On Wednesday, July 16, 2014 1:48:38 PM UTC-4, Ed Pawlowski wrote:
That's for sure. I wonder if anything else was going on, like
the passenger giving BS to the cop. It still would not justify
giving him a DUI.
I have problems with cops arresting people for DUI under those
circumstances too. If I were the cop, and I found someone safely
pulled over, sleeping in the car, in a reasonable place, I'd give
them a pass. If they fell asleep at a light, or were parked somewhere
they clearly did not belong, that would be different. I think it's
better to send a message that if you're drunk and make a mistake, but
realize it, you have a safe way out. If you know you're going to get
a DUI for going to a rest stop, it really leaves the best option for
everyone off the table.
Not DUI. but care and control while under the influence.
If you are going to "sleep it off" in your car, make sure you hide
the key OUTSIDE of the vehicle. A friend's wife was sitting in the car
after an accident and was charged with "care and control". If she had
given someone else thekeys, or stayed out of the car, they would not
have been able to charge her and make it stick. No proof she was the
driver, and no proof she had been drinking before the accident.
On Wednesday, July 16, 2014 5:44:59 PM UTC-4, email@example.com wrote:
One has to wonder how you can be charged with "care and control while
under the influence" with no proof that you had been drinking before
the accident. Did she have a bottle with her and claim that she had
been drinking *after* the accident?
I haven't read all 98 posts, but this doesn't seem right. If G
increases, that is, a steeper hill, d increases, and that is fine.
But if f increases, the formula shows that d also increases, and that
seems clearly wrong. The higher the coefficient of friction, that is,
the more friction there is between the road and the tires**, the sooner
the truck stops. A very low f would be like melting ice. A very high
f would be like a road made of big rocks, maybe.
At any rate f and G are parellel in the formula, even though high values
of each have the opposite effect. Maybe there should be a negative
**And what about between the brake pads and the rotors? Or are we
assuming the wheels are locked up? I didn't think that was the case.
On Saturday, September 6, 2014 6:12:24 PM UTC-4, micky wrote:
You also must have been MIA, this is an old thread.
Actually if either G or f increase, then d will *decrease*. Both G and f
are in the denominator.
The higher the coefficient of friction, that is,
I think there is a negative sign, depending on whether the truck is going up
the grade or down it. For the formula to be correct, it would appear that
a positive grade is one where the truck is going uphill. If the grade is
negative, ie truck going downhill, then when the grade cancels out the friction,
the stopping distance becomes infinity. Beyond that, the formula doesn't work.
If the downhill grade exceeds the friction, it would give a negative stopping
distance, but that isn't right, the truck would never stop, so it doesn't work
for neg denominators.
I would think the formula is based on the wheels being locked up. How
close it is to reality, even then, IDK.
You are correct. I pointed that out in one of the other posts in this
thread. I don't blame you for not reading them all.
The use of the formula relies on the user understanding that the
correct sign must be used for GRADE.
The formula only deals with the friction, f, at the tire/road
interface. The formula can be used to solve for any of it's variables
based on knowing the other variables and plugging them in to get a
solution. It also needs to be kept in mind that the way it's often
used, and how its being used here, one simply makes an assumption that
is said to be the f for the road. That's a bit simplistic because
there is no "f' for the road. f only has meaning relative to what
ever material is the other half of the interface. In this case it's a
tire. And different tires can produce different f's, sometimes
significantly different. Also, as you mention, there is the issue of
whether the brakes are locked. If they are locked and the tires
skidding you have a different f value then what you would get/use if
the tires were not skidding. F also varies with the speed of the
things moving against each other at the interface, so the f value is
an average of all the f's as the speed goes from the starting speed
down to zero.
Or are we
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