slow drivers

brat

Gilbert2 wrote: »

What?

It is not what suites them at all, you wouldn't drive like that in a test and give that reason because you would get a fail.

They have passed the test, their skill level is at a minimum standard at the very least.

And that means to not drive without due care, which is what a slow driver is doing if causing problems for other road users.

If you are the average driver, you would probably not pass the current driving test, so I'm not sure why you want to use the test pass standard as the benchmark for careless or inconsiderate driving.

If a driver wants to do 40 on an NSL road, then let him. He's perfectly entitled. Get past when it's safe, or don't. That's up to you. There are not many roads in the country where such a motorist would hold you up for more than a couple of minutes, so it's not worth twisting your knickers over.

RichardD1970

Joe_Horner wrote: »

No, but can you prove that 1+1=2? (Hint: No you can't - but that doesn't stop it being obviously true)

Yes you can,

The proof starts from the Peano Postulates, which define the natural numbers N. N is the smallest set satisfying these postulates:

P1. 1 is in N.
P2. If x is in N, then its "successor" x' is in N.
P3. There is no x such that x' = 1.
P4. If x isn't 1, then there is a y in N such that y' = x.
P5. If S is a subset of N, 1 is in S, and the implication
(x in S => x' in S) holds, then S = N.

Then you have to define addition recursively:
Def: Let a and b be in N. If b = 1, then define a + b = a'
(using P1 and P2). If b isn't 1, then let c' = b, with c in N
(using P4), and define a + b = (a + c)'.

Then you have to define 2:
Def: 2 = 1'

2 is in N by P1, P2, and the definition of 2.

Theorem: 1 + 1 = 2

Proof: Use the first part of the definition of + with a = b = 1.
Then 1 + 1 = 1' = 2 Q.E.D.

Note: There is an alternate formulation of the Peano Postulates which replaces 1 with 0 in P1, P3, P4, and P5. Then you have to change the definition of addition to this:
Def: Let a and b be in N. If b = 0, then define a + b = a.
If b isn't 0, then let c' = b, with c in N, and define
a + b = (a + c)'.

You also have to define 1 = 0', and 2 = 1'. Then the proof of the
Theorem above is a little different:

Proof: Use the second part of the definition of + first:
1 + 1 = (1 + 0)'
Now use the first part of the definition of + on the sum in
parentheses: 1 + 1 = (1)' = 1' = 2 Q.E.D.

:rotfl:

ILW

RichardD1970 wrote: »

Yes you can,

The proof starts from the Peano Postulates, which define the natural numbers N. N is the smallest set satisfying these postulates:

P1. 1 is in N.
P2. If x is in N, then its "successor" x' is in N.
P3. There is no x such that x' = 1.
P4. If x isn't 1, then there is a y in N such that y' = x.
P5. If S is a subset of N, 1 is in S, and the implication
(x in S => x' in S) holds, then S = N.

Then you have to define addition recursively:
Def: Let a and b be in N. If b = 1, then define a + b = a'
(using P1 and P2). If b isn't 1, then let c' = b, with c in N
(using P4), and define a + b = (a + c)'.

Then you have to define 2:
Def: 2 = 1'

2 is in N by P1, P2, and the definition of 2.

Theorem: 1 + 1 = 2

Proof: Use the first part of the definition of + with a = b = 1.
Then 1 + 1 = 1' = 2 Q.E.D.

Note: There is an alternate formulation of the Peano Postulates which replaces 1 with 0 in P1, P3, P4, and P5. Then you have to change the definition of addition to this:
Def: Let a and b be in N. If b = 0, then define a + b = a.
If b isn't 0, then let c' = b, with c in N, and define
a + b = (a + c)'.

You also have to define 1 = 0', and 2 = 1'. Then the proof of the
Theorem above is a little different:

Proof: Use the second part of the definition of + first:
1 + 1 = (1 + 0)'
Now use the first part of the definition of + on the sum in
parentheses: 1 + 1 = (1)' = 1' = 2 Q.E.D.

:rotfl:

That does assume that the first and second number 1 are of equal value.

Ich_2

There's no court in the land that's going to convict soneone doing 40 in a 60 NSL for inconsiderate driving!

Though I do seem to recall a driver caught doing 65 in lane 3 of the M3 was convicted of Driving without due care and attention.
I do know of JCB drivers threatened with prosecution for not pulling over to let a line of cars pass them.

Can't post any proof of these though

ILW

Ich wrote: »

Though I do seem to recall a driver caught doing 65 in lane 3 of the M3 was convicted of Driving without due care and attention.
I do know of JCB drivers threatened with prosecution for not pulling over to let a line of cars pass them.

Can't post any proof of these though

That would be about driving in the wrong lane rather than speed.
Lanes 2 and 3 are for overtaking.

brat

Gilbert2 wrote: »

Can you prove this statement?

No HGV driver would be convicted of doing 40mph in the S/C NSL, because that's the max speed, and he's far more of an obstruction on the road than Mrs Miggins in her Honda Jazz.

So if a court won't convict an HGV driver for obstructing the road while doing the legal max of 40mph, there is no case to offer against a car driver doing the same speed.

So, proved!

martinthebandit

brat wrote: »

If a driver wants to do 40 on an NSL road, then let him. He's perfectly entitled. Get past when it's safe, or don't. That's up to you. There are not many roads in the country where such a motorist would hold you up for more than a couple of minutes, so it's not worth twisting your knickers over.

I don't think anyone is arguing that he is not entitled, rather I am asking for the reasons why they chose to be so inconsiderate to other drivers.

If I could understand why, then maybe I wouldn't consider them to be inconsiderate. IYSWIM

new_owner

About 143 accidents a year are directly caused by slow drivers, according to figures for 2009 from the Department for Transport. Quote from Telegraph

brat

Ich wrote: »

I do know of JCB drivers threatened with prosecution for not pulling over to let a line of cars pass them.

Can't post any proof of these though

Nor can I, but I have successfully reported a few (three or four) resulting in guilty pleas, with just one case requiring proving at court.

brat

new_owner wrote: »

About 143 accidents a year are directly caused by slow drivers, according to figures for 2009 from the Department for Transport. Quote from Telegraph

Given that there are something like 2 million accidents each year that's a tiny proportion. I'd love to know how the causation factors are directly attributable to driving too slowly, or if it's down to the other driver's incorrect anticipation.

Incidentally, many fatal accidents are caused at slow speed, not usually by people driving too slowly.