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17-12-2009, 03:28 PM | #61 | |||
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I cant believe I am disagreeing with a Physics doctorate but I am.. If you assume a perfectly inelastic collision, there is no difference between hitting a unyielding brick wall OR an identical car of the same mass travelling toward you at the same speed. Its conservation of momentum - and you can take that to the bank! |
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17-12-2009, 03:31 PM | #62 | |||
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If you were driving 100km/h an ultra light car into a B double, and you went backwards at 100km/h, then it is equivalent to a collision into a solid wall at 200km/h. Result is that the B double is the safer vehicle to be in purely because of mass. Force is equal to mass x acceleration. If both vehicles have the same mass and crumple zones, then the closing speed is 200km/h, but the effect is a negation of the speed of the opposing vehicle. Therefore mass is equal, acceleration is equal (both are negative), so the force of each cancel each other out - so it is not a 200km/h collision. The force is the same as a brick wall. Landcruiser vs fiesta - the landcruiser will not completely stop - it will still be travelling at some speed after the collision until it stops due to the alteration to rolling resistance (bent wheels etc). It will have the equivalent of a lesser impact than the fiesta. The fiesta however will be going backwards as a result of the collision, so the force upon it is greater than 100km/h. Try plugging some rudimentary numbers into the F=MA equation and see what you get. If the forces are equal, then the acceleration figures are skewed in favour of the cruiser - less deceleration due to the weight. BTW degrees in quantum physics are fine in theory, but mechanical physics is where it's at for this type of question. Perhaps we have an engineer here to answer it.
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XA coupe 8.8sec @ 150mph http://www.fordforums.com.au/showthr...coupe+drag+car BA GT-P for the shed Mustang GT for the other half E3 chubsport - fully fat (and slow), sitting there waiting for me to get sick of it and sell it. BA XR6T for a daily NT Pajero for the bush XB 4 door project- swallows a BF xr6 turbo My dad is a generous bloke. He gave away his dead car batteries free of charge.... |
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17-12-2009, 03:45 PM | #63 | ||
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A couple of websites for those who do not mind a bit of physics reading over their coffee
http://www.physicsclassroom.com/mmed...ntum/cthoi.cfm http://www.physicsclassroom.com/mmed...ntum/cthoe.cfm www.physicsclassroom.com/mmedia/#momentum Explains some of the problems raised here
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XA coupe 8.8sec @ 150mph http://www.fordforums.com.au/showthr...coupe+drag+car BA GT-P for the shed Mustang GT for the other half E3 chubsport - fully fat (and slow), sitting there waiting for me to get sick of it and sell it. BA XR6T for a daily NT Pajero for the bush XB 4 door project- swallows a BF xr6 turbo My dad is a generous bloke. He gave away his dead car batteries free of charge.... |
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17-12-2009, 04:22 PM | #64 | ||||
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335 S/C GT: The new KING of Australian made performance cars.. |
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17-12-2009, 04:52 PM | #65 | ||
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^Correct because you are looking at a single frame of reference.
In the Car vs Car scenario you can imagine it as one car being stationary and the other travelling at 200kph (as that is the relative velocity). While yes the other car offers come extra deformation due to crumple zones to absorb the energy it is not a huge amount at such speeds. Conservation of momentum holds true as in the Car v Wall scenario the wall has no velocity whereas Car vs Car, both cars have identical momentum. GeckoGT I can see the point Brazen is trying to make. If a 1200kg light car hits a wall at 60kph it is impacting with much lower kinetic energy then a 1800kg car hitting a wall at the same speed. Thus the smaller car requires much less energy to impart the same amount of damage to it as the larger car would (seems common sense). But you are right in that it doesn't particularly matter as in the real world collisions are not planned. This is just my understanding, I dont profess to be an expert and I would love to be proven wrong.
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17-12-2009, 05:01 PM | #66 | ||
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Looking at it from an energy point of view, in the two-car head on collision there is twice as much energy at the time of impact, but there is two cars involved so there is twice as much body work to absorb that energy. So each car still only has to absorb the energy of one impact with an immovable wall.
The crumple zones and time period of the accident are doubled to counteract the doubling of speed and energy. |
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17-12-2009, 05:30 PM | #67 | ||
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WOW, Bogan Uni is going strong.
A 100km/h head-on crash with identical cars is equal to a 100km/h crash into a non-deformable wall
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17-12-2009, 05:37 PM | #68 | ||
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Where this is being confused.
Twice the energy or force but twice the absorbsion. In the single car there is X energy absorbed by the car crumpling against the wall. In the dual car the is 2 * X energy being absorbed by 2 cars crumpling against each other. Part of the energy is dissapated squishing one car, the rest by the other. If the cars are identical then the absorbtion will be identical and therefore half of the total. If one of the cars had a yellow S on its cape then the other car would have twice the splat. As opposed to quantam physics, in the mechanical world 1 + 1 always equals 2 even for very large, very small or indeterminate values of 1. If the cars are not identical or the wall is deformed in any way as I said in my first post, the result could be anything...... |
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17-12-2009, 05:46 PM | #69 | |||
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17-12-2009, 05:46 PM | #70 | ||
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Actually I just thought of another way of describing this.
Imagine the wall is between the two cars and is infinitely thin. Each car will crumple as it hits the wall with exactly the same force but in opposite directions. Remove the wall. What happens? |
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17-12-2009, 06:20 PM | #71 | |||
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335 S/C GT: The new KING of Australian made performance cars.. |
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17-12-2009, 07:36 PM | #72 | ||
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what an amusing thread, or rather, what amusing responses!!!!
It's all got to do with the kinetic energy of the moving object which gets dissipated in a crash. Here's the theory (from what I can remember from high school and uni). Kinetic Energy (KE) = 1/2 x mass (m) x velocity squared (v^2) If a GTP weighing 2000kg drives at 100 km/h (27.8 m/s), it's Kinetic Energy is 1/2 x 2000 x (27.8)^2 = 772,840 Joules. If it hits an immovable object, then in the crash you dissipate 773 kJ of energy through your GTP (assuming no energy is absorbed by the immovable object) Now, if 2 GTPs travelling in opposite directions at 100 km/h each hit head on, then each will be carrying 773 kJ of kinetic energy which is dispensed equally among them, so a total of 1546 kJ of energy divided by 2 cars = 773 kJ each ... same as hitting an immovable object in one car. However, if a GTP is travelling at 200 km/h (55.6 m/s), your Kinetic Energy is 1/2 x 2000 x (55.6)^2 = 3091 kJ, or four times the kinetic energy of travelling 100 km/h (double the speed = four times the KE). So, if you hit an immovable object at 200 km/h, you have to dissipate 3091 KJ of energy through one GTP. This goes to show that two cars colliding head on at 100 km/h each is not the equivalent of hitting a solid wall at 200 km/h, actually, you're 4 times better off!!!!! Drive slow people, Merry X'mas |
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17-12-2009, 07:48 PM | #73 | |||
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I understand what he is saying and you are right it is logical. But the point is that I have tried to put across is that the ANCAP does not assess the damage to the vehicle, it rates the injury causing forces applied to the occupants in the vehicle (2 adults in the front, a 3 year old and a 18 month old, both in approved restraints). Thus it assesses the car ability to absorb a known force. No one buying a small car is going to say "I would like a small car, I only need a small car, but if I get hit by a 4wd it will do lots of damage so I better buy the 4WD". Just like those that are looking at large sedans don't opt for a Hummer because they might get hit by a truck. In crash protection, people want to know that they will be protected in an average crash with minimal chance of injury. Considering the average crash happens at less than 40 km/h and is normally front to rear type crashes, the ANCAP testing does very well in giving a decent indication of what is going to happen if you buy it and are involved in an accident.
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17-12-2009, 09:01 PM | #74 | |||
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Conservation of momentum would mean the wall has to transfer the energy for any rebound. If the wall is totally inert, immoveable, invulnerable etc this can't happen, so because the momentum must be conserved and the only way it can go is into whatever is anchoring the wall = the planet. If we place the same caveats on the cars and say they don't have any absorption and momentum must be retained, they are travelling at 100kph initial speed, but after butting heads and bouncing off each other they have a final velocity of -100kph. Put that into the V-V1 formula (100 --100) and you get 200 kph for the energy function of each. Deduct whatever you want for absorption after that. Last edited by Wally; 17-12-2009 at 09:14 PM. |
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17-12-2009, 09:18 PM | #75 | ||
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What we really need to know is >
If you put both cars back to back on the equator, and then they both drive away from each other at 100kph - how long would it take before they run into each other ?
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17-12-2009, 09:20 PM | #76 | |||
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17-12-2009, 10:32 PM | #77 | |||
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The OPs question is a bit more complex then time/distance.
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17-12-2009, 10:44 PM | #78 | ||
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^ correct - It was a small diversion to break the trend of the over complication of this thread.
After those 8 and a bit days the cars would collide as if they were hitting an immovable object !!
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17-12-2009, 10:56 PM | #79 | |||
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Newtons laws don't apply. The wall has no reaction:- force of wall = 0. You could go so far as to say as there is no reaction (no initial moment and no final moment) , no force is applied, so therefore no collision at all. |
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17-12-2009, 11:12 PM | #80 | ||
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Um, obviously from by brief skimp on this thread, no one has watched Deathproof!
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17-12-2009, 11:19 PM | #81 | |||
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18-12-2009, 12:14 AM | #82 | ||
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here is the physics behind this question with conservation of momentum and energy in an inelastic collision:
The answer is that a car of the same mass travelling at the same speed (say 1m/s) is the same as hitting a very heavy solid wall at 1m/s. car 1 moving 1m/sec hitting car 2 moving 1m/sec in opposite direction m1 v1 + m2 v2 = (m1+m2) v3 100* 1 + 100* -1 = 200* v3 v3 = 0 (final velocity =0 show both car come to a stop) loss KE of each car = 0.5*m*(v1^2-v3^2) = 50 ---------------------------------------------------------------------------------------------- car 1 moving 1m/sec hitting a heavy (10,000kg) solid wall m1 v1 + m2 v2 = (m1+m2) v3 100* 1 + 10000* 0 = 10100* v3 v3 = 0.00990099 (final velocity = approx. 0 show car1 almost come to a stop) loss KE of car1 = 0.5*m*(v1^2-v3^2) = 49.99509852
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18-12-2009, 08:07 AM | #83 | |||
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we did this in first year mech eng, hitting identical cars head on at the same speed is the same as hit a wall at the speed indicated. The other car adds its inertia to the crash = mv2 Ie mv2=mv2 |
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18-12-2009, 08:12 AM | #84 | |||
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18-12-2009, 08:19 AM | #85 | |||
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wall does not need to be padded as it has to return ALL of the energy back into the car and the speed is the same NOT 200kph Don't you guys see that when they collide they they impart all their energy onto each other and when they hit the wall (immovable object) they get all of their own energy back |
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18-12-2009, 08:51 AM | #86 | ||
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As far as I can tell, the issue of disagreement here is the point of reference.
From the driver's perspective, when he's driving towards the brick wall at 100km/h, his speed relative to the wall is 100km/h. When he hits the wall, the force applied by the wall on the car is enough to cause the car to accelerate (negative) to 0km/h in the distance of whatever the crumple zone is. From the driver's perspective, when he's driving towards the vehicle approaching from the opposite direction, his speed relative to the other vehicle is 200km/h, if both vehciles are doing 100km/h in directly opposite directions (in the same plane approaching each other). When the two vehicles collide, from the driver's perspective, his speed relative to the other vehicle is initally 200km/h at the beginning of the collision, and 0km/h at the end of the collision. If both vehicles are of equal weight, then the force applied by the opposing vehicle is sufficient to stop the driver's vehicle in the same distance as when he hit the wall. So, relative to the wall, the driver was doing 100km/h and stopped in the distance of his cars crumple zone. Relative to the other vehicle, the driver was doing 200km/h and still stopped in the distance of his cars crumple zone. I would much rather hit the wall if I had to choose. Most people are looking at this from the perspective of an observer standing on the side of the road watching a car hit the wall or two cars collide. That's fine, but in the scenario of two cars colliding head on, the forces viewed from the perspective of the observer are not the forces acting on the driver because the observer see's the first car with negative acceleration forces from 100km/h to 0km/h and the second car with negative acceleration from 100km/h to 0km/h as effectively two separate events, where as to the driver of either vehicle, the collision is very definitely a single event. |
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18-12-2009, 09:49 AM | #87 | |||
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My issue is with the more common car on car crashes, Im not carte blanche stating that a smaller car is worse off hitting a large car, I am saying that the crash rating themselves are flawed due to the vehicles differing masses. Remember E = 0.5mv2 so the vehicles mass is certainly an issue in determining the energy of a crash between two vehicles. This issue is not tested in an immovable object crash as the wall is exerting the exact opposite of the cars force in the crash. So an Ancap crash of a Fiesta has a lower kinetic energy than an Ancap crash of a Landcruiser, so I disagree that each crash test is a test which is equivalent. Remember that a crash with an immovable object is the equivalent of a crash with a car with the same mass and speed from the opposite direction. So a Fiesta crash test is the equivalent of a Fiesta hitting another Fiesta head on, likewise a Landcruiser test is the equivalent of a Landcruiser hitting another Landcruiser head on - you can start to see its not a level playing field as a Fiesta is having a crash test with a lot lower kinetic energies involved. The Ancap tests basically penalise a car for its mass. My way would to solve it would be testing cars up against a movable block. This block could represent the weight of the average car on the road travelling at the same speed. All of a sudden a Fiesta would have to work a lot harder to get a good rating and a Landcruiser would have to work a lot less, some people would say thats not fair but in reality real life crashes are seldom fair and would usually involve cars of different masses. |
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18-12-2009, 09:54 AM | #88 | ||
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Brazen, I completely understand what you are saying.
Small car vs Big car, as has been mentioned earlier is like having two same size cars head-on travelling at different speeds. The small, or slower car, will decelerate more than the big, or faster, car. I think you'll find though the offset frontal accident is the most common accident and in that test ANCAP use a deformable object. So it may be that they're one step ahead of you. |
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18-12-2009, 10:03 AM | #89 | |||
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The difference in speed of the first car at the beginning of the event to the end of the event in 100km/h NOT 200km/h. Same for the second car. The collision point is fixed so it does not matter what it is hitting the differential is the same. The relative view of the observer or either car is irrelvent. The only thing that matters is the state of each car before and after the event. Both doing 100km/h then both stopped. If all the cars and the wall were on a giant conveyer (and not taking off ) doing 2000km/h when the events happened would the damage be more or less? If both cars are on the surface of the earth doing several hundered km/h as it rotates and whatever else as we are big banging about the universe would the damage be more or less?. If the event were to be viewed as one car doing 200km/h running into another that is stopped then at the end of the event the combined mess would have to be doing -100km/h as the second car must be doing 100km/h less than it was at the beginning...... |
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18-12-2009, 10:10 AM | #90 | ||||
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