stevei wrote:
It's not a trap, your logic is incorrect. I've described a scenario where speed has a systematic effect on the outcome. These are the only scenarios that are relevant. Out of all the possible scenarios, positions of vehicles, speeds etc, if you were to count up the number of scenarios where the 70mph driver couldn't stop, and the number of scenarios where the 90mph driver couldn't stop, the 90mph driver would hit the lorry more times, because there are more possible starting locations where he will be unable to stop. This says everything about the risk, indeed it's the very definition of it - the number of scenarios where a particular outcome will occur out of all the possible scenarios.
You've desribed one tiny subset of all possible scenarios.
I'm not disputing that the risk is greater at a higher speed. What I am disputing is the degree by which it is greater - according to them.
By their reckoning, risk is proportional to stopping distance - which is braking distance plus reaction distance, and so the relative risk works out at something approaching the square of the ratio of the two speeds. But this does not take into account the probability of being at a certain place at a certain time.
With probability taken into account, the risk is proportional to the stopping time, or more accurately, the sum of reaction time and half the braking time - which makes the RR something approaching the ratio of the two speeds.
The half the braking time is a bit confusing at first glance, but in fact that's the time it would take to cover the braking distance at the pre-incident speed. And it's that which determines the risk. Anything that happens once you've started reacting to an incident (the apparent 'slowing down' of time), is already
fait accompli and no longer plays a part in the risk.
How it works mathematically is as follows:
Stopping distance is reaction distance plus braking distance, so
D = ( s * r ) + ( ( s * s ) / ( 2 * g ) )
where: s = pre-incident speed, r = reaction time and g = deceleration.
As a worked example, the stopping distance D from 30mph (13.33m/s) assuming a deceleration of 0.9g (8.82m/s/s) and reaction time of 0.6s is:
( 13.33 * 0.6 ) + ( 13.33 * 13.33 / ( 2 * 8.82 ) ) = 18.078m
and the stopping distance from 60mph (26.66m/s) is:
( 26.66 * 0.6 ) + ( 26.66 * 26.66 / ( 2 * 8.82 ) ) = 56.312m
So, according to them, the relative risk at 60mph compared to 30mph is:
56.312 / 18.078 = 3.11
But the laws of probability tell us that the probability of being between two points at a particular time is proportional to the time spent between the two points, ie inversely proportional to speed.
So modifying the first equation gives us:
( ( s * r ) / s ) + ( ( s * s ) / ( 2 * g * s ) ) = r + ( s / ( 2 * g ) )
At 30mph we get a figure of:
0.6 + ( 13.33 / ( 2 * 8.82 ) ) = 1.356
which, incidentally, is the time taken to cover 18.078m at 30mph
and at 60mph we get:
0.6 + ( 26.66 / ( 2 * 8.82 ) ) = 2.112
So the risk ratio is actually:
2.112 / 1.356 = 1.557
which is half of what they say.
This is not difficult to prove, if proof is needed, by writing a fairly simple computer simulation as follows:
1) Simulate a stream of vehicles passing a fixed point P at an interval of V. V can be fixed (say every 3 seconds) or random
2) Simulate hazard appearing in the road at an interval of H. H can be fixed (say every 10 seconds) or random. The point at which the hazard appears is A, which can be fixed, or random (say up to half a mile either side of point P)
At this point I must add that it makes no difference whether V, H or A are fixed or random, just as long as at least one of them is random.
Also, they should be chosen to be valid for the whole range of interest - for example it makes no sense to make V one second if, at the highest speed being considered, the stopping time is two seconds, as this will result in a collision in every case.
3) At every interval H, calculate whether there's a vehicle approaching point A, and within its stopping distance of point A. If this is true, record a collision. You could, at this point, also calculate the impact speed and record it.
4) Run this for as long as is necessary to get a good spread of random values (suggest in the order of a million times)
5) Divide the number of collisions by the number of hazards and the number of vehicles passed point P. This will give the number of collisions per hazard per vehicle.
6) Run the simulation again with a different vehicle speed. The risk ratio will be the ratio of the two results obtained.
You can try different variations of the simulation, just as long as you're careful not to artificially skew the results. You could even add other parameters if you so choose.
Cheers
Peter