[Fluffy]

dissapointing first run , rotation has a too big impact on the movement direction after a hit, hits don't translate enough into rotation and worst of all, my system is highly undeterministic, different runs yield very different results.

EDIT: Now I think of it I didn't imply any physics at all, just put it in a way that would work and not tresspass and see what happens so I shouldn't be dissapointed lol. But the first try worked so great that I though I was a genious for a moment, that was just random though, the second run showed better

[Fluffy]

New stupid question

If I throw a steel bar at me at it hits me on its middle it will hit me with its full mass

however if it hits me on one of its extremes, some of the bar will just keep going when the bar rotates and the blow won't be as tough, so my speed after the hit will be lower than before. Or so I think, correct me if wrong please.

Any smart way to calculate what % of mass is hitting? what if the bar is a sphere instead? what if its an orthoedrum?

[hotShot]

It's not a percentage of the mass.

Setting:

You stand there and the steel bar thrown at you, has it's ends parallel to the line between your sholders.

Locking from your position, the left end of the steel bar passes you, while it's right end hits your left shoulder.

The steel bar will start to rotate and your chest will take the same amount of angular momentum (so hat the combined angular momentum stays 0.), your body will also take some of the bars momentum, but a (big) part of it, is used to deform your body and crush your bones.

The steel bar will fly further with a reduced speed.


So if you want to simplify it to a percentage you have to take a percentage of the momentum.

Since the mass of the steel bar does not change, the change in momentum (mass * speed)^{thanks helene_t} leads to a change of speed.

You get the amount of energy available for deformation, by calculating the changes in kinetic energy ( mass * speed²)
.

[helene_t]

If you can assume either an elastic collision or an unelastic one you can compute how much energy you absorb under the different scenarios. But OK, this is probably a partially elastic collision.....

Hoshot: momentum is mass*velocity, mass*speed^2 is 2*(kinetic energy). But you know that of course ....

[phil_20686]

Fluffy, on 2010-October-27, 04:15, said:

New stupid question

If I throw a steel bar at me at it hits me on its middle it will hit me with its full mass

however if it hits me on one of its extremes, some of the bar will just keep going when the bar rotates and the blow won't be as tough, so my speed after the hit will be lower than before. Or so I think, correct me if wrong please.

Yes. Again you must look only at conservation of angular momentum and linear momentum. The most natural way to get an estimate is to assume that the human absorbs no angular momentum (ie the human doesn't start spinning), if you do this you should find that it is identical problem to the sphere and pellet, only with different moments of inertia.

[Fluffy]

thx guys I am starting to understand the things now, sadly I understand it on my own terms.

What I am tryin to solve is this:

assuming elastic collision, modeling me as a vertical bar (X) standing at the beginning. the bar Y starts horizontally and moves horizontally and hits Bar X right in the middle, and the collision instant they both form a beatifull cross. Both bars have constant density on their volume.

Assuming the bar Y hit me with its middle, no rotation happens, and I can calculate my speed at the end, wich we will name V, because X takes some ammount of energy from the hit, lets call this energy E

but if Y hit me lets say, at 2/3 of its lenght (1/6 away from its centre) then, can we calculate bar X new velocity? assuming bar X won't spin, just translate.

X could be a bar, a sphere, or whatever, but the hit direction goes through its gravity center so no rotation will happen. However I assume the farther we move from Y's gravity center, the less energy/speed X takes. Is it easilly calculable? is it independant of X and Y masses? (assuming we already know V and E)

In my dreams I'd like to hear someone telling me, if Y's hit position is 1/6 of its mass centre then V1 and E1 will be equal to (N/M)*V and (N/M)*E where N is blablabla and M is blablabla

[phil_20686]

Fluffy, on 2010-October-27, 08:26, said:

thx guys I am starting to understand the things now, sadly I understand it on my own terms.

What I am tryin to solve is this:

assuming elastic collision, modeling me as a vertical bar (X) standing at the beginning. the bar Y starts horizontally and moves horizontally and hits Bar X right in the middle, and the collision instant they both form a beatifull cross. Both bars have constant density on their volume.

Assuming the bar Y hit me with its middle, no rotation happens, and I can calculate my speed at the end, wich we will name V, because X takes some ammount of energy from the hit, lets call this energy E

but if Y hit me lets say, at 2/3 of its lenght (1/6 away from its centre) then, can we calculate bar X new velocity? assuming bar X won't spin, just translate.

X could be a bar, a sphere, or whatever, but the hit direction goes through its gravity center so no rotation will happen. However I assume the farther we move from Y's gravity center, the less energy/speed X takes. Is it easilly calculable? is it independant of X and Y masses? (assuming we already know V and E)

In my dreams I'd like to hear someone telling me, if Y's hit position is 1/6 of its mass centre then V1 and E1 will be equal to (N/M)*V and (N/M)*E where N is blablabla and M is blablabla

If you demand that no energy is lost the problem is normally simpler,, provided it does not become over-constrained .

Its generally a useful trick to find observers frame of reference that maximises the symettry, in this case that means at the centre of the horzontal bar in its centre of mass frame. That leads to the easiest callculation:

consider an observer at the centre of the horizontal bar, originally P = m*v, L = m*v*1/6*L

after P = -mv' + MV L = -mv'L/6+I*W

These are two linear equations between 3 unknown and will have a one parameter family of solutions. These solutions correspond to all the possible energetic outcomes. Fixing an energy condition, of either CoE or any given fraction lost, will fix you to one (or occasionally two) points on the family of solutions.

Note that while this method works for a goo first approximation for elastic or mostly elastic collisions, as soon as you start losing a lot of energy that is roughly equivalent that the collision takes a finite time, after which the vertical bar will not rebound straight and things get a lot more complicated very quickly.

[Fluffy]

oh *****, here comes friction lol, its impossible to do a perfect model, I just hope mine is practical and doesn't look very odd.

[helene_t]

But the question was how much damage the object does to Gonzalo's buddy, wasn't it? Then we should not assume that no energy is lost since the damage is related to the amount of energy absorbed by Gonzalo's buddy.

Ideally we want to know how much energy is absorbed by fragile and vital organs like brain and heart. Energy absorbed by his clothes is no problem. But obviously that would require a very detailed description of the scenario to compute these things.

[Fluffy]

nah, that was just an example, my body is not important hehe. Seriously, on my model objects are solid-rigid undestroyable (if that word exists in english). But I am taking away some energy depending on some factors, will see what do I use that energy for later, maybe it is for some cool explosion or body amputation

[Fluffy]

Here is the problem:

I throw a spinning ball into a suface vertically, the ball doesn't bounce vertically back, but somehow its speed "rotates" towards (or against) the spin. Why?

I think of 2 answers:

-Because when the ball hits the surface, there is a friction that creates a force that opposes the spin, this force slows down the speed (rotation) but also speeds the gravity center of the ball towards/off the spin (translation)

-Because when the ball hits, the hit takes a finite time, and while the ball its at it, its vertical axis rotates, and the bounce goes on the new axis direction.


Any of them is correct?

[phil_20686]

Fluffy, on 2010-October-29, 07:33, said:

Here is the problem:

I throw a spinning ball into a suface vertically, the ball doesn't bounce vertically back, but somehow its speed "rotates" towards (or against) the spin. Why?

I think of 2 answers:

-Because when the ball hits the surface, there is a friction that creates a force that opposes the spin, this force slows down the speed (rotation) but also speeds the gravity center of the ball towards/off the spin (translation)

-Because when the ball hits, the hit takes a finite time, and while the ball its at it, its vertical axis rotates, and the bounce goes on the new axis direction.


Any of them is correct?

Both are correct. A truly elastic collision necessarily takes no time (time in contact means distortion of the shape means friction means in-elasticity), and hence there can be no frictional force. After all, the change in momentum is the integral of force over time. A collision that takes no time cannot feel a frictional force. I think you will find that if you calculate it this kind of collision is always inelastic, but, that is hard. The imposition of elastic collision may be technically inconsistent, but it can still give you an accurate approximation for two rigid bodies colliding (think spinning ball bearing on a marble floor). Its actually a very difficult problem I think. Suppose put my observer at the point where the centre of the ball will be at the collision, it bounces away at an angle, naively you might assume that neither in bound nor outbound trajectory has any angular momentum as they have no impact parameter, but the spin speed has been decreased, so angular momentum has decreased. In actuality, the ball rolls a little while in contact with the ground and this created an impact parameter and makes the problem soluble. For me the most "natural assumption" is to assume complete friction along the ground so that the time in contact can be related to the distance that the ball rolls, this will allow a conservation of angular momentum to proceed. Also, in practice the ground might carry away momentum energy so we fix our coordinates to the ground (insuring ground speed = zero in all frames) and assume that the effects of being in an accelerating frame are completely negligible. This will also give you an estimate of how much the ball will deform on impact (after all by symmetry it must take roughly half the time for top of the ball to stop moving downwards under a force given by hooks law), and you can use this to provide an estimate of how much energy will be lost here. (think like damping in a spring....)

[Fluffy]

yeah, I was suspecting this, the higher the speed at wich the ball collides, the bigger the contact time and the bigger the friction, so friction depends on the vertical speed, this messes my calculations even more hehe.

I've been researching through webs, and I am happy to see nobody has a perfect model, all are aproximations, so I feel a bit better now.


Another problem, if a ball is rolling though a surface, the hitting point at each moment is non-moving, so there is no slide, no slide means no friction, so whatever the surface's friction coeficent I give it, it won't make the ball stop. Is this correct?

[Fluffy]

regarding friction, does low friction dominate high friction or the contrary? I mean, if you put the highest friction surface against the lowest one, will it be maximum friction, no friction or something in between?.

I am tinking of sandpaper vs humid ice but I dunno what they'd do.

Regarding elasticity, I have the same problem question, inelastic collisions I think lose a lot of energy beacause one of the objects vibrates, and I guess if one object is highly inelasic it will dominate elastic ones (golf ball against a sofa for example)

[phil_20686]

Fluffy, on 2010-October-29, 11:14, said:

yeah, I was suspecting this, the higher the speed at wich the ball collides, the bigger the contact time and the bigger the friction, so friction depends on the vertical speed, this messes my calculations even more hehe.

I've been researching through webs, and I am happy to see nobody has a perfect model, all are aproximations, so I feel a bit better now.


Another problem, if a ball is rolling though a surface, the hitting point at each moment is non-moving, so there is no slide, no slide means no friction, so whatever the surface's friction coeficent I give it, it won't make the ball stop. Is this correct?

There are at least 3 kinds of friction in this example:

1) Rolling resistance: this comes from the fact that neither the ball nor the ground can be perfectly smooth, thus there will be some vertical movement - on some microscopic level both the ground and the sphere will ahve cross sections like the edge of a saw.

2) Air resistance

3) Distortion of the ball itself. - think of a tyre, the bottom compresses under the weight of the car, as the wheel rotates the bid under compression which means the tyre undergoes stretching. The more distortion the more friction, that is why an under pressure tyre becomes warm when you drive on it.

For something like a polished marble on a steel surface both 1 and 3 can be made entirely negligible. Two will normally not be negligible as the speed grows, but you can evacuate the chamber or something if you are doing a thought experiment.

[phil_20686]

Fluffy, on 2010-October-31, 07:59, said:

regarding friction, does low friction dominate high friction or the contrary? I mean, if you put the highest friction surface against the lowest one, will it be maximum friction, no friction or something in between?.

Regarding elasticity, I have the same problem question, inelastic collisions I think lose a lot of energy beacause one of the objects vibrates, and I guess if one object is highly inelasic it will dominate elastic ones (golf ball against a sofa for example)

I'm pretty sure that the first question is very case specific. E.g metal can slide smoothly over metal so you might think it has "low" friction, but if you put metal over stone it will appear as if the highest friction (stone) is dominating. But should I add glycerine to the stone and slide the metal over it I will get an almost zero friction surface - so then the "smoothest" surface is dominating. Perhaps when adding a lubricant you should regard it as 3 surfaces : stone-lubricant, lubricant-lubricant, and lubricant-steel. This brings up the general case that when you have multiple sliding layers atop one another, and you but a transverse force onto the top one, then they will slide at the surface of lowest friction. - this is what makes it possible to water-plane in a car with relatively little water on the road.

I could probably think of a better example, but I am pretty sure that the friction at an interface is a very complicated thing to work out. If you make the assumption that they are polished and rigid, such that the only friction comes from a kind of saw tooth effect forcing the objects to go up and down as they move along (think two saw toothed blades atop one another, but one with smaller teeth), then it is clearly dominated by the "smoother" of the two, but in practice in such a situation one or the other might rapidly erode/gouge the other. This is at least something one might check - if you get strips of metal and polish one side both sides or no sides, and slide them over each other do you expect the one polished to be equal to the two polished and much less than no polished, or one polished close to no polished and far from both polished? Experiment is the test of all knowledge

I think elasticity it is more the case of "domination" by the more elastic. However, the elasticity of a substance can vary greatly depending on the speed of impact - ever seen that ceramic that is like play dough, you can mould it in your hands, but it is easily hard enough to deflect a bullet, and if you bit it with a hammer it will shatter like glass. This is an extreme example of the "speed of sound" effect. Basically, the rate at which a substance can macroscopically deform to get out of the way is limited by the speed of sound in that material. That is why air resistance charateristics for fighter jets change dramatically at the speed of sound. in that case the density of the plane is so much more than the density of the air that it can force it out of the way by main force. Water is my favourite example of this: if you sink a ship it will vanish under the sea smoothly with almost no friction. If you smash an asteroid into the sea at greater than 3000 m/s (speed of sound in water) the sea will be entirely unable to get out the way, and the asteroid will be forced to vapourise the volume of water that it moves through (you can get the force from cross section * velocity * energy to vapourise 1m^3 water = force*velocity by CoE) and the back reaction will almost certainly vapourise/disintegrate the asteroid.

Thus we have several important caveats: if you have measured the elasticity at a velocity appropriate for the collision you intend to model, then you should get domination by the most elastic.

[Fluffy]

Thank you so much phil, I've been moving forward. Things are going slowly, but at least they keep going, I will probably have to change the core of my collision calculation because it slows down the program too much.

If anyone cares here are some examples that show more or less how it works

Attached File(s)

•  proyecto.zip (462.98K)
  Number of downloads: 1

[Fluffy]

New question, 2 cylinders, with the same mass, same radius, but different height, they are hit with identical bullets on the middle of its height (but not along gravity center) making them spin (over their circumferences) and move perpendiculary to the height.

They have same mass but different height, assume constant density along its volume, will they have same speed and rotation after the hit?


New examples at

www.angelfire.com/sk2/flufbbteams/proyecto.rar

[gwnn]

same speed and rotation, because the moment of inertia is the same (0.5*mr²), I would say. of course the question is, how exactly does the bullet interact with the surface? does it matter that the longer cylinder has a smaller density? I would say no.