Colliding Ovals: Exploring the Identical Outcomes of 4 Collisions

  • Thread starter Michel_vdg
  • Start date
In summary: I'm assuming all the collisions conserve energy. If that's not the case than the...different center-of mass-impact-speed might have an effect.
  • #1
Michel_vdg
107
1
Hi,

I Have set up 4 collisions between two Ovals (AA,BB,CC,DD) and I would like to know if the outcome will be identical for all 4 of them.

First of, when looking at this orange collision that is going to take place, one can draw straight vertical lines between the two centers ...

attachment.php?attachmentid=67307&stc=1&d=1394114608.png


… and if you fly along with the lower oval, and draw it out; than it looks as if they are approaching each other in a straight line:

attachment.php?attachmentid=67308&stc=1&d=1394114608.png


Next, look at the 4 collisions (AA,BB,CC,DD) in the image below, than you can draw 4 times the same visual thing as the image above, when flying each time along with the lower Oval ABDC, be it for collision AA, BB, CC or DD.

The right frame of reference to calculate collisions in is center of mass frame ... and when looking at this from a 'set' view from the top, than the outcome would be 4 x times different, but is it also 4 times different? ... cause they are all coming from different directions, and are differently orientated before they collide, and inertia-wise ... so my guess would be that the out comes are in fact also really different each time, although they approach each other visually in the same way.

attachment.php?attachmentid=67310&stc=1&d=1394115793.png


A friend gave me this explanation (see quote below), but it doesn't seem to be right ... so that's why I'm posting it here, just to be sure.

Looking forward to hear your opinion(s), thanks!

m.

ABCD are all physically the same collision. In each case, you can transform to a reference frame in which they approach not along the black arrows, but along the orange ... yellow arrows. This transformation is precisely just subtracting out the average velocity.

The outcomes must be the same physically, because the situation is the same physically. The way the particles behave cannot depend on the camera you choose to observe them with.

I should add, the thing that makes ABCD _appear_ to be different collisions is that you not only subtract their average velocity before calculating the collision, but you add it back on afterwards. So A appears to bounce in a primarily right-wards direction, because the average velocity in the case of A is primarily pointing rightwards, and so on.
 

Attachments

  • StraightLine.png
    StraightLine.png
    9.4 KB · Views: 447
  • Staight_Contact_vectors.png
    Staight_Contact_vectors.png
    3.3 KB · Views: 431
  • Collision_ABCD.png
    Collision_ABCD.png
    26 KB · Views: 464
Physics news on Phys.org
  • #2
Your friend is right. The quote is correct
 
  • #3
dauto said:
Your friend is right. The quote is correct

But if you replace the 'mirrored' lower half with a flat plane than 'C' would be smacked onto that plane; while 'A' would 'shoot' straight through. So what's the difference with that?
 
  • #4
No, A would also reflect off the plane
 
  • #5
The one factor that might be different between the collisions is how fast the collision happens. But you did not mention that at all so I assume it is not important for that particular question.
 
  • #6
dauto said:
The one factor that might be different between the collisions is how fast the collision happens. But you did not mention that at all so I assume it is not important for that particular question.

Mh, yes, the feeling I have is that 'C' is moving more powerful (faster?) in the Vertical direction vs. the Horizontal direction, and visa versa for 'A'. Let's say that they are all moving at the same speed towards the 'rendezvous point ' (black arrow).

I'm adding an image to show what I meant with 'smack' vs. 'shoot through':

attachment.php?attachmentid=67328&stc=1&d=1394127081.png
 

Attachments

  • Flat_Shoot.png
    Flat_Shoot.png
    6.5 KB · Views: 400
  • #7
Yes, if the magnitude of their speeds are all the same than the component of the speeds along the vertical will be different. That doesn't change the fact that it is best to model all these collisions in the center of mass frame. The collisions will all look the same except that some will happen faster than others. I'm assuming all the collisions conserve energy. If that's not the case than the different center-of mass-impact-speed might have an effect.
 
  • #8
dauto said:
Yes, if the magnitude of their speeds are all the same than the component of the speeds along the vertical will be different.
So all 4 collisions (AA, BB, CC, DD) are different if V(black arrow)=1

btw, they are all placed in a circle, so that should have indicated that they're all moving with V(black arrow)=1 towards the center of the collision. (It wasn't exactly so in the first image, so I've made a newer version where this is the case.)

btw if (x,y) -> V(black arrow) ≠ 1 than I couldn't have made the illustration this way, and the distribution should be looking different, no?

attachment.php?attachmentid=67332&stc=1&d=1394129813.png
dauto said:
That doesn't change the fact that it is best to model all these collisions in the center of mass frame.
ok.

dauto said:
The collisions will all look the same except that some will happen faster than others.
Faster or slower ... now they all 'look the same', but what if V(black arrow) = 1

Shouldn't they than not look the same?
dauto said:
I'm assuming all the collisions conserve energy. If that's not the case than the different center-of mass-impact-speed might have an effect.

'might' that's a vague answer, sorry for being so picky :smile: but I've added a new 'mirrored' image where the outcomes are totally different, the orange pair is moving further horizontally ahead while the green ones are almost stationary in the same spot, that IS a big difference, so all in all the outcomes aren't the same at all, is this correct?

attachment.php?attachmentid=67333&stc=1&d=1394129813.png
 

Attachments

  • Centered_Velocity.png
    Centered_Velocity.png
    32.1 KB · Views: 474
  • mirrored.png
    mirrored.png
    19.7 KB · Views: 437
Last edited:
  • #9
Michel_vdg said:
I've added a new 'mirrored' image where the outcomes are totally different, the orange pair is moving further horizontally ahead while the green ones are almost stationary in the same spot, that IS a big difference, so all in all the outcomes aren't the same at all, is this correct?
What physics did you use to determine that there is a big difference? What equation did you solve to determine the different outcomes?

Sure you can draw a picture, but that shows your artistic ability, not the way nature works.

Is the collision elastic (they bounce off each other undamaged) or plastic (they stick together)?
 
  • #10
DaleSpam said:
What physics did you use to determine that there is a big difference? What equation did you solve to determine the different outcomes?

Sure you can draw a picture, but that shows your artistic ability, not the way nature works.

Is the collision elastic (they bounce off each other undamaged) or plastic (they stick together)?

Doesn't really matter; if you set two cases to the most extreme angles 180° (AA) with velocities (-1,0) and (-1,0); and 90° (BB) with velocities (0,-1) and (0,1); than you would have particles that either never *collide* and fly next to each other horizontally (AA), versus those that collide 100% head on (BB).
 
Last edited:
  • #11
DaleSpam said:
What physics did you use to determine that there is a big difference? What equation did you solve to determine the different outcomes?

Ok, Ignore my previous comment, my friend has given his equations and method that I'll quote below, but first I'll explain the whole concept and why this question in the first place.

--

The goal is to simulate a gas made up of slippery frictionless ellipsoids that can slide over each other during interaction. Sort of like these billiard ball-like soap bubbles:

http://www.800million.org/collisions/billiard_1.jpg [Broken]

Because this kind of fictitious particle, with the sliding interaction, is nearly impossible to figure out with actual physics equations, the idea was to use 2 programs:
  1. Collision editor: where it's possible to preset the outcome of different kinds of collisions intuitively by hand.
  2. Simulator: that uses the presets from the Collision editor as out come for each collision, to generate this gas.

So when 2 particles collide in the Simulator the outcome is what is defined in the 'Collision editor'.

I'm adding a couple of images of:

1. Collision editor:

http://www.800million.org/collisions/collision_editor.png [Broken]

--

2. Simulator:

http://www.800million.org/collisions/simulator.JPG [Broken]--

And here is a scheme of what kinds of presets there should be, for a set particle 'A' and an in-flying particle 'B' that changes position relative to 'A', and where things could be mirrored. The 'in between' results should be calculated by interpolation.

http://www.800million.org/collisions/CE_1.png [Broken]

-------------Now here's the suggested method to solve this:

Notation for vectors: v.x is the x component of the vector v

So, let's say we have two particles with velocities (vectors) v1 and v2 at positions (vectors) r1 and r2:

dR = r1-r2
dV = v1-v2

In the center-of-mass frame, we want to calculate what 'theta_in' was. To do this, we calculate the angle associated with dV:

theta_0 = angle(dV) (angle is atan2(dV.y,dV.x) in C, or atan(dV.y/dVx) mathematically)

This angle is subtracted from all other angles in order to compute in the center-of-mass frame.

The next step is to figure out the angle of impact. This is the angle between dR and dV. This angle is theta_in:

theta_in = angle(dR) - angle(dV)

Then we calculate the two 'phi' angles - the tilts. These are defined (in the CoM 'Collision editor' at least) as an angle between the tilt angle and the angle of dR.

So:

phi1 = angle(particle1) - angle(dR)
phi2 = -(angle(particle2) - angle(dR))

(I think this is right, but I haven't had a chance to test this in detail)

So now we have theta_in, phi1, phi2. We use the interpolator formula to find theta_out, alpha, and beta.

The interpolator works as follows - for each thing we want to estimate, we sum over all presets and weight the presets according to a weight function:

d = |theta_in - theta_in_preset| + |phi1 - phi_1_preset| + |phi2 - phi_2_preset|

weight = 1/(1+(d/d0)^2)

Here 'd0' defines the sharpness of the interpolation.

What I mean by a weighted sum is that we compute total_weight = sum_i weight_i over each preset, and weighted_variable = sum_i weight_i * variable_i over each preset. The weighted sum is then: variable = weighted_variable / total_weight.

One note on adding and subtracting angles. Since angles 'wrap around', one must do the following to do z=x-y where z,x,y are angles.

z0 = x-y

if (z0>180 degrees) z=z0-360 degrees
else if (z0<-180 degrees) z=z0+360 degrees
else z = z0

The reason is that we want to find the smaller of the two delta-angles between the two angles we're subtracting. Imagine if we're finding the distance between 5 degrees and 10 degrees. One arc goes straight from 5 degrees to 10 degrees without passing 0; the other arc goes all the way around on the negative side, from 5 through 0/360 and all the way around to 10. We want to always use the smaller of those two angles.

Anyhow, now that we have the theta_out, alpha, beta for the collision we can figure out the leaving angles in the CoM frame and use that to find the leaving velocities v1' and v2'.

v1' = (v1+v2)/2 + alpha * |dV| * [cos(theta_0 + theta_out), sin(theta_0 + theta_out)]
v2' = (v1+v2)/2 - alpha * |dV| * [cos(theta_0 + theta_out), sin(theta_0 + theta_out)]

That concludes the collision. For added accuracy, we allow a collision to endure for an amount of time proportional to dR dot dV/|dV|^2, where 'dot' is the dot product and the denominator is dV^2 because it should be 'distance/velocity' and we have an extra factor of velocity in the numerator.

--

That's it. Looking forward hear your opinion. btw any suggestions to tackle this problem are welcome.
 
Last edited by a moderator:
  • #12
Sim in 2D, right?
The whole reason using ovals would seem to be to look at the spins and effect of moments because of the different ways contact between ovals can impact each other at different points around the circumferences, hence causing different moments and spins. But I'm not seeing that your ovals are spinning? ...they will be after collision, so your phi angle tilts are going to be changing..
Once they are spinning you'll need to know where on the ovals collision contacts occur so that impact calculations can include a component of their moments of inertia as well as their approach speeds of centers of mass.

I think this means that as the centers of mass of individual ovals approach each other you'll need to know their phi angles at the instant of contact, which sounds like it means each oval is going to have an oscillating boundary with respect to its direction of travel, and when two ovals' centers of mass approach close to the length of the long axis of the ovals (half of each of the two) where contact may be immanent, the phi angles would need to be "projected forward" to both predict a contact time (step) and contact orientation of the ovals, and specific location of contact on each oval circumference to compute relative moments, etc... sounds pretty complicated, but think you need this to assign the impact of collision, its distance and direction from the centers of mass and the resulting effects on the ovals' spins, as well as the directions and speeds of their centers of mass.

This might be a routine that detects an immanent collision (or keeps track of all the ovals' rotation conditions for each step) and uses a sub to step "ahead" in time to determine what the relative orientations and contact points will be, then sending those values to the main engine for the contact step to calculate moments at contact, resulting phi angles and angular rotation speed after contact, and resulting directions and speeds of the centers of mass after contact.
 
  • #13
bahamagreen said:
Sim in 2D, right?

Yes.

3D perhaps one day, first 2D ...
bahamagreen said:
The whole reason using ovals would seem to be to look at the spins and effect of moments because of the different ways contact between ovals can impact each other at different points around the circumferences, hence causing different moments and spins.

No, 'spin' shouldn't be included. The reason for this, is when you add 'spin' to the outcome of the collision, than the orientation of the ovals could be anything based on their outgoing velocity, and the time of when their next collision takes place ... also more slippery is more spin(?), or less spin(?) ... so it would probably be fairly meaningless to start presetting the outcome of collisions with 'spin', cause the gas would be chaotic (no control). Thus the orientations should be locked in set directions when the particles leave the collision, based on how they collide. One of the main points also is to get a grip on the number of unique ingoing (preset) collisions to get a fair level of smoothness.

The actual reason of this concept is to look for possible packing formations sort of like what you can see in the images below, but they are generated thanks to the bulk-volume-interaction of rods (oval particles). So by using a more exotic type of particle/interaction some results might be created with far less particles.

http://www.800million.org/collisions/GRNLR_1.jpg [Broken]... I guess this first response, exclusion of spin, effects the rest of your comment here:
bahamagreen said:
... But I'm not seeing that your ovals are spinning? ...they will be after collision, so your phi angle tilts are going to be changing..
Once they are spinning you'll need to know where on the ovals collision contacts occur so that impact calculations can include a component of their moments of inertia as well as their approach speeds of centers of mass.

I think this means that as the centers of mass of individual ovals approach each other you'll need to know their phi angles at the instant of contact, which sounds like it means each oval is going to have an oscillating boundary with respect to its direction of travel, and when two ovals' centers of mass approach close to the length of the long axis of the ovals (half of each of the two) where contact may be immanent, the phi angles would need to be "projected forward" to both predict a contact time (step) and contact orientation of the ovals, and specific location of contact on each oval circumference to compute relative moments, etc... sounds pretty complicated, but think you need this to assign the impact of collision, its distance and direction from the centers of mass and the resulting effects on the ovals' spins, as well as the directions and speeds of their centers of mass.

This might be a routine that detects an immanent collision (or keeps track of all the ovals' rotation conditions for each step) and uses a sub to step "ahead" in time to determine what the relative orientations and contact points will be, then sending those values to the main engine for the contact step to calculate moments at contact, resulting phi angles and angular rotation speed after contact, and resulting directions and speeds of the centers of mass after contact.
 
Last edited by a moderator:
  • #14
Michel_vdg said:
'might' that's a vague answer, sorry for being so picky

That's because the properties are vague. If your ovals won't spin and energy is conserved than their shape doesn't really matter at all (as far as the outcome of the collision) and all the collisions in the first post will indeed produce identical outcomes. Either way, I strongly recommend that you program you collision generator in the center of mass reference frame.
 
  • #15
dauto said:
That's because the properties are vague.

What is vague about the properties? It are particles with spin=0 that collide, coming from different directions. I agree that the outcome will be unnatural, because it will be manipulated by the program, but that's irrelevant regarding my question.

What my question is about is if in a natural setting the outcome would be different for AA, BB, CC and DD? ... and yes spin, change of momentum and energy loss may all be considered during the collision. This is important to know because if the outcome would be for each case the same, than only one 'preset' can be used for a lot of collisions; otherwise there need to more presets for each of those unique cases.

dauto said:
If your ovals won't spin and energy is conserved than their shape doesn't really matter at all (as far as the outcome of the collision) and all the collisions in the first post will indeed produce identical outcomes.

The shape is important, because it will be the heart of each collision, and serving like an oval roundabout redirecting particles into a specific direction, depending on how they are orientated towards each other and how they make contact. Therefor it should be a perfectly normal and natural until the point that they start exiting the collision which is in fact something else.
 
  • #16
Michel_vdg said:
[*]Collision editor: where it's possible to preset the outcome of different kinds of collisions intuitively by hand.
Intuitively is fine, but then you are just doing some computer programming or a video game. You are not doing physics. If you want to do physics then you need to actually solve for this physically, not intuitively. If you just want to do a video game then there is nothing to discuss here in Classical Physics.

Your idea for presetting outcomes is a good idea. That way you can solve the equations in advance and plug in the results as needed. It trades memory for computation speed. As has been suggested earlier, your outcomes table should be written in the center of momentum frame. That will reduce the number of entries by a lot. The process would be:

Transform to center of momentum frame
Lookup collision results
Update velocities
Transform back to original frame
 
  • #17
Michel_vdg said:
No, 'spin' shouldn't be included.
If you do not include spin then either
a) The collisions must be the same as a collision of a spherical object (i.e. the forces must be exerted along the line joining the centers of mass, regardless of the point of contact)
b) You will violate the conservation of angular momentum

You cannot both conserve angular momentum and have forces which do not pass through the center of mass.
 
  • #18
DaleSpam said:
Intuitively is fine, but then you are just doing some computer programming or a video game. You are not doing physics. If you want to do physics then you need to actually solve for this physically, not intuitively. If you just want to do a video game then there is nothing to discuss here in Classical Physics.
Until a certain point it is Classical Physics and that's the part I'm questioning here.

So what I now have found out regarding that is:

The relative velocities that they are moving towards each other is different for each case (colored vertical arrows), although the velocity and time that they move towards the rendezvous point is for each case the same (black arrows); and because of this difference in velocity the normal reaction force is different so the torque is different for each case.

A second point was that if you would look only at these relative velocities, one wouldn't know that there's a possible difference in angle of approach, and for this someone told me:

To calculate the collisional impulse all you need is
* the orientation of the objects relative to the tangent line
* the relative velocities of the points on the body that are going to make contact (it doesn't matter what combination of linear and angular velocity causes this).
That is, the key variable is the velocities of the contact points, not of the centers of mass.
 
  • #19
Michel_vdg said:
That is, the key variable is the velocities of the contact points, not of the centers of mass.
Not if the spin is to remain 0. You cannot have it both ways.
 
  • #20
DaleSpam said:
Not if the spin is to remain 0. You cannot have it both ways.

Why not? I mentioned ...

"Until a certain point it is Classical Physics and that's the part I'm questioning here."

So I guessed one can figure out what different types of collisions there are in Classical Physics by using the 'not centers of mass method', and than manipulate those results and create presets, this part has of course no longer anything to do with Classical Physics. Why would it be impossible to make the jump between the two in a Collision_editing program?
 
  • #21
It certainly isn't impossible to program. All sorts of video games have non-realistic physics engines. There is nothing keeping you from writing a program which violates the conservation of angular momentum, as you have described and as you seem determined to do.

It simply isn't material for this forum.
 

1. What is the purpose of studying colliding ovals?

The purpose of studying colliding ovals is to understand the behavior and outcomes of collisions between two objects with identical properties, such as size and mass. This can provide insights into larger scale collisions, such as those between planets or galaxies.

2. How do you conduct experiments on colliding ovals?

Experiments on colliding ovals can be conducted using computer simulations or physical models. In computer simulations, the properties of the two ovals can be adjusted and the collisions can be observed and analyzed. Physical models can be created using objects with identical properties and measuring the outcomes of their collisions.

3. What are the factors that affect the outcome of colliding ovals?

The outcome of colliding ovals is affected by several factors, including the initial velocity and angle of the ovals, the elasticity of the objects, and any external forces acting on the ovals during the collision. The mass and size of the ovals also play a role in determining the final outcome.

4. Can colliding ovals produce different outcomes?

No, colliding ovals with identical properties will always produce the same outcome. This is due to the conservation of momentum and energy, which dictate that in a closed system, the total amount of momentum and energy must remain constant.

5. What real-life applications can be derived from studying colliding ovals?

Studying colliding ovals can have several real-life applications, such as understanding and predicting the outcomes of collisions in sports, traffic accidents, or industrial processes. It can also provide insights into the formation and dynamics of celestial bodies, such as asteroids and planets.

Similar threads

  • Advanced Physics Homework Help
Replies
1
Views
784
  • Special and General Relativity
2
Replies
67
Views
4K
  • Introductory Physics Homework Help
Replies
34
Views
597
  • Introductory Physics Homework Help
Replies
7
Views
1K
Replies
43
Views
3K
  • High Energy, Nuclear, Particle Physics
Replies
2
Views
2K
Replies
6
Views
2K
  • Special and General Relativity
Replies
1
Views
512
  • Special and General Relativity
Replies
25
Views
787
  • Special and General Relativity
Replies
14
Views
923
Back
Top