> .05@ #bjbj22 .HXX>>>>>>>R<TR[R(zzzzVVV[ [ [ [ [ [ [$U\R^D[>WKJVWWD[>>zz+Y[YYYWX>z>z[YW[Y(YZIZe>>ZzF@mWX6ZZ<o[0[ZI_Y@I_ZRR>>>>I_>Z$V0 W"Y+WGWVVVD[D[RRdYRRCritical speed on a cambered road
(mgcos(
mv2/R
(mv2/R)cos( mgsin(
(
mgcos(
mg
Available force is (mgcos( + mgsin(
This will supply the component of the centripetal force = (mv2/R).cos(
The critical speed is given by
(mgcos( + mgsin( = (mv2/R).cos(
or EMBED Equation.3
Example: A bend has a radius of curvature of 100 metres and a camber of 15%. What is the fastest speed at which the bend can be taken?
(Assume that the vehicle follows the curvature of the road, a = 7 ms2)
EMBED Equation.3
( = a/g = 7/9.81 = 0.714
Vc = 29.1 ms1 = 65.1 mph
Measuring curvature
Method A:
D
(
Angular change in road direction = (o
Distance between straight sections = D
Therefore:
D/(2(R) = (/360
and R = 180D/((()
Method B:
chord Middle offset
h
L
(Rh)
R
(L/2)2 + (R  h)2 = R2
L2/4 + R2  2Rh + h2 = R2
2Rh = L2/4 + h2
R = (L2/8h) + h/2
Note: The percentage error in R is equal to the percentage error in h. Therefore h should be as large as possible and at least 0.3 metres.
CRITICAL SPEED TYRE MARKS
1 There should be at least two tyre marks visible from the outside wheels on the curve which should show lateral striations.
2 Measurements of R should be made from the front outside tyre mark.
Aligning boards should be used where possible.
3 The chord length L should be 15 metres or more in length and h should be greater than 0.3 metres.
4 The yaw of the rear wheels should not exceed half the track width. The rear wheels should track outside the front wheels.
5 Measure the gradient and crossfall.
Momentum calculations
In any collision, the total linear momentum of the vehicles just before impact equals the total linear momentum just after impact.
Momentum does not change into energy. It is independent of the damage sustained.
Momentum does not change into angular momentum. It is independent of any subsequent vehicle spin.
Momentum is conserved when the impact is offset or even glancing. It applies to any kind of collision.
Momentum is a vector quantity. It has magnitude and direction.
The momentum of a vehicle is given by (mass of vehicle) x (its velocity)
Example:
A car of mass 1000 kg travelling at 30 mph eastwards collides directly head on with a van of mass 1500 kg travelling at 50 mph westwards.
What will happen?
(assume an inelastic collision)
Precollision:
momentum of car = 1000 x 30 kg.mph
momentum of van = 1500 x 50 kg.mph
Postcollision:
Let immediate postcollision velocity be U mph
momentum of vehicles =
(1000 + 1500) x U kg.mph
Conservation of linear momentum gives
(1000x30)  (1500x50) = (2500xU)
U = 18 mph
Immediately after impact, both vehicles will be travelling westwards at 18 mph.
Another approach . . . .
In the example, the common centre of gravity prior to impact is always situated 1500/2500 = 60% of the distance from the car to the van.
common C of G
100 miles
30 20 50 miles
1 hour later
12
42 18
60 miles
Consider the two vehicles 100 miles apart. The common centre of mass is 60 miles from the car.
After 1 hour, the vehicles are 20 miles apart with the car 30 miles further east.
After 1 hour, the common centre of mass is 12 miles from the car or (30+12) = 42 miles from the original position of the car.
Thus, the common centre of mass has moved westwards by (6042) = 18 miles in one hour.
After the collision, the common centre of mass will continue to move westwards at 18 mph.
IMPLICATIONS:
Example:
Hill brow Greenhow
Volvo
Blubberhouses
Gouge Rover
A high speed collision in a layby!
Speed of Rover deduced from flight of bonnet.
Gouge mark indicates the impact point.
Common centre of mass has moved towards Greenhow after the collision.
Common Centre of mass has moved further into the layby after the collision.
Both cars have the same mass in this case.
The Rover has been travelling towards Greenhow at between 50 and 60 mph and has veered towards the layby. The Volvo has been travelling towards Blubberhouses at about 40 mph.
Rover driver sues North Yorkshire Council for placing a layby in a dangerous position and not maintaining it.
Collisions at significant angles
m1v1
north
(2
(1 m2v2
u1
m1 east
u2
m2
Momentum northwards:
Before impact = m2u2
After impact = m1v1cos(1 + m2v2cos(2
Thus, m2u2 = m1v1cos(1 + m2v2cos(2
Momentum eastwards:
Before impact = m1u1
After impact = m1v1sin(1 + m2v2sin(2
Thus, m1u1 = m1v1sin(1 + m2v2sin(2
v1 and v2 can be estimated from the postcollision movements of the vehicles.
m1 and m2 can be determined from manufacturer's data.
(1 and (2 can be determined by direct measurement.
The only unknowns are then u1 and u2.
Example: m1 = 1000 kg, m2 = 1200 kg, (1 = 45o, (2 = 60o, car1 moves 5 metres after impact, car2 moves 8 metres after impact. Both cars remain fully braked on a dry road surface with a = 7 ms2.
v1 = 18.7 mph v2 = 23.7 mph
u2 = 22.9 mph u1 = 37.9 mph
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