Merged DDWFTTW in a “Real” Wind

[humber]

RP,

I was just thinking about your question about self starting when I was passing through an airport a few days ago. They had the "escalator like" people movers which run on a flat surface. I recall Denver and Atlanta airports as having examples.

And you "understand" relative motion?

If you were in the car heading for the stationary truck, would you apply the brakes? If you were in the stationary car, with the truck heading towards you, would you put your foot on the brake?

 

[humber]

Yes, I can point you to the part where you agreed with him. You simply missed it. If you go back and read it again you'll see that you were in violent agreement that the relative velocity was 120mph.

It is. But there is no closing velocity, and I say that there isn't whereas Spork says there is. Have you comprehended that much?
It's that erroneous idea of closing velocity that make for the TM. Don't you worry, both the TM and the trial will soon be buried.

 

[3point14]

Don't you worry, both the TM and the trial will soon be buried.

Crikey, well that's a weight off my mind. Now I can devote all my time to worrying about the end of the world in 2012.

 

[windgrins]

Ball on a circular treadmill (to Spork)

I'd have to look at it again, but my recollection is that it imitated typical orbital dynamics. If you look at the behavior of a spacecraft in a local gravity-aligned frame, it can seem pretty bizarre (and makes it difficult for astronauts to control them for docking, etc.). When you hit forward thrusters you go "up" (away from Earth). Reverse thrusters make you go down (toward earth). Down thrusters will send you forward, and up thrusters will send you back.

So if you're all lined up for docking, and hit forward thrusters to do the deed, you end up missing the target. In a true inertial frame it all makes sense. But in the local frame, the craft appears to go in circles as a result of a single thrust.

Good observation but I'm not quite clear on the forces. I am reasonably familiar and understand your answer wrt to orbiting spacecraft. The component in the treadmill test I didn't see was what provides a force that keeps the ball from starting to roll with the surface. When the ball is held on long enough to come up to rotational speed, I can see that there is very little wind drag (other than entrained skin friction drag) and there is no R-bar (centrifugal) force since the ball is stationary over the ground (roughly). The circular motion over the ground reference would seem to generate an ever increasing radius circle but there may be something about the dynamics that provide a counteracting force that are not obvious (to me).

From the discussion and the video, I couldn't tell if this was a long term steady state behavior or just a transient condition until the ball fell off. This is why I was also a bit annoyed at Ynot's comments about trying to catch us with his varying speed treadmill. It was actually and interesting side problem and I could have learned something from that. I might have to do that experiment myself.

This is reason the "flame" doesn't do me much good as it obscures the actual knowledge transfer. On the other hand, forums aren't run just for me and everyone has their reasons for being here and the cost for admission is only time so I remain mollified!

Thanks for the comment Spork and if you do have insight, I'd like to hear it.

We may have a tough time convincing Ynot to re-shoot the video under steady state conditions but I'm not above asking him as I'm really lazy!

Ynot. What do you think if you are out there? An opportunity to do some good with your equipment. Let us know.

Regards,

WindGrins

 

[humber]

There is

[latex]${\vec v}_{\rm rel} = {d \over dt} \left( \vec x_1 - \vec x_2 \right)$[/latex] and
[latex]s_{\rm clos} = {d \over dt} \left| \vec x_1 - \vec x_2 \right|[/latex].

But [latex]$|{\vec v}_{\rm rel}| \neq s_{\rm clos}$[/latex]. In the situation described, [latex]$|{\vec v}_{\rm rel}| = 2v \equiv 120{\rm mph}$[/latex], while [latex]$s_{\rm clos}=2v \left[2vt/\sqrt{(2vt)^2+b^2} \right]$[/latex] (b is the distance between the cars at t=0 when they pass each other).

Gee, Sol, you are a wizz.
First you write an equation around the "relative velocity" ( which is also the differential velocity, with a common mode velocity of the surface of the planet, but we are going to make that zero, aren't we now?), and the same equation, relabeled "closing velocity"

You then write an equation that says the closing velocity becomes zero when the distance, b = 0,
But oddly enough, that equation produces a zero when t = 0 or v = 0.
but also 2V, and some other values dependent upon b or t...etc...etc...
Meaning that if you are right, the relative velocity is zero too, but only if t = 0, which gives you a problem for your first two equations....etc...etc...
Good work.
ETA:
You may get away with that, if the velocity can be expressed in terms of "c".

 

[zooterkin]

humber admitted to us long ago that he has Alexia. He is not able to read your post with more care.

If true, this would suggest that a text-based forum is not the best place for him to be arguing with people, since he can never be sure he understands what they are saying. Of course, getting that point across to him here is also going to be hard...

 

[humber]

Crikey, well that's a weight off my mind. Now I can devote all my time to worrying about the end of the world in 2012.

I am here to help, 3.14. It won't be your world that is significantly affected so you need not worry. I hope Sol's tenure is not up for re-appraisal.

 

[humber]

If true, this would suggest that a text-based forum is not the best place for him to be arguing with people, since he can never be sure he understands what they are saying. Of course, getting that point across to him here is also going to be hard...

I wrote that in a PM to Spork for a number of reasons, including to see if he would mention it. He posted that PM, and knowingly (or thought he did) ridiculed someone having an unfortunate disorder. I wonder if NALSA and Google and Joby know about that unacceptable publicity.
He will soon learn what that means.

 

[zooterkin]

I wrote that in a PM to Spork for a number of reasons, including to see if he would mention it. He posted that PM, and knowingly (or thought he did) ridiculed someone having an unfortunate disorder. I wonder if NALSA and Google and Joby know about that unacceptable publicity.
He will soon learn what that means.

So, is it true, or not? Do you have alexia?

 

[quarky]

RP,

I was just thinking about your question about self starting when I was passing through an airport a few days ago. They had the "escalator like" people movers which run on a flat surface. I recall Denver and Atlanta airports as having examples.

If those were turned up in speed just a bit, one could probably get the cart set on the entrance end and before it fell off exit end, it should self start and run back up and run off the entrance end. (the entrance and exit being defined by the way people normally use it.) This would provide the appropriate tailwind condition. If any of our demo cart people live near these airports, It might be an interesting (albeit predictable) video.

For those of us who do understand the correct use of inertial reference frames, it wouldn't add much, but a very cute demo, nonetheless.

Regards,

WindGrins

Are you ready for the 'slinky on the escalator' experiment?

 

[humber]

So, is it true, or not? Do you have alexia?

Funny how I am either seen as a master troll, or an idiot. That makes me wonder about the capacity of those who say that, to discriminate between fact and fiction, and why that may apply to following the cart.

 

[sol invictus]

Gee, Sol, you are a wizz.
First you write an equation around the "relative velocity" ( which is also the differential velocity, with a common mode velocity of the surface of the planet, but we are going to make that zero, aren't we now?), and the same equation, relabeled "closing velocity"

Gibberish.

You then write an equation that says the closing velocity becomes zero when the distance, b = 0,

No. Read it again.

But oddly enough, that equation produces a zero when t = 0 or v = 0.

There's nothing odd about it.

ETA:
You may get away with that, if the velocity can be expressed in terms of "c".

This has nothing to do with relativity. That formula follows from trivial geometry - it took me much longer to type in than it did to figure out (approximately 30 seconds in my head, most of which were devoted to factors of 2 and 1/2 which cancel).

 

[humber]

Are you ready for the 'slinky on the escalator' experiment?

According to my experience, all airport moving walkways are behind passport control, so security is likely to jump on anyone with a cart and video camera.
At Tullamarine Airport in Melbourne, there is one moving walkway that is fun. If you de-focus, and let your eyes see the walls, you get a sensation of moving backwards, that is so convincing, you try and right yourself.

BTW. The speed of the wind can be measured using an acoustic anemometer, so no moving parts to produce error.

The airport belt is a lesson in taking what you see to be fact.
What if I were to deliberately build a TM to fool you? You will no doubt know of Maglev trains.
I could make a small version of that, with the track hidden under the belt. The cart then moves as a linear motor. I could tell you that the cart is either moving relative to the air, or the belt, but the magnetic field passes through the belt, so the cart is actually moving relative to magnets and so the ground. That magnetic field is analogous to gravity. Is the TM cart moving relative to the belt, ground or air?

 

[Dr. Keith]

I put an open challenge to Spork to prove that he has the skill to support his claims which are not to be found in any text book.

That he can't do that, says a lot.

Like all of the TM tests that have been ignored.

But since I claim no degree you won't discuss this with me?

A bit elitist, but I was really hoping you could help me out.

You post tens of times a day and still haven't been able to address why a neutrally buoyant balloon would travel at a different speed than the medium and how that difference is calculated. If it is so minor just pint me to a reference that addresses this condition. Surely meteorologists must have tables to calculate wind speed based on measured balloon speed or something of the like.

 

[AvalonXQ]

According to my experience, all airport moving walkways are behind passport control, so security is likely to jump on anyone with a cart and video camera.

I know at least one set of moving walkways between the parking garage and the terminals in the airport closest to me. I'm sure if you looked carefully you could find plenty of available moving walkways for this experiment.
And, honestly, you wouldn't even have to do it under the table. Talk to the facilities manager at the airport and ask to use a walkway after hours. I think you'd be surprised how reasonable people can be.

 

[ThinAirDesigns]

I know at least one set of moving walkways between the parking garage and the terminals in the airport closest to me.

I've too have had no trouble finding moving walkways at airports that were not behind security. Our local airport has several.

The problem is not finding one, but rather with finding one that moves fast enough. Our best model will run down to something in the range of 4mph and the fastest walkway I have found only moves around 3mph (most are close to 2mph).

I have thought several times about custom building a device that would work down in the 2mph range (it would just take a custom prop) but I likely won't until I'm old and bored. Self-starting on a straight treadmill is really the only thing the device has never demonstrated (though Ynot's turntable is a good as that to all but the hidiots).

JB

 

[windgrins]

A long winded but careful response to Humber

Humber,

I'm impressed. These are nearly coherent replies but still wrong. But I applaud your coherency. I take a great deal of care to try to understand your comments so please do me the same service, as you tend to reply to a perception of the comments that's not close to what is intended. A careful reading and a little bit of thought before replying would be appreciated. I've read a lot of your posts and these same types of error and misconceptions appear throughout so please take a look-carefully. The problem with this style is that it starts to look like a beginners physics text and this is probably not appreciated by the posters as brevity is appreciated.

Incorrect, and the same folksy mistake. At no time is there a relative speed, anything other than 120mph. When at 100m that is true, at 1m and 1mm.

The distance between the vehicles falls as they approach, but the rate of change is constant. That is, measuring that change in distance over time, gives the relative speed and that is constant. There can be no "closing speed" and "relative speeds", that become more closely aligned as the cars approach, And no, there is no "sweet spot, when d =0 .

The misunderstanding here Humber is that I and Spork actually know what relative velocity mean and his statement is correct. The "closing speed" is used in various areas to discuss the idea of how fast it appears that the relative speed between two objects is changing. This is a semantics issue and not an issue in understanding. So even though the cars pass at 120 mph, the instantaneous distance between them is not increasing or decreasing at that point so the "closing speed" is 0mph. On Collision courses the closing speed always is the relative velocity of the two objects. On non-collision courses between two objects, the closing speed is a function of velocity of the objects and the path they are on and are not numerically equal except at infinity so this was why Spork made his comment to highlight this subtlety.

The others concerns balloons that accelerate with zero time, which implies infinite force, and that NB balloons have no gravitational potential energy.

With regards to this issue, I think you missed the point of Spork's video in the car. For a NB balloon, the balloon occupies a space that air would normally have occupied. The air weighs the exact same as the balloon that is occupying the space. This is why it is neutrally buoyant. Once this is understood the issue becomes clear: It's an old Jedi mind trick of one's own brain. The brain perceives the balloon as an object and normally objects have mass and react to f=ma. In this case the overall object is just a separated region of the medium with the overall same mass. Since it was not obvious, I'll point this out. The reason Spork and I were discussing the mass properties is that even though the region (balloon) has the same weight as the air it is displacing, it doesn't have the same mass distribution. This is why the balloon in the car demonstrated the "torque effects" illustrated. So even though the car completely simulated the "gust" effects in the wind via acceleration/deceleration and curves (which is all just acceleration if it is represented as a vector), The overall balloon maintained its relative position in the airmass but generated torques illustrating the existence of the acceleration. So, the point was that the NB Balloon is behaving just like the air that would normally occupy the same volume except for the mass property effects shown. The balloon nor the air in the car reacted instantaneously but it reacted at the same speed as the medium which I believe to be the speed of sound in the media. But since it reacts at the exact same speed of the media in bulk, for the purposes of the discussion, it reacts immediately-same as the surrounding media. This does not imply infinite force contrary to what you stated. This was the subtlety of Spork's video and I'm sure it was hoped by the presenters that "bright" students watching this would observe all these effects and learn something. I can't make the claim to be "bright" but I sure learned something.

With regards to potential energy, the NB balloon has no potential energy with respect to the ground because it occupies a volume containing the equivalent mass of air as the balloon. When someone moves the balloon, a volume of mass of air would move in to take its place. Since the air has the same mass as the balloon, any change in height of the NB balloon is accompanied by an opposite change in height of an airmass of the same size so m(balloon)gh=-m(air equiv)gh. Therefore if one moves the balloon up or down there is no change in PE. This is the subtlety that we didn't want to bludgeon most readers with as it is normally assumed it is understood. When questions are asked, it makes sense to explain-or as the phrase goes "Otherwise we'll be here all day!" Sheesh-we sure exceeded that one so maybe its worthwhile after all being initially concise-lesson learned on my part.

The point about the two cars, it that Spork mistakes differential speed ( two things moving in opposite directions) and two things moving in the same direction, and that mistake appears in the TM.

From an observer in space or even altitude, the surface of the planet and a car may be moving in the same direction, but the vehicle may be moving forwards faster. (That is more than an illusion, but real, and would be necessary say, if the driver wanted to circumnavigate the Equator, in the direction of rotation, but do so in less that 24 hours. )
At no time does the Earth move backwards relative to the car, at any speed including zero. If the car is mentally stopped, or in your parlance, viewed from the frame of the high attitude observer, then the ground still moves forwards but under the car. If the ground is mentally stopped, the car moves forwards.

OK except that Spork really did understand the issue and didn't misunderstand anything. He actually answered wrt to what was stated.

We have just witnessed the Carteers being corrected on their notions of relative KE, but that is only part of the error of thinking that A hitting B, is the same as B hitting A, based solely on the idea that the relative velocities are the same. From that error, comes scoffing about "most people" thinking that a car hitting a truck, is different from the same truck hitting the car at the same speeds.
"From the frame of our stationary truck", the car is inevitably approaching the at speed. Do you think that at the point of impact, and prior to that, the driver may have applied the brakes and for good reason?

"From the frame of the stationary car" with the truck approaching, would you have your foot on the brake or not?

You have not "corrected" the carteers as your understanding of this is wrong.

Here are the basics:

You seem to think that Kinetic Energy is some sort of intrinsic absolute property of a mass. Your error is the following: KE is only defined for a body with respect to an IRF. The same body moving has two different KEs for two translating reference frames. It is not an intrinsic property. You keep marching out the KE=1/2mv*2 argument but this formula intrinsically assumes a reference frame (which the authors generally neglect to mention because they assume it is obvious.) If one views the same object from a different IRF, it has an actual different KE with respect to that frame in that the KE is a computed scalar property from the vector velocities wrt to that frame and this has nothing to do with it being a non-linear or polynomial that can't be added. In general for any field scalar property, one can translate the underlying vector components to the new target reference frame and recompute the scalar property in the reference frame. Some scalar properties are the same in the two reference frames and some are not. For instance, pressure is a scalar that generally is consistent from one frame to the next but KE is not. It all depends upon how the scalar property is defined over the field.

The proper way to compute KE for any IRF is as stated earlier: Determine the vector velocity of the object in the desired frame and then compute KE=1/2mv*2 with the :"v" in this equation being the "v" determined for this reference frame. It's not the same v as any other reference frame so one will get two different numerical values for the two frames for the same object at the same point in time and space. If one moves from one frame to another, one can compute the KE in the target frame by translating the vector velocity from the inital frame to the target frame and then computing the KE as described. Or one can simply measure the vector velocity in the target frame and compute the KE using 1/2mv*2 (measured v from the target frame) and the answer computed will be the same if there are no errors in the calculation. That is the KE in the initial frame is numerically (and actually) different for the object in question in the second IRF. If you will read this carefully, you will potentially clear up an enormous amount of your confusion.

With regards to the question of brakes. If I couldn't shift into reverse due to time constraints, I would put my foot on the brake to mitigate the ensuing motion that the cars steering wheel is about to partake in wrt to the driver. Brakes are better than no brakes here to mitigate the damage to the driver but not to the car. The car will be worse for wear!

This really is your chance to gain some understanding and not just bluster and insult and post back quickly defending your point. It took me quite a time to write this and I don't intend to do this on future posts but please read it carefully. If you want to refute it, please do so with the same care as it was written. As I don't consider it my job to teach you physics, I think you should recognize I'm going to quite a bit of effort here.

Regards

WindGrins

 

[zooterkin]

Funny how I am either seen as a master troll, or an idiot. That makes me wonder about the capacity of those who say that, to discriminate between fact and fiction, and why that may apply to following the cart.

It's a straightforward question, and the fact you choose to dodge it could be read a number of ways. Do you, as it seems you once claimed, actually have alexia?

 

[windgrins]

What Windgrins would do with an approching colliding truck.

And you "understand" relative motion?

If you were in the car heading for the stationary truck, would you apply the brakes? If you were in the stationary car, with the truck heading towards you, would you put your foot on the brake?

I would shift into reverse and try to back up real fast!

Windgrins.

 

[humber]

Gibberish.
No. Read it again.
There's nothing odd about it.
This has nothing to do with relativity. That formula follows from trivial geometry - it took me much longer to type in than it did to figure out (approximately 30 seconds in my head, most of which were devoted to factors of 2 and 1/2 which cancel).

It took me less time (nano-seconds) to know what you were going to do.

You have invented "closing distance", which works for the single value t= 0
There is no such thing as "closing velocity" in this context.

V = dx/dt is a constant.

The distance, x, isn't.

Your equation works when v = 0, and the cars are then stationary, and perhaps parallel parked. Squaring the velocity -v changes the direction, so you can say they are traveling at 60mph in parallel for all t, or some other whatever that falls out of the churn.