This document discusses how vehicle acceleration is affected by load transfer during acceleration. It contains the following key points: 1) Vehicle acceleration is limited by the coefficient of friction between the tires and the driving surface. For front-wheel drive vehicles, the maximum tractive effort is equal to the coefficient of friction multiplied by the static reaction force at the front wheels minus the transfer of load to the rear wheels during acceleration. 2) For rear-wheel drive vehicles, the maximum tractive effort is equal to the coefficient of friction multiplied by the static reaction at the rear wheels plus the transfer of load to the rear wheels. 3) An example calculation is shown to find the maximum acceleration of a 1.8 tonne

1. Vehicle Dynamics
V e h i c l e A c c e l e r a t i o n – E f f e c t o f L o a d
T r a n s f e r
Dr. Budi Waluyo, MT
5 Oktober 2023
Fungky Dyan Pertiwi, MT

2. V e h i c l e A c c e l e r a t i o n

3. Front wheel drive
The maximum tractive effort (Te) that can be applied is governed by the coefficient of friction ⍵ between
the tyre and the driving surface.
The coefficient of friction
⍵ =
tractive effort
force pressing the surfaces together
= Newton/Newton
This means that
Te = ⍵ × force pressing surfaces together
Tractive
effort (Te)
F=m.a
Rf Rr

4. Static reaction at the front wheels – load transfer
= > RF − F. h/b
So…… Maximum Tractive Effort for front wheel drive:
Te = ⍵.(RF − F. h/b)
Te
so…..
Te = ⍵.(RF − F. h/b)
 F = ⍵.(RF − F. h/b), because F = m.a, so…..
 m.a = ⍵.(RF − m.a. h/b)
 m.a = ⍵.RF - ⍵.m.a. h/b
the front wheels – load
transfer
coefficient of friction
Te = F = m.a

5. Maximum Acceleration – rear wheel drive
In this case, maximum tractive
effort
=>Te = µ[RR + Fh/b]
=>m.a = µ[RR + Fh/b]
=>m.a = µ[RR + m.a.h/b]

6. 
Four wheel drive – fixed
The drive is now transmitted through all four points of
contact of the tyres on the driving surface;
The force pressing down on the surface is the weight
of the vehicle mass × gravitational constant mg
The maximum tractive effort
Te = μ.mg.

7. Example
A vehicle of weight = 1.8 tones has wheelbase of 2.6 m and its Centre of Gravity is
0.7 m above ground level and midway between the axles. If the coefficient of
friction between the tires and the road is 0.7, calculate the maximum acceleration
that is possible on a level surface:
(a) when the vehicle has front wheel drive;
(b) when the vehicle dimensions are the same but rear wheel drive is used.
Take g = 9.81m/s2

8. Solution
Known
m = 1800 kg
b = 2.6 m
h= 0.7 m
⍵ = 0.7
x = y = 2,6/2 = 1.3 m (midway between the axles)
g = 9.81 m/s2
Asked :Max acceleration when the vehicle has front wheel drive
1. Front wheel drive
2. Rear wheel drive
Solution
RF = RR = m. g / 2 = (1800 x 9.81)/2 = 8829 Newton
 By substituting in (m.a = ⍵.RF - ⍵.m.a. h/b)
 So……..

9. 1. For Front wheel drive
m.a = ⍵.RF - ⍵.m.a. h/b
m.a + ⍵.m.a. h/b = ⍵.RF
a.(m + ⍵.m. h/b ) = ⍵.RF
a =
⍵.RF
m + ⍵.m. h/b
a =
⍵.RF
m. (1 + ⍵.h/b)
a =
0.7 x .8829
1800. (1 + 0.7x 0.7/2.6)
= 2.89 m/s2
2. For Rear wheel drive
m.a = ⍵.RR + ⍵.m.a. h/b
m.a - ⍵.m.a. h/b = ⍵.RR
a.(m - ⍵.m. h/b ) = ⍵.RR
a =
⍵.RF
m − ⍵.m. h/b
a =
⍵.RF
m. (1 − ⍵.h/b)
a =
0.7 x .8829
1800. (1 − 0.7x 0.7/2.6)
= 4.23 m/s2

10. Thank You for your attention………………… !!!!!
Continue working on structured assignments….