# Angle between the orientation of a moving object and its velocity

Discussion in 'Physics & Math' started by Pete, Nov 23, 2011.

1. ### TrippyALEA IACTA ESTStaff Member

Messages:
10,890
To illustrate this:
Consider this diagram:

Here we have three paths connecting P to P'. The black path, the green path, and the blue path all have the same displacement even though the cover different distances.

3. ### TrippyALEA IACTA ESTStaff Member

Messages:
10,890
All the equations in the world are useless if you do not understand the fundamentals like the difference between distance and displacement, and how it's relevant in this scenario.

5. ### TrippyALEA IACTA ESTStaff Member

Messages:
10,890
It's fine, we get it, straight forward vector addition is to complicated for you to follow (that is, after all, what the diagram illustrates).

Correct.
V[sub]t[/sub] is the tangential velocity.
V[sub]i[/sub] is the instantaneous velocity.

Haven't you learned anything about me yet?

Of course I can. Whether or not I will is the question. Well, that and whether or not you will understand it.

Taken in the classical limit.
V is the translational velocity of the axel as the wheel rolls along the ground.
V[sub]x[/sub] is the component of the instaneous velocity paralell to the ground.
V[sub]y[/sub] is the component of the instantaneous velocity perpindicular to the ground.
$\omega$ is the angular velocity of a point on the surface of the wheel as it rolls (as viewed by a 'stationary' observer - eg a camera or a light source).
$\alpha$ is the angle of rotation as measured from the vertical.
And $\psi$ is the angle between the instaneous velocity and the ground.
Rotational kinematics gives us V[sub]t[/sub] as:
$V_t = \omega r$
In the classical limit, V[sub]x[/sub] is the horizontal component of V[sub]t[/sub] combined with the horizontal motion of the wheel.
$V_x = V+\omega r cos(\alpha)$
And V[sub]y[/sub] is simply the vertical component of V[sub]t[/sub]
$V_y=\omega r sin(\alpha)$
Pythagoras tells us that:
$V_i=\sqrt{V_x^2 + V_y^2}$
And trig tells us that:
$\psi=tan^{-1}(\frac{V_y}{V_x})$
I generally dislike using the tan function, however, it illustrates the point most effectively.
Which is to say that:
$\psi=tan^{-1}(\frac{\omega r sin(\alpha)}{V+\omega r cos(\alpha)})$
Now, what does this tell us?
$\psi$ is defined by our chosen frame, which in this case happens to be the rest frame of the ground, but would also be the rest frame of the camera (or the light), as is $\alpha$.
Clearly then, $\psi$ can only be zero when $sin(\alpha)$ is zero, which only happens when $\alpha=0$ or $\alpha=\pi$ radians.

I have an inkling there was soemthing else I was going to go into, however, that's all I have time for at the moment.

7. ### TachBannedBanned

Messages:
5,265
You don't use $V_i$ anywhere

...that you were asked to calculate the angle between $V_t$ and the instantaneous tangent to the circle. In the axle frame, that angle is clearly zero. What did you do? Certainly not what you were asked to do. Do you want to continue with this charade? Can you, at least , pause for a few minutes and look at what pete and I are doing to get an idea about what you actually need to calculate?

Last edited: Nov 24, 2011
8. ### TrippyALEA IACTA ESTStaff Member

Messages:
10,890
Just incidentally:

Not the claim I made, in fact, counter to the claim I made:

Once again...

9. ### RJBeeryNatural PhilosopherValued Senior Member

Messages:
4,222
This discussion of microfacets on the cylinder is exactly why a matte surface (be it flat, round etc) cannot hide Doppler effects as it moves between a primary light source and a camera. I hear, very generally, the same thing coming from AN, Pete and Trippy, so I'm attempting to give my perspective with both visuals and math. (For Tach, this also addresses the "moved goalposts" of our debate, which apparently sit on casters

)

For a mirror whose orientation is parallel to its velocity moving between a single light source and a camera, the camera will ALWAYS detect the frequency emitted from the light source. This has been beat to death.

However, for a matte surface, the angle of incidence to the plane of the surface does not necessarily equal the angle of reflection away from it. This is due to the rough surface inherent in a matte material. The angle of incidence is equal to the angle of reflection against whatever the surface orientation happens to be at that point.

If we have $\phi_1$ = $\frac{\pi}{4}$ and $f_0$=550 THz, the plane of the matte object moving at $\frac{V}{c}$=.5 will "see" the light source as being roughly 305 THz.

In the case above, however, $\phi_2$ = $\frac{\pi}{8}$. This is trivially evidenced by the fact that a matte white wall is visible from all angles, regardless of the source of light. Therefore, using the formulas laid out here, $f_{s'}$ = 595 THz.

Summary:
$f_0$ = 550 THz
$f_{matte}$ = 305 THz
$f_{s'}$ = 595 THz

Conclusion:
Any surface with microfacets whose orientation is not tangential to its velocity from the frame of the camera will induce Doppler effects upon the light reflected to that camera. This includes mirrored washers, mirrored cylinders, and matte objects of all types.

Messages:
5,265

11. ### Neddy BateValued Senior Member

Messages:
2,514

Let's imagine that Tach is standing in front of an oncoming steamroller. In classical physics, the cross-section of the roller is a circle, the steamroller has a forward velocity component, and Tach get's plowed over. However, in relativistic physics, the cross-section of the roller is not a circle. Applying Tach's ridiculous logic (quoted above), the forward velocity component of the steamroller would be zero, and the roller would just spin in place. So Tach is claiming that he would be safe standing in front of an oncoming steamroller!!!!

12. ### TachBannedBanned

Messages:
5,265
Nope, this is not what I am saying, you don't even begin to understand what I am saying.

You whiffed again. You can stop embarrassing yourself now.

You also had your shot at the problem and you self-destructed just the same. Instead of wasting space and bandwidth with your trolling , why don't you let me and pete sort things out?

13. ### RJBeeryNatural PhilosopherValued Senior Member

Messages:
4,222
There's no statute of limitations on truth. If you disagree with the math please be explicit about the nature of your contention.

Messages:
5,265

15. ### RJBeeryNatural PhilosopherValued Senior Member

Messages:
4,222
Done, see you there, but please be mindful to be explicit about any issues you see with my analysis.

16. ### TrippyALEA IACTA ESTStaff Member

Messages:
10,890
And?
I do, however, prove that V[sub]i[/sub]≠V[sub]t[/sub], as well as proving that $\psi$≠0 (except under 'special conditions).

No. Here is what you said:
You explicitly asked me to demonstrate that V[sub]i[/sub] and V[sub]t[/sub] were not co-linear, and that is precisely what I have done.

The charade is yours, and yours alone.

Let's recap shall we?

This is precisely what I did, except I took the classical limit, rather than the relativistic approach.

This is precisely what I did. The vector described by $\psi$ and V[sub]i[/sub] precisely describes the instantaneous motion of one of your microfacets relative to the ground (in the ground frame) when one considers a wheel rolling along the ground.

This is precisely what I have done by demonstrating that V[sub]i[/sub] the instaneous velocity of the microfacet, and V[sub]t[/sub] the tangential velocity of the microfacet are not co-linear in the ground frame.

So I have done everything that you have agreed I need to do, and therefore proven you wrong. All that's left is your trying to worm your way out of it because of the amount of emotional currency you've invested in being right.

Incidentally, surely even you can see that this diagram:
Is just a re-treatment of this diagram:
The differences being that I have included the motion of the axel relative to the ground, as well as including the various component vectors.

And that this:
Which is just another way of saying this:
$tan(\psi)=\frac{\omega r sin(\alpha)}{V+\omega r cos(\alpha)}$
Is just the classical limit of this:
Because $|\vec{V_p}|=\omega r$
And $\psi = \theta'_A$
And $\alpha = \theta_A$

Or perhaps you should get Pete to explain it to you.

And now I have Pink Floyd: Pigs (Three different kinds) running through my head.

Last edited: Nov 24, 2011
17. ### Neddy BateValued Senior Member

Messages:
2,514

I know what you've been saying. You've been saying that the instantaneous velocity vectors of all points on the rim of a rolling wheel are tangent to the rim. That is quite wrong, but it is what you've been saying.

18. ### RJBeeryNatural PhilosopherValued Senior Member

Messages:
4,222
Hey bro we can squabble all day long about our thought experiments but nothing ends the ambiguity like real-world results. :xctd:

19. ### TachBannedBanned

Messages:
5,265
$V_i$ is the tangent to the cycloid, the trajectory of the point on the circumference calculated in the ground frame. What you have been asked to do is to calculate the angle between the velocity $V_t$ and the tangent to the surface , not between $V_t$ and $V_i$. If you read my exchange with pete as to what is being calculated you would not have wasted your time. Besides, you would have learned the correct derivation, including the relativistic case. As such, you wasted your time but thanks for playing. Would you now let me and pete finish our discussion? We are down to disagreeing on one formula.

20. ### PeteIt's not rocket surgeryRegistered Senior Member

Messages:
10,167
Don't confuse the discussion, Tach. The scenario described explicitly describes a flat surface, not any microfacets on a curved surface.
Please don't hijack this thread (too late) by trying to discuss a different problem.
No.
That is not the situation described in the opening posts.
The surface A is at rest in S, with angle $\theta_A$ between the surface and the x-axis.

Please try to focus on the problem at hand, don't be confused by other discussions.

Changing the scenario and changing definitions doesn't qualify as "fixing a modelling error."

Good idea.

21. ### AlphaNumericFully ionizedRegistered Senior Member

Messages:
6,702
Your reply to me which Neddy quotes shows you don't understand what I'm saying. My example didn't even need to consider the wheel, the plane mirror is enough.

A mirror extended in the x direction and moving in the x direction in frame S will have non-zero y' component if you boost to S' via a LT in a direction with non-zero y component. The mirror is then extended in the x' direction but can have velocity in both the x' and y' direction.

Your claim is false. How many ways do you need to be told by how many people with how much mathematics?

22. ### TachBannedBanned

Messages:
5,265

Then you did not address the issue at all. What you have is the angle between a stationary flat surface and the velocity of a point on the rim. Therefore your derivation does not address the problem at all.

I think that you are backpeddalling now that your solution has been shown to be wrong. We have been talking about the microfacets making up the rim for quite awhile, now you are changing your story?

...which makes your solution quite useless, as pointed out.

I am , this is why I gave you the correct solution, a few days ago and this is why I point out the error in your solution. If you don't want to solve the problem for the circular rim, just say so and we are done.

For me, the scenario was always the same, find out how the angle between tangent to moving microfacet and point velocity transforms from the axle frame in the ground frame. This is the solution I provided you with.

Ok, open a debate between the two of us, it is long overdue.

Last edited: Nov 24, 2011
23. ### TachBannedBanned

Messages:
5,265
True, so what? I posted the exact formulas eons ago. The point the contention is whether the velocity and the tangent to the rim remain colinear when boosted from the axle frame to the ground frame.

Based on what you post, you clearly do not even understand my claim despite explaining it to you (now) for the third time.