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

[humber]

Ummm.... what is it?

Please explain, because to the rest of us "balloons pass left to right w.r.t the observer" and "observer passes the balloons right to left" sound strangely similar...

It seems clear enough, but OK.

Wind (real)
Two balloons tethered against the wind from blowing left to right say velocity 'v'. Assume they are horizontal, and their motion is simple linear.
Observer rabbit is distance 'n' from said balloons.
Balloons simultaneously released.
When they pass the rabbit it will see that they are distance 'x' apart.

Treadmill wind (fictive)
Two balloons tethered to device attached to belt, beltspeed 'v'
Observer rabbit is distance 'n' to the right of attached devise, and also on the belt.
Balloons simultaneously released
When the rabbit passes by the balloons, it will see that they are distance 'x' apart.
X will be larger in the real wind. No numbers are really needed, because that difference is arbitrarily large.

I have 3 more (new) when you have done with that.
Can't tell the difference between real and treadmill wind?

 

[humber]

Great. I look forward to seeing your demonstration on YouTube.

By Monday.

 

[humber]

Nice drawing,I must say.

Yes

And we seem to be getting closer.
I'm going to assume that the difference you refer to is the difference in distance between the balloons as they pass by the bunny?

Mrs Rabbit, yes.

This difference is not arbitrarily large, as I can state the exact limits of the difference.

I can make it otherwise.

For a small treadmill the boundary layer does not reach up to the balloons so they will arrive at the same time.

Boundary assumption correct. Just as well I didn't include boundary flow!

The wind driven balloons will not. The difference will then be:

[latex]\small{
$ \int_{t_{snip}}^{t_{pass}} \int_{h_1}^{h_2}
\frac{\partial}{\partial h}} V_{wind}(h,t)
dh dt [/latex]

where h₁ and h₂ are the heights of the balloons above the surface. V_wind is the boundary layer windspeed, and t_pass is the time the first balloon passes the bunny.

In the case of an infinitely long treadmill or trolley, the boundary layer thicknesses and velocity gradients will be about equal, making the difference 0.
This calculation assumes 2D laminar horizontal flow

Those details are irrelevant, but if you wish.

.
Anyone, feel free correct me if I've made a mistake. I do do that.

What's good for the goose....(paraphrasing)
"let me assure you this is quite correct. It's simply that you do not really understand relative velocities, nor basic physics and until you do...".

For a small treadmill the boundary layer does not reach up to the balloons so they will arrive at the same time.

Boundary assumption correct. Assume constant homogeneous flow. Just as well I didn't include that!
No, try again.
No, not correct. But thanks for the proof that it will be

For a small treadmill the boundary layer does not reach up to the balloons so they will arrive at the same time.... The wind driven balloons will not.

If you insist. I win already.

 

[John Freestone]

sure. I was about ot tell you thta the "quote button" didn't do anything on my machine, but it turns out it's just fricken slow.

Yeah, I thought after I posted that, you're gonna hate me as you wait for the server. It varies, maybe with traffic, but it's one of the negatives about this forum (there are lots of positives).

Yep - you are right, and the need to add those fictitious forces one way we can tell we are not in an inertial frame of reference. I think it's true to say (in a very strict and pedantic sense at least) that there are no frames of reference in the "real universe" that are truly inertial, at least with respect to Newtonian mechanics. All we can ever hope for with Newtonian mechanics is to get something that is "approximately inertial". This might not be a problem with General Relativity, but perhaps I just need to go and read those Wikipedia pages yet again.

Thanks for those links, Clive. I keep thinking that I should read a bit more about what ref frames actually are instead of assuming I know. Obviously you're off into more advanced territory here with rotating frames and "fictitious forces", and the simplest of explanations is enough to undertand the equivalence of treadmill testing, etc. In fact more that that. I think it has a deeper effect, this relativity of velocities and related quantities, if you play with the idea in your personal experience. Our own frame of reference is the one we're physically bound to, yet we step out of it in imagination all the time, and there is no 'correct' frame. I mean, when I drive somewhere, and think about what's happening, I jump out of my own frame and see it from the Earth's frame. I don't think I'm dragging my destination to me by the wheels of my car. Put it together with the up and down directions, and it's even more enlightening. The Earth is just somewhere next to me, and I'm using the fact that it's stuck to me by some weird force to use my Earth-grabber wheel-turny thing to drag my destination to me!

OTOH, it seems that it's not quite equivalent, because of the masses and accelerations. If I use my frame, the earth's surface accelerates the same amount as I do if I measure from the Earth's frame, but there is something artificial about that, because my mass is tiny compared to the earth, so for the applied force, I accelerate much more - I want to say "in real terms". I don't know what to make of that. It obviously sounds dangerously humberistic, and I know that for little carts and treadmills and airflows of the types we're discussing, such differences don't apply, or do they? No, I suppose because in neither frame (earth or belt) are we actually concerned with the movement of the earth or anything with great mass. It's the air and the cart and a surface.

My background knowledge is a bit too sketchy to make it all fit properly yet, I think. Presumably, if I tried to work out my engine's power output or fuel consumption by reference to how it can move the Earth around, something would be way off. An ant could shift the planet 'miles'. But I'm doing it again - writing about it when I should be reading!

Anyway, we're getting way off topic...

Well, if you are Pedantic (the overwhelming desire to correct people when they are wrong), it's still better than being Humdantic (The overwhelming desire to correct people when they are RIGHT!!)

That's better. That's a humberdinger.

BTW, what? No comeback from "negative dragons"? I don't know why I bleedin' bother!

 

[John Freestone]

It seems clear enough, but OK.

Wind (real)
Two balloons tethered against the wind from blowing left to right say velocity 'v'. Assume they are horizontal, and their motion is simple linear.
Observer rabbit is distance 'n' from said balloons.
Balloons simultaneously released.
When they pass the rabbit it will see that they are distance 'x' apart.

Treadmill wind (fictive)
Two balloons tethered to device attached to belt, beltspeed 'v'
Observer rabbit is distance 'n' to the right of attached devise, and also on the belt.
Balloons simultaneously released
When the rabbit passes by the balloons, it will see that they are distance 'x' apart.
X will be larger in the real wind. No numbers are really needed, because that difference is arbitrarily large.

I have 3 more (new) when you have done with that.
Can't tell the difference between real and treadmill wind?

Once again I think you're just stating something that the rest of the world would dispute utterly. No numbers are really needed except zero. x=0 [ETA: I mean the difference between x in the first example and x in the second will be zero]. You didn't even say how the balloons are positioned with respect to each other!] How could x be different? You don't specify what dimensions these distances are in, but it doesn't matter anyway, due to the equivalence of frames of reference.

But yeah, let's have 3 more why not?

 

[humber]

Dang!! This is so classically humber I thought it was humb

Yes, it is quite embarrassing for you, when many of have replied, and then quoted him in support. He is quite useful, as I told him.

 

[humber]

Once again I think you're just stating something that the rest of the world would dispute utterly. No numbers are really needed except zero. x=0 [ETA: I mean the difference between x in the first example and x in the second will be zero]. You didn't even say how the balloons are positioned with respect to each other!] How could x be different? You don't specify what dimensions these distances are in, but it doesn't matter anyway, due to the equivalence of frames of reference.

But yeah, let's have 3 more why not?

Not the answer, but put numbers in if you like.

If I give you more than one, you won't be able to sleep, and you know how cranky you get when that happens.

ETA:

That's better. That's a humberdinger.

BTW, what? No comeback from "negative dragons"? I don't know why I bleedin' bother!

That's why.

 

[sol invictus]

Wind (real)
Two balloons tethered against the wind from blowing left to right say velocity 'v'. Assume they are horizontal, and their motion is simple linear.

What do you mean, "assume they are horizontal"? Do you mean at the same height above the belt? Or do you mean the opposite, as in your sketch (same horizontal starting position but different heights)?

The only thing that can make a difference in x is a difference in wind speed at the heights of the balloons. It's true that a finite sized treadmill will have a slightly different vertical windspeed profile, as we've all said all along - but that difference will become smaller and smaller as the belt gets bigger and bigger. Is that what you're trying to say - that you agree with us all now?

 

[John Freestone]

This thread has developed into an exploration of Humber's perception of the validity of classical physics, and no matter how much he writes, no-one can get any more understanding of his theories.
It would help If he would respond to particular questions. I'd love to know if he finds his methods work in practice, for him. That is, if he has a practical problem involving physical objects in the conventional physical world, involving other people outside (who could validate that they experience the same world), can he work out and predict the outcome of events. It seems to most of us that it would be very dangerous to get involved, since the force and mass interactions could affect our bodies, limbs etc.
Are there some basic experiments we could agree on which involve the use of our bodies rather than only external objects like carts etc, which would give us real motivation to be correct? Would Humber survive, could we tell?
I'd like him to predict this one and explain why. If two cars accelerate toward eachother then maintain a constant velocity, then launch off a ramp into the air, and collide mid air, will they get damaged before hitting the ground? Would Humber try this?

Yes, or if he could predict which way a piece of paper would bend if he waved it through the air. Then, perhaps after empirical verification if he likes, he could move on to attaching bits of tissue paper to a couple of lolly sticks stuck through a large piece of card and wave that about. This series of experiments in the real world are designed, of course, to demonstrate the direction of a boundary layer w.r.t. a moving surface travelling through a fluid medium, which he seeeeeems to suggest is opposite to how it flows when the medium moves over the surface.

This is one of his main objections to the use of the treadmill as an equivalent inertial frame and proof of DDWFTTW, that the boundary layer is going the wrong way. Testing presumably isn't required, since he is so sure of this theory.

Asking a bit more about the theory leads nowhere either - like the fact that if his theory were true, it would mean that other surfaces would act in the same manner. An aeroplane's wing (or just a car body, to keep lift and such things out of the argument) would have monotonic flow w.r.t. it when in a windtunnel, but suddenly find a reverse flow at the boundary layer when moving through still air.

For the treadmill example, I even asked for help in deciding where these opposite flows are divided from each other (I haven't even got into what the result of such sheer would be!), but no actually numbers or any other actual data is required, apparently for the same reason as the arbitrary difference in x for the rabbit above.

 

[humber]

Yes, or if he could predict which way a piece of paper would bend if he ......

I'll let you know when you can speak for me, John.

What a load of self congratulatory dross. It's a toy cart on a belt. The only matter of interest at all, is that you all can't see that.

 

[humber]

What do you mean, "assume they are horizontal"? Do you mean at the same height above the belt? Or do you mean the opposite, as in your sketch (same horizontal starting position but different heights)?

The only thing that can make a difference in x is a difference in wind speed at the heights of the balloons. It's true that a finite sized treadmill will have a slightly different vertical windspeed profile, as we've all said all along - but that difference will become smaller and smaller as the belt gets bigger and bigger. Is that what you're trying to say - that you agree with us all now?

The heights are the same. Heth'etheth has the circumstances correct.

 

[GregLondon]

OK, multi-quote either doesn't work for me or is taking more than five minutes, so I did a single quote and have to post Humber's statements manually:

humber:

I mention the maximum power theorem,

humber:

I'm an electrical engineer, and that's an electrical engineering term. wikipedia agrees:

http://en.wikipedia.org/wiki/Maximum_power_theorem

In EE, all it means is that if you have an amplifier with an internal resistance of 4 ohms, then if you want to transfer maximum power to the load, it should be 4 ohms too.

Do you have a URL to the Maximum power theorem you're talking about?

No, really, I'm trying to understand what you mean by "maximum power theorem".


humber:

Power is not measured like that.

Huh? Power is work per time. Work is force times distance.

Power (watts)=force(newtons) * distance(meters) / time (seconds)

If you can extract 10 watts of power from a windmill, you can lift 1 kg at 1 vertical meters per second, or half a Kg at 2 m/s, or 2 kg at one-half m/s or whatever, it's all 10 watts.

Do you still disagree with this, Humber? Or can we agree that power translates to motion?

Oh, and I thought of a better electro-generator cart design.

Did you read this one, humber? Do you see it working or do you disagree?

Also, out of curiosity, what do you do for a living?

I was asking so I could google "maximum power theorem" and add in whatever career title you have or had to see if there's a specific meaning that isn't electrical engineering related. If you're retired, what did you used to do for your career?

 

[GregLondon]

In my work,

What line of work was that again? So, I'll assume you're not retired. Are you an aero guy? mech? physics? mathematics? Do you design stuff or test it?

 

[GregLondon]

Greg, better check your last big post, it appears that you suddenly went from producing 12 watts to 24 watts with the same relative generator speed.

Ooh, pooperdoodles. Does the "edit" button go away after some period of time? Cause I can't seem to find the button on that post. There's an edit button on teh post I just put up, but not the one from yesterday.

Yeah, it should be 12 watts, always 12 watts. not sure how that happened.

oh well.

 

[John Freestone]

The heights are the same. Heth'etheth has the circumstances correct.

Look, the only reason I have to speak for you (and it was only suggesting you do an experiment and what you might learn is up to you) is because you won't ever explain what causes the effect you assert will be observered. Is it the difference in boundary layer speeds due to the different drag between surface and air in the two situations? If so, big deal. They are minor arbitrary differences of scale, dealt with.

If the 'circumstances are the same', which I think you stated in the original, then the outcome is the same. You have to specify exactly WHAT differences there are for a different outcome. Same circumstances = same outcome.

Does the boundary layer go forwards w.r.t. the treadmill, as you asserted before? Is that the cause of the difference? Because that is quite simply wrong (again, you must be very used to it now), or if it isn't, then tell me why they test aerodynamic flow over surfaces in windtunnels, for surfaces that are meant to go through the air - moving, like a treadmill surface, driven through the still air. The boundary layer would be reversed. Air on a plane's wing would go forwards over it. No answer from you on that. Why? Because it's garbage.

Or is the bunny seeing something different for some other reason? You can't just state an effect and then refuse to agree to any objections we raise, without having said what you think will cause the difference.

 

[Christian Klippel]

Do you design stuff or test it?

Hello Greg,

nope, he simply makes up stuff as he goes. One could also call it "fantasizing".

Greetings,

Chris

 

[H'ethetheth]

Wow, that is a hackneyed piece of text if ever I've seen one. No less than three replies to the same sentence, two of which nearly identical.

I can make it otherwise.

Please do.

Boundary assumption correct. Just as well I didn't include boundary flow!

You did incluse it implicitly. And also, many people have told you before that this is the only appreciable difference between the treadmill and the road.

What's good for the goose....(paraphrasing)
"let me assure you this is quite correct. It's simply that you do not really understand relative velocities, nor basic physics and until you do...".

The things you're bringing up here were discussed miles back in the DDWFTTW thread. I asked you about boundary layer flows long ago, and you said the differences between the treadmill and the road had nothing to do with boundary layer thicknesses. Now they do?

No, try again.
No, not correct. But thanks for the proof that it will be

What? What does this refer to? What does it even mean? You know what? Never mind.

If you insist. I win already.

Ah, the old win-by-blatant-assertion method. Very nice.

Bye.

 

[John Freestone]

Ooh, pooperdoodles. Does the "edit" button go away after some period of time? Cause I can't seem to find the button on that post. There's an edit button on teh post I just put up, but not the one from yesterday.

Yeah, that's one of the negative things about the forum - or it has disadvantages anyway. It's meant to stop people saying stuff and then backtracking, but I'm not sure it's really of much worth, and it is very annoying when you want to alter a genuine mistake or remove wrong information for good reasons. I can't remember how long it is, only a matter of an hour or so I think.

Negative dragons.

 

[sol invictus]

The heights are the same. Heth'etheth has the circumstances correct.

"The heights are the same" - it's truly comical how bad you are at expressing yourself. Almost as if it's intentional....

Anyway, in that case Heth's perfectly correct analysis applies (although you don't need to write the result as an integral over h), and you are (of course) wrong. I'm afraid that your assertion to the contrary, supported as it is with zero evidence or reasoning, isn't very compelling.

But what's nice about this is that it almost verges on a concrete claim. So tell us, humber - what do you need to do to make the difference large? Be careful, because someone may actually test this....

 

[RossFW]

did I mention-

Cavitation in Gas? Citation?

Relevence of Clay pipes and fibreglass thingies to ridiculous terms like "Negative drag?"

Notice how Greg came on, was very polite, made some very clear (well, to well people) explanations, but now thinks you are a Whacko?