Force calculation for chassis
VD-Rev Mechanics
Geometric consideration
The most important part of motorcycle designing is determining the geometry of the chassis.
It results in the overall behavior of the motorcycle. The geometry determines the overall
stiffness, aerodynamics, rider position and comfort, and also affects handling,
manoeuvrability and responsiveness. The parameters that primarily dictate the geometry
include:
1. Castor angle (Rake angle)
2. Trail
3. Wheelbase
4. Steering offset
These parameters form the basis of our motorcycle geometry. The inter dependency of
these parameters are given by the following equations:
   1. a=an/cosϵ, where ‘a’ is the ground trail, ‘ϵ’ represents the rake or caster angle and ‘an’
      is the normal trail
   2. bn=(p + a) cos ϵ, where ‘bn’ is the rear normal trail and ‘p’ is the wheelbase
   3. an=Rf sin ϵ -d, where ‘Rf’ is the front wheel radius and ‘d’ is the steering offset each
        parameter affects the performance and handling of the motorcycle differently thus
        there is no optimal value for each parameter. The type of geometry required depends
        on the characteristics expected for the motorcycle by the rider. Through numerous
        iterations we arrived at the optimal geometry for our motorcycle.
Each parameter affects the performance and handling of the motorcycle differently thus
there is no optimal value for each parameter. The type of geometry required depends on the
characteristics expected from the bike:
1. Castor angle: Castor angle is the angle between the center axis of the steering head and
the vertical center axis of the front wheel. The caster angle varies according to the type of
motorcycle: from 19° (speedway) to 21-24° (competition or sport motorcycles), up to 27-34°
(touring motorcycles). From a structural point of view, a very small angle causes notable stress
on the fork during braking. Since the front fork which is of a beam like structure is rather
deform able, both flexion ally and torsion ally. Hence, small changes in the values of the angle
will lead to drastic changes in stress experienced and therefore greater deformations, which
can cause dangerous vibrations in the front assembly (oscillation of the front assembly around
the axis of the steering head, called wobble). Castor angle without an offset in a motorcycle
geometry is the sole contributor to the trail which dictates the steering characteristic.
2. Trail: The primary function of trail is to provide the desired steering characteristic. It can be
classified into real and ground trail. Ground trail is the distance between the center of tire
contact point and the point where the steering axis meets the ground. Front and rear tires
have their respective trail values. Real or mechanical trail is measured at right angles to the
steering axis. Real trail is reduced from the ground trail by the cosine of the rake angle. For
practical purposes we use real trial while designing the geometry of a motorcycle. If the value
of the trail were negative (the contact point in front of the intersection points of the steering
head axis with the road plane) and considering that friction force is always in the opposite
direction of the velocity of slippage, a moment around the steering head axis that would tend
to increase the rotation to the left would be generated. One can observe how friction force F
would amplify the disturbing effect, seriously compromising the motorcycle’s equilibrium.
This is the reason why we avoid having a negative trail value. Since the load on the front
wheels is high due to the weight of the motorcycle (weight distribution of the protype is 55%-
45%), the choice of a small trail lowers the value of the torque that the rider must apply to
the handlebars to execute a given maneuver which gives a competitive edge in motorsport
but a balance has to be achieved. As the trail is decreased the directional stability of the
vehicle decreases as at higher speeds which is not ideal in motorsports as the safety of rider
is vital when travelling at high speed.
3. Wheelbase:
In general, an increase in the wheelbase, assuming that the other parameters remain
constant, leads to:
1. An unfavorable increase in the flexional and torsional deform-ability of the frame. These
parameters are very important for maneuverability (frames that are more deform able make
the motorcycle less maneuverable).
2. An unfavorable increase in the minimum curvature radius, since it makes it more difficult
to turn on a path that has a small curvature radius, in order to turn, there must be an
unfavorable increase in the torque applied to the handlebars.
3. A favorable decrease in transferring the load between the two wheels during the
acceleration and braking phases, with a resulting decrease in the pitching motion.
4. A favorable reduction in the pitching movement generated by road unevenness
5. A favorable increase in the directional stability of the motorcycle.
The value of the wheelbase varies according to the type of motorcycle. It ranges from 1200
mm in the case of small scooters to 1300 mm for light motorcycles (125 cc displacement) to
1350 mm for medium displacement motorcycles (250 cc displacement) up to 1600 mm, and
beyond, for touring motorcycles with greater displacement. Wheelbase is also influenced by
the components that need to be fit inside the chassis. Considering the effects of above
parameters, we conducted multiple iterations to arrive at the ideal vehicle geometry for our
prototype.
                               (Figure 1) Geometry of a motorcycle
                             [COSSALTER, Vittore. Motorcycle dynamics.
                                      2n edition. 2006. Pag.4]
Squat characterization
In the rear suspension the squat ratio ‘R’ is an important parameter to consider as it decides
how the rear suspension reacts to the different moments generated, both static and dynamic.
(Figure 3) Squat geometry
[COSSALTER, Vittore. Motorcycle dynamics.
2n edition. 2006. Pag.246]
We define the squat ratio R as the ratio between the moment generated by the load transfer
and the moment generated by the sum of the chain force and the driving force:
R = (Ntr L cos φ)/(S L sin φ + T L sin(φ – n))
Where,
Φ= swingarm angle
L=swingarm length
S= moment generated by driving force
T= moment generated by chain force
Ntr= load transfer
n is given by
n= arcsin[(L sin Φ + yp-(rc – rp))/Lc)
where, yp is the vertical coordinate of the drive sprocket shaft rp and rc represent the radii of
the drive sprocket and the rear sprocket, b represents the vertical distance between the axis
of the sprockets, Lc is the length of the straight-line section of the chain. Expressing the load
transfer as a function of the driving force, the ratio is a function of only the geometric
characteristics.
R= (h*cos Φ)/(A[sin Φ + Rr*sin(Φ-n)/rc)
According to the squat ratio three cases can occur:
1. Point A lies on the straight line of the load transfer, that is σ=τ ; in this case R = 1. During
the thrust phase there are no additional moments operating on the swinging arm, so the
suspension spring is no longer stressed compared with the static condition.
2. Point A lies under the straight line of the load transfer, that is σ<τ ; in this case R >1 : the
moment generated by the resultant force Fr causes a compression of the spring in addition
to the one created by the static load.
3. Point A lies above the straight line of the load transfer, that is σ>τ ; in this case R<1 :The
moment generated by the resultant force Fr causes the extension of the spring.
The squat ratio ‘R’ can also be expressed as
                                         R=tan τ/tan σ
Where,
1. τ (load transfer angle)
2. σ (squat angle)
Detailed static load cases
As dynamic analysis was not conducted a factor of safety of 1.5 - 2 is favorable as per previous
experience to compensate for fatigue and to compensate for random factors a factor of safety
of around 2 was considered appropriate.
The following forces act on a motorcycle:
- The weight mg acts at its center of gravity.
- The driving force T, which the ground applies to the motorcycle at the contact point of the
rear wheel.
- The vertical reaction forces Nf and Nr exchanged between the tires and the road plane.
    Parameter (Unit)                         Symbol           Value
    Mass of the vehicle (Kg)                 md               220
    Wheelbase (mm)                           A                1345
    Castor angle (degree)                    E                23
    Offset (mm)                              d                30
    Weight distribution (%) (F-R)                             55-45
    COG to rear (mm)                         Gx               739.75
    COG height (mm)                          Gy               420 - 480
    Max velocity of vehicle (km/hr)          Vmax             100
    Radius of rear wheel (mm)                Rr               306
    Radius of front wheel (mm)               Rf               293
    Radius of rear sprocket (mm)             Rcrown           113.788
    Max power (W)                            P                2000
    Swingarm inclination angle (degree)      phi              6.04
    Angular velocity (m/s)                   Ω                22.22
    Radius of curvature (m)                  Rc               40
                                 EXTERNAL FORCES
1. Maximum acceleration
Knowing that during the maximum acceleration the front wheel or the normal on the front
wheel will be approximately neglected.
Nf = 0
Nr = Ndr-Tdr = md*g
Tdr = md*g*(A-Gx)/Gy
2. Max braking
Nr = 0
Nf = md*g
Tdf = md*g*Gx/A
                                    INTERNAL FORCES
1. Max acceleration
Resolving forces across x and y direction,
Tdr-F1+Fcb*cos(phi)-Fcu*cos(phi)=0      - (1)
Nr-F2+Fcb*sin(phi)-Fcu*sin(phi)=0       -(2)
Vmax = 100 Km/hr = 27.777m/s
Vmax = Wwheel*Rr
Wwheel = Vmax/Rr = 27.777/0.306.9 = 90.50 rad/s
Vchain = Wwheel*Rcrown = 90.50*0.113788 =10.29 m/s
Fcb = W/Vchain = 2000/10.29 = 194.19 N
Fcu<Fcb
Therefore, Fcu = 150 N
Solving equation we get,
F1 = 3081 N
F2 = 2163 N
2. Max braking
AB=0.2929m
BC=0.5435m
CD=0.160m
Tdf*AB+F2*BC-BD*F1=0
F2*CD+Nf*BD*sin (23)-Tdf*(AB-BD)*cos (23) =0;
Solving equations we get,
F1 = 5525
F2 = 6512
3. When in a curve
           i = 384.60 mm
           j = 529.35 mm
           Fcentrifugal = (md*Ω^2)/Rc = 2716 N
           ΣMC=0
           Fcentrifugal(j) – Fsteering axle(i+j)=0
           Solving equations we get,
           Fsteering axel = 1573 N
Different Gy      External                                  Internal
with respective   Forces in                                 forces
Forces            N                                         in N
    Gy(420-
    478mm)                                 Max               Max
  (Increasing)    Max acc                 braking            acc                  Max braking   When in a curve
                     Nr         Tdr         Nf        Tf      Fcb      F1    F2    F1    F2     Fcen   Fsteering
     420            2158       3110        2158      1187    197.6     3154 2163 5525 6512 2716          1573
     430            2158       3038        2158      1187    197.6     3081 2163 5525 6512 2716          1573
     440            2158       2969        2158      1187    197.6     3012 2163 5525 6512 2716          1573
     450            2158       2903        2158      1187    197.6     2946 2163 5525 6512 2716          1573
     460            2158       2840        2158      1187    197.6     2883 2163 5525 6512 2716          1573
     470            2158       2778        2158      1187    197.6     2823 2163 5525 6512 2716          1573
     478            2158       2733        2158      1187    197.6     2776 2163 5525 6512 2716          1573