1. i
EVALUATE VEHICLE PERFORMANCE ...........................................................................................1
TASK 1.................................................................................................................................................1
First and Intermediate Ratios of the Gearbox .............................................................................1
Maximum Gradeability of the Vehicle..........................................................................................2
Road Speed When Changing from 3rd
to 4th
Gear ......................................................................3
3rd
Gear Road Speed...................................................................................................................3
TASK 2.................................................................................................................................................5
TASK 3.................................................................................................................................................7
Power to Weight Ratio.................................................................................................................7
Number of Gears .........................................................................................................................8
Range Change and Splitter Boxes ............................................................................................10
Vehicle Body Shape...................................................................................................................11
Aerodynamic Styling ..................................................................................................................14
Trailer Height and Design..........................................................................................................16
BIBLIOGRAPHY ...............................................................................................................................17
CONVERSION TABLE.....................................................................................................................A-1

2. i
Abbreviations
Cd Drag Coefficient
EU European Union
FDR Final Drive Ratio
GR Gear Ratio
HGV Heavy Goods Vehicle
Kmh Kilometres per hour
Kg Kilogram
KN Kilo newton
KW Kilo Watt
LPPV Light Protected Patrol Vehicle (Foxhound)
m/s Metres per second
PM Protected Mobility (Vehicle)
RPM Revolutions per Minute
Ra Air resistance
Rg Gradient Resistance
Rr Rolling Resistance
TE Tractive effort
TR Tractive Resistance

3. 1
EVALUATE VEHICLE PERFORMANCE
This assignment will cover calculations of vehicle gearing and its gradeabilty based on them, it will
establish the parameters of tractive effort versus tractive resistance and how they interrelate. The
third part of this assignment will focus on the choice of vehicle gearboxes and the impact of
aerodynamic resistance on Heavy Goods Vehicles (HGV). I will set myself a period of 3 weeks to
complete the first two tasks and use the remainder of the time available to research aerodynamic
issues as this is not something I have looked at in detail.
TASK 1
1. The data given in table 1 below is for an experimental vehicle that could be manufactured,
based on the information given we will calculate the first and intermediate ratios of the gearbox, the
vehicles maximum gradeability and gradeability in top gear and the road speed change that will
occur when going from 3rd
to 4th
gear. I will use student notes and formula books (Greer, 1989) for
equations as these are relevant to the tasks.
Table 1. Vehicle Information
First and intermediate ratios of the gearbox
2. Assuming geometric progression of gear ratios (a,ar,ar2
…..or n,nz,nz2
…..)1
. The gear ratio or
n in 5th
gear is n5=1:1, therefore n5 = n1 z4
. To conduct this equation we require our z value, this
is the engine speed ratio, the engine speed ratio is the area of engine rpm between maximum
torque and maximum power, this is also known as the engine operating speed range. It is
calculated by:
𝑧 =
𝑟𝑝𝑚 𝑎𝑡 max𝑡𝑜𝑟𝑞𝑢𝑒
𝑟𝑝𝑚 𝑎𝑡max𝑝𝑜𝑤𝑒𝑟
2 𝑧 =
1540
2100
= 0.73
n5 = n1 x z4
or 1 = n1 x 0.734
Therefore n1 = n5 / z4
which is 1 / 0.734
= 3.52:1
n2 = 3.52 x 0.73 = 2.57:1
n3 = 3.52 x 0.732
= 1.88:1
n4 = 3.52 x 0.733
= 1.37:1
n5 = 3.52 x 0.734
= 0.999:1 or 1:1
1
(Greer, 1989)
2
(Defence Schoolof Electronic and Mechanical Engineering)
Mass 15tonne Final drive ratio 5.248:1
Max Power 165KW @2100rpm Efficiency of other ratios 90.50%
Max Torque 710Nm @ 1540rpm Transmission Efficiency 92%
5th gear ratio 01:01 Wheel rolling radius 0.328m
Vehicle Information

4. 2
3. To confirm our figures we can check that our r or z figure is correct by calculating:
𝑟 𝑜𝑟 𝑧 = 4√
1
3.52
= 0.73
4. These figures would be the ideal gear ratios to use in the gearbox based on the data given,
however, the gears must also fit inside the gearbox and if a constant mesh gearbox is to be used
than gears must be found that can all fit while running on the same central shafts, therefore the
actual gear ratio’s used may be altered to suit.
Maximum gradeability of the vehicle
Table 2. Vehicle Information
5. Gradeability is the amount of incline that a vehicle can drive up dependant on the torque and
gear ratios it is using, it is calculated by:
𝑀𝑎𝑥 𝑇𝑜𝑟𝑞𝑢𝑒 𝑥 𝑆𝑎𝑓𝑒𝑡𝑦 𝑓𝑎𝑐𝑡𝑜𝑟 𝑥 𝐺𝑒𝑎𝑟 𝑟𝑎𝑡𝑖𝑜 𝑥 𝐹𝑖𝑛𝑎𝑙 𝑑𝑟𝑖𝑣𝑒 𝑟𝑎𝑡𝑖𝑜 𝑥
𝑉𝑒ℎ𝑖𝑐𝑙𝑒 𝑓𝑜𝑟𝑐𝑒 ( 𝑁) 𝑥 𝐷𝑟𝑖𝑣𝑖𝑛𝑔 𝑡𝑦𝑟𝑒 𝑟𝑎𝑑𝑖𝑢𝑠
3
6. This equation includes a safety factor to ensure that we are always working within safe limits
and won’t design a vehicle that is dangerous to use. By using a safety factor of 95% we can
calculate maximum gradeability in first gear and gradeability in top gear as:
𝑀𝑎𝑥 𝑔𝑟𝑎𝑑𝑒𝑎𝑏𝑖𝑙𝑖𝑡𝑦 =
710 𝑥 0.95 𝑥 3.52 𝑥 5.248 𝑥 0.905
147150 𝑥 0.328
= 𝟎. 𝟐𝟑𝟑𝟔
𝐺𝑟𝑎𝑑𝑒𝑎𝑏𝑖𝑙𝑖𝑡𝑦 𝑖𝑛 𝑡𝑜𝑝 𝑔𝑒𝑎𝑟 =
710 𝑥 0.95 𝑥 1 𝑥 5.248 𝑥 0.92
147150 𝑥 0.328
= 𝟎. 𝟎𝟔𝟕𝟓
7. By taking these figures and comparing them against a conversion table at Annex A we reach
a tanƟ (%) figure of 22.5% in first gear and 5.5% in top gear, this shows that by using a higher
gear ratio we have lost the advantage of a lower gear and the tractive effort available to overcome
the tractive resistance of the gradient is insufficient, the driver must therefore shift down to a lower
gear.
3
(Defence Schoolof Electronic and Mechanical Engineering)
lowest gear ratio 3.52:1
transmission mech
efficiency (top gear)
0.92
top gear ratio 01:01
transmission mech
efficiency (other
0.905
max engine torque
(NM)
710 vehicle weight 147150
safety factor 95% driving wheel size 0.328
final drive ratio 5.248:1
Vehicle Data

5. 3
Road speed when changing from 3rd
to 4th
gear
Table 3. Vehicle Information
8. As the vehicle drives it will accelerate through the rev range until a point where it needs to
shift up to the next gear ratio, by shifting up a gear we will move down into the max torque area to
enable us to accelerate away in a new gear ratio that allows higher speed, this is shown below.
To calculate the change in speed we require the following equations:
𝑅𝑜𝑎𝑑 𝑤ℎ𝑒𝑒𝑙 𝑠𝑝𝑒𝑒𝑑 =
𝐸𝑛𝑔𝑖𝑛𝑒 𝑠𝑝𝑒𝑒𝑑 𝑎𝑡 𝑚𝑎𝑥𝑖𝑚𝑢𝑚 𝑝𝑜𝑤𝑒𝑟
𝐺𝑒𝑎𝑟 𝑟𝑎𝑡𝑖𝑜 × 𝑓𝑖𝑛𝑎𝑙 𝑑𝑟𝑖𝑣𝑒 𝑟𝑎𝑡𝑖𝑜
4
𝑅𝑜𝑎𝑑 𝑠𝑝𝑒𝑒𝑑 = (
𝑅𝑜𝑎𝑑 𝑤ℎ𝑒𝑒𝑙 𝑠𝑝𝑒𝑒𝑑 × 𝜋 × 𝐷
60
) × 3.6
3rd
gear road speed
𝑅𝑜𝑎𝑑 𝑤ℎ𝑒𝑒𝑙 𝑠𝑝𝑒𝑒𝑑 =
2100
1.88 × 5.428
= 212.85 𝑟𝑝𝑚
𝑅𝑜𝑎𝑑 𝑠𝑝𝑒𝑒𝑑 = (
212.85 × 𝜋 × 0.656
60
) × 3.6 = 26.32km/h
9. The above assumes that the driver is making a gear change as soon as the vehicle gets to
its maximum power point, however the vehicle could continue driving until engine maximum rpm,
despite being passed maximum power there would still be an increase in road speed, E.g:
𝑅𝑜𝑎𝑑 𝑤ℎ𝑒𝑒𝑙 𝑠𝑝𝑒𝑒𝑑 =
2500
1.88 × 5.428
= 244.99 𝑟𝑝𝑚
𝑅𝑜𝑎𝑑 𝑠𝑝𝑒𝑒𝑑 = (
244.99 × 𝜋 × 0.656
60
) × 3.6 = 30.294km/h
10. The speed at which the vehicle changes from 3rd
to 4th
gear occurs at 26.32 km/h, when the
gear change is made the vehicle is shifted into a higher gear ratio, in this instance 1.37, when this
happens the engine rpm will drop down due to the increase in resistance now offered by the higher
gear ratio. By transposing our equation for road wheel speed and assuming the road wheel speed
does not change at the point of gear change we can calculate the effect on engine speed (n) of this
gear change:
𝐹𝑜𝑟𝑚𝑢𝑙𝑎 𝑡𝑟𝑎𝑛𝑠𝑝𝑜𝑠𝑒𝑑 𝑖𝑠 𝑛 = 212.84 (1.88 𝑥 5.248) = 2100𝑟𝑝𝑚
4
(Defence Schoolof Electronic and Mechanical Engineering)
3rd gear ratio 1.88
4th gear ratio 1.37
wheel rolling diameter 0.636
engine max power 2100rpm
engine max torque 1540 rpm
Vehicle Data

6. 4
If gear ratio is changed to 1.37:
𝑛 = 212.84 (1.37 𝑥 5.248) = 1530.27𝑟𝑝𝑚
11. We can now see that by shifting up a gear we have gone from the point where our engine is
making maximum power (2100) to close to the point where our engine is making maximum torque
(1540), we now have maximum torque available for acceleration.

7. 5
TASK 2
12. The graph below contains information on a vehicles performance, by reading the graph we
can calculate max road speed, gradeability and tractive effort (TE). Again calculations are taken
from student notes and a formula book (Greer, 1989).
13. Maximum road speed of the vehicle. This is where the tractive effort and tractive
resistance meet, using the highest gear ratio of 5th
and the tractive resistance based on a gradient
of 0% we reach a figure of around 115 km/h.
Graph 1. Tractive effort and resistance v Speed5
14. The gradeability of the vehicle in 1st
gear. This is where the tractive effort in that gear ratio
is at its maximum, looking at the graph we can see that 1st
gear has a tractive effort curve that
reads at 23.5 KN at its maximum, the TE required to climb a 35% gradient is around 24KN,
therefore technically in this instance we do not have sufficient TE to do this, our maximum
gradeability is probably around 34%, reading from the graph we would say that maximum
gradeability is 30%.
15. Gear ratio to negotiate a 10% gradient. For this we would need to find a gear that has
enough TE to overcome the TR present, at 10% this ranges from 7.8KN at 20 km/h to 9KN at
130km/h, the TE in 3rd
gear is less than 7 KN so this gear ratio would be too high to allow the
vehicle to climb, the maximum TE in 2nd
gear is 13 KN so this gear ratio gives us enough TE to
climb the hill but also an extra 5 KN at 17km/h should the gradient increase, we would also be able
to use this extra TE to accelerate past a slower moving vehicle up to a speed of about 37.5km/h ( a
line drawn down from the TE curve would cut the 10% TR line at about 37.5 km/h).
5
Assignment Unit 25 Paper 11

8. 6
16. We could also use 1st
gear which has close to 24 KN of TE available but at a much reduced
speed, it can be seen that TE in lower gears drops of greatly for a small increase in speed while at
higher gear ratios (4 and 5) the curve drop is much shallower, this is due to where in the rev range
the engine produces most torque and this is dependent on engine design.
17. Surplus tractive effort. Travelling at 80km/h in 5th
gear on a 0% gradient we see from the
graph that approximately 1.8 KN is required but our available TE is around 3 KN, therefore we
have 1.2 KN surplus tractive effort. If we were travelling at 40 km/h on a 5% gradient in 3rd
gear
than we have a TR of 4.25 KN while our gear ratio and speed gives us a TE of 7 KN, our surplus
TE is 7 – 4.25 = 2.75 KN.
18. Acceleration. To work out the acceleration at 60 km/h on a 0% gradient in 5th
gear with a
vehicle with a mass of 3.6 tonnes (3600 kg). TE available is TE-TR = (3 – 1) 2 KN, mass of vehicle
is 3600 kg.
𝐹 = 𝑀𝑎𝑠𝑠 𝑋 𝐴𝑐𝑐𝑒𝑙𝑒𝑟𝑎𝑡𝑖𝑜𝑛. 𝑇ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒 𝐴𝐶𝐶𝐸𝐿𝐸𝑅𝐴𝑇𝐼𝑂𝑁 =
𝐹𝑂𝑅𝐶𝐸
𝑀𝐴𝑆𝑆
=
2000𝑁
3600𝐾𝐺
= 0.556 𝑀/𝑆2
19. Based on the performance curves I would recommend that this engine be suitable for a small
to medium HGV, it offers a reasonably high speed in top gear and the tractive effort available in
gears 1 and 2 allow for steep hills to be climbed but also provide sufficient surplus power to
accelerate past slower moving vehicles. The spread of 5 gears would also be indicative of a small
truck such as a Leyland Daf 4 tonne.

9. 7
TASK 3
Power to weight ratio
20. This can be defined as the power developed divided by the weight of the laden vehicle. It is
based on the maximum brake power of the vehicle divided by its weight, the weight used is
normally the curb weight of the vehicle which is the vehicle weight minus the weight of drivers/crew
or additional load, but other ratios such as combat weight may be used to give a more accurate
figure. It is calculated by:
𝑃𝑜𝑤𝑒𝑟 𝑡𝑜 𝑊𝑒𝑖𝑔ℎ𝑡 𝑟𝑎𝑡𝑖𝑜 =
𝐵𝑟𝑎𝑘𝑒 𝑝𝑜𝑤𝑒𝑟
𝑊𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑡ℎ𝑒 𝑣𝑒ℎ𝑖𝑐𝑙𝑒 (𝑙𝑎𝑑𝑒𝑛)
6
21. As a comparison of these ratios we can look at 2 Army vehicles, the Ford Jeep light vehicle
from the 1940s and the Foxhound Light Protected Patrol Vehicle (LPPV) HGV from 2010, both
these vehicles are small combat vehicles used for reconnaissance and utility tasks and designed to
carry about four soldiers. I have used the combat weight of the vehicles as this provides a more
realistic power to weight ratio of the vehicles in actual service.
22. The Ford Jeep has a mass of 1530kg and a power output of 44.74 KW7
.
𝑃𝑜𝑤𝑒𝑟 𝑡𝑜 𝑊𝑒𝑖𝑔ℎ𝑡 𝑟𝑎𝑡𝑖𝑜 =
44.74 𝐾𝑊
1.530 tonnes
= 𝟐𝟗. 𝟐𝟒 𝑲𝑾 𝒑𝒆𝒓 𝒕𝒐𝒏𝒏𝒆
23. The Foxhound Light Protected Patrol Vehicle (LPPV) has a mass of 7500 kg and a power
output of 145 KW8
.
𝑃𝑜𝑤𝑒𝑟 𝑡𝑜 𝑊𝑒𝑖𝑔ℎ𝑡 𝑟𝑎𝑡𝑖𝑜 =
145 𝐾𝑊
7.5 𝑡𝑜𝑛𝑛𝑒𝑠
= 𝟏𝟗. 𝟑 𝑲𝑾 𝒑𝒆𝒓 𝒕𝒐𝒏𝒏𝒆
Fig 1. Ford Jeep9
Fig 2. Foxhound LLPV10
24. This illustrates that although the Foxhound has over three times the power of the Ford Jeep,
because we have increased the weight nearly 5 times as much we have a much lower power to
weight ratio, the relationship between power and weight is not linear. A poor power to weight ratio
poses a particular problem when climbing a gradient, power is needed to drive the vehicle uphill,
gradient resistance and tractive effort required to overcome it is proportional to the weight of the
6
(Defence Schoolof Electronic and Mechanical Engineering)
7
(Ware, 2010)
8
(Genral Dynamics Land Systems, 2014)
9
(Wikipedia, 2014)
10
(Genral Dynamics Land Systems, 2014)

10. 8
vehicle and the steepness of the gradient, therefore a vehicle with a poor power to weight ratio has
less power in reserve to tackle the climb.
25. To counter the problem of low power to weight ratio we must first accept that the extra weight
will mean a generally slower vehicle, increasing engine power is possible but the increased weight
of a larger engine and chassis to hold/house it will mean we will only ever create slight
improvements. By knowing the gradients the vehicle is likely to climb along with its weight and
power the correct gearing can be produced for it.
Number of gears
26. Gears are required in order to allow engine torque to be multiplied at the road wheel, this is
crucial for accelerating, the first gear ratio must be low enough that the torque can overcome the
resistance offered by the vehicle and the ground. The trade-off is speed, a gear ratio (GR) low
enough to move the vehicle forward such as 4:1 will mean the engine is spinning quicker than the
road wheel (with further final drive reduction (FDR) of 5:1), therefore the engine will reach max rpm
while the vehicle is only travelling slowly. For example
Engine 4000 rpm / (GR) 4 = 1000rpm / (FDR) 5 =wheel spinning at 200rpm / 60 = 3.3 revs per
second
Wheel diameter of 0.5m means distance of 1.57m per revolution
1.57 x 3.3 =5.18 m/s or 18.66 km/h
27. So, if we need to increase speed we need to shift up to the next gear, however, we require
torque to accelerate further, max torque will occur at a set rpm and as explained earlier rpm will
drop on gear upshifts dependant on the gear ratio, from para 14:
𝑅𝑃𝑀 (𝑛) = 212.84 (1.88 𝑥 5.248) = 2100𝑟𝑝𝑚
If gear ratio is changed to 1.37:
𝑛 = 212.84 (1.37 𝑥 5.248) = 1530.27𝑟𝑝𝑚
28. It is therefore clear that if a heavy vehicle with a low revving diesel engine with a max torque
high up in the rev range, has a gearbox with few gears, then the upshifts will result in the engine
speed dropping below an acceptable torque figure and further acceleration will be sluggish (known
as torque recovery). To remedy this we require more gear ratios to ensure that each upshift
results in only a minor drop in rpm, the downside will be an increase in power losses through the
extra gears.
29. In order to minimise journey times it is beneficial to drive in top gear at the quickest speed
possible, this has the negative effect of increasing fuel consumption. Consider the graph below,
this displays the torque, power and specific fuel consumption (SFC) of an example vehicle, the
engine operating speed range lies between max torque and max power, SFC is at a minimum from
around peak torque to a point 200-400rpm further on the positive torque rise curve. This band is
known as the economy speed range and if engine rpm is kept within this range then maximum fuel
efficiency will be realised.

11. 9
Graph 2. Example vehicle data graph11
30. Let us consider that our vehicle has just 5 gears rather than the 8 indicated, in 5th
gear and
staying in the economy speed range we can achieve around 35 km/h, if we go to max engine rpm
we could achieve around 50 km/h but with massively increased fuel consumption. If we now “re-fit”
our 8 speed gearbox we can achieve around 70 km/h but for the same fuel consumption as 35 in
5th
. By increasing the number of gears we can increase our speed while still remaining within our
economy speed range.
31. Example:
5th
gear inside economy range (2000 rpm) = SFG (kg/kWh) X Power (KW) X 1 hour
= 0.20 kg/kWh X 225 KW X 1 hour
= 45 kg/h
Relative density of fuel = 0.8
45 / 0.8 = 56.25 Litres per hour at 2000 rpm
Speed is 35 km/h for 8 hours = 280 km travelled plus 450 litres of fuel used
However, if we increase the number of gears but remain in the economy range at 2000 rpm:
Speed is now 70 km/h x 8 hours = 560 km/h plus 450 litres of fuel used
32. So it is clear to see that by increasing the number of gears we can travel further and faster
but remain within the most economic range, this is key for HGV’s. The limiting factor on number of
gears is likely to be the size of the gearbox. A standard gearbox will usually contain 4, 5 or 6 gears
but by using a splitter or range change box we can double that.
11
(Heisler, Advanced engine technology, 1995)

12. 10
Range change and Splitter boxes
33. As discussed previously an issue with HGV’s that have very poor power to weight ratios is
that when we change up a gear we must accept a drop in engine rpm, this drop in rpm could put
our rpm below that were maximum torque is produced. As we require that torque to accelerate
away we will find that with large changes between gearbox ratios our torque recovery or
acceleration will be sluggish. To counter this and increase fuel economy splitter or range
gearboxes can be fitted.
34. Splitter gearbox. In this arrangement the gear ratios are spread out wide as per a
conventional gearbox while the 2 speed auxiliary gearbox at the front has one gear in direct ratio
and the second gear is either a step up or step down ratio. The ratio is chosen so as to “split” the
main gearbox ratios in half with a typical ratio being 1.25:1. As well as the normal gearstick to
select gears there would be another means to select either the high or the low ratio of the splitter
gearbox. (fig 3)
Figure 3. Splitter gearbox12
35. With this gearbox the procedure from moving off to top gear would be to select first gear and
low on the splitter box, then as speed increases select high on the splitter box, further acceleration
then requires the driver to select 2nd
gear and low on the splitter box. This is then repeated
through acceleration or, on deceleration, continually alternating between high and low on the
gearbox as shown in the figure 4.
Figure 4. Splitter gearbox gear selection13
12+12
(Heisler, Advanced engine technology, 1995)

13. 11
36. While this system addresses the requirement for a reduction in the ratio difference between
gears in a conventional gearbox it clearly is a somewhat complicated procedure and could cause a
grinding of the gears if the wrong high/low splitter ratio is selected, an improvement on this system
is the range change.
37. Range change. In this setup the gearbox has gear ratios set close together, the auxiliary
gearbox at the back has a 2 speed box with one gear again in direct ratio to the gearbox and one
at a ratio slightly larger than the gearbox gear ratio spread (fig 5).
Figure 5. Range change gearbox14
38. The procedure for gear change with this arrangement would be to engage first gear and low
on the range change box, accelerating further and shifting up gears through 2,3,4,5 etc until top
gear is reached, high ratio is then selected and the sequence is repeated through 1,2,3,4 and 5
only now they are in reality 6,7,8,9 and 10.
39. The principal difference in design between range change splitter gearboxes is that the splitter
box “splits” the gear changes between high and low for each individual gear by having auxiliary
gears before the gearbox, whilst the range change uses the gear progression as normal with the
auxiliary gears changing the output from low to high after the gearbox. Certain specialised HGV’s
such as those that carry very heavy loads or climb severe gradients may be fitted with a third
auxiliary box which would allow for very low or crawler gears.
Vehicle Body Shape
40. When a vehicle drives forwards its speed will be dictated by many factors such as vehicle
power, gearing, and rolling and gradient resistance. Another form of resistance is from the air, air
resistance (Ra) is calculated as:
𝑅𝑎 =
1
2
𝜌𝐴𝑉2 𝐶𝑑15
14
(Heisler, Advanced engine technology, 1995)
15
(Defence Schoolof Electronic and Mechanical Engineering)

14. 12
41. Where 𝜌 is air density, A is the projected frontal area, V is velocity of the air relative to the
vehicle and Cd is the drag coefficient of the vehicle shape. As A, Cd and 𝜌 are relatively fixed it
can be seen that as velocity increases the air resistance will increase to the square of the velocity.
Although HGV’s are travelling at slower speeds then cars the large frontal area adds considerably
to the air resistance. An increase in air resistance can lead to an increase in fuel consumption.
“Aerodynamic drag is responsible for 35-40% of a 40t lorry’s fuel consumption. Reducing
aerodynamic drag therefore offers a very promising way to reduce fuel consumption and GHG
emissions from HGVs.”16
42. At low speeds Ra will be minimal and other resistances such as gradient or rolling
resistances will have more of an impact, however as speed increases the increase in velocity
together with the large frontal area will cause Ra to become the dominant resisting force. Consider
the example below with air density of 1.23 Kg/m3
and Cd of 0.65:
Fig 6. HGV Frontal area17
Fig 7. Cd values of vehicles18
At 25 Km/h (6.9m/s) 𝑅𝑎 =
1
2
1.23(2.55𝑥4) 𝑥6.92 𝑥0.65 =194.13 N
At 100 Km/h (27.78m/s) 𝑅𝑎 =
1
2
1.23(2.55𝑥4) 𝑥27.782 𝑥0.65 =3146.68N
43. By increasing our speed by a factor of 4 we have caused an increase in air resistance of over
16 times the resistance at the slower speed. As we don’t really want to reduce our velocity and as
air density is (relatively) fixed we are left with trying to alter the frontal area of the vehicle and
decreasing its Cd.
44. Looking at it in closer detail air resistance can actually be classified as a drag force acting
on the vehicle and this can be broken down into the following areas:
a. Forebody Drag. Caused by the vehicle displacing the air molecules in front of it, the
faster the vehicle moves the less time there is for the air molecules to move out of the way
and so more energy is required to move them, this energy comes from the kinetic energy
created by the motion of the vehicle. This can be reduced with smooth surfaces.
b. Skin Friction Drag. Air consists of layers of molecules and those at the surface of the
vehicle attach themselves to it and move at the same speed as the vehicle, the next layer
above it moves a little slower, and this then repeats again and again until a point where the
vehicle no longer has an effect on the air, as each layer moves over the lower one shear
16,17,18
(Dings, 2012)

15. 13
forces are created, the sum of these is the skin friction drag. The volume in which these
layers act is called the boundary layer
c. Pressure or Base Drag. As the air separates from the rear of the vehicle it cannot fill
up the area behind it quickly enough, this creates an area of turbulent vortices known as
eddies which create an area of low pressure, the pressure acting against the front of the
vehicle is now higher than that at the back and causes a vacuum or suction effect on the
vehicle.
45. Of note now is that air resistance is dependent not just on the frontal area, but the sides, top
and back as well. In order to reduce air resistance on a HGV we can add some of these devices:
a. Rounded corners on the cab edges. This allows air to flow easily over the vehicle and
thereby reduce the drag coefficient. The effect of an increased radius on the corner is to
encourage the air flow to remain attached to the body rather than separating from it.
b. Ensure the cab and body are of equal height. This is to reduce the frontal area that air
will act on and reduce the disruption of the air flow. In order to maximise the load carried it
will be likely that the trailer height will be the maximum permissible by law, this is likely to
place it higher than the cab, in order to minimise this a deflector can be fitted above the cab
to push the airflow over the trailer and reduce the difference in height.
c. Ensure a minimal gap between the cab and trailer. This will reduce the vortices
between cab and trailer and reduce the effect of crosswinds being “caught” in between the
gap, it can be reduced by fitting extended seals between cab and trailer which remain flexible
to allow for the vehicle turning.
d. Fit trailer skirts. In order to reduce any crosswind that may sweep under the vehicle and
create an increase in drag coefficient. These can however prove problematic for
inspection/servicing of the vehicle.
Fig 8. Effect of aerodynamic devices19
46. The effects of these items on air flow are illustrated in figure 8 and show the smoothing of air
flow lines over the cab and trailer, the reduction in the drag coefficient is illustrated below in figure
9 which also illustrates how these devices effect the “yaw angle”, this is caused as a result of
19
(Heisler, 2002)

16. 14
crosswinds which can raise the drag coefficient, by knowing the direction of wind and the direction
of the vehicle a vector diagram can be drawn from which the angle of yaw can be calculated.
Fig 9. Effect of aerodynamic devices on Cd and Yaw angle20
Aerodynamic Styling
47. In order to reduce air resistance further we could alter the design of the cab and trailer itself.
Consider the effect of introducing a more curved structure to the cab as evaluated by a Transport
and Environment research paper by FKA automotive research, this is a justifiable source as it is an
independent research company acting on behalf of EU legislators and relates to this assignment.
Shown below is a computer simulated model of the air resistance on the front of a standard and
potential truck (fig 10+11).
Fig 10. Standard HGV21
Fig 11. Improved HGV22
48. By increasing the cab length (currently legislated at 2.35m) by 80 cm and allowing for a more
aerodynamic design incorporating a rounded and aerodynamic frontal area has led to a reduction
in aerodynamic drag by 12%, this is evidenced by the reduced high pressure zone on the improved
design.
49. As discussed under pressure drag we must also address the resistance caused as a result of
the trailer, because of its role the trailer’s most likely shape will be rectangular with an end
perpendicular to the body (a square back configuration), and this has the effect of increasing
pressure drag.
20
(Heisler, 2002)
21+20
(Dings, 2012)

17. 15
50. Pressure drag is caused by the airflow separating from the body at the back of the vehicle, a
highly aerodynamic shape such as a teardrop will mean that airflow separation occurs smoothly at
the back, the square back of the trailer prevents this and so flow separation is not smooth and
pressure drag is roughly equal to the height of the trailers end (base area wake), this creates a
large negative pressure area which has the effect of increasing drag.
51. In order to reduce this effect we can alter the shape of the trailer itself and/or the rear end
configuration, two options are to create a trailer that is physically more tear drop shaped (fig 12),
or, to fit a device at the back known as a boat tail which consist of angled boards at the back of the
vehicle (fig 13).
Fig. 12 Teardrop design23
Fig 13. Boat tail design24
52. The effect of the teardrop design is clear to see, because the airflow remains attached to the
trailer for longer it has a reduced area of negative pressure indicated by the white area between
the flow lines, note also the improved flow between cab and trailer due to the improved shape of
the trailer between it and the cab. The effect of a boat tail is to encourage the air to flow down and
fill the void created easier and so has a similar effect to that of the teardrop in that it reduces the
negative pressure at the back of the truck.
53. While both these designs have shown reductions in drag they are not without their
disadvantages, principle among these are the requirements for a trailer to be a practical item for
carrying loads. For the teardrop design to be aerodynamically efficient it should follow the fineness
ratio where the ratio between its length and height should be between 2 and 4 with the widest part
about a third of the total length from the front (consider the tear drop trailer and cab as one unit),
this therefore places design constraints on the height of the rear as the design requires an initial
increase in height to make the tear shape, height is regulated by law and so a genuine increase in
carrying capacity may prove difficult to achieve.
54. The constantly changing height could also make fitting and securing loads difficult while the
reduction in size of the rear doors can make loading/unloading operations difficult. The boat tail
would seem to negate these downsides however EU laws place strict rules on lengths, heights and
overhangs of vehicles and this could be prohibitive in this case. Further considerations of inter-
operability with other driving cabs must be taken into consideration
55. The ultimate driving goal in improving aerodynamics will be in the value for money it
represents, fuel will make up a large amount in the cost of fleet running and while reducing
aerodynamic drag is an appealing aspect it may not reduce fuel costs in proportion to the cost of
the design ideas. While the reduction of emissions is attractive it will not be fully considered until
legislation either enforces it or, allows it by authorising design alterations as alluded to by the
Transport and Environment FKA report from para 52.
23+24
(Carter, 2011)

18. 16
Trailer Height and Design
56. A critical aspect in aerodynamic drag is the height of the trailer in relation to the height of the
cab, this is known as the t/c ratio. Although the cab is a relatively poor aerodynamic shape it does
contain devices to improve the air flow over and around it, a standard box trailer has few of these
devices, therefore if the trailer height is much larger than the cab height the flow of air will be
hindered by these large area.
57. By dividing the trailer height by the height of the cab gives us the t/c ratio. Wind tunnel tests
have shown that a t/c ratio of less than 1.2 has a negligible effect on drag coefficient (Cd), however
after 1.2 Cd rose considerably due to this increase in surface area (fig 14). The increase is a
combination of the air hitting the surface area of exposed trailer and also that some of the air hitting
the trailer will be directed down between it and the cab, this increases negative pressure.
Fig 14. Effect of t/c ratio on Drag coefficient (Cd)25
58. This increase of Cd amounts to 0.23 which is considerable, the t/c ratio is stopped at 1.5 as
this represents the maximum trailer height allowed by law. In order to reduce this problem most
trucks are designed with a set trailer it can pull and these designs would ensure the t/c ratio to be
1.2 or less. However fleet managers may need to task cabs to pull or backhaul different trailers
depending on the needs of the company, in this instance a cab could be required to pull a trailer
outside of its ideal t/c ratio.
59. To counter this a cab roof deflector can be utilised, this was discussed briefly in para 49.
The deflector will direct the air to flow up and over the air gap between cab and trailer and flow
further down the trailer (fig 15). An adjustable deflector will allow the altering t/c ratios for different
trailers to be accounted for. As discussed previously deflectors are affected by crosswinds which
can sometimes increase Cd.
Fig 15. Effect of deflector on directing airflow26
25+26
(Heisler, 2002)

19. 17
Conclusion
60. The design of vehicles has changed considerably over time, it is still an evolving process
which changes in-line with our understanding of the physics of forces and the development of
materials to make vehicles lighter and stronger. As task 3 has shown a big push for altering
vehicle design comes from fuel efficiency savings and legal requirements for it.
61. This assignment covers a wide range of technical difficulties in vehicle design, by drawing on
lessons covered within the science modules I was able to produce the calculations required.
Because of this I was able to complete tasks 1 and 2 quickly and ahead of schedule which then
gave me more time to consider the effects of aerodynamic resistance, this is a large feature of
vehicle design and I have addressed how much more important it becomes as vehicles travel at
quicker speeds. I had initially underestimated how much work would be required for task 3 but
internet research provided me with an excellent report from the transport and environment agency
which helped a great deal.
62. This assignment was enjoyable to complete, the area of aerodynamics is very interesting to
me and is something I will look to study further at degree level, in hindsight I would’ve liked to have
carried out some computer aided drawing to show the effect of changing a vehicle shape to reduce
drag.
BIBLIOGRAPHY
Carter, M. (2011). Aerodynamics in HGV's.
Defence School of Electronic and Mechanical Engineering. (n.d.). Engine and Vehicle Design and
Performance Student Notes Version 1.
Dings, J. (2012, February). www.transportenvironment.org. (J. Dings, Ed.) Retrieved November 17,
2014, from transportenvironment.
Genral Dynamics Land Systems. (2014, November 11). Retrieved from General Dynamics Land
Systems: http://www.gdls.com/index.php/products/other-vehicles/ocelot
Greer, A. &. (1989). Tables, Data and Formulae for Engineeers & Mathematicians. Cheltenham:
Stanley Thomas.
Heisler, H. (2002). advanced vehicle technology (2 ed.). Oxford: Butterworth Heineman.
Sully, F. (1974). Motor Vehicle Craft Studies 380 (Vol. 1). London: Butterworth & Co LTD.
Ware, P. (2010). Military Jeep. Yeovil: Haynes Publishing.
Wikipedia. (2014, December 16). Retrieved from Wikipedia: http://en.wikipedia.org/wiki/Jeep
Figures
All figures taken from the sources quoted.
Tables
Except where stated all tables produced by the author.
Annex:
A. Conversion table.

20. Annex A
(Student
Handout) to
Unit 25
Dated 15 Dec 14
CONVERSION TABLE