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Vancouver Auto Chat 2016 VAC Community Head Moderator: Raid3n

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Old 10-26-2011, 07:35 PM   #26
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Originally Posted by EvoFire View Post
By your definition, you are still wrong.

If a car hits an immovable wall at 120mph, the car takes 100% of the hit, whereas if a 120mph car hits a 0mph car, the energy is spread into the 2nd car. Car =/= wall

On top of that, a 60mph car hitting a 60mph car means they are taking damage for their speed respectively. Sure one may take a bigger chunk of the hit than the other depending on circumstance, but its still two entities all the same.

Moreover, if you took physics in highschool (this is taught in Physics 11) It takes more energy to make an object more faster. It is not linear. Just check out Musky's math.
U guys suck at physics n logic
Musky calculation is wrong because it is using the wrong frame of reference to measure the speed n energy

Back to high school physic.... Just think of speed relativity?
The totally energy of a collision would be the same whether it is 120->0 or 60-><-60
The difference is how the energy distribute between the 2 objects will be different depending on whether it is a car or a brick wall
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Old 10-26-2011, 07:36 PM   #27
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Holy shit......scary because I use to drive a ford focus
but not at 200km? any car will be destroyed like that.
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Old 10-26-2011, 07:41 PM   #28
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Originally Posted by Ferra View Post

Speed and energy are both relative
your equation is looking at the vehicle speed from an OUTSIDER perspective...(which has nothing to do with the two cars colliding with one another)

since you are measuring the damages & energy on the vehicles, the correct frame of reference should be on the vehicles and not on you.
ITT people need to go back to high school

go back to your 2-dimensional kinematics and energy conservation equations. these are like the first 2 equations that you learn in high school. and look at how v^2 + v^2 =/= (2*v)^2, which shows that velocity increases energy by a square factor


Quote:
Originally Posted by Ferra View Post
U guys suck at physics n logic
Musky calculation is wrong because it is using the wrong frame of reference to measure the speed n energy

Back to high school physic.... Just think of speed relativity?
The totally energy of a collision would be the same whether it is 120->0 or 60-><-60
The difference is how the energy distribute between the 2 objects will be different depending on whether it is a car or a brick wall
learn to walk before you run. Both cars are running on the same reference plane (ground) as the observer. Total energy in those two collisions are NOT the same. There is a reason why there is a square term in velocity. I can assure you the sole moving car will take more damage even if a stationary car replaces the brick wall.

Last edited by Zyzz; 10-26-2011 at 07:52 PM.
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Old 10-26-2011, 08:11 PM   #29
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ITT people need to go back to high school

go back to your 2-dimensional kinematics and energy conservation equations. these are like the first 2 equations that you learn in high school. and look at how v^2 + v^2 =/= (2*v)^2, which shows that velocity increases energy by a square factor
The energy distribution to mass and the absolute distance traveled to come to a complete stop are also factor.

Assuming a 1300kg car hit a wall, stopping from 100kph to 0 in 0m, the impact on the 1300kg mass is around 5000 tons of force to the car. The energy needed to stop the moving object completely in such distance.

Same car hitting a stopped car, which moves forward upon impact (let's say 10m), the number becomes 5 tons.

Two cars crashing makes no difference. It depends on each car's speed and mass to calculate their forward momentum and the force needed to stop each other.
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Old 10-26-2011, 08:13 PM   #30
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Originally Posted by Zyzz View Post
ITT people need to go back to high school

go back to your 2-dimensional kinematics and energy conservation equations. these are like the first 2 equations that you learn in high school. and look at how v^2 + v^2 =/= (2*v)^2, which shows that velocity increases energy by a square factor




learn to walk before you run. Both cars are running on the same reference plane (ground) as the observer. Total energy in those two collisions are NOT the same. There is a reason why there is a square term in velocity. I can assure you the sole moving car will take more damage even if a stationary car replaces the brick wall.
You should think this through before you start replying...

you are trying to measure the kinetic energy of the COLLISION between the 2 cars
and you need an inertia frame of reference when you use the 1/2mv^2 formula... You do not use the ground as ur frame of reference because the collision has nothing to the ground...the collision speed & energy is only relative to the 2 cars' speed.

You would use the "ground" as ur frame of reference only if the two cars are being smashed into the ground.... (i.e. one car smashing into the ground at 120mph produce more energy than 2 cars smashing into the ground at 60mph)

just think of speed relativity...i.e. there is NO WAY to tell whether you are moving toward an object @ 120mph or an object is moving toward you @ 120mph or if you and the object are moving toward each other at 30/90, 60/60, 20/100mph....etc

tl;dr
the total energy of the collision between 2 objects will be the same whether is is a 120mph->0mph or 60mph-><-60mph.
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Old 10-26-2011, 09:00 PM   #31
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Ok think of it this way

2 inputs enter a system (a black box), each with the energy going into the system with the energy E, with speed v. The energy obtained by the system is 2*E = E1

Now a single input enter into the system with 2*v1, with the energy E2

U will find that E2 is greater than E1 by a factor of 2. Note that E1 here already includes the energy from both cars. Therefore the output of the system (the damage that incurs on the car) is greater than the first scenario.

No just no. You DO use the ground as the reference point in this case. If we go by your point of view, where we take the car as the reference point, I will never be able to push/move a box, because according to Newton's law, the box is exerting an equivalent force on my hand. You seem to be confused as to where in which scenario to take the reference points.

Last edited by Zyzz; 10-26-2011 at 09:34 PM.
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Old 10-27-2011, 07:07 AM   #32
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Ok think of it this way

2 inputs enter a system (a black box), each with the energy going into the system with the energy E, with speed v. The energy obtained by the system is 2*E = E1

Now a single input enter into the system with 2*v1, with the energy E2

U will find that E2 is greater than E1 by a factor of 2. Note that E1 here already includes the energy from both cars. Therefore the output of the system (the damage that incurs on the car) is greater than the first scenario.

No just no. You DO use the ground as the reference point in this case. If we go by your point of view, where we take the car as the reference point, I will never be able to push/move a box, because according to Newton's law, the box is exerting an equivalent force on my hand. You seem to be confused as to where in which scenario to take the reference points.

you are trying to measure the energy of the collision here....and the energy of the collision is dependent upon the relative speed of the objects...NOT how much energy they carry relative to the ground/earth/solar system/etc

energy is relative in the exact same way speed is relative....

Just give it some thought...
Using the same example, what if this time the ground is moving?? or the collision happened inside a moving train?)
The energy calculation will be different using the the ground as a frame of reference but obviously the result of the collision will be the same regardless of whether the ground is moving or not.

I understand how you are thinking about the problem and you are missing the point....you are not trying to find the energy of the two objects relative to the ground or that is contained relative to a closed system.
You are trying to find the energy of the two objects' collision.

Another example you can think of: This time, instead of head on collision, what if a 120mph object hit a 119mph object both moving in the same direction?? You "ground reference" would indicate a shitload of energy when the actual energy of the collision is minimal due to the small relative speed.


Not sure how this has anything to do with newton's law here
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Old 10-27-2011, 09:37 AM   #33
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So I am wrong, the tv show is wrong, the physicists are wrong, and the whole world is wrong, and you are correct.

I think you should write a report to be submitted for Nobel prize. I give up on you.
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Old 10-27-2011, 11:32 AM   #34
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So I am wrong, the tv show is wrong, the physicists are wrong, and the whole world is wrong, and you are correct.

I think you should write a report to be submitted for Nobel prize. I give up on you.
Good choice lol
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