Does anyone know what the braking distances might be on a 44 tonne 12809 cc 2015 Mercedes Actros 2545 Wheel Plan 3 axle + 3 axle artic, on the flat, in good conditions or the best way I might be able to find out? Fully loaded or with no load (just curtain sided trailer)
Thanks
Go buy a ball of string undo it and trow it then cut it when it stops and repeat
A few times and measure the cut lengths 
Many variables can give various answers
At what speed? Start with a my of about 0.7
Take it for a drive on the M25 or the roadworks on M1, youāll find out in about 10 minutes
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Methinks somebody is trying to build a case (against somebody else).
Too many variables - tyres, road surface, liquid load and weather etc.
Guess there might be nominal values at some German laboratory or other.
If you hit something from behind I donāt think youāll get away with it by blaming the lorry.
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In theory you can work it out somewhere near since it must be capable of meeting 50% efficiency on a brake test for MOT and 50% efficiency is 1/2G . However you really need to know the speed and the coefficient of friction of the road surface among other factors.
What I can tell you is that I did a Tapley Test on 4x2 unit and tandem trailer on Thursday with the following results on smooth dry concrete. GTW30500kg. Speed 25 km/h Peak deceleration 60%G Average deceleration 44%G Stopping distance 5.8 M Average brake efficiency 45%. These are figures from a controlled emergency stop as done on a driving test, they are not the result of a panic stop. This particular test was done with the unit drive axle only weighing 6505 kg which flagged as insufficient weight on the Roller Brake Test carried out on the same combination and load the same day. The Tapley Test conducted with the trailer brakes disconnected indicated - as expected- that as the trailer which was at near maximum bogie weight the tractor unit could have contributed more if it had been nearer to its design GVW. RBT figures Unit 49% all lock, Trailer 58% all lock. A similar tractor unit loaded to its GVW produced 67%G all lock on RBT.
If you want the formulae to calculate this then here they are:
Assuming 1/2g acceleration, and rounding we get -5m/s.
Initial velocities conversions.
10 kph to 10,000 m/hr or 10,000 m/60x60 secs > 2.8m/sec approx.
Stopping from say 50kph (30 mph) is 19m/s so about 4 seconds.
Av speed is 8.5m/s so 34m travelled.
Open to correction.
My thoughts exactly. I think its likely to be the old ābad workman blaming the toolsā situation
You would also need to factor in real life reaction time on top of any manufacturers measurements conducted under controlled conditions. Thatās going to be fairly challenging to do post real-life incident.
Iāve seen some figures from the Transport Research Laboratory, which did a study on increased reaction times. Short version of that study was that: a standard reaction time was measured at 1 second for a normal driving situation. This could be increased to about 1.47 seconds if the driver was (for a variety of reasons) not concentrating normally.
Less than half a second? Not much is it? But at 56mph that equated to an increase in stopping distance of 13.1 metres.
In more usable terms, almost the same as the length of a trailer.
So that increased distance would cover three medium-sized family cars (based on a Ford Focus measuring 4.38m). In a RL situation, thatās anywhere between 3 to a potential 15 casualties resulting from that 0.47 second increase over the 1.0 second standard time.
EDIT: If weāre not done playing with numbers, we could consider the peak impact force of a 44 tonner hitting something at 56mph. We can measure that as 27.5 Million Newtons, but in more understandable everyday terms we can equate that to 2,804 Tonnes.
Agreed. Just playing with some rough numbers, and not offering it as a good basis to drive to on real roads.
Ans roughly using the same assumptions as I did earlier, we get a stopping distance at 90k/hr of about 169 metersā¦
Plus the thinking distance 
Well worth asking then given a reaction time of 0.47 seconds for the driver of a 44tonne HGV travelling at 90kph.
If the driver of the focus travelling at the same speed has the same reaction time to something falling off a lorry also at 90 kph in front also of him without bouncing becasue he maintained a 2 second following distance and manages to stop dead safely (did he). How far in distance and time should the 44t lorry have been behind to stop safely?
Donāt ask me the answer!
Thatās easy: It is inversely proportional to 2Ļ.r divided by NA (this is to say, Avogadroās number, which is as you know, 6.02214076Ć10Ā²Ā³ molā»Ā¹ )
But as a general rule, itās half the value of the amount of wood a woodchuck would chuck if a woodchuck could chuck wood.
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No complicated mathematics required really. Whatever you consider the safe distance to be, double it. Youāll thank me.
I think the general consensus is that that suggestion might have been required at some earlier point in time. OP is oddly quiet following his initial enquiry
Itās Saturday. He/she could be doing productive things such as being in the pub or maybe tailgating Honda Jazz drivers with blue hair on the M25. Who knows.
Saturday now. Original post was made three days ago
Agh, the post was asking about stopping distance which has lots of variables. Then add in reaction time! Highway Code reaction is accepted as being unrealistic it just happens to match the original stopping distances they used.
Beware of the use of kph when the calcs using g are in metres per seconds per second.
And of course, these calcs are mass independent- the greater issue is the tyre-road co-efficient, and on a dry road it would be calculated closer to 0.7-0.75 mu. For quick calls try using half and full g as a deceleration rate and basic 1,1.5 and 2.0 s reaction time
A picture paints a thousand words (at least), so hereās a two thousand word essay on braking distances
The lorry is being lifted into the air by the car in the second picture
That car contains the adult driver and three children, all unsurprisingly deceased.
