[DML] Re: crash tests / safety
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[DML] Re: crash tests / safety



[Note to Moderator; my minor was in physics, hardly a physicist but 
perhaps it will suffice: I tried to be more to the point here, as 
well as give reference sources, I promise this is my last post on the 
topic] 

How to read a long post: the First bit is some quick quotes, second 
bit re-explains the math, third bit demonstrates-via-example the 
two 'debated' scenarios (rigid barrier vs. head-on) and compares them 
for vehicles of identical mass and speed.

Firstly: some sources

Quote from http://www.safercar.gov/info.htm 
"Vehicles are crashed into a fixed barrier at 35 miles per hour 
(mph), which is equivalent to a head-on collision between two similar 
vehicles each moving at 35 mph. " 

Quote from http://tinyurl.com/ycx3lb  (NHTSA)
"As mentioned above, offset rigid barrier collisions
simulate collisions between vehicles of the same weight, or
collisions into structures."

a good book worth reading if you desire to have a basis for 
stating 'engineering facts' on lists..
"Vehicle Crash Mechanics"
By Huang Huang, Matthew Huang


Second-Bit: the requisite math

>>the crash speed is cumulative. IE: if barrier = 40
>>mph, head-on = 80 mph = 40 mph + 40 mph.
You can NOT add them together like that! Momentum is a VECTOR Ugh!  

To figure out how much Kinetic Energy (KE) is converted into 
heat/damage/sound/etc we need to answer a four questions?

1st Question: What is the final velocity of the two objects (now 
stuck together after impact, note: assuming completely inelastic 
collision).  
1st Answer: Vf = (m1v1 + m2v2)) / (m1 + m2)

2nd Question: what is the initial KE (BEFORE collision)
2nd Answer: Ki = (1/2) m1v1^2 + (1/2) m2v2^2

3rd Question: what was the final KE (After collision)
3rd Answer: Kf = (1/2) (m1 + m2) Vf^2

Final Question: How much KE was lost? (note: not all the loss can be 
attributed to destructive forces)
Final Answer: Kloss = Ki-Kf

 

Third-Bit: using the above formulas to answer the principle 
question: "if two cars collide head-on @ x MPH , is the KE loss 
(destructive forces) the same as if one of those cars had hit a RIGID 
barrier?"

---
Scenario A: "Two Cars collide Head-on" Assuming v=(40mpg ~ 18m/s), 
m=1200kg

Final Velocity = Vf = [(1200kg) (18m/s) + (1200kg) (-18m/s)] / 
(1200kg + 1200kg) = 0m/s = 0mph
KE Before Collision = Ki = (1/2) (1200kg) (18m/s)^2 + (1/2) (1200kg) 
(-18m/s)^2 = 388800J
KE After Collision = Kf = (1/2) (1200kg + 1200kg) (0m/s)^2 = 0J
KE Lost during Collision = 19634400J ? 0J = 19634400J

Now, you may not agree on how we divide the KE between the two cars, 
so lets compute the initial KE of both cars BEFORE the collision 
(after the collision they are both @ 0J KE)

KE car1 before collision  = (1/2) (1200kg) (18m/s)^2 = 194400J
KE car2 before collision  = (1/2) (1200kg) (18m/s)^2 = 194400J

KE Car 2 after collision = 0J
KE Car 2 after collision = 0J



Therefore, both cars `experienced' 194400J of destruction EACH.
---

---
Scenario B: "One car collides with a rigid barrier" For ease of 
comparison the barrier will be `car2' with zero velocity and high 
mass.  Assuming v=(40mpg ~ 18m/s), m of car=1200kg, m of barrier = 
120000000kg.

Final Velocity = VF = ((1200kg)*(18m/s)+(120000000)*(0m/s))/
(1200kg+120000000kg) = ,00018m/s ~ 0m/s
KE of Car1 before Collision = (1/2) (1200kg) (18m/s)^2 = 194400J

KE of Barrier before Collision = (1/2) (1200kg) (0m/s)^2 =  0
KE of system after Collision = (1/2) (1200kg + 120000000kg) (0m/s)^2 
= 0J

KE Loss of Car1 = 194400J
KE Loss of Barrier = 0J
---

**Conclusion: KE loss for Car-1 is the same regardless of scenario A 
or B.


As other people noted, it is the conversion of KE into destructive 
energies over TIME that results in the Severity Index of the crash, 
therefore the more your car can collapse,, the less YOU have to. 


-Nate
(11501)





--- In dmcnews@xxxxxxxxxxxxxxx, "Cameron, Peter" <cameron@...> wrote:
>
> ***** Moderator's Note *****
> The trouble with physicists is there is never one 
> around when you need one. Until we can find one
> or somebody finds their college physics text and
> looks up the equations, consider this thread 
> closed.







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