24 watts when it's really 12 watts and you refuse to actually say "Yeah, its 12 watts, I was wrong when I said 24", then I have no way of knowing if you still think its 24 or if you've realized your mistake, understood the problem, and have correctly figured out that its 12 watts.
F...

And if I can't tell if you've corrected a mistake, then I have no way of knowing if we're ever making any progress. And I like to make progress. So, you're avoidance of coming out and admitting when you're wrong has removed one of my big incentives to engage in conversation with someone.
Progress is not made through this method. Assuming all "mistakes" are deception or evasion, should be left at the church door. It is your lack of understanding and willingness to learn that are the obstacles. If you want to save face, stick it in the sand. Enough.
"If at first you don't succeed, fail again, fail better" (Beckett). Get it?

For me, it's not just about equations and formulas and being right, its about being willing to be wrong, learn from that, and move forward. We're not movign forward. Or if we are, I have no way of knowing. If you say "I used the wrong term", then at least I know we're making some sort of progress. If not, then you're basically asserting that "maximum power theoem" has something to do with newtonian objects moving around, and it doesn't.
It's also about understanding, and accepting that process will entail making mistakes. Talk to the hand.

well, then we've just demonstrated that they are inertial frames.
An inertial frame is a frame where Newtons three laws hold true.
That is so, but also true for the conservation of energy. Just like the ground.
Just because the laws hold, does not mean that it is a new frame of .
reference. A logical, not physical error.

They hold true for a windmill on the ground in a 10 mph wind, and they hold true for a windmill on a treadmill movign at 10 mph.
They would. Like an engine on bus, works like an engine on a dyno.

In either case, the windmill produces the same amount of power. So Newton's laws apply to both.
Only if all parameters are the same. For example, to run the belt-generator, requires that the windmill first be accelerated to windspeed. That is KE that is not in the stationary generator. There will be some loss that is not recoverable when the motor is tuned off and the KE returned.
"Steady State" is misused to cover these sort of inconvenient facts.

If the windmill produced more power in one scenario than another, then we'd have to accoutn for an extra force. If the treadmill was accelerating, then that would cause the windmill to generate more power, which would mean it isnt' an inertial frame. But newton's three laws hold true in both cases, so their both inertial frames.
Also for a windmill simply being driven by a belt. Losses are losses, gains are gains, the books will balance.

except you said "I am not interested in the cart. yo-yo's are a distraction." So, we've shown how the cart is an inertial frame using the power generated by a windmill, and then you say the cart wont' work. But then you're not interested in the cart.
You have reiterated an error.

Perhaps you can see why I might be having a hard time understanding what it is you wnat to talk about here.
Inertial frames as applied to the treadmill, is BS. That is what I am telling you. Not about Einstein, hobos in cars, inertial frames or equivlancy, but this specific application and abuse of those precepts.

Ah, so, the treadmill isnt' the problem. THe cart is the problem. The power is generated by the difference between the wind and the ground.
No, the treadmill is the problem. Please do not tell me what I think, Greg.

If you have a standard windmill planted in the ground and the wind isn't moving relative to the ground, if there is no difference between wind and ground, then the windmill can generate no power.
Nobel prize sighted.

(2) The cart cannot go faster than teh wind if there is no wind over the ground. Because there is no difference in velocity between wind and ground, so there is no power to be extracted.
Field's medal sighted.

The cart can only extract power if there is a difference between wind and ground.
Yes.

And because the cart uses a simple mechanism to extract power based on the difference, it doesn't matter how fast the cart is going, there is always a difference between wind and ground speed.
Energy cannot come from the ground. See your own remark (2)
Why is there no energy when there is no wind?

If the cart is on a treadmill indoors, and the treadmill is moving at 10 mph, then the cart has power available equal to the cart stationary in a 10 mph tailwind. Windmill on the ground in a 10 mph wind. Windmill on a treadmill moving at 10 mph. Either way it has power available because there is a difference between the air and ground.
There is no wind on the treadmill that is independent of the belt. So, yes, you could say that is "ground power" but that is a substitute for the real wind's power. In reality, there is no motor driving the road.

Because of the way the prop is geared to the ground wheel, it will extract power based on the difference betwen wind and ground, no matter what the speed of the cart is.
Between the wind and ground. Period.
Why do you not see wind-generators on tracks? Siemens have some very bright engineers, and they seem to have missed that golden opportunity to make history.
 
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Why do you not see wind-generators on tracks?


Hey check it out! humber asked his first intelligent question.

The reason... complexity and cost. Windmills on tracks or turntables can in fact harness significantly more energy than their stationary counterparts.
 
OK, boundary layer measurement and determining whether we are in equivalent frames or not.
Last chance. All others, my diagrams, or draw you own.

Go to the imaginary hardware store. Pick up some 1"x1" moulding. We want to build a tower of sorts. It will have a base that is 1 foot wide and 1 foot deep. And we want it to stick up into the air as far as we think there may be a boundary layer difference. Just to be crazy, lets make it 6 feet tall.
I may tell you where to stick it.

We need some kind of wind measuring device. A bunch of them actually. spinning anenometers, or simple hanging plates that get deflected by the wind. Or little propellers tied to little generators which connect to voltmeters. I don't care. We want them spaced evenly along the height of the tower. Put one on the tower every inch.
No. Too vague.

Now, put the tower on the ground, and have a 10 mph wind blowing on it. The velocity right at the ground will be zero. Then lets say that the velocity measured equals the height in inches above the ground. That the boundary layer effect is 1 mph per inch, with the ground always having stationary air.

height above ground = measured airspeed.
0 = 0
1 inch above ground => 1 mph measured windspeed
2 inch above groudn => 2 mph measured windspeed.
3 inch above ground => 3 mph measured windspeed.
...
9 inch above ground => 9 mph measured windspeed.
10 inch above ground -> 10 mph measured windspeed
11 inch above ground -> 10 mph measured windepseed
12 inch above ground -> 10 mph measured windepseed
13 inch above ground -> 10 mph measured windepseed
...
6 feet above ground -> 10 mph measured windspeed.

I don't think this is an accurate representation of boundary layer effects, but it's easy to work with the numbers.

OK, now, lets take our tower and place it on the treadmill. We'll put the treadmill indoors so there is no wind. And we'll set the treadmill to turn at 10 mph. What do we measure?

0 = 0
1 inch above treadmill=> 1 mph measured windspeed
2 inch above treadmill=> 2 mph measured windspeed.
3 inch above treadmill=> 3 mph measured windspeed.
...
9 inch above treadmill=> 9 mph measured windspeed.
10 inch above treadmill-> 10 mph measured windspeed
11 inch above treadmill-> 10 mph measured windepseed
12 inch above treadmill-> 10 mph measured windepseed
13 inch above treadmill-> 10 mph measured windepseed
...
6 feet above treadmill-> 10 mph measured windspeed.

And then the tower crashes off the end of the treadmill and falls to teh ground.

Everyone agree with this so far?
Perhaps? How do you make the tower standstill on the belt?

OK, now, go to the imaginary store. Go into the toy section, and pick up a nice remote controlled truck. You want something big and powerful enough to carry the tower on top of it and have enough juice to move at 10 mph, and be able to do it long enough to make some measurements.
No. remote truck are self powered. Try an elephant, they have good windsense.

<snip>
A diagram, and the effort to draw it may simplfy your experiment, and allow you to better see your own errors.
 
Hey check it out! humber asked his first intelligent question.

The reason... complexity and cost. Windmills on tracks or turntables can in fact harness significantly more energy than their stationary counterparts.

Cost is not so relevant. They are along term capital investment. Even a small gain is advantageous because that is real income.
Any gain, should it not be possible to incorporate it in a static design, will not be because of the motion, but better use of the wind.
Outside LA. you can see such a farm, and the different types of windmill used according to location on the hills.
 
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blah, blah, blah. Find your own. Don't get hysteresis-terical.
You have NO IDEA what you are talking about, and are trying to bluff, Captain.
I know quite a lot about rheology, so your failure is 100% guaranteed. Over.


In other words, you can't produce a single example of the term "Negative drag" in terms of aerodynamics, or "Cavitation" in regards to gas, so you will have a hissy fit.

Grandious terms you bandy about, then get shirty when you get called on them.

Pathetic.
 
In other words, you can't produce a single example of the term "Negative drag" in terms of aerodynamics, or "Cavitation" in regards to gas, so you will have a hissy fit.

Grandious terms you bandy about, then get shirty when you get called on them.

Pathetic.

No, you still don't have a clue, and are trying to make much of a definition. Got ya Google. Take a look for any of those terms. You can even find cavitation for air, if you look. It won't mean anything to you of course, Captain. There is a cretin certain amount of "negative credibility" when a pilot writes about the stubbornness of other pilots, only in order to soil himself.
Over.
 
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Humber, I realise there's something new to distract you - or rather, quite old now - in the form of Michael's videos, but if you wouldn't mind answering the questions, please, I'm sure we'll have the misunderstanding cleared up in no time, and you will, of course, be proved right. I just want to get to the bottom of it.

Is the land-based 8 m/s to the right above or below the 6 m/s? (above/below)

Does it go at 8 m/s right to left above the treadmill, w.r.t. the still air/room/earth? (Y/N...it goes at). Thanks.
Any thoughts? I'm not interested in Greg's new scenario, like you asked me not to post any new scenarios. I want you to answer these simple questions about the one we were discussing.

Sorry, Greg, I'm sure it's fine when I get round to reading it.
 
Is that really humber's paddle wheel argument? Why does he use a different relative speed of the boundary layer over the surface when the surface is a belt instead of the ground? Is this one of those "lets solve a different problem and get a different answer and claim that the problems are different" sort of things?

And why does he screw up so badly in the lower left panel? Are there to many vectors for him to keep track of?

Several pages back, Humber posted a fairly detailed "analysis" of his boundary layer claims. There were two pretty obvious errors: He started off with beltspeed = windspeed, then figured the boundary layer on the belt relative to the ground instead of the belt, then used the wrong sign to transform from ground to belt reference, and came up with the boundary layer moving faster than the belt in the same direction. Like all of Humber's other "proofs" it proved only that he simply cannot grasp a frame of reference other than the ground.
 
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