Discussion
Something that occured to me t'other evening on the Watford link road...
If you have an NSL dual carriageway with a roundabout on it, but the main road has a one lane "bypass" to enable traffic not needing the roundabout to skirt it, what is the limit on the bypass bit? If the whole section is NSL, does the one lane bit revert to a 60, then back to a 70 when the bypass rejoins the main road after the roundabout, or would that need additional signing? Come to think of it, could there be additional signing, given it would be the same sign?!
If you have an NSL dual carriageway with a roundabout on it, but the main road has a one lane "bypass" to enable traffic not needing the roundabout to skirt it, what is the limit on the bypass bit? If the whole section is NSL, does the one lane bit revert to a 60, then back to a 70 when the bypass rejoins the main road after the roundabout, or would that need additional signing? Come to think of it, could there be additional signing, given it would be the same sign?!
Consider this (A590 Cumbria good example)you travel along a road which is dual carriageway (70). It then reverts to single carraigeway (60)and then miles later back to dual(70). How many of you seem signs leaving DC -60 - and on single 70 ?
If it doesn't fall within the definition of a dual and not covered by an Order and without street lighting then its 60.
DVD
>> Edited by Dwight VanDriver on Tuesday 2nd August 07:17
If it doesn't fall within the definition of a dual and not covered by an Order and without street lighting then its 60.
DVD
>> Edited by Dwight VanDriver on Tuesday 2nd August 07:17
hornet said:
Something that occured to me t'other evening on the Watford link road...
If you have an NSL dual carriageway with a roundabout on it, but the main road has a one lane "bypass" to enable traffic not needing the roundabout to skirt it, what is the limit on the bypass bit? If the whole section is NSL, does the one lane bit revert to a 60, then back to a 70 when the bypass rejoins the main road after the roundabout, or would that need additional signing? Come to think of it, could there be additional signing, given it would be the same sign?!
Surely it would be a dual carridgeway? As it was serperated by a central barrier..
Just floating 7db understand.....
a to b = one limit(X)
b to c = different limit (Y)
then average speed a to c is X squared + Y squared equals a to c average speed unsquared.
If this result is greater that any one of the limits on a to b or b to c then someone has been speeding?
DVD
(OK Fif I know I fluffed Maths at Grammar)
a to b = one limit(X)
b to c = different limit (Y)
then average speed a to c is X squared + Y squared equals a to c average speed unsquared.
If this result is greater that any one of the limits on a to b or b to c then someone has been speeding?
DVD
(OK Fif I know I fluffed Maths at Grammar)
Dwight VanDriver said:
Just floating 7db understand.....
a to b = one limit(X)
b to c = different limit (Y)
then average speed a to c is X squared + Y squared equals a to c average speed unsquared.
If this result is greater that any one of the limits on a to b or b to c then someone has been speeding?
DVD
(OK Fif I know I fluffed Maths at Grammar)
That is quite OK by me, I made an utter hollyhocks of latin. Opening bat for the cricket team though...
Should I make my way to Edgbaston?
Dwight VanDriver said:
Just floating 7db understand.....
a to b = one limit(X)
b to c = different limit (Y)
then average speed a to c is X squared + Y squared equals a to c average speed unsquared.
If this result is greater that any one of the limits on a to b or b to c then someone has been speeding?
DVD
Not sure what your formula is supposed to solve, DVD (looks like Pythagoras to me). If:
X = 30 mph
Y = 40 mph
X^2 + Y^2 = 900 + 1600 = 2500
2500 unsquared (square root?) = 50
So average speed is 50 mph!?
hornet said:
Something that occured to me t'other evening on the Watford link road...
If you have an NSL dual carriageway with a roundabout on it, but the main road has a one lane "bypass" to enable traffic not needing the roundabout to skirt it, what is the limit on the bypass bit? If the whole section is NSL, does the one lane bit revert to a 60, then back to a 70 when the bypass rejoins the main road after the roundabout, or would that need additional signing? Come to think of it, could there be additional signing, given it would be the same sign?!
The dual carriageway - which ist genuine dual carriageway mit the barrier un the verge separating each two lane carriageway ist 70 mph.
If there ist no central barrier - ist 60mph - und so I would take the single bit to be 60 mph.
A65 DVD, mein Liebchen, ist Steviebababes' fave hunting ground after M6 at Shap bridge und Ings
. Ist a singel carriageway - but a lot read as dual carriages because in places - ist two lanes each way - but ist only paint in centre - und ist thus a single carriageay. Ist a good little earner apparently 
"If it doesn't fall within the definition of a dual and not covered by an Order and without street lighting then its 60."
DVD
Yes indeed. I remember reading about Don finding this out at cost to his driving licence some time ago...
>> Edited by princeperch on Wednesday 3rd August 19:08
Observer2 said:
Just floating 7db understand.....
a to b = one limit(X)
b to c = different limit (Y)
then average speed a to c is X squared + Y squared equals a to c average speed unsquared.
If this result is greater that any one of the limits on a to b or b to c then someone has been speeding?
DVD
Not sure what your formula is supposed to solve, DVD (looks like Pythagoras to me). If:
X = 30 mph
Y = 40 mph
X^2 + Y^2 = 900 + 1600 = 2500
2500 unsquared (square root?) = 50
So average speed is 50 mph!?
Perhaps it was a triangular roundabout?
Best wishes all,
Dave.
Dwight VanDriver said:
Just floating 7db understand.....
I think the maths has gone a bit wonky there, but if you know the distance a-b and the speed limit you know the minimum legal time a-b, and similarly for b-c. Add these times together and you get the minimum legal time a-c. If you drive a to b to c in less time, you must have broken the law at some point.
I think that the only safe thing to say is that if the average speed a-c is over the higher of the limits a-b and b-c then someone has been speeding.
Without knowing more about the relative distances involved, it would be impossible to say more.
Since in our example, we are talking about a very short section of higher limit road, this would be a big distortion.
Two maths degrees, and I'm reduced to this?
Without knowing more about the relative distances involved, it would be impossible to say more.
Since in our example, we are talking about a very short section of higher limit road, this would be a big distortion.
Two maths degrees, and I'm reduced to this?
WildCat said:Wildy - the number of lanes in either direction is immaterial - there are some dual carriageways where either or both sides have a single lane ... and many that have three (or more) lanes - Streaky
The dual carriageway - which ist genuine dual carriageway mit the barrier un the verge separating each two lane carriageway ist 70 mph.
7db said:
I think that the only safe thing to say is that if the average speed a-c is over the higher of the limits a-b and b-c then someone has been speeding.
Without knowing more about the relative distances involved, it would be impossible to say more.
Since in our example, we are talking about a very short section of higher limit road, this would be a big distortion.
Two maths degrees, and I'm reduced to this?
Distances ARE fundamental.
A to B, then B to C. Speed limits are X and Y respectively. We won't work with speed, but with time taken to cover A to C.
A to B at the limit should take time = (B-A)/X (call it t1)
B to C at the limit should take time = (C-B)/Y (call it t2)
So if total time is less than t1+t2, then the driver has been speeding. In fact, if junctions are involved you've a small amount of leeway because you'll have to slow down...
Depending on enforcement, there may be a partial answer if you're JUST having fun...do what the French do on the peage's...stop for a break mid-way!
PS - TWO maths degrees?!?
What was wrong with the first one?!? And personally I found one more than enough!!! 
havoc said:Something doesn't add up! - Streaky
I think that the only safe thing to say is that if the average speed a-c is over the higher of the limits a-b and b-c then someone has been speeding.
Without knowing more about the relative distances involved, it would be impossible to say more.
Since in our example, we are talking about a very short section of higher limit road, this would be a big distortion.
Two maths degrees, and I'm reduced to this?
Distances ARE fundamental.
A to B, then B to C. Speed limits are X and Y respectively. We won't work with speed, but with time taken to cover A to C.
A to B at the limit should take time = (B-A)/X (call it t1)
B to C at the limit should take time = (C-B)/Y (call it t2)
So if total time is less than t1+t2, then the driver has been speeding. In fact, if junctions are involved you've a small amount of leeway because you'll have to slow down...
Depending on enforcement, there may be a partial answer if you're JUST having fun...do what the French do on the peage's...stop for a break mid-way!
PS - TWO maths degrees?!?
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