Crash Talk Here

Now that's funny!!!!

There is another forum for "Other Cars - Especially Hybrids."

 

Jim,

These questions are considerably more difficult because we have to make assumptions about the front ends of the cars and how they deform. For example, I can think of ways of modeling the front ends of the cars in case 1 to make either result occur. But regardless of how they deform two things are true: in case 1, at any given point in time both cars will have the same deceleration, even if their front ends deform differently. This is because both cars must have the same instantaneous decelerative force (for every force, there's an equal force going the other way), and both have the same mass (F = ma). In case 2, the SUV will have exactly half the deceleration as the Prius for the same reasons above (this is also what 200volt said).

1) There are two cases I can think of. First, if the second car's front end is made of silly putty, then the Prius will experience a low decelerative force for an extended period of time (probably killing the silly putty car's driver). In this case the peak deceleration of the Prius is low. Second, if the second car is made of granite, then when the Prius crashes into it, there's the same amount of kinetic energy that needs to be dissipated as if the Prius had crashed into another Prius. However, there's only one crumple zone to absorb that energy. Instead of splitting between two Prii, the one Prius has to absorb twice as much energy as before. I think no matter what, this is less safe for the Prius driver. Both the silly putty and the granite car would have low crash safety scores.

In the case of hitting the granite car, if you model the front end of the Prius as a linear "one-way" spring that deforms until absorbing energy E (where E is the kinetic energy of 1 Prius going 35mph), it compresses sqrt(2E/k) distance at a maximum force of sqrt(2Ek), where k is the spring constant. If it must absorb 2E energy, it compresses sqrt(4E/k) distance with a maximum force of sqrt(4Ek). Since acceleration is proportional to force, the peak deceleration is a factor of sqrt(2) higher. But it could also be much worse: what if the front end were designed only to absorb energy E, and after that the engine gets shoved into your face?

2) This one is also tricky, but it makes the analysis easier if we assume both front ends are identical. In that case, both front ends will crush safely and symmetrically until 2E of the kinetic energy is absorbed, leaving E kinetic energy left. If the front ends become much stiffer after the crush zone is used up, then the force felt by each car spikes, and the deceleration felt by each driver jumps, but again the Prius driver feels twice the deceleration as the SUV driver the entire time. If we revert to the "one-way" spring model and give both vehicles identical front ends, then the springs would deform symmetrically because they feel equal compression forces during the entire collision. Each vehicle then absorbs energy 3E/2, giving a peak deceleration in the Prius a factor of sqrt(3/2) higher than the 2 Prius example.

I think the conclusions I can draw are these:

1. The stiffer the other guy's car, the worse it is for you. This is true whenever you hit a body-on-frame truck or SUV because the frame is very stiff.
2. The heavier the other guy's car, the worse it is for you. First, there is more energy that needs to be absorbed, and second, if the heavy car were designed to have good crash safety, it would likely have a stiffer front end to absorb the higher kinetic energy of the vehicle. Doubly bad for you.
3. The heavier your car, the better it is for you. The peak force is divided by your mass to get your peak deceleration.

 

Plasm,

I have to compliment you on your ability to think things through clearly... and explain them in a clear and concise manner... and then (most importantly)... give a meaningful conclusion. Nice job.

Your conclusions exactly match my preconceptions... but your thought process was more methodical.

Thanks again.

/Jim

 

I decided to try analyzing the effect of a 3200 lb vehicle colliding with a 5920 lb vehicle; elastic and inelastic collision to see what the effect would be on an unrestrained driver. I assume that the semi-inelastic collision was 75% inelastic and 25% elastic. Follows here:

The challenge: What happens when a 3200 lb Prius collides head on with a 6000 lb vehicle. Herein an attempt.

Vehicle #1 – 3200 lb Prius (2890 +310 lbs fuel/pax).
Vehicle #2 – 5920 lb Expedition (5610 + 310 lbs pax/fuel)

m1 = 3200 lbf-s^2/ft = 100 slugs , m2 = 5920 lbf-s^2/ft = 185 slugs
V1 = 40 f/s (27.3 mph), V2 = -40 f/s

Find: velocity of the center of mass (Vcm) : Vcm = Sum of momentum/ total mass.

Momentum = mV. ; Sum mV = 100 sl*40 f/s – 185 sl*40 f/s = -3400 sl-f/s
Then Vcm = (-3400 sl –f/s)/285 sl = -11.93 fps.
So, the center of mass is moving rt to lf at 11.93 f/s.

Elastic Collision: Equation ; 2* Vcm – V (before) = V (after)

V1 = 2*(-11.93) – 40 = -63.86 f/s...............V2 = 2*(-11.93) – (-40) = +16.14 f/s
Del V1 = - 103.9 f/s................................... Del V2 = +56.14 f/s
That is, rebound velocity change is 85% greater for the perfectly elastic Prius than for the Expedition..

Before/After Momentum: Conservation of momentum means that total momentum should be constant. Check totals

...................Prius..........Expedition.............Total
....................mV..............mV.....................mV
before....... 4000sl-f/s -7400sl-f/s = -3400sl-f/s

after........ -6386sl-f/s +2985sl-f/s = -3400 sl-f/s

Inelastic Collision: Assume each vehicle’s crush zone collapses in 0.1 second.
Assume (jus’ b’because) that 75% of momentum is absorbed, the remaining 25% of momentum being pseudo-elastic. Velocity of the cm (Vcm) needs to be recalculated.

Vcm (inelastic) = .25*Vcm (elastic) = .25*(-11.93) = - 3 f/s

Impulse-Momentum Exchange: I = Impulse = F*t = momentum =mV.

For 2 vehicles I = F*t = m1*V1 + m2*V2. The collision occurs over 0.1 seconds, from impact to vehicle rebound. Velocity also changes during that time. Let’s call the velocity change del V and the time interval del t .

Velocities after the collision are now:
2*Vcm – V(before) = V(after) ;...V1 = 2*(-3)-40 = -46 f’/s,...V2 = 2*(-3) –(-40) = +34 f/s.

del V = V(final) – V(initial):...del V1 = -46-40 = -86 f/s;...del V2 = +34 – (-40) = 74 f/s.

Solve for Force of Collision in g’s: F = Sum of momentums/ del t

m = W/g = 32*slug lbf-s^2/f ; m1 = 3200 lbf-s^2/f, m2 = 5920 lbf-s^2/f.

Then F = (3200*(-86) lbf-s + 5920*74 lbf-s )/0.1 sec = 1,628,800 lbf.

Acceleration: From F = ma, get a = F/m. Determine acceleration for each vehicle based on the force of the collision.

Prius: a = F/ m = 1.63x10^6 lbf / 3200 lbf-s^2/f = 509 f/s^2. g’s = 509 f/s^2/ 32 f/s^2 = 15.9
Expedition: a = 1.63x10^6 lbf/ 592 lbf-s^2/f = 275 f/s^2. g’s = 275 f’s^2 / 32 f/s^2 = 8.6

If each driver were unrestrained, the Prius driver would hit the steering wheel and windshield at nearly 16 g’s! The expedition driver – under 9 g’s!

 

So the expedition has nearly twice the mass... and nearly half the G's. Is this relationship linear? Would a vehicle with 3X the mass experience 1/3 of the G's?

/Jim

 

Depends. On.....speed of each vehicle, actual inelastic vs. elastic % of the crash, etc. Besides, this was a problem made up out of whole cloth. Real world kinematics are much more complicated than the simple assumption of 75% of contact duration was inelastic and the remaining 25% purely elastic. I did that just to get some numbers that COULD be ballpark.

It does follow that as the inequality of the masses gets larger, given a similar set of assumptions, one would expect a 3 to 1 weight advantage (of larger vehicle) to result in 33% to 40% of the smaller vehicle's g-loading.

 

Hi Denny, thanks for the analysis. You highlight a model in which the front ends of the cars deform with uniform force over time and space versus my model of linearly increasing force with crumple distance. I do not have the expertise to say which is more realistic, but I'd bet it's a mixture of the two plus some nonlinear stuff. Here are my thoughts (please do not take offense, I am merely trying to help/clarify):

Momentum is always conserved whether the collision is elastic or inelastic, so Vcm should be the same regardless. Elasticity only modifies how much kinetic energy is recovered after the collision. Notice how your inelastic V2 = +34 f/s is greater than in the elastic experiment V2 = +16 f/s, which doesn't make sense. The SUV should had a more negative V2 after a more inelastic collision.

Here you take the sum of the change in momentum over both vehicles and divide by del t. I believe you should only take the change in momentum for one of the vehicles and divide by del t to get the impact force because both vehicles feel the equivalent force in opposite directions (Newton's 3rd) and have their momentums changed equivalently.

Here you compute the acceleration of each vehicle by taking force divided by mass. This is the acceleration felt by the car, not by the driver. Imagine a collision where the driver is unrestrained. For a while after the collision has begun, the driver's velocity doesn't change while the dashboard is decelerating. By the time the driver hits the dashboard, the driver's velocity must change extremely rapidly to match the new lower velocity of the dashboard, resulting in an instantaneous deceleration much higher than that felt by the car. The above method would compute the acceleration felt by a perfectly restrained driver that feels only the acceleration that the car experiences.

I could have made mistakes in my analysis, so please comment as you like!

 

For a perfectly restrained driver, it should be perfectly linear because the force felt by each vehicle is the same. Then you just use a = F/m. If m triples, a is divided by 3.

 

Plasm – your comments were appreciated. Rather than reposting all you typed, I’ll respond by revising and commenting as I go along.

**Continue after Elastic collision momentum analysis

Before/After Momentum: Conservation of momentum means that total momentum should be constant. Check totals

................Prius...................Expedition .................Total
.................mV .......................mV........................ mV
before.... 4000sl-f/s ............-7400sl-f/s ..........= -3400sl-f/s

after....... -6386sl-f/s ...........+2985sl-f/s .........= -3400 sl-f/s

- Re: Inelastic collision comment. To wit; Kinetic energy change and conservation of momentum. I must admit I haven’t done any of this kinetics stuff since the late 70’s. My old befuddled mind needs this exercise in cobweb removal. Obviously, as you point out, the Vcm remains constant (conservation of momentum)whilst the velocity of the 2 vehicles must change relative to the cm, and in proportion to the ratio of the two masses. That wrong assumption affected all else – except frame of reference in the driver’s compartment. A FBD could have prevented that – if I can remember how to do ‘em.

REVISE - Inelastic Collision: Assume each vehicle’s crush zone collapses in 0.1 second. Assume (jus’ b’because) that 75% of KE is absorbed, the remaining 25% of KE preserved during a pseudo-elastic end of collision. Velocity of the vehicles at end of inelastic portion needs to be recalculated.

Speed of Prius = 40f/s and Expedition = - 40f/s -> before contact.

If each car were the same mass then the velocity at end of elastic collision
would be identical. But ratio of mass is m1/m2 = 100/185 = 0.54. By iteration(wag too):
V1 = +3 f/s & V2 f/s = -20 f/s

del V1/del V2 = (+3 – (+40))/(-20 – (-40)) = -37/20 = -1.85 = -1/0.54

Velocities after the collision are now:
2*Vcm – V(before) = V(after) ; V1 = 2*(-12) -3 = -27 f/s,
V2 = 2*(-12) – (-20) = -4 f/s.

Conservation of Momentum Chk: At completion of inelastic contact;

P = m1V1+m2V2 = 100*3+185*(-20) = 300-3700 sl-f/s= -3400 sl-f/s.
So, momentum is in order! If V1 = -27 and V2 = -4 are plugged in the momentum remains = –3400 sl-f/s.

REVISE: Collision force:
Impulse-Momentum Exchange: I = Impulse = F* del t = m delV.
Concur, as you pointed out – each vehicle feels identical force.
For 2 vehicles I = F1*del t = m1*delV1 = F2*del t = m2*delV2. The collision occurs over 0.1 seconds, from impact to vehicle rebound.

F1 = m1(del V1)/del t = F2 = m2(del V2)/del t, where del t =0.1sec

F1 = 3200(-37)lbf-s / 0.1 s = F2 = 5920(20)lbf-s/ 0.1 s = 1.184 E6 lbf

Acceleration: From F = ma, then a = F/m. Determine acceleration for each vehicle based on the force of the collision.

Prius: a = F/ m = 1.184x10^6 lbf / 3200 lbf-s^2/f = 370 f/s^2
g’s = 370 f/s^2/ 32 f/s^2 = 11.5

Expedition: a = 1.184x10^6 lbf/ 5920 lbf-s^2/f = 202 f/s^2
g’s = 202 f’s^2 / 32 f/s^2 = 6.3

If each driver were (REVISED) “restrainedâ€, the Prius driver would be subjected to around 11 g’s! The expedition driver – around 6 g’s!

Nuff, eh! :blink: