Harness Bar Question (Why its needed)
05-25-2018, 12:03 PM
#21
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St. Jude Donor '12-'13-'14-'15-'16-'17
"for every force there is an equal and opposite force"
If two cars with similar mass and speed collide at 70mph it is just a 70mph crash it is not the same as a 140mph crash.
Kinetic energy is based on the square of the velocity, so a car moving at twice the speed would have 4 times the velocity. Momentum is the mass x velocity so if both cars have similar mass and velocity... it is a 70 mph crash it is not the same as a car moving twice the speed @140mph.
If two cars with similar mass and speed collide at 70mph it is just a 70mph crash it is not the same as a 140mph crash.
Kinetic energy is based on the square of the velocity, so a car moving at twice the speed would have 4 times the velocity. Momentum is the mass x velocity so if both cars have similar mass and velocity... it is a 70 mph crash it is not the same as a car moving twice the speed @140mph.
So, you contend that a car crashing into a solid, immovable object at 70 mph is exactly the same as crashing into something weighing a couple thousand pounds moving at 70 mph towards you? Are you seriously saying that? If so, you are dead wrong.
05-25-2018, 12:05 PM
#22
To be specific I am saying that two cars (both traveling 70 mph) colliding head on into each other is NOT the same as one car hitting a wall at 140 mph.
Change my mind, prove me wrong. .
Last edited by c0ntract_thrilla; 05-25-2018 at 12:11 PM.
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franknbeans (05-26-2018)
05-25-2018, 12:23 PM
#23
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St. Jude Donor '12-'13-'14-'15-'16-'17
Determine how much mass the crashed object contains. For example, consider a 2,000-pound car. On Earth, there are 2.2 pounds for every kilogram (kg) of mass, so:
Mass of car = 2,000 pounds ÷ 2.2 kg/pound = 909.1 kg
Determine the acceleration, or deceleration, involved in the crash. Imagine that the car was traveling as 27 meters per second (m/s)--roughly 60 miles per hour--when it hit a wall, coming to a complete stop in 0.05 seconds--5 hundredths of a second. To calculate the acceleration, simply divide the change in speed by the time it took to change.
Acceleration of the car = (0 m/s - 27 m/s) ÷ 0.05 s = -540 m/s2
Note: the negative sign on the acceleration indicates that it was deceleration that occurred, and is not important when calculating the net force involved.
Use Newton's Second Law to calculate the net force involved in the crash.
Force = mass x acceleration = 909.1 kg x 540 m/s2 = 490,914 Newtons (N)
The car exerts a force of 490,914 N on the wall, which is roughly equivalent to 550 times the car's weight. That's for one car crashing into a wall. If two cars going the opposite direction crash into each other, there would be almost 1,000,000 Newtons of force expended. That force is absorbed by the cars and whatever is in them.
One car hitting a wall is not the same as two cars hitting head on at the same speed as the car that hit the wall, at least in terms of energy involved, which is what matters.
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rb185afm (05-25-2018)
05-25-2018, 12:43 PM
#24
According to Newton's Second Law of Motion, the force, in Newtons, that an object exerts on another object is equal to the mass of the object times its acceleration. How can this be applied to calculating the forces involved in a crash? Keep in mind that acceleration is an object's change in speed over time. Objects involved in crashes usually decelerate--the numerically negative form of acceleration--to a stop. Calculating the amount of force involved in a crash is as simple as multiplying the mass of the crashing object by its deceleration.
Determine how much mass the crashed object contains. For example, consider a 2,000-pound car. On Earth, there are 2.2 pounds for every kilogram (kg) of mass, so:
Mass of car = 2,000 pounds ÷ 2.2 kg/pound = 909.1 kg
Determine the acceleration, or deceleration, involved in the crash. Imagine that the car was traveling as 27 meters per second (m/s)--roughly 60 miles per hour--when it hit a wall, coming to a complete stop in 0.05 seconds--5 hundredths of a second. To calculate the acceleration, simply divide the change in speed by the time it took to change.
Acceleration of the car = (0 m/s - 27 m/s) ÷ 0.05 s = -540 m/s2
Note: the negative sign on the acceleration indicates that it was deceleration that occurred, and is not important when calculating the net force involved.
Use Newton's Second Law to calculate the net force involved in the crash.
Force = mass x acceleration = 909.1 kg x 540 m/s2 = 490,914 Newtons (N)
The car exerts a force of 490,914 N on the wall, which is roughly equivalent to 550 times the car's weight. That's for one car crashing into a wall. If two cars going the opposite direction crash into each other, there would be almost 1,000,000 Newtons of force expended. That force is absorbed by the cars and whatever is in them.
One car hitting a wall is not the same as two cars hitting head on at the same speed as the car that hit the wall, at least in terms of energy involved, which is what matters.
Determine how much mass the crashed object contains. For example, consider a 2,000-pound car. On Earth, there are 2.2 pounds for every kilogram (kg) of mass, so:
Mass of car = 2,000 pounds ÷ 2.2 kg/pound = 909.1 kg
Determine the acceleration, or deceleration, involved in the crash. Imagine that the car was traveling as 27 meters per second (m/s)--roughly 60 miles per hour--when it hit a wall, coming to a complete stop in 0.05 seconds--5 hundredths of a second. To calculate the acceleration, simply divide the change in speed by the time it took to change.
Acceleration of the car = (0 m/s - 27 m/s) ÷ 0.05 s = -540 m/s2
Note: the negative sign on the acceleration indicates that it was deceleration that occurred, and is not important when calculating the net force involved.
Use Newton's Second Law to calculate the net force involved in the crash.
Force = mass x acceleration = 909.1 kg x 540 m/s2 = 490,914 Newtons (N)
The car exerts a force of 490,914 N on the wall, which is roughly equivalent to 550 times the car's weight. That's for one car crashing into a wall. If two cars going the opposite direction crash into each other, there would be almost 1,000,000 Newtons of force expended. That force is absorbed by the cars and whatever is in them.
One car hitting a wall is not the same as two cars hitting head on at the same speed as the car that hit the wall, at least in terms of energy involved, which is what matters.
You didn't do a second calculation for 140 mph which is 4 times the velocity with the same mass which is more kinetic energy.
Last edited by c0ntract_thrilla; 05-25-2018 at 12:59 PM.
05-25-2018, 06:17 PM
#25
"for every force there is an equal and opposite force"
If two cars with similar mass and speed collide at 70mph it is just a 70mph crash it is not the same as a 140mph crash.
Kinetic energy is based on the square of the velocity, so a car moving at twice the speed would have 4 times the velocity. Momentum is the mass x velocity so if both cars have similar mass and velocity... it is a 70 mph crash it is not the same as a car moving twice the speed @140mph.
If two cars with similar mass and speed collide at 70mph it is just a 70mph crash it is not the same as a 140mph crash.
Kinetic energy is based on the square of the velocity, so a car moving at twice the speed would have 4 times the velocity. Momentum is the mass x velocity so if both cars have similar mass and velocity... it is a 70 mph crash it is not the same as a car moving twice the speed @140mph.
Of course that was 55 years ago when I was dozing thru High School Physics!
I was at Irwindale Raceway in late 60s/early 70s when I saw a Mustang
Driver ( I think Les Ritchey ) have his head CUT entirely off on the track
with his helmet still in place on his head.
Since then I have always mounted a Anti-Submarine Belt from my harness to the floor
slightly under the seat.
Last edited by GTX JOHN; 05-25-2018 at 10:18 PM.
05-25-2018, 06:37 PM
#26
Although the two-car crash doubles the speed, the energy the crash is transferred to twice the mass resulting in a crash that looks like just one car hitting a wall at 70 mph vs one car hitting a wall at 140 mph.
You didn't do a second calculation for 140 mph which is 4 times the velocity with the same mass which is more kinetic energy.
You didn't do a second calculation for 140 mph which is 4 times the velocity with the same mass which is more kinetic energy.
If identical cars A and B crash perfectly square into each other driving identical speeds, Both cars should stop moving. As if car A is hitting a non-movable object because their momentum cancel each other out perfectly.
Now if Car B was moving at 90, faster than Car A going 70 (and you're in car A). Then, the crash felt by A will be 70 +20 (the difference).
Same if Car B has more mass (say a semi), now you're in big trouble. Likely Car B will barely lose momentum and transfer it all to Car A.
Last edited by BrunoTheMellow; 05-25-2018 at 06:42 PM.
05-26-2018, 10:39 AM
#27
Le Mans Master
Thread Starter
Dudes, the maximum force generated here is the deceleration of the car parts from 70mph to 0 within the length of the car. This assumes it hits an immovable object (giant steel die, stone wall, whatever).
Yes, that's the same as hitting an oncoming car at 70mph if everything is perfectly aligned and opposite, which is, of course, impossible in the real world but the same in "theory" if you could stage it.
And I'm sure there are ways to double the speed and halve the mass of the oncoming car and get the same result.
Don't worry about the speeds, worry about the actual deceleration and then it's much easier to grok in your head.
FWIW, as an aside, the front bumper encounters WAY higher forces than the rear bumper, as the rear bumper decels over the course of 16 feet while the front bumper has to stop "right now".
Yes, that's the same as hitting an oncoming car at 70mph if everything is perfectly aligned and opposite, which is, of course, impossible in the real world but the same in "theory" if you could stage it.
And I'm sure there are ways to double the speed and halve the mass of the oncoming car and get the same result.
Don't worry about the speeds, worry about the actual deceleration and then it's much easier to grok in your head.
FWIW, as an aside, the front bumper encounters WAY higher forces than the rear bumper, as the rear bumper decels over the course of 16 feet while the front bumper has to stop "right now".
Last edited by davepl; 05-26-2018 at 10:41 AM.
