Yeah? Guess what I'm thinking now!
Answer: That is what you are thinking.
Next thought: DOH!
Split Thread The validity of classical physics (split from: DWFTTW)
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Answer: That is what you are thinking.
Next thought: DOH!
sol invictus
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jj, just in case you don't know what he's referring to, it's a book on meteorology I linked to earlier. In a section describing the motion of balloons in the wind, it says precisely the opposite of what humber claims it says here.
Can we calculate the magnitude of the flux of humber's lies?
That's exactly the same as saying that if you give a canoe a push on a dead calm lake, it will make a bow wave all the way across the lake because it can't ever reach waterspeed. Amazing!
Well, at least you're on the right forum. All you have to do is prove this or your equivalent statement that I quoted and you'll be a millionaire! Congratulations!
Sure... if you buy into that nonsense spouted by Galileo, Newton, Einstein, and every high schooler that passed their first physics mid-term.
Dan O.
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With ideal collisions and particles all the same size, it's as if the particles don't even collide but pass right through each other (complete exchange of momentum). The model boat is then just colliding with each successive particle and sending that particle forward with just under 2x the boats current speed.
Since this is 1-d, there can be no turbulence and with elastic collisions, all the energy and momentum are conserved.
Playing with the model, I found a geometric barrier that will slow the boat to 1/2 it's initial velocity, accelerate the barrier to exactly match the boats velocity (like an inelastic collision) and put all the excess energy and momentum into a single lightweight particle.
I'd like to believe you're right about humber, and I've even waffled back and forth on the issue, but in the end I fear that humber is exactly what he appears to be.
Hopefully I'll be proved wrong in the end. Maybe humber will admit to being a grad student in psychology, doing a research project on the behavior of crowds when each individual is somewhat anonamous. Or maybe he has a bet with one of the lunatics on another forum to see which of them can maintain the ruse the longest.
Unfortunately, one things I've learned is that things are usually no deeper than they appear.
http://books.google.com/books?id=Zj...X&oi=book_result&resnum=9&ct=result#PPA219,M1
The book quite clearly states that the balloon travels at the mean speed of the wind. The concern is that the dynamic behaviour of the balloon will not allow it to track the variations around that mean windspeed.
For example, if the balloon is too massive, the wind may not be able to accelerate the balloon to track the wind, and if it has too much momentum, it may not slow appropriately. The concern the is not one of the balloon's velocity, but of acceleration.
Perhaps the wind is 50km/hr, +/-2km/hr. A balloon traveling at say 45km/hr, yet capable of accelerating +/2km/hr ;not instantaneously but within specified time (equation 12.8), then the balloon is satisfactory for the purposes. It does not have to be "stationary w.r.t the wind" to be of use.
(1) If the balloon spends any time accelerating, its average velocity must be below that of the wind. Intermittent changes in the wind's peak velocity are used to accelerate the balloon from its lower mean value, so that the changes may be tracked.
(2) If that acceleration is possible, then there must be net force upon the balloon. If sufficient force were constantly available from a steady wind, the question of acceleration and the time it takes to do so, would not be an issue.
(3) The author then makes the simplifying assumption that the balloon is capable of windspeed, (the worst case condition for available force needed to accelerate the balloon), then goes to show that there is enough force to track the variation within the desired time. Equation 12.8, provides the time constant.
Equation 12.7 shows that the balloon's velocity Vb, is less the Vwind, and that difference is proportional to the derivative of Vb. The "steady state" solution, Vb = Vwind, is merely the simplification that the author used as a premise.
(4) That book shows that rapid acceleration depends upon the balloon's characteristics. Low mass, yet large area. That is the intent. To show that this form is capable of the desired performance, and not that all objects can to the same. Quite clearly they cannot.
(5) Given that a balloon is a low mass, large area device, yet does not quite reach windspeeed, it is difficult to see how heavier devices of indeterminate area and drag coefficient, could do the same. This would seem to deny the claim that all objects can reach windspeed. Acceleration is necessary to the claimed mean-speed, showing that the net force upon the balloon cannot be zero.
This book does not support these claims;
(a) Raleigh's formula shows that all objects with a non-zero Cd can reach windspeed
(b) Windspeed has been reached for the cited example.
(c) The net force upon the balloon is zero, or could be so at windspeed.
Well said. The very contrary of learning.
Some can see when they are being deceived, Clive.
Those names, so gratuitously wasted, and upon you.
Support or explanation for your conclusion please, Mender
sol invictus
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Which is precisely what everyone has been telling you, and you have been denying for the last 3,000 posts.
If the balloon is massive, it will take longer to accelerate to the new speed - just as everyone has been telling you. That doesn't mean the wind won't be able to accelerate it.
Nonsense. If the wind slows, the balloon will be left moving faster than the wind. If it speeds up, the balloon will be left moving slower. Its time-averaged velocity could be faster or slower.
Incomprehensible gibberish, and the rest of your post is more nonsense that's far too boring to even read.
The book directly contradicts your claim that the balloon will travel slower than the mean speed of the wind. It even calculates (as I already did in this thread) how long it will take for the balloon the reach windspeed if for some reason it's moving at a different speed.
Humb and humber...
sol invictus
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That should give you an average force proportional to v2, then. There's one factor of v because the momentum transferred per particle is proportional to v, and there's another because the number of particles the thing collides with per unit time is also proportional to v. So I don't see how you can get an exponential out of that.
John Freestone
Graduate Poster
No answer to my question again, humber? Can't describe your mystery force that keeps a balloon at below windspeed, or explain why that happens while it also somehow manages to travel at the mean speed of the wind? Of course you can't. They are mutually exclusive conditions, for a start, one true (the one you denied), the other false. Having made yet another schoolboy error that clearly defies Newton, it has all been bluster from that moment on. Not being able to admit you were wrong is a very dismal character trait.
I'm not sure why you think you'll make better progress in classical mechanics by pretending it's all just like electricity. I don't see any relevance of the analogue so I haven't bothered clicking on your links. I know people remember stuff in odd ways. Maybe you learn history by playing it on the piano. The electricity thing isn't working, though. It might be confusing you. Something is.
Pseudomorphism is another invitation to go off on a pointless tangent, whilst pretending to have great expertise in some related area of knowledge. None but the stupidest initiates to your antics would be persuaded by it.
Well, that's great, humber. You have posted a link to a book about weather balloons that not only assumes that the balloon will travel at windspeed (and explicitly states that assumption), but gives a fine calculation of how long it will take the balloon to approach windspeed when there is a change in windspeed (e.g. a gust). The example (in section 12.3) that they give shows a 100 g balloon, and they calculate that the time constant for it to get to windspeed in a 1 m/s wind; the result is 0.45 seconds (which does not mean that it will reach windspeed that fast, but that is the number that can be used to determine how long it will take to get to any given speed that it arbitrarily close to windspeed).
If you think that the balloon will wind up at a speed lower than windspeed, tell us what that "terminal velocity" will be in this example. And I will tell you how long it will take to get to a speed that is faster than that.
If you think that the balloon will wind up at a speed lower than windspeed, tell us what that "terminal velocity" will be in this example. And I will tell you how long it will take to get to a speed that is faster than that.
No. The author acknowledges that the balloon cannot actually reach windspeed, when Vwind and Vb equated. Later, it is shown that the calculation for the acceleration time to holds even for winspeeds as low as 1m/s, but not that that speed is attained.
In fact, the quadratic equation (V-Vb)^2 is simplified to V-Vb. Care to support that?
No, the aim is to track both increases and decreases within the specified time. Anyway, this approach does not support your claim even in principle.
Of major concern to the novice white water canoeist, is getting down to the bottom of the run in one piece, and so they tend to follow the lower velocity course of the mean flow.
In time trials, the canoeist uses faster water that is created by obstacles, so as to accelerate the canoe beyond the mean water flow. However, even the mean speed is not achievable in a steady flow.
The analogy here is that of a river with waves. A canoe can exploit those waves to achieve a higher velocity than the mean speed. So can the balloon. The variation in speed allows it it gain energy and therefore velocity by exploiting acceleration not available in a steady flow.
I have told you this before. Wind variation is readily exploitable, and not at all surprising. Your claim, and the only one of possible interest, is steady flow.
I'll liven it up for you. I'll make you a joke.
Nice try. I have already said that the book makes good that claim, but denies your claim that it actually travels at windspeed. It contradicts you.
Your calculation is false, that is the point. This example is not comparable for at least one reason. The balloon equation is valid if, and only if, there is a velocity difference to exploit, and that condition exists for at least as long as the time it takes to accelerate that body. How long is "steady state"?
So you say.
I agree completely. But who is actually being deceived and who are those that can see they are being? You're being rather ambiguous again!
Here's a simple question that I hope you'll answer for me. It may seem a rather odd thing to ask but I believe that an open and honest answer from you can only help to support your position - which I'll assume for now is that you are in fact acting in good faith, and not deliberately setting out (in general at least) to deliberately deceive or frustrate others who have contributed to this thread.
So here's the question. Have you come to the realisation that any part (at all) of your previous understanding of physics was incorrect as a result, directly or indirectly, of anything written in this thread and its parent? For extra credibility, can you tell us what it was that you used to incorrectly believe and how you see that now? (Obviously something of some significance and relevance to the cart and/or treadmill rather than a very tiny detail in more or less unrelated area will probably carry more weight.)
Now, it could be that your honest answer to the above is "no". But I think I would find that somewhat surprising and therefore harder to believe than hearing you say something like, "Yes, I was wrong about some things, and here is an example....", and then proceed to show us something you have learned.
I'm happy to do the same if that seems more equitable. Just ask! However I don't think people will be nearly as interested in what I might have to say about my own knowledge or lack of it as what you might have to say about yours given the circumstances.
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That is essentially so, JJCote. It was not my argument, but Sol_invictus' support for his abuse of Rayleigh's formula for steady flow.
The point is moot. The question is for steady wind or water. Balloons can exploit wind variation to increase their velocity to the mean speed. I expect that some devices could do better.
No news there for you, I hope, JJcote.
When seen that way, it makes you wonder why he would support it, and trip over himself in the process.
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