Angle between the orientation of a moving object and its velocity

[James R]

Err, no you begged for extra "chancies" , yet you failed to produce an answer.

I didn't beg for anything.

I asked that you consider being reasonable and accept a post that was 12.5 hours over the agreed time limit. When you decided to engage in some of that "arsing" you're so fond of, I closed the thread as per your request.

I provided no analysis of your many errors because you wimped out of the debate early.

Still sour grapes? Grasping at straws?

I accuse you of plagiarising the Sfarti article.

How do you respond?

 

[Trippy]

This can't be since $$\vec{V_t}=\vec{\omega}x\vec{r}$$ in the frame of the axle, so you can't have the same expression in the frame of the ground.

Here's why you're wrong.
First, at T0 we define the ground reference frame:

Then at T1

We measure the point P1 being the location of the axle, and P2 being the point on the cirfumerence of the wheel.

Then at T2

We repeat the same measurements, all using the rods and clocks from the reference frame we defined at T0.

Now, if we want to be extra careful, we can trace the path of the cycloid in our reference frame, because that gives us additional information that might be useful.

Now, without leaving our reference frame, we can measure the change in angle between the blue line and the ground, which by dumb luck in this case appears to be approximately $$2\pi$$.

P1 still has a velocity perpendicular to the blue line in our reference frame, it just exists as a vector component of the V[sub]i[/sub].

Original grid from Source, repoduced and edited without permission, not for personal gains, under fair use blah blah.

 

[Trippy]

Or, to put it another way:

The grid is the ground frame.
The black circle rolls from left to rightm without slipping, along a surface represented by the green line.
In the ground frame the blue line represents the path traced by a point on the surface of the wheel.
In the ground frame the first brown line describes the tangent at that point.
In the ground frame the yellow line represents the displacement vector of the point, as it moves from the first location to the second location.
In the ground frame the velocity calculated is necessarily paralell to that.
In the ground frame The teal line represents the component of that vector that is perpendicular to the tangent plane (which neccessarily implies there is also a component paralell to the tangent plane in the ground frame.

Clearly, in the ground frame, the teal vector is not zero, and so clearly and observer in the ground frame will measure doppler shifting from the wheel.
Source for the cycloid (again, without permission, not for profit, so on and so forth).

 

[Pete]

Sorry, it refutes your post 2.

No, it refutes your mistaken assumption of what it seems you thought this thread was about.

I did, I knew it was not true

Without even reading it properly, you knew it wasn't true.
That's why this is taking so long.

I was under the impression that you were dealing with the microfacets on the rim, as I did in the "Discussion" forum. So, now you want to open a debate thread on the SAME subject? This is a monumental waste of time, I thought that you were planning in earnest to solve the microfacet problem so we could put the zero Doppler effect to bed once and for all.

Yes, this is a step toward solving the microfacet problem.

It is totally useless for the purpose of the studying the Doppler shift.

I think it is both useful in itself, and a useful step toward resolving the doppler problem.

Now I see why we were having the disagreement, you were using a fixed surface while I solved the problem for the moving surfaces (microfacets).

You made an unwarranted assumption and didn't read the opening posts, leading to a big waste of time.
If you would kindly read post 2 without assumptions you'll see that it is all correct.
Then move on to post three, and we can consider how it relates to the wheel facets.

 

[Tach]

Ok,

Time to put this away for good.
1. In the frame co-moving with the axle, the trajectory of a point on the circumference is:

$$x=r cos(\omega t)$$
$$y=r sin(\omega t)$$

where $$\omega$$ is the angular speed of the wheel and r is its radius.

The tangential speed of a particle $$\vec{v_p}$$ is identical to the speed of the tangent plane in any point $$\vec{v_p}=\vec{v_t}=(-r \omega sin(\omega t), r \omega cos (\omega t))$$, a well known fact. So,the angle between $$\vec{v_p}$$ and $$\vec{v_t}$$ is ZERO everywhere in the axle frame.

2. In the ground frame

The axle moves with speed $$\vec{V}=(V,0)$$ so:

$$\vec{v'_p}=\vec{v'_t}=(\frac {V-r \omega sin (\omega t)}{1-Vr \omega sin(\omega t) }, \frac{r \omega/ \gamma(V) cos(\omega t)}{1-Vr \omega sin(\omega t)})$$, c=1.

so, not only that the two speeds are identical (of course they are), their angle is ZERO in the ground frame AS WELL. This is true for both relativistic and "classical" case (pay attention, Trippy). It is also true whether the wheel motion has slippage or not. The "no slip" situation (much touted as being "key" by AN) is just a particular case of the above, $$r \omega=V$$:

$$\vec{v'_p}=\vec{v'_t}=(\frac{V(1- sin (\omega t))}{1-V^2 sin(\omega t) }, \frac{V/\gamma(V) cos(\omega t)}{1-V^2 sin(\omega t) })$$

Of course, the same holds, the zero angle in the axle frame transforms into a zero angle in the ground frame.

The END.

 

[Trippy]

I'm fairly sure that there is an elemnt of circular logic:

Ok,

Time to put this away for good.
1. In the frame co-moving with the axle, the trajectory of a point on the circumference is:

$$x=r cos(\omega t)$$
$$y=r sin(\omega t)$$

where $$\omega$$ is the angular speed of the wheel and r is its radius.

The tangential speed of a particle $$\vec{v_p}$$ is identical to the speed of the tangent plane in any point $$\vec{v_p}=\vec{v_t}=(-r \omega sin(\omega t), r \omega cos (\omega t))$$, a well known fact. So,the angle between $$\vec{v_p}$$ and $$\vec{v_t}$$ is ZERO everywhere in the axle frame.

2. In the ground frame

The axle moves with speed $$\vec{V}=(V,0)$$ so:

$$\vec{v'_p}=\vec{v'_t}=(\frac {V-r \omega sin (\omega t)}{1-Vr \omega sin(\omega t) }, \frac{r \omega/ \gamma(V) cos(\omega t)}{1-Vr \omega sin(\omega t)})$$, c=1.

so, not only that the two speeds are identical (of course they are), their angle is ZERO in the ground frame AS WELL. This is true for both relativistic and "classical" case (pay attention, Trippy). It is also true whether the wheel motion has slippage or not. The "no slip" situation (much touted as being "key" by AN) is just a particular case of the above, $$r \omega=V$$:

$$\vec{v'_p}=\vec{v'_t}=(\frac{V(1- sin (\omega t))}{1-V^2 sin(\omega t) }, \frac{V/\gamma(V) cos(\omega t)}{1-V^2 sin(\omega t) })$$

Of course, the same holds, the zero angle in the axle frame transforms into a zero angle in the ground frame.

The END.

Not that it matters too much, because you're still wrong, as has been demonstrated time and again. There's an element you've been neglecting in your considerations of the axles frame. Do you know what that is? Would you like me to tell you?

The movement of the observer relative to the wheel.

Look, the simple fact is:
$$\vec{V} = \frac{dS}{dt}$$
So as long as $$\vec{S}$$ can be demonstrated to have a component perpendicular to the tangent, then it follows that $$\vec{V}$$ also has one as $$\theta_V$$ is determined by $$\theta_S$$ which is true for all observers of all microfacets with the exception of those observers observing microfacets that lie on the line connecting their location to the location of the axle.

 

[Tach]

I'm fairly sure that there is an elemnt of circular logic:


Not that it matters too much, because you're still wrong, as has been demonstrated time and again. There's an element you've been neglecting in your considerations of the axles frame. Do you know what that is? Would you like me to tell you?

The movement of the observer relative to the wheel.

Look, the simple fact is:
$$\vec{V} = \frac{dS}{dt}$$
So as long as $$\vec{S}$$ can be demonstrated to have a component perpendicular to the tangent, then it follows that $$\vec{V}$$ also has one as $$\theta_V$$ is determined by $$\theta_S$$ which is true for all observers of all microfacets with the exception of those observers observing microfacets that lie on the line connecting their location to the location of the axle.

No, Trippy. It is really The End.

 

[Trippy]

No, Trippy. It is really The End.

No, it isn't, not even close.

Repeating the assertion does not counter the argument presented.

This is now about your inability to understand a straightforward vector addition.

 

[OnlyMe]

No, it isn't, not even close.

Repeating the assertion does not counter the argument presented.

Arguing this issue with Tach is a kin to debating evolution with a creationist. Neither, seem to be interested in anything but thier own narrow argument.

 

[Trippy]

Arguing this issue with Tach is a kin to debating evolution with a creationist. Neither, seem to be interested in anything but thier own narrow argument.

Yeah, I mean... The real world disagrees with him. In the real world, we can measure the specular reflection of GPS satelite signals off a lake surface, and measure a range of doppler shifts, when compared to the direct signal recieved by the elevated antenna. Which agrees with my interpretation, rather than Tachs.

 

[RJBeery]

...but...but Pauli said it was true!

 

[Trippy]

...but...but Pauli said it was true!

And on the same page Pauli also said that if the mirror has any motion perpendicular to the plane of the mirror, doppler shifting would occur. Which in itself implies that if the observer has any motion perpendicular to the mirror in the mirrors co-moving frame, then doppler shifting will also occur.

Tachs fundamental mistake, well, one of them anyway, is he proves that the angle between the instanaeous velocity of a particle on the rim of the wheel, and the tangential velocity is zero in the co-moving frame of the axle of the wheel (Duh! I could have told him that!) and then argues because it is zero there, it must be zero in all frames. But what he has failed to account for is that if the wheel is moving in an observers rest frame, then the observer must be moving in mirrors co-moving refrence frame. And so if the motion of the mirror is the salient consideration in the observers rest frame, then the motion of the observer is what we must consider in the mirrors co-moving frame, not the motion of the mirror.

But it's worse than that, and even more fundamental than that, because for the reasons I demonstrated, but Tach dismissed out of hand, even a co-moving observer would measure doppler shift that is dependent on on the angle from their line of sight, because on one side of the line of sight, if the observer was to measure the distance between them and a point on the surface of the wheel, they would find it move towards them, and on the other side of their line of sight, it would be moving away from them.

 

[RJBeery]

And on the same page Pauli also said that if the mirror has any motion perpendicular to the plane of the mirror, doppler shifting would occur. Which in itself implies that if the observer has any motion perpendicular to the mirror in the mirrors co-moving frame, then doppler shifting will also occur.

Tachs fundamental mistake, well, one of them anyway, is he proves that the angle between the instanaeous velocity of a particle on the rim of the wheel, and the tangential velocity is zero in the co-moving frame of the axle of the wheel (Duh! I could have told him that!) and then argues because it is zero there, it must be zero in all frames. But what he has failed to account for is that if the wheel is moving in an observers rest frame, then the observer must be moving in mirrors co-moving refrence frame. And so if the motion of the mirror is the salient consideration in the observers rest frame, then the motion of the observer is what we must consider in the mirrors co-moving frame, not the motion of the mirror.

But it's worse than that, and even more fundamental than that, because for the reasons I demonstrated, but Tach dismissed out of hand, even a co-moving observer would measure doppler shift that is dependent on on the angle from their line of sight, because on one side of the line of sight, if the observer was to measure the distance between them and a point on the surface of the wheel, they would find it move towards them, and on the other side of their line of sight, it would be moving away from them.

This is all very well put. Tach's problem has more to do with a character flaw (immaturity/lack of integrity) rather than simply being wrong in this instance. I mean, we're all wrong at some point...

 

[RJBeery]

...not that I would know!

 

[Trippy]

This is all very well put. Tach's problem has more to do with a character flaw (immaturity/lack of integrity) rather than simply being wrong in this instance. I mean, we're all wrong at some point...

Thanks, I take pride in being able to express the same concepts equally well using diagrams, prose, and maths. It's one of the things that I agree with Fraggle Rocker on, a good scientist should be able to communicate their point clearly, which requires the use of all three tools.

I'm not going to call Tach immature, nor am I going to suggest that he lacks integrity. In that regard, I'm going to simply restate what I have already said a couple of times, the problem in that rergard is that Tach has too much emotional currency invested in being right - he's been too rude to too many people about their skills, and his backed himself into the position where admitting error looks increasingly untenable.

 

[RJBeery]

I'm not going to call Tach immature, nor am I going to suggest that he lacks integrity. In that regard, I'm going to simply restate what I have already said a couple of times, the problem in that rergard is that Tach has too much emotional currency invested in being right - he's been too rude to too many people about their skills, and his backed himself into the position where admitting error looks increasingly untenable.

Again, this is a mature analysis of the situation. OTOH, I don't claim to be mature, but I do question the integrity of someone that changes their posts after those posts have been commented upon, or moves the goalposts of a discussion immediately after they realize that they're about to lose a debate on some point.

 

[Trippy]

Again, this is a mature analysis of the situation. OTOH, I don't claim to be mature, but I do question the integrity of someone that changes their posts after those posts have been commented upon, or moves the goalposts of a discussion immediately after they realize that they're about to lose a debate on some point.

I can understand that.
Personally, if I have something to add, or correct, I've gotten into the habit of checking the thread first. If there have been replies since I posted, I will post it as an addendum, quoting my original post, otherwise I will simply make the changes. Sometimes I even remember to check the thread before I save the changes.

There's a couple of reasons I do this. First, it irritates me when people do it to me. Second, I've noticed that changes under these circumstances tend not to get noticed.

Of course, all of this has the consequence that I tend to ignore changes made to a post after I have replied to it. Unless there is something that obviously stands out as new, at which point I will reply to that seperately.

 

[Farsight]

Thanks, I take pride in being able to express the same concepts equally well using diagrams, prose, and maths. It's one of the things that I agree with Fraggle Rocker on, a good scientist should be able to communicate their point clearly, which requires the use of all three tools.

Well said Trippy.

 

[Trippy]

Well said Trippy.

You need all three, using only one or two is handicapping yourself, and making yourself prone to the kind of oversights we see in this thread.

 

[Tach]

No, it isn't, not even close.

Repeating the assertion does not counter the argument presented.

No need to continue to dig yourself deeper. Notice that the cheerleader squad has become quiet, you are left in the company of the likes of :Farsight, RJBeery and OnlyMe. This should tell you something.

This is now about your inability to understand a straightforward vector addition.

But this is exactly what the second set of formulas is, if you looked closely you would have recognized the Lorentz transforms for velocity. In the past you have accused me of needing remedial "calcalus", now you claim that I don't know vector addition, this is getting embarrassing (and not for me). If you do not understand the explanation, why don't you ask?