Auto insurance ripoff by GEICO

In article >,
"Rod Speed" > wrote:

<snip>

> > this equation completely describes the situation
> > and constant acceleration is the best case scenario
> > for minimizing the forces the tires must resist.
>
> Like hell it does when you just ignore crumpling which is what
> absorbs most the mometum at the higher impact speeds.
>
> That's why cars are now designed to crumple, fool.

I doubt you can learn anything, but I'll try anyway:

<http://www.skytran.net/09Safety/10sfty.htm>

Note the simple relationship between speed, crumple distance and
acceleration.

"ecelerating at 6"g's" is NOT the same as crashing! A crash from 100
MPH into a solid object that results in say a 4 foot of vehicle front
compartment deformation means the passengers inside (if securely
fastened in) also decelerate from 100 MPH to zero MPH in 4 feet!"

Note that they don't make any mention of your nonsense about crumpling
somehow magically dissipating the momentum. They are presenting the best
case scenario and their figures match exactly with my formula.

Here's an example:

Crumple distance, d: 5 feet

Speed, v: 30 mph, 44 feet/second

So: 0 = v^2 + 2ad

a = -(v^2)/2d
= -(44^2)/10
= -193.6 feet/sec/sec

Or, since the acceleration due to gravity (g) is 32 feet/sec/sec,

a = -6.05g

Just as their website says.

Does that help you get your feeble brain around the subject?

LOL

--
Alan Baker
Vancouver, British Columbia
"If you raise the ceiling 4 feet, move the fireplace from that wall
to that wall, you'll still only get the full stereophonic effect
if you sit in the bottom of that cupboard."

In article
>,
"Rod Speed" > wrote:

> "Alan Baker" > wrote in message
> ...
> > In article >,
> > "Rod Speed" > wrote:
> >
> >> "Alan Baker" > wrote in message
> >> ...
> >> > In article >,
> >> > keith > wrote:
> >> >
> >> >> On Tue, 26 Apr 2005 01:49:39 +0000, Alan Baker wrote:
> >> >>
> >> >> > In article >,
> >> >> > ?hadoaeraith -id> wrote:
> >> >> >
> >> >> >> Alan Baker wrote:
> >> >> >>
> >> >> >> > In article >,
> >> >> >> > Alan Baker > wrote:
> >> >> >> >
> >> >> >> > <snip>
> >> >> >> >
> >> >> >> >>> >> Pity what was actually being discussed was the effect of the
> >> >> >> >>> >> stationary car HAVING THE BRAKES ON AT THE TIME OF
> >> >> >> >>> >> THE COLLISION ON THE DISTANCE IT MOVES FORWARD
> >> >> >> >>> >> AND WHETHER THAT CAN AVOID RUNNING INTO THE
> >> >> >> >>> >> CAR IN FRONT OF THE STATIONARY CAR.
> >> >> >> >>>
> >> >> >> >>> > And it *does* move forward.
> >> >> >> >>>
> >> >> >> >>> Depends on the accident detail. If the speed of the moving
> >> >> >> >>> car is low enough, not necessarily if its weight is similar.
> >> >> >> >>>
> >> >> >> >>> > Momentum doesn't just disappear.
> >> >> >> >>>
> >> >> >> >>> Doesnt need to, most obviously if when the moving car
> >> >> >> >>> just bounces back off the stationary one, or the crumpling
> >> >> >> >>> of one or both absorbs the momentum, stupid.
> >> >> >> >>
> >> >> >> >> Bouncing back makes it *worse* for the stopped car, not better
> >> >> >> >>
> >> >> >> >> Study "Conservation of Momentum.
> >> >> >> >>
> >> >> >> >>>
> >> >> >> >>> > Having the brakes on doesn't magically make that speed go
> >> >> >> >>> > away.
> >> >> >> >>>
> >> >> >> >>> It does however increase the likelyhood of the moving car
> >> >> >> >>> bouncing
> >> >> >> >>> back or the crumple zones crumpling with lower speed collisions,
> >> >> >> >>> stupid.
> >> >> >> >>
> >> >> >> >> Do the math.
> >> >> >> >
> >> >> >> > I have to apologize for asking you to do the math; it's clearly
> >> >> >> > beyond
> >> >> >> > you.
> >> >> >> >
> >> >> >> > So you think that crumple zones will save the situation, do you?
> >> >> >> >
> >> >> >> > Let's examine that:
> >> >> >> >
> >> >> >> > Two cars, same mass, the rear travelling at 10 mph, the front one
> >> >> >> > stationary and -- let's be generous -- capable of crumpling 4 feet
> >> >> >> > (far
> >> >> >> > more than real life is likely to grant you).
> >> >> >> >
> >> >> >> > So if the rear car is to be stopped completely in 4 feet without
> >> >> >> > the
> >> >> >> > front car moving, the acceleration necessary to do so must be less
> >> >> >> > than
> >> >> >> > the locked tires can provide.
> >> >> >> >
> >> >> >> > Maximum tire deceleration: -0.8g, -25.6 feet/sec/sec
> >> >> >> >
> >> >> >> > If we assume constant acceleration (unrealistic, but any other
> >> >> >> > assumption makes for larger peak acceleration)
> >> >> >> >
> >> >> >> > v(initial): v: 10 mph, 14.667 feet/second
> >> >> >> > V(final): V: 0 feet/second
> >> >> >> > distance: d: 4 feet
> >> >> >> >
> >> >> >> > V^2 = v^2 + 2ad
> >> >> >> >
> >> >> >> > Solving:
> >> >> >> >
> >> >> >> > 0 = (14.667)^2 + 8a
> >> >> >> >
> >> >> >> > a = -215.094/8 = -26.886 feet/sec/sec
> >> >> >> >
> >> >> >> > IOW, at best, the deceleration necessary is greater than the tires
> >> >> >> > can
> >> >> >> > provide. Even with the unrealistic (and generous) assumptions of 4
> >> >> >> > feet
> >> >> >> > of crumple space and a constant decelerating force.
> >> >> >> >
> >> >> >> > The front car is going to get push along.
> >> >> >>
> >> >> >> It's conservation of energy and your unrealistic hypothetical
> >> >> >> hyperbole
> >> >> >> does not take into account absorption factors (which do not require
> >> >> >> "4
> >> >> >> feet
> >> >> >> of crumple space"). But please do carry on. It's entertaining.
> >> >> >
> >> >> > I invite you to show the the math that reaches a different
> >> >> > conclusion.
> >> >> >
> >> >> > v(f)^2 = v(i)^2 + 2ad; it tells the story.
> >> >>
> >> >> *Energy* is conserved, but kenetic energy is only conserved in an
> >> >> inelastic collision, which automobiles certainly aren't.
> >>
> >> > It doesn't *matter*.
> >>
> >> Corse it does.
> >>
> >> > We're looking at *kinematics* with this.
> >>
> >> Pig ignorant waffle.
>
> > Can you even *define* kinematics?
>
> Even you should be able to bull**** your way out of
> your predicament better than that pathetic effort, fool.
>
> >>> Under constant acceleration,
>
> >> Soorree, aint seen in the situation being discussed.
>
> > Anything else just makes the situation worse.
>
> Wrong. As always.
>
> > If the acceleration isn't constant, then any distance
> > traveled under the average must be made up for by
> > distance *over* the average acceleration; leading to
> > a higher peak than staying on the average would be.
>
> Irrelevant to what speeds and weights of the
> moving car cause the stationary car to slide, fool.
>
> >>> this equation completely describes the situation
> >>> and constant acceleration is the best case scenario
> >>> for minimizing the forces the tires must resist.
>
> >> Like hell it does when you just ignore crumpling which is what
> >> absorbs most the mometum at the higher impact speeds.
>
> >> That's why cars are now designed to crumple, fool.
>
> > The crumpling doesn't make the force any different.
>
> Thanks for that completely superfluous
> proof that you've never had a ****ing clue.
>
> > If the moving car stops in x feet, then that is going to be y force.
>
> Duh. Pity thats irrelevant to what happens to the stationary car, fool.
>
> > The minimum y can be is for constant acceleration.
> > So we can figure the minimum acceleration necessary
> > for any combination of speed and crumple space we please.
>
> Pity aint relevant to whether the stationary car will slide or not, fool.
>
> > Once we know the acceleration,
>
> We dont. Because the effect of the crumple aint
> readily quantifiable by plucking numbers out of your arse.
>
> > we can relate that to the maximum acceleration
> > which a tire can generate on asphalt
>
> Soorree, THERE AINT NO ACCELERATION WITH
> THE TIRES OF THE STATIONARY CAR, FOOL.

No. There is a *force*. A force which can be directly related to an
acceleration by noting that the force which will stop a car a certain
deceleration is going to be exactly the same as the force required to
stop another car striking yours from behind under the same rate of
deceleration.

Get it? If your car can skid to a stop, generating 0.8g (a reasonable
figure; look it up), then another car with the same mass and stopping in
a distance of a little more than 4 feet will require the same force.

>
> > and see whether or not there is going to be sufficient
> > traction to keep the stationary car stationary.
>
> Pathetic, really.

--
Alan Baker
Vancouver, British Columbia
"If you raise the ceiling 4 feet, move the fireplace from that wall
to that wall, you'll still only get the full stereophonic effect
if you sit in the bottom of that cupboard."

Alan Baker > wrote in message
...
> Rod Speed > wrote
>> Alan Baker > wrote
>>> Rod Speed > wrote
>>>> Alan Baker > wrote
>>>>> ?hadoaeraith -id> wrote:
>>>>>> Alan Baker wrote:
>>>>>>> Alan Baker > wrote:

>> >> >> >>> >> Pity what was actually being discussed was the effect of the
>> >> >> >>> >> stationary car HAVING THE BRAKES ON AT THE TIME OF
>> >> >> >>> >> THE COLLISION ON THE DISTANCE IT MOVES FORWARD
>> >> >> >>> >> AND WHETHER THAT CAN AVOID RUNNING INTO THE
>> >> >> >>> >> CAR IN FRONT OF THE STATIONARY CAR.
>> >> >> >>>
>> >> >> >>> > And it *does* move forward.
>> >> >> >>>
>> >> >> >>> Depends on the accident detail. If the speed of the moving
>> >> >> >>> car is low enough, not necessarily if its weight is similar.
>> >> >> >>>
>> >> >> >>> > Momentum doesn't just disappear.
>> >> >> >>>
>> >> >> >>> Doesnt need to, most obviously if when the moving car
>> >> >> >>> just bounces back off the stationary one, or the crumpling
>> >> >> >>> of one or both absorbs the momentum, stupid.
>> >> >> >>
>> >> >> >> Bouncing back makes it *worse* for the stopped car, not better
>> >> >> >>
>> >> >> >> Study "Conservation of Momentum.
>> >> >> >>
>> >> >> >>>
>> >> >> >>> > Having the brakes on doesn't magically make that speed go away.
>> >> >> >>>
>> >> >> >>> It does however increase the likelyhood of the moving car bouncing
>> >> >> >>> back or the crumple zones crumpling with lower speed collisions,
>> >> >> >>> stupid.
>> >> >> >>
>> >> >> >> Do the math.
>> >> >> >
>> >> >> > I have to apologize for asking you to do the math; it's clearly
>> >> >> > beyond
>> >> >> > you.
>> >> >> >
>> >> >> > So you think that crumple zones will save the situation, do you?
>> >> >> >
>> >> >> > Let's examine that:
>> >> >> >
>> >> >> > Two cars, same mass, the rear travelling at 10 mph, the front one
>> >> >> > stationary and -- let's be generous -- capable of crumpling 4 feet
>> >> >> > (far
>> >> >> > more than real life is likely to grant you).
>> >> >> >
>> >> >> > So if the rear car is to be stopped completely in 4 feet without the
>> >> >> > front car moving, the acceleration necessary to do so must be less
>> >> >> > than
>> >> >> > the locked tires can provide.
>> >> >> >
>> >> >> > Maximum tire deceleration: -0.8g, -25.6 feet/sec/sec
>> >> >> >
>> >> >> > If we assume constant acceleration (unrealistic, but any other
>> >> >> > assumption makes for larger peak acceleration)
>> >> >> >
>> >> >> > v(initial): v: 10 mph, 14.667 feet/second
>> >> >> > V(final): V: 0 feet/second
>> >> >> > distance: d: 4 feet
>> >> >> >
>> >> >> > V^2 = v^2 + 2ad
>> >> >> >
>> >> >> > Solving:
>> >> >> >
>> >> >> > 0 = (14.667)^2 + 8a
>> >> >> >
>> >> >> > a = -215.094/8 = -26.886 feet/sec/sec
>> >> >> >
>> >> >> > IOW, at best, the deceleration necessary is greater than the tires
>> >> >> > can
>> >> >> > provide. Even with the unrealistic (and generous) assumptions of 4
>> >> >> > feet
>> >> >> > of crumple space and a constant decelerating force.
>> >> >> >
>> >> >> > The front car is going to get push along.
>> >> >>
>> >> >> It's conservation of energy and your unrealistic hypothetical hyperbole
>> >> >> does not take into account absorption factors (which do not require "4
>> >> >> feet
>> >> >> of crumple space"). But please do carry on. It's entertaining.
>> >>
>> >> > I invite you to show the the math that reaches a different conclusion.
>> >>
>> >> You dont need any math for it to be obvious to anyone with a
>> >> clue that crumpling will dissipate some of the momentum with
>> >> the higher collision speeds, and anyone but a pig ignorant
>> >> fool realises that it isnt practical to quantify what momentum
>> >> is dissipated in crumpling, because that varys so much from
>> >> car to car and collision to collision.
>>
>> > The forces involved don't just disappear, sunshine.
>>
>> Never said they did, arseshine.
>>
>> > When the moving car hits the rear of the
>> > stationary one, it exerts a *force* on it.
>>
>> Duh.
>>
>> > According to Newton, the stationary car exerts a force of equal
>> > magnituded but in the opposite direction on the moving car.
>>
>> Duh.
>>
>> > It can only remain stationary if there is some other
>> > force exerted on it; in this case by the tires.

>> Pity about the crumpling you are ignoring, arseshine.

> I'm not ignoring it.

Corse you are with your silly 'math'

> Do you think the car crumples under no force at all?

Nope, but your silly 'math' clearly doesnt include that.

> If it did, where would the force that stops the moving car come from?

Pathetic, really.

>>> But if the force that must be opposed is
>>> greater than what the tires can supply,

>> You aint established that it is with the 10mph you waved around.

> Go ahead and explain where my analysis is in error.

Been there, done that. You just lie about that and fool absolutely no one at
all.

> Use actual figures

Not even feasible with crumple, ****wit.

> and the appropriate math.

> Get your teacher to help...

Even you should be able to bull**** your way out of your
predicament better than that pathetic effort, pig ignorant ******.

>>> then the car's going to move, isn't it?

>> Corse it is, but you aint established at what speed that occurs
>> no matter how desperately you wanked with irrelevant numbers
>> and its completely irrelevant to what was actually being
>> discussed, WHETHER IS BETTER TO STOP WITH THE BRAKES
>> ON THAN NOT AS FAR AS GETTING RAMMED INTO THE CAR
>> IN FRONT OF THE STATIONARY CAR IS CONCERNED, you ****wit.

>>> By knowing how much distance the car is stopping in,

>> You dont know that.

> We *defined* it for that problem.

Even you should be able to bull**** your way out of your
predicament better than that pathetic effort, pig ignorant ******.

> Change the distance for the less and
> you change the forces for the greater.

Duh.

>>> we can calculate the accelerations involved, from the acceleration
>>> and the mass of the vehicle, we can get the force involved.

>> Duh.

>>> But also from the mass of the vehicle, along with the acceleration
>>> due to gravity and the coefficient of friction for rubber on asphalt,
>>> we can reduce the whole thing to talking about accelerations.

>> No you cant when you dont know how much
>> of the momentum is absorbed by crumpling.

> It can't change the fact that the car has to be decelerated

Duh.

> and that there must be a force exerted on the stationary
> car equal to the force that it is exerting on the moving car.

Duh.

> That force is then transfer to the ground by the tires.

Wrong. Some of it gets dissipated in the crumpling, ****wit.

> It isn't lessened by the crumpling; *all* of it must be reacted through the
> tires.

Thanks for that completely superfluous proof that you've never had a ****ing
clue.

>>>> WHAT MATTERS IS THAT YOU ARE BETTER OFF WITH THE
>>>> BRAKES APPLIED IN THE STATIONARY VEHICLE AS FAR AS
>>>> THE RISK OF BEING RAMMED INTO THE CAR IN FRONT OF
>>>> THE STATIONARY VEHICLE IS CONCERNED, REGARDLESS
>>>> OF WHETHER THE STATIONARY VEHICLE IS MOVED OR NOT.

>>> Nobody's disputing that completely obvious point.

>> You've desperately wanking about complete irrelevancys then.

> And yet you can't actually present any math to dispute me.

Dont need any math. Anyone but a pig ignorant fool
realises that quite a bit of kinetic energy is dissipated
in the crumpling at the higher collision speeds, ****wit.

Even someone as stupid as you should be able to grasp that
with an immovable barrier being crashed into, ALMOST ALL
the kinetic energy gets dissipated in the crumpling with a
head on collision, if someone was actually stupid enough
to lend you a seeing eye dog and a white cane.

>>>> None of your desperate masturbating with useless numbers
>>>> plucked out of your stupid pig ignorant arse are gunna change that.

>>>>> v(f)^2 = v(i)^2 + 2ad; it tells the story.

>>>> Wrong. As always.

>> > Show *how* I'm wrong, bright boy. Or is that beyond you?

>> Done that on the FACT that that does NOT 'tell the story', fool.

> It tells the story for the best case you can get.

Wrong. It IGNORES THE KINETIC ENERGY DISSIPATED IN
THE CRUMPLING AT THE HIGHER SPEEDS, ****WIT.

> Any other deceleration scenario results in a higher peak value.

Thanks for that completely superfluous
proof that you've never had a ****ing clue.

>>> Bottom line: that is the formula that relates final velocity with
>>> initial velocity, acceleration and distance for constant acceleration.

>> Pity you dont know the distance.

> It was defined

Nope, you just plucked that stupid number out of your arse.

> and defined generously. No real car is going to crumple 4 feet.

Duh. Pity that number is completely useless on the kinetic
energy that gets used by the crumpling, ****wit.

>>> Anything other than constant acceleration is going to be worse
>>> for the vehicle trying to stay still, because the distance traveled
>>> at less than what constant would have been will have to be made
>>> up for by distance traveled at *more* than constant, and all we
>>> need to know is whether the acceleration necessary to stop the
>>> vehicle is more or less than that that can be provided by the tires.

>> And you dont know that when you cant quantify the crumpling, fool.

> I don't have to.

Corse you do.

>>>> You must actually be that thick. Its too convincing to be an act/troll.

>>> And yet you're the one who can't do the math...

>> Lying again. I just keep rubbing your silly little nose in
>> the FACT that you have just desperately wanked with
>> numbers that you have plucked out of your arse that
>> arent even relevant to the situation being discussed.

>> So thick that you cant even grasp that your stupid numbers
>> are completely irrelevant to the situation being discussed.

> So present some that are relevant.

Not even feasible with crumpling, ****wit.

> Show us all your superior understanding.

Been there, done that. Everyone is ****ing
themselves laughing at your predicament, ****wit.

In article >,
"Rod Speed" > wrote:

> > So present some that are relevant.
>
> Not even feasible with crumpling, ****wit.

You keep hiding behind "crumpling".

It doesn't matter whether it is crumpling or springs or cheetos.

If a vehicle is going to stop from "v" speed in "d" distance, then the
formula for constant acceleration, "a" (the one which provides the
lowest possible acceleration) is:

a = -(v^2)/2d

Period.

And for two vehicles of equal mass, that means that the force that the
stopped vehicle experiences (and which must be resisted by the tires, or
else the freebody diagram is unbalanced and there *will* be movement) is
the same as the force that the tires would provide decelerating that car
at the same rate.

If a is larger than the maximum traction available, then the stationary
car *will* move.

Period.

--
Alan Baker
Vancouver, British Columbia
"If you raise the ceiling 4 feet, move the fireplace from that wall
to that wall, you'll still only get the full stereophonic effect
if you sit in the bottom of that cupboard."

"Alan Baker" > wrote in message
...
> In article
> >,
> "Rod Speed" > wrote:
>
>> "Alan Baker" > wrote in message
>> ...
>> > In article >,
>> > "Rod Speed" > wrote:
>> >
>> >> "Alan Baker" > wrote in message
>> >> ...
>> >> > In article >,
>> >> > keith > wrote:
>> >> >
>> >> >> On Tue, 26 Apr 2005 01:49:39 +0000, Alan Baker wrote:
>> >> >>
>> >> >> > In article >,
>> >> >> > ?hadoaeraith -id> wrote:
>> >> >> >
>> >> >> >> Alan Baker wrote:
>> >> >> >>
>> >> >> >> > In article >,
>> >> >> >> > Alan Baker > wrote:
>> >> >> >> >
>> >> >> >> > <snip>
>> >> >> >> >
>> >> >> >> >>> >> Pity what was actually being discussed was the effect of the
>> >> >> >> >>> >> stationary car HAVING THE BRAKES ON AT THE TIME OF
>> >> >> >> >>> >> THE COLLISION ON THE DISTANCE IT MOVES FORWARD
>> >> >> >> >>> >> AND WHETHER THAT CAN AVOID RUNNING INTO THE
>> >> >> >> >>> >> CAR IN FRONT OF THE STATIONARY CAR.
>> >> >> >> >>>
>> >> >> >> >>> > And it *does* move forward.
>> >> >> >> >>>
>> >> >> >> >>> Depends on the accident detail. If the speed of the moving
>> >> >> >> >>> car is low enough, not necessarily if its weight is similar.
>> >> >> >> >>>
>> >> >> >> >>> > Momentum doesn't just disappear.
>> >> >> >> >>>
>> >> >> >> >>> Doesnt need to, most obviously if when the moving car
>> >> >> >> >>> just bounces back off the stationary one, or the crumpling
>> >> >> >> >>> of one or both absorbs the momentum, stupid.
>> >> >> >> >>
>> >> >> >> >> Bouncing back makes it *worse* for the stopped car, not better
>> >> >> >> >>
>> >> >> >> >> Study "Conservation of Momentum.
>> >> >> >> >>
>> >> >> >> >>>
>> >> >> >> >>> > Having the brakes on doesn't magically make that speed go
>> >> >> >> >>> > away.
>> >> >> >> >>>
>> >> >> >> >>> It does however increase the likelyhood of the moving car
>> >> >> >> >>> bouncing
>> >> >> >> >>> back or the crumple zones crumpling with lower speed collisions,
>> >> >> >> >>> stupid.
>> >> >> >> >>
>> >> >> >> >> Do the math.
>> >> >> >> >
>> >> >> >> > I have to apologize for asking you to do the math; it's clearly
>> >> >> >> > beyond
>> >> >> >> > you.
>> >> >> >> >
>> >> >> >> > So you think that crumple zones will save the situation, do you?
>> >> >> >> >
>> >> >> >> > Let's examine that:
>> >> >> >> >
>> >> >> >> > Two cars, same mass, the rear travelling at 10 mph, the front one
>> >> >> >> > stationary and -- let's be generous -- capable of crumpling 4 feet
>> >> >> >> > (far
>> >> >> >> > more than real life is likely to grant you).
>> >> >> >> >
>> >> >> >> > So if the rear car is to be stopped completely in 4 feet without
>> >> >> >> > the
>> >> >> >> > front car moving, the acceleration necessary to do so must be less
>> >> >> >> > than
>> >> >> >> > the locked tires can provide.
>> >> >> >> >
>> >> >> >> > Maximum tire deceleration: -0.8g, -25.6 feet/sec/sec
>> >> >> >> >
>> >> >> >> > If we assume constant acceleration (unrealistic, but any other
>> >> >> >> > assumption makes for larger peak acceleration)
>> >> >> >> >
>> >> >> >> > v(initial): v: 10 mph, 14.667 feet/second
>> >> >> >> > V(final): V: 0 feet/second
>> >> >> >> > distance: d: 4 feet
>> >> >> >> >
>> >> >> >> > V^2 = v^2 + 2ad
>> >> >> >> >
>> >> >> >> > Solving:
>> >> >> >> >
>> >> >> >> > 0 = (14.667)^2 + 8a
>> >> >> >> >
>> >> >> >> > a = -215.094/8 = -26.886 feet/sec/sec
>> >> >> >> >
>> >> >> >> > IOW, at best, the deceleration necessary is greater than the tires
>> >> >> >> > can
>> >> >> >> > provide. Even with the unrealistic (and generous) assumptions of 4
>> >> >> >> > feet
>> >> >> >> > of crumple space and a constant decelerating force.
>> >> >> >> >
>> >> >> >> > The front car is going to get push along.
>> >> >> >>
>> >> >> >> It's conservation of energy and your unrealistic hypothetical
>> >> >> >> hyperbole
>> >> >> >> does not take into account absorption factors (which do not require
>> >> >> >> "4
>> >> >> >> feet
>> >> >> >> of crumple space"). But please do carry on. It's entertaining.
>> >> >> >
>> >> >> > I invite you to show the the math that reaches a different
>> >> >> > conclusion.
>> >> >> >
>> >> >> > v(f)^2 = v(i)^2 + 2ad; it tells the story.
>> >> >>
>> >> >> *Energy* is conserved, but kenetic energy is only conserved in an
>> >> >> inelastic collision, which automobiles certainly aren't.
>> >>
>> >> > It doesn't *matter*.
>> >>
>> >> Corse it does.
>> >>
>> >> > We're looking at *kinematics* with this.
>> >>
>> >> Pig ignorant waffle.
>>
>> > Can you even *define* kinematics?
>>
>> Even you should be able to bull**** your way out of
>> your predicament better than that pathetic effort, fool.
>>
>> >>> Under constant acceleration,
>>
>> >> Soorree, aint seen in the situation being discussed.
>>
>> > Anything else just makes the situation worse.
>>
>> Wrong. As always.
>>
>> > If the acceleration isn't constant, then any distance
>> > traveled under the average must be made up for by
>> > distance *over* the average acceleration; leading to
>> > a higher peak than staying on the average would be.
>>
>> Irrelevant to what speeds and weights of the
>> moving car cause the stationary car to slide, fool.
>>
>> >>> this equation completely describes the situation
>> >>> and constant acceleration is the best case scenario
>> >>> for minimizing the forces the tires must resist.
>>
>> >> Like hell it does when you just ignore crumpling which is what
>> >> absorbs most the mometum at the higher impact speeds.
>>
>> >> That's why cars are now designed to crumple, fool.
>>
>> > The crumpling doesn't make the force any different.
>>
>> Thanks for that completely superfluous
>> proof that you've never had a ****ing clue.
>>
>> > If the moving car stops in x feet, then that is going to be y force.
>>
>> Duh. Pity thats irrelevant to what happens to the stationary car, fool.
>>
>> > The minimum y can be is for constant acceleration.
>> > So we can figure the minimum acceleration necessary
>> > for any combination of speed and crumple space we please.
>>
>> Pity aint relevant to whether the stationary car will slide or not, fool.
>>
>> > Once we know the acceleration,
>>
>> We dont. Because the effect of the crumple aint
>> readily quantifiable by plucking numbers out of your arse.
>>
>> > we can relate that to the maximum acceleration
>> > which a tire can generate on asphalt
>>
>> Soorree, THERE AINT NO ACCELERATION WITH
>> THE TIRES OF THE STATIONARY CAR, FOOL.

> No. There is a *force*. A force which can be directly related
> to an acceleration by noting that the force which will stop a
> car a certain deceleration is going to be exactly the same
> as the force required to stop another car striking yours
> from behind under the same rate of deceleration.

Wrong. Take an example of the stationary vehicle being a fully
loaded semi with the brakes on, and the moving vehicle being
a light car. Even with quite high speeds of the moving car,
THE SEMI WONT BE MOVED FORWARD AT ALL.

ALL THE KINETIC ENERGY OF THE CAR WILL BE DISSIPATED
IN THE CRUMPING OF THE CAR AROUND THE BACK OF THE SEMI.

THERE WILL BE NO ACCELERATION OF THE SEMI AT ALL.

NONE, ZERO, ZILTCH.

And with a lighter stationary vehicle, at some speed the stationary
vehicle will be slid along the road, AND IT DOESNT ACTUALLY
MATTER A **** AT WHAT SPEED OF THE MOVING VEHICLE
THAT HAPPENS WITH THE QUESTION BEING DISCUSSED,
WHETHER ITS BETTER TO HAVE THE BRAKES ON WHEN STOPPED.

> Get it?

Nothing to 'get', ****wit.

> If your car can skid to a stop, generating 0.8g (a reasonable figure;
> look it up), then another car with the same mass and stopping in
> a distance of a little more than 4 feet will require the same force.

Pathetic, really.

ITS COMPLETELY IRRELEVANT TO WHAT MATTERS,
WHETHER THERE WILL BE ENOUGH KINETIC ENERGY
LEFT AFTER THE CRUMPLING, IF ANY, TO OVERCOME THE
RESISTANCE OF THE TIRES ON THE ROAD TO DRAGGING.

>>> and see whether or not there is going to be sufficient
>>> traction to keep the stationary car stationary.

>> Pathetic, really.

He's right, as always.

Alan Baker > wrote in message
...
> Rod Speed > wrote

>>> this equation completely describes the situation
>>> and constant acceleration is the best case scenario
>>> for minimizing the forces the tires must resist.

>> Like hell it does when you just ignore crumpling which is what
>> absorbs most the mometum at the higher impact speeds.

>> That's why cars are now designed to crumple, fool.

> I doubt you can learn anything, but I'll try anyway:

Even you should be able to bull**** your way out of your
predicament better than that pathetic effort, clown.

> <http://www.skytran.net/09Safety/10sfty.htm>

> Note the simple relationship between speed,
> crumple distance and acceleration.

COMPLETELY IRRELEVANT TO WHETHER THERE IS ENOUGH
KINETIC ENERGY LEFT AFTER THE CRUMPLING TO OVERCOME THE
DRAG OF THE TIRES OF THE STATIONARY VEHICLE ON THE ROAD.

Reams and reams of desperate wanking with completely
irrelevant numbers flushed where they belong.

In article >,
?hadoaeraith -id> wrote:

> Alan Baker wrote:
>
> > In article >,
> > ?hadoaeraith -id> wrote:
> >
> >> Alan Baker wrote:
> >>
> >>> In article >,
> >>> Alan Baker > wrote:
> >>>
> >>> <snip>
> >>>
> >>>>> >> Pity what was actually being discussed was the effect of the
> >>>>> >> stationary car HAVING THE BRAKES ON AT THE TIME OF
> >>>>> >> THE COLLISION ON THE DISTANCE IT MOVES FORWARD
> >>>>> >> AND WHETHER THAT CAN AVOID RUNNING INTO THE
> >>>>> >> CAR IN FRONT OF THE STATIONARY CAR.
> >>>>>
> >>>>> > And it *does* move forward.
> >>>>>
> >>>>> Depends on the accident detail. If the speed of the moving
> >>>>> car is low enough, not necessarily if its weight is similar.
> >>>>>
> >>>>> > Momentum doesn't just disappear.
> >>>>>
> >>>>> Doesnt need to, most obviously if when the moving car
> >>>>> just bounces back off the stationary one, or the crumpling
> >>>>> of one or both absorbs the momentum, stupid.
> >>>>
> >>>> Bouncing back makes it *worse* for the stopped car, not better
> >>>>
> >>>> Study "Conservation of Momentum.
> >>>>
> >>>>>
> >>>>> > Having the brakes on doesn't magically make that speed go away.
> >>>>>
> >>>>> It does however increase the likelyhood of the moving car bouncing
> >>>>> back or the crumple zones crumpling with lower speed collisions, stupid.
> >>>>
> >>>> Do the math.
> >>>
> >>> I have to apologize for asking you to do the math; it's clearly beyond
> >>> you.
> >>>
> >>> So you think that crumple zones will save the situation, do you?
> >>>
> >>> Let's examine that:
> >>>
> >>> Two cars, same mass, the rear travelling at 10 mph, the front one
> >>> stationary and -- let's be generous -- capable of crumpling 4 feet (far
> >>> more than real life is likely to grant you).
> >>>
> >>> So if the rear car is to be stopped completely in 4 feet without the
> >>> front car moving, the acceleration necessary to do so must be less than
> >>> the locked tires can provide.
> >>>
> >>> Maximum tire deceleration: -0.8g, -25.6 feet/sec/sec
> >>>
> >>> If we assume constant acceleration (unrealistic, but any other
> >>> assumption makes for larger peak acceleration)
> >>>
> >>> v(initial): v: 10 mph, 14.667 feet/second
> >>> V(final): V: 0 feet/second
> >>> distance: d: 4 feet
> >>>
> >>> V^2 = v^2 + 2ad
> >>>
> >>> Solving:
> >>>
> >>> 0 = (14.667)^2 + 8a
> >>>
> >>> a = -215.094/8 = -26.886 feet/sec/sec
> >>>
> >>> IOW, at best, the deceleration necessary is greater than the tires can
> >>> provide. Even with the unrealistic (and generous) assumptions of 4 feet
> >>> of crumple space and a constant decelerating force.
> >>>
> >>> The front car is going to get push along.
> >>
> >> It's conservation of energy and your unrealistic hypothetical hyperbole
> >> does not take into account absorption factors (which do not require "4 feet
> >> of crumple space"). But please do carry on. It's entertaining.
> >
> > I invite you to show the the math that reaches a different conclusion.
> >
> > v(f)^2 = v(i)^2 + 2ad; it tells the story.
>
> Your asinine equation implies a VW bug hitting a stationary VW bug will
> displace the stationary VW bug the same exact distance as an 18-wheeler
> hitting a stationary 18-wheeler. Put another way, if you must engage in
> public mathturbation please set up your problem set correctly. Here's a
> hint:

No, it doesn't. Because I said up front that the calculations were for
two vehicles of equal mass.

See:

> >>> Two cars, same mass, the rear travelling at 10 mph, the front one

Don't you feel stupid now...

>
> c1 c2
> +-(x)->(K)+-(xp)->
> | |
> W W
>
> a = ???v/???t
> v = ???x/???t
> x = x(0) + v(0)t + ??at^2
> K = ??mv^2
> W = mg
>
> Where a = acceleration, v = velocity, t = time, x = displacement, xp =
> projected impact displacement, m = mass, g = gravity, W = weight, and K =
> kinetic energy. Furthermore, when computing projected impact displacement,
> you must also take into account angle of impact, crush factor (which is
> dependent upon car make/model/year and speed at impact), road conditions,
> and tire friction (which in and of itself is dependent upon vehicle weight,
> center of gravity, tire width, and tire tread). These factored together
> provide a means to determine approximate force distribution between c1 &
> c2, resulting in projected displacement. I'm sure I left a few things out
> which you'll no doubt correct. Should you so choose, please do get some
> fresh air first, m-kay?

Certainly where the vehicle masses differ one must do a more detailed
calculation, but I was deliberately keeping it simple to illustrate the
point:

Even at moderate speeds (10 mph) with generous crumple distance to keep
acceleration low (4 feet) if two cars of equal mass collide, the
stationary car is going to get moved.

--
Alan Baker
Vancouver, British Columbia
"If you raise the ceiling 4 feet, move the fireplace from that wall
to that wall, you'll still only get the full stereophonic effect
if you sit in the bottom of that cupboard."

In article
>,
"Rod Speed" > wrote:

> Alan Baker > wrote in message
> ...
> > Rod Speed > wrote
>
> >>> this equation completely describes the situation
> >>> and constant acceleration is the best case scenario
> >>> for minimizing the forces the tires must resist.
>
> >> Like hell it does when you just ignore crumpling which is what
> >> absorbs most the mometum at the higher impact speeds.
>
> >> That's why cars are now designed to crumple, fool.
>
> > I doubt you can learn anything, but I'll try anyway:
>
> Even you should be able to bull**** your way out of your
> predicament better than that pathetic effort, clown.
>
> > <http://www.skytran.net/09Safety/10sfty.htm>
>
> > Note the simple relationship between speed,
> > crumple distance and acceleration.
>
> COMPLETELY IRRELEVANT TO WHETHER THERE IS ENOUGH
> KINETIC ENERGY LEFT AFTER THE CRUMPLING TO OVERCOME THE
> DRAG OF THE TIRES OF THE STATIONARY VEHICLE ON THE ROAD.
>
> Reams and reams of desperate wanking with completely
> irrelevant numbers flushed where they belong.

Sigh. A thought experiment.

Imagine a completely rigid car with a separate crumple box on the back
of it. For the sake of argument, this box is designed so that it
decelerates a vehicle that strikes it at a constant rate until it is
completely crushed. The box can crush precisely 4 feet before it is
completely solid and will crush no more.

Now: When the moving car strikes the rear of the box, it exerts a force
on the box, right? And one can calculate that force by knowing the
acceleration of the vehicle and its mass, correct? That same force that
the moving car exerts on the box is transfer

--
Alan Baker
Vancouver, British Columbia
"If you raise the ceiling 4 feet, move the fireplace from that wall
to that wall, you'll still only get the full stereophonic effect
if you sit in the bottom of that cupboard."

Alan Baker > wrote in message
...
> Rod Speed > wrote

>>> So present some that are relevant.

>> Not even feasible with crumpling, ****wit.

> You keep hiding behind "crumpling".

You keep desperately attempting to bull**** your way out of
your predicament and are clearly fooling absolutely no one at all.

> It doesn't matter whether it is crumpling or springs or cheetos.

Wrong with the kinetic energy dissipated with the higher speed collisions.

> If a vehicle is going to stop from "v" speed in "d" distance,
> then the formula for constant acceleration, "a" (the one
> which provides the lowest possible acceleration) is:

> a = -(v^2)/2d

> Period.

All completely and utter irrelevant to how much of the
kinetic energy gets dissipated in crumpling, ****wit.

Clearly ALMOST ALL OF IT IS WITH AN IMMOVABLE OBJECT.

> And for two vehicles of equal mass, that means
> that the force that the stopped vehicle experiences
> (and which must be resisted by the tires,

Wrong when much of its dissipated by the
crumpling with higher speed collisions, ****wit.

> or else the freebody diagram is
> unbalanced and there *will* be movement)

Thanks for that completely superfluous proof
that you've never ever had a ****ing clue.

> is the same as the force that the tires would
> provide decelerating that car at the same rate.

Thanks for that completely superfluous proof
that you've never ever had a ****ing clue.

The resistance to the car being slid across the road
has NOTHING to do with how effective the brakes
are as slowing the car down when its moving.

> If a is larger than the maximum traction
> available, then the stationary car *will* move.

Duh.

> Period.

Keep desperately digging that hole you are in, ****wit.

You'll be out in china any day now.

In article >,
Garth Almgren > wrote:
>Around 4/25/2005 9:10 AM, Matthew Russotto wrote:
>
>> I just got a ticket in Philadelphia. Well, actually, drove away from
>> it while the revenuer was writing it up. Where the hell was I supposed to
>> park to unload 50" x 40" artwork, when all the nearby spots fit into
>> the above categories?
>
>How about the loading zones you mentioned, or is unloading in a loading
>zone also prohibited? Wouldn't surprise me if that were the case...

One loading zone was blocked by construction, the other by someone
parked in it. It was Saturday; I believe the latter loading zone was
actually legal to park.

I have gotten a ticket for unloading in a loading zone before, though
it wasn't in Philadelphia. Fought it in court, judge called me a liar
in so many words.
--
There's no such thing as a free lunch, but certain accounting practices can
result in a fully-depreciated one.