Subduction Zone
Muse
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Wow, neat article sol invictus. What do those numbers work out to in regular terms. How much momentum can you store? Is it related to the electron drift of the current?
Split Thread The validity of classical physics (split from: DWFTTW)
- Thread starter Dan O.
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Wow, neat article sol invictus. What do those numbers work out to in regular terms. How much momentum can you store? Is it related to the electron drift of the current?
These forums allow you to use LaTeX if you are familiar with that. I am not, but will simply copy some LaTeX markup to demonstrate (and try it out for the first time!)
[LATEX]
\begin{align}
E &= mc^2 \\
m &= \frac{m_0}{\sqrt{1-\frac{v^2}{c^2}}}
\end{align}
[/LATEX]
That was generated by putting the following "raw text" (copied from this wikipedia article) between the appropriate tags ([latex] and [/latex]) like this.
Code:
[PLAIN][latex][/PLAIN]
\begin{align}
E &= mc^2 \\
m &= \frac{m_0}{\sqrt{1-\frac{v^2}{c^2}}}
\end{align}
[PLAIN][/latex][/PLAIN]
ETA: There's a whole thread about the LaTeX option here.
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A great deal of the damage will have been done by the release of the energy stored mechanically within the core.
Perpetual Student
Illuminator
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AHA!!! Thanks.
[latex]\int \limits_a^bx \mathrm dx [/latex]
It appears to work just like aurora. Now, I'm looking forward to having a reason to use it.
sol invictus
Philosopher
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As much as you want, limited only by the intensity of the fields you can create. You can store quite a lot of momentum in strong fields, as you can see from that picture.
No, nothing to do with it. Remember, light (and all other forms of EM radiation) carries momentum. That light could propagate half way across the universe and exert a pressure on something. The momentum is stored in the fields themselves.
sol invictus
Philosopher
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- Oct 21, 2007
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So humber - are you going to keep lying? Did you read the page from that book I linked to - the one that says that balloons travel at windspeed, and calculates how rapidly they return to it when the wind changes (and miraculously get the same solution I already showed you)?
The simplifying assumption is first made that the balloon is close to windpeed, then it is shown to be capable of following that wind. That is, there is enough force to accelerate the object to track the wind at constant percentage of the wind, but does not actually reach winsdpeed.
As the formula shows, in order to maintain that speed a constant force is required, whereas you claim that to be zero.
The force is required to accelerate the balloon, and so maintain a fixed, (V- Vb) depends upon drag, cross-sectional area, and the density of the medium. This is quite different from your calculation, and (u) is still not zero.
There is no doubt that some objects will get closer to wind or waterspeed than others, but force will still be needed to maintain that speed, whereas you have that force as zero.
Also, you claim is that all objects will get to waterspeed, for all values of c(u)>0.
Claim denied. Are you going to continue to lie, I mean, be be evasive?
ETA:
If you like, I can use a balloon to show that the treadmill is false.
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I would like you to show us the equation and calculate what is the constant percentage of the wind that the balloon will reach. Feel free to assume any fuction you prefer for describing the drag as a fuction of speed and assume any velocities and other quantities you need as examples in your calculations.
Again humber comes up with the claim F<>ma.
What the formula F=ma shows, is indeed that a force is recuired to accelerate, it is not required to maintain speed. In fact to maintain speed, it is indeed required that there is no force acting on the body.
What humber writes on this thread is truly astonishing. The ONLY thing that mitigates the fact that every single thing he says is dead wrong is the fact that he really says so incredibly little despite using so many words. I'd put his SNR at about -80 dB, and when you do get signal you realize the noise was more meaningful.
Tracking balloons are used to follow the wind, to have short enough acceleration times to follow the changes. They travel at the mean speed of the wind, not the peak (maximum) speed of the wind.
It's even in the text.
Nope.
Do you think that force on the left hand side may be what is called the force of the wind? Do you think then that increasing the mass of the balloon, increases that force?
The force is not zero at windspeed. If the balloon is held stationary w.r.t the ground there is the full force of the wind upon it, but there is also an equal restraining force. When released, the force of the wind can accelerate that body, because the restraining force is no longer there. In the ideal case, the balloon would be accelerated to windpeed, but the force of the wind can no longer accelerate w.r.t the surrounding air, but the force remains. This is also applies for terminal velocities below windspeed.
There is still the constant force of the wind, driving the balloon forward. You seem to say the force is zero because the force is zero with "reference to itself ".
When the balloon is stationary, drag couples the wind to the balloon. As the balloon accelerates, that drag goes to zero, but that does not mean the force is zero. Drag exists while there is a velocity difference, but the force remains.
Wind is a moving mass, and when at windspeed, the balloon becomes essentially part of it. How do you get to windspeed in this manner?
Be more like the air, like a balloon, but even they can't quite do it. The drag is never quite zero, so 100% windspeed is not possible.
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Is that what I'm thinking now, is it Sol, huh is it?
No, -80dB means that the noise is 80dB lower than the signal. DOH!
That balloon article will do just fine when comparisons are made to your treadmill. Quite helpful, really. Too bad that you didn't notice that huh34 is wrong.
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Hello humber.
You seem to given up the ghost when it comes to responding to my post #2744? (In the original thread.) I'm still interested in hearing your thoughts on that though if I can persuade you to reconsider.
Clearly your input has been the main reason this thread and its parent have been as popular and long-lived as they have. I can't imagine how much time you've put into it. But I for one would really like to gain some real understanding of your point of view when it comes to the treadmill. If a back-to-basics "lesson" from you is what is needed, then I'm willing to listen.
You seem to given up the ghost when it comes to responding to my post #2744? (In the original thread.) I'm still interested in hearing your thoughts on that though if I can persuade you to reconsider.
What are the "connections" you mention here? Is it anything to do with the "holding with the hand" business in the following quote (from later in the same post)?
I'm afraid I couldn't really work out what the "holding with the hand" issue might be about. All I could think of was that this was something to do with me leaning over the equator with the cart in my hand and placing it on the ground in the northern hemisphere. Note that I am not trying to make an argument in favour of the cart going downwind faster than the cart. It's just a thought experiment where we end up with two identical carts running side by side at the same speed in the same "real wind" on equally "real ground". What else do we need to know in order to be able sure that they will perform identically (regardless of whether that is to slow down or accelerate)?
I'm also keen to hear the details of the thought experiment you mentioned. Mender also expressed interest in this so that makes at least two of us.
Yes please to this also!
Clearly your input has been the main reason this thread and its parent have been as popular and long-lived as they have. I can't imagine how much time you've put into it. But I for one would really like to gain some real understanding of your point of view when it comes to the treadmill. If a back-to-basics "lesson" from you is what is needed, then I'm willing to listen.
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Let the record show that humber accidentally wrote something that is quite correct. On average they do travel at exactly the mean speed of wind, not at some fixed percentage of that mean.
I don't understand the question. If the wind is moving from left to right in relation to the balloon, the force exerted on the balloon is directed from left to right. You can think of it as a "pushing" force on the left hand side of the balloon or as well as a "pulling" force on the right hand side. Makes no differece.
No. Increasing the mass of the balloon decreases the acceleration. For example assume a force of 1 N from the wind pushing the balloon. If the balloons mass is 1 kg, it will accelerate at the rate of 1 m/s2. (F=ma, you see.) If we increase the mass to say 2 kg, the 1 N force will accelerate the balloon at the rate of 0,5 m/s2.
Wich way is this nonzero force of yours pointing at? If the wind moves 10 m/s FLTR and the balloon moves 10 m/s FLTR, is the forze from left to right or vice versa?
Newtons first: "A body continues to maintain its state of rest or of uniform motion unless acted upon by an external unbalanced force."
That means, if the balloon is moving at a constant velocity, there is no net force acting on it. If there is such a force, the balloon will accelerate (F=ma, you see)
IN THE SAME FREAKING POST?
Humber, I claim that a body with nozero mass can move with a constant speed in relation to the ground with zero external net force acting on it. In fact I claim this is the only situation, where the body will move at constant speed, in any other situation the body will accelerate of decelerate. Do you or do you not agree?
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In humberverse not true:
v(t) = s'(t) and a(t) = v'(t) = s"(t)
Where v(t) is the first derivative of the position function and a(t) is the first derivative of the velocity function. Also note it is the second derivative of the position function!
I think that this must be among the things that huber totally fails to understand. In humberverse acceleration and speed are not directly coupled in time. Therefore in humberverse F=m(v+a). An object moving at a constant speed needs a force F=mv to be able to keep moving. If the force exerted on an object is less than required for constant speed, the object will decelerate (a<0). If the force is greater than required for constant speed, the object will accelerate (a>0).
v(t) = s'(t) and a(t) = v'(t) = s"(t)
Where v(t) is the first derivative of the position function and a(t) is the first derivative of the velocity function. Also note it is the second derivative of the position function!
I think that this must be among the things that huber totally fails to understand. In humberverse acceleration and speed are not directly coupled in time. Therefore in humberverse F=m(v+a). An object moving at a constant speed needs a force F=mv to be able to keep moving. If the force exerted on an object is less than required for constant speed, the object will decelerate (a<0). If the force is greater than required for constant speed, the object will accelerate (a>0).
John Freestone
Graduate Poster
If it remained below windspeed, there would be wind moving past it. That wind would exert the force known as drag, which would accelerate it. As it accelerated, the drag would decrease, but still remain if the object were moving slower than the wind, so that would accelerate the object a little more, increasing its speed to nearer that of the wind and reducing the drag. That is how the speed increases (WRT the ground; decreases the difference between the object and the air) as an exponential curve. The only point at which the drag reduces to zero, and therefore the acceleration (that's when something goes faster than it was before) reaches zero, is when the object is moving at windspeed. From certain strict mathematical consideration points of view, you could argue that that never happens, but we have covered that point - it continues to approach that, it gets so close as to be indistinguishable, real world fluctuations are much more important and cancel out that pure mathematical truth, etc. It would even be mathematically correct to say that if I put an object, initially stationary, into a perfect fluid flowing at 10 m/s, the object will never reach 10 m/s, because the whole system won't. The stream will be slowed a little by the drag just as the object is accelerated by the drag. The momentum will be maintained. However, again that's just a ridiculously tiny effect. If you throw a ball at 10 m/s, you don't argue that it won't go the distance calculated because the earth will move backwards by a small amount. But once again I've given you another fact you can use later to wriggle out of what you were actually saying, and remind us that you said it wouldn't go 100% of windspeed, which is true!
For that to be true, and yet Newton to be right, there would have to be another force holding an object in a fluid back from reaching fluid speed. A net force of zero, as claimed, does indeed equate to the maintenance of a constant speed. What is the other force cancelling out the drag, which will accelerate the object at lower than fluid speed? Simple question, not answered. I have stated before that the only force applicable is the force of drag between fluid and object.
Parsing that is tricky. It seems to state the same, that force is required to cause acceleration and thus maintain a constant velocity. Wrong.
Once again, this says Newton was wrong, or we have a mystery force.
The problem is so trivial I have to admit I've been skimming some of the maths, but I gather c(u) is the coefficient of friction, or how draggy something is. Right? So if something has absolutely no drag coefficient, and we drop it in a stream, the stream can't get hold of it and accelerate it. There is no force of drag. (No-one has yet invented such a material, unfortunately.) If there is just a tiny weeny bit of draggability it will be accelerated a tiny weeny bit. That will continue (even though, as always, the tiny weeny force will reduce even more as the object accelerates) until the force is zero. When is that? It is when the object is going at waterspeed.
So far your only refutation of that relies on denying Newtonian mechanics. But why change the habit of a lifetime?
And that will be how many wrong demonstrations you've either done or, more usually, not done? But hey, don't let me stop you. It will no doubt be just as amusing as the others.
You claim that acceleration is needed to maintain a fixed speed?
MAY THE FORCE BE WITH YOU, HUMBER!
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John Freestone
Graduate Poster
...and is the very point we Newtonianists have been making all along. They travel at the (mean, if you like) speed of the wind! Yes, finally, you have got it. All you have to do now is backtrack to the stated problem, which involved an ideal, constant speed of wind, and then its mean speed is the same as its speed! Well done.
Oh dear. I spoke too soon.
Well just think what happens if you put a couple of tons extra weight in your car. Does that increase the power of the engine, or reduce your acceleration?
WHY NOT? WHAT IS IT? WHAT FORCE ACTS ON A BODY FLOATING IN THE WIND AT WINDSPEED? And please, can we keep away from coriolis forces, quantum forces, gravitational forces.... and stick to the lateral mechanical forces. What force is there at windspeed that is going to slow the object down? That is all you have to answer.
Correct. That is why it doesn't move anywhere. The net force is zero.
Yes.
Correct-ish. In the ideal case it would approach it as an exponential curve approaches a particular value. In the real world case, it does real world complicated things. But interestingly, you have again repeated the original point that we Newtonianists were making. Well done. Now say, "So you were right after all, sorry."
Oh crap
If the force remains, why can't it "accelerate [the object] w.r.t. the surrounding air"? That is what force does, unless all the forces aren't calculated and so it isn't the net force (remember how you do a force diagram and work out the net force to decide on the acceleration of things)? Once again, you have a mystery force to tell me about, or you have to explain why a force remains that can't accelerate something.
Parsing problems arise again, but 'terminal velocity' is a misleading term, I think. It usually applies to the velocity gained when a constant force (e.g. gravity) is applied to an object in a fluid (usually an object falling through still, or vertically motionless, air). This is an interesting case, almost the reverse of the balloon/boat problem, where an object is accelerated by another force (indeed, it could be the boat problem when driven by a motor) until that force matches the drag. It is therefore the velocity something (like a person) reaches when falling through the air. If you could miraculously suck all the mass out of a person as they fell, in fact, you'd find that they slowed (due to that drag force) until they stopped falling...i.e. were going windspeed...i.e. not going. The same is true of the boat with its motor cut. It slows to waterspeed, whatever that is and in whatever direction that is, due to the force of drag. But look at me waffling on talking correct physics.
This kind of statement (esp. the first sentence) demonstrates why you might possibly be wrong, as opposed to just lying. You have failed to get your head round the fact that a body in uniform motion continues to move at a constant velocity unless an external force is applied. The balloon does not need to be driven forward in the wind constantly. It is accelerated by force only: force does not maintain constant velocity, it buggers it up. Once the object's molecules are travelling along side-by-side with the molecules of the air, there are no more collisions between them, there is no more drag caused by them rubbing up against each other (er, kind of). They just make friends and chill and watch the clouds. It's an easy mistake to make, for a 6 year old or someone who has lived isolated from modern societies, say in the Amazon jungle. It is the kind of mistake people made all the time before the Enlightenment.
Of course, again, there is another way to look at it that makes your statement right: the pressure of the molecules on the downwind side of the balloon does exist. There is a force impacting the balloon pushing it forward. The only problem with that view is that it ignores the equal pressure on the other side, and to decide on the acceleration, you have to calculate the net force.
The whole problem can be thought of in terms of little balls (molecules) moving in space. Just try it that way. They impact your object on one side, bounce off it, and cause it to accelerate. They impact a little slower, but continue to impact that side until the whole sherbang is gliding along through space. You do know that an object moving through space continues to maintain its speed without being pushed, don't you? I think Newton said something about that.
No, it doesn't. You're bound to muddle yourself if you start thinking that drag couples one thing to another. You used to keep saying things were coupled to other things regarding the treadmill. It's misleading.
Hurrah!
God help us. DRAG is the name of a KIND OF FORCE. It happens to be the ONLY force in the system. Therefore when drag goes to zero, force goes to zero.
Hurrah!
You keep saying the truth, but your ignorance remains.
Hurrah!
By acceleration due to drag-force.
Can't diagnose that sentence, but it's ill.
I refer you to my earlier waffle concerning the mathematical world and the real world. You're right and you're wrong. In the real world the drag, the object's speed, the acceleration and the windspeed are all changing. If they don't mess about too much, balloons move at windspeed and immersed dead fish move at riverspeed, and floating boats on rivers maintain some balance depending on the windspeed above the waterline and the riverspeed below the waterline. Thus has it always been.
sol invictus
Philosopher
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- Oct 21, 2007
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- 8,613
Well, it's true that to maintain a fixed and non-zero v-vb one needs a propulsive force which is equal in magnitude and opposite in direction to Fdrag. Absent such a propulsive force, Fdrag is the only force acting and the velocity v will rapidly approach vb (how rapidly depends on the mass and shape of the object, but all objects behave that way, and none can achieve a steady state at a velocity other than that).
This is all totally obvious to everyone... except our hero Forrest Hump, of course.
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John Freestone
Graduate Poster
But here you're talking about needing a constant propulsive force to overcome drag - like my car's engine needs to be running to keep moving forward - surely? In the balloon situation there is no propulsive force except drag. Drag is not the enemy of constant velocity, but its cause. As soon as u becomes non-zero, a drag arises. That's the bit humber doesn't make sense of - he lets drag go to zero, and says there's still a force of propulsion.
We could argue (and I guarantee humber now will) that there is a constant propulsive force provided by the difference in pressures between one area and another, which is what causes the wind in the first place. That would be a further obfuscation, however. It would not allow his earlier assertion that the balloon maintains a speed significantly less than windspeed, and anyway, he has agreed that it does maintain its average speed, which is the same, so we've won.
John Freestone
Graduate Poster
Forest Hump. 
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