And I will use logic, and thus go wrong with confidence.

1000kg car weight
20m/s
2 cars.
f=mv^2
f=1000kg * 20m/s * 20m/s
f=4000
00kgm^2/s^2, or 400000joules
x2 = 800000joules energy to disipate in the dual car crash

1000kg car weight
40m/s
f=1000kg * 40m/s * 40m/s
f=1600000joules to disipate in the single car crash

In the equation, shouldn't the rate of speed be doubled, since the closing rate is the sum of the 2 car's speed?

I'm not sure, but that's the basis of my view.

Frank

Would the same physics apply if both cars were on a conveyor-belt, one car going in the direction of the belt's movement, one going in the opposite direction?

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I don't think so, Frank. The energy in each car is independent of the energy in the other, unless you have to consider the relative speeds of the two cars...

And, Cap'n, what is the energy of the car if it's on a conveyor belt which is running at 88fps and the car is faing against the direction of the belt's travel doing 60mph? <img src="/forums/images/graemlins/evil.gif" alt="" />

I think that's my point. It would seem to me that each car is experiencing a 100mph contact with the other car. Anyway, accepting your math, we're left with an unresolved question:

If a car crashes into a wall at 50mph and the force results in pushing the front of that car rearward by 3'..... and we state the force required as "X"..... and another car crashes into the same wall at 100mph and the force results in pushing the front of the car rearward by 6'..... and we state the force required as "2X". Then we crash 2 cars going 50mph into each other and find that the front of each car has been pushed rearward by 3' - which we say requires "X"/car......... and state that only "X" force has occurred..... when we can see the result of "X" to each car..... then what did the damage to the second car? Aren't we missing another "X"?

Frank

Zero, if that 60 MPH is relative to the conveyor, and not the Earth. <img src="/forums/images/graemlins/butwiggle.gif" alt="" />

Now instead of a car, let's substitute a Cessna 182 at half its rated load capacity.

Nerds!


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I like my simple farm boy logic better than all that math. <img src="/forums/images/graemlins/wink.gif" alt="" />

I was never puzzled at all by the airplane one, but the car one has some interesting ramifications. The damn plane only flies when the airspeed over the wing is sufficient to create lift, period, end of story.

The car collisions involve whether relative speed or absolute speed is the proper frame of reference. The more I think about this, the less sure I am. I would be more confident if the test was two cars hitting at 50mph each vs. one at 100mph and one stopped, taking the differing "give" of the struck objects out of the equation.

1) What's indicated wind speed and direction?
2) Is the engine running? If so, at what rpm?
3) What's the diameter and pitch of the prop?
4) 230HP Continental or 235 Lycoming Turbo?
5) What's the coefficient of friction of the wheel bearings?
6) Is it tricycle gear or a tail dragger? I guess no floats or skis if it's on a conveyor?
7) If it's a tail dragger, is it a skid or a wheel?
8) Wheel pants or open?
9) Retractable gear or fixed?
10) Child seats or not?

No matter, they're all good - but I really, really like the old original '50s fat fuselage square rudder variants.

Frank

Pooey on a 182, RG or not. I want a 210T with the real big door...