train crossing truck driver
in [Aussie Drivers]
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From: Daryl Walford on 11 Jun 2007 22:00 Clockmeister wrote: > "veritas" <veritas(a)coldmail.con> wrote in message > news:Gr5bi.11793$wH4.1190(a)news-server.bigpond.net.au... >> Daryl Walford wrote: >>> There are many "alternatives" that can prevent a repeat of last Tuesday's >>> crash, better roads, better management of road transport which may be as >>> simple as paying drivers a lot more so there won't be a driver shortage >>> which causes people to work excess hours and we can put more freight onto >>> trains but none of those measures will completely eliminate the >>> possibility of it happening again. >>> As I said previously even the very best drivers can have momentary lapses >>> in concentration and thats all it takes. > > It took more then a momentary lapse of concentraction by the looks of it, > more like a reckless act. > > It would only take a short concentration lapse to reduce the stopping distance enough to where he had no hope of stopping, he had a 14 tonne payload so his gross weight would be approx 34 tonne and that takes a bit to stop from 100kph. I very much doubt that we will ever know with 100% certainty exactly what caused the crash and we both may be partly correct, maybe he wasn't concentrating enough, saw the train then after realizing he had no hope of stopping he tried to beat it, either way we only only speculating. Daryl
From: John Hudson on 11 Jun 2007 23:02 "reg-john" <al(a)fddfd.com> wrote in message news:qwwai.10993$wH4.8742(a)news-server.bigpond.net.au... > 60,000 times more damage? did you proofread this post? > I've had a bit of a look round the net and it is true. The calculation is axle load to the fourth power. This means that a 40 tonne truck will do yes, 60,000 times more damage than a one ton car. This refers to bitumen roads. If a car does sfa damage and you multiply it by 60,000 then it's still sfa. What one really needs to know is what this means in dollar terms. regards, huddo
From: hoot on 12 Jun 2007 01:47 "John Hudson" <huddo(a)bigpond.net.au> wrote in message news:f1obi.12624$wH4.5379(a)news-server.bigpond.net.au... > > "reg-john" <al(a)fddfd.com> wrote in message > news:qwwai.10993$wH4.8742(a)news-server.bigpond.net.au... >> 60,000 times more damage? did you proofread this post? >> > > I've had a bit of a look round the net and it is true. The calculation is > axle load to the fourth power. This means that a 40 tonne truck will do > yes, 60,000 times more damage than a one ton car. This refers to bitumen > roads. If a car does sfa damage and you multiply it by 60,000 then it's > still sfa. What one really needs to know is what this means in dollar > terms. > regards, > huddo > John, could you tell me where to find this info. I'm having trouble understanding what you actually get (what units) when you raise the weight to the fourth power. 40^4 = 2560000 and 1^4 = 1 so the truck ends up 2.56 million times higher than the car. A 20ton truck is only 160000 time higher. But what does it actually refer to? Do i need to divide by the number of axles? Why is it raised to the foruth? How do the numbers generated translate into damage? What is the SI unit for damage? And how many tons of wight equals one unit of damage? What is damage? a pot hole? Cracking? Rippling? Does the cost of damage just refer to the cost of repair or should it take into account lost revenue caused by road interuption or extra fuel/wear and tear/time cost for driving around the damage? If damage is related to wieght, why is the truck damage not 40 times the car damage? If a car does ten dollars worth of damage does a truck really do six hundred thousand dollars worth. (or $25,600,000 worth depending on the figure used)? Also if we had dedicated truck lanes would they really cost sixty thousand times as much to maintain? How much does a bus weigh? How much more damaged do Bus Lanes get? I ask these questions seriously, but i don't expect you to answer them, if you could point me to where you found your info i'll have a look for myself. I agree about needing to know what it means in dollar term but can't see a way to express these figures in dollar terms without a quantifier. There is much more info needed to validate these figures. Just generating numbers from a formular is a long way from an engineering proof. And without proof or validation, using the figures as one of the reasons to remove trucks from the road seems a little half baked. I'm willing to pay for all the theoretical damage in the world, as long as i can use my theoretical dollars. H.
From: reg-john on 12 Jun 2007 05:49 "Noddy" <dg4163@(nospam)dodo.com.au> wrote in message news:466be091$0$59820$c30e37c6(a)lon-reader.news.telstra.net... > > "reg-john" <al(a)fddfd.com> wrote in message > news:o7Oai.11367$wH4.7252(a)news-server.bigpond.net.au... > >> this is how it will all eventually be, but until it is you will get >> companies pushing their drivers. and expect the cost of most goods to >> increase in line with the large costs incurred in complying 100% with >> safety regulations. > > The sooner it's made law the better. would you then be in favour of the same system for cars? after all if every speed infraction can be recorded, a lot less people will do so. > > -- > Regards, > Noddy. >
From: Noddy on 12 Jun 2007 06:24
"reg-john" <al(a)fddfd.com> wrote in message news:P_tbi.12883$wH4.11831(a)news-server.bigpond.net.au... > would you then be in favour of the same system for cars? Absolutely. > after all if every speed infraction can be recorded, a lot less people > will do so. Indeed. To that end, I'd like to see a lot of the revenue (if not all of it) generated by speed cameras used to put a large number of police on our roads so the visible deterrent actually works rather than using the "speed kills" hype as a revenue raiser. -- Regards, Noddy. |