Road Road LoadsLoads
vehicles. Low racers incur relatively little load shift, while upright bicycles may experience a significant amount.
Equations 7-31 are valid for both positive and negative accelerations (increas- ing and decreasing speed, respectively). As noted, human-powered vehicles generally are not capable of achieving high positive accelerations under most conditions. Braking does produce large negative accelerations, and the load shift from rear to front axle may be very significant for vehicles with high or forward centers of gravity. In extreme cases, this can lead to vehicle pitchover—the vehicle flips forward, usually sending the rider over the handlebars in a dangerous acci- dent. Upright bicycles and tadpole tricycles tend to be more prone to braking pi- tchover. Tadpole trikes often have a center of gravity closer to the front axle than the rear (large b) while upright bikes have a high cg (large h).
To be strictly complete, the drive torque reaction acting on the vehicle frame also contributes somewhat to the load shift. Under heavy load, there can be a sig- nificant torque at the rear wheel. This is particularly true at very low speed when the operator is exerting a large force on the pedals. The frame reacts against the wheel torque, which also unloads the front wheel. This is a small effect, and may generally be neglected.
Braking Braking
Peak braking forces on human-powered vehicles can be much larger than peak acceleration forces from hard pedaling. The ability to stop in a short distance Figure 7-7
Figure 7-7 Free body diagram used to find the longitudinal load shift during braking
Design of Human-Powered Machines Design of Human-Powered Machines
Example 7-3 Example 7-3
A 77 kg cyclist rides a 14 kg long wheelbase recumbent bicycle. The wheelbase is 1.6 m, while the center of gravity position and height are 56 cm and 75 cm respectively. Compare the front wheel load fraction for riding at constant speed on level ground with that for 0.5 G braking on a 6% down grade.
Solution:
For riding on level ground, the front wheel load is (77 14) 9.91 .56 312 1.60 f mgb W N L + × × = = =
The corresponding weight fraction is
312 .35 893 f f f f W W f f W W = = =
This is a typical value for long wheelbase vehicles, which tend to have lightly loaded front wheels. For braking on a 6% downslope, the front wheel load and the corresponding front wheel weight fraction is:
893 ( ) (.56 .05 .75 .6 .75) 547 1.60 547 0.61 893 f x f f W W h Gh b A N L W f W = − − = + × + × = = = =
Note that both the grade term and the deceleration term are positive, un- like the first Equation 7-31. This is because the grade is negative (downhill) and the acceleration in negative (slowing).
The increased load on the front axle is rather significant in this example, although the rear wheel is not fully unloaded. It is instructive to look at the braking and down hill components separately, to see which has the greater effect. A total of 234 Newtons was transferred to the front axle, of which 25 N were due to the down grade and 209 N were due to braking. This illustrates how important braking can be on vehicle loads.
Road Road LoadsLoads
while maintaining control is an important safety attribute of any vehicle. Good braking performance is particularly important when riding in traffic in order to avoid collisions. Even with good brakes, most upright bicycles require a longer stopping distance than a typical automobile with the same initial speed. Stopping power for recumbent bicycles is often superior to a comparable upright bike, due to the lower CG height and the greater distance between the CG and the front axle. For either type, the dangers of inadequate stopping distance must still be addressed during the design stage.
Like acceleration, the maximum braking force is limited by either loss of trac- tion or by shifting so much weight to the front wheel that the rear wheel is lifted off the ground. The first case results in reduced braking performance and possible loss of control, either of which can lead to an accident. In good traction condi- tions, the forward weight shift may cause the rider to be thrown over the handle- bars, potentially sustaining serious head or neck injuries.
The longitudinal load shift from the rear wheel to the front wheel always oc- curs during braking. The pitchover threshold, at which the vertical force on the rear wheel just goes to zero, limits braking when traction is good. This can be found from Equation 7-31 by setting the rear wheel load to zero and solving for the acceleration. Assuming level ground, the term Gh may be omitted, giving the following braking limit:
,Unload rear wheel
( ) x L b A h − = − (7-32)
The negative sign simply indicates that the acceleration is negative, that is, the vehicle is braking. Some skillful riders can brake hard, lifting the rear wheel off the ground while maintaining control of the vehicle. More often this results in a hazardous accident in which the rider is thrown over the handlebars. While it is possible to exceed this limit on recumbent vehicles, it is more common with upright bicycles and tadpole trikes.
In some cases, traction is not sufficient for a vehicle to pitch over. In this case, the vehicle skids prior to unloading the rear wheel. Usually, the rear wheel begins to skid first, as the vertical load is reduced. Once skidding starts, braking is ad- versely affected and the deceleration is decreased. This usually precludes both skidding and pitchover from occurring. However, if the front brakes have good traction and the CG is forward, the rear wheel can skid and, with increased front braking, the vehicle can still pitch over. Tadpole trikes are more prone to this than many other vehicles.
Design of Human-Powered Machines Design of Human-Powered Machines
On level ground, traction limited deceleration is found by noting that F B,max = m
m mg. Applying Newton’s second law givesmm mg = – ma x. Dividing through by mg,
and solving for the deceleration gives the skid limit for braking.
A x,skid = –mm (7-33)
This equation differs from that for traction-limited acceleration, Equation 7-23, only in sign. The maximum deceleration (in G’s) is just equal to the braking coeffi- cient. Both limiting braking equations are reproduced here for clarity.
, , Pitchover Limited: Traction Limited: x MAX x MAX L b A h A
−
= −
= −µ
(7-34)Due to the load transfer during deceleration, it is unusual for most vehicles to lose traction on both axles at the same time. With bicycles, the rear wheel generally loses traction first, due to the reduced vertical load. Maximum braking performance will occur at a deceleration that just keeps the rear wheel in contact with the ground or just keeps the wheels from skidding. Exceeding these limits can increase stopping distance and risk loss of control of the vehicle.
Braking Performance Braking Performance
Braking performance may be measured either by the time or the distance to come to a full stop from a given initial speed. Equation 7-17 can be used to ana- lyze stopping distance by setting the drive force to zero and assuming a constant deceleration. Assume that the braking force is constant during the deceleration. Also assume a constant grade and that the rolling resistance is independent of velocity. The relevant equation is then
{ F B + F RR + mgG} + F AERO = ma x (7-35)
The terms in the curly braces can be taken as constant retarding forces, desig- nated FD for convenience. The drag force cannot be combined with the remaining forces because it is dependent on the squared velocity of the vehicle, and hence does not remain constant during the braking process. However, the magnitude of the drag force is small compared to the braking force, particularly at stops made
Road Road LoadsLoads
from low to moderate speeds. Under these conditions, drag can be neglected with little loss of accuracy.
For simplicity, initially assume a stop made from a low initial speed such that aerodynamic drag can be neglected. With F AERO = 0 and this simplification, and as-
suming that F D is constant, the deceleration will also be constant. The distance
and time required to stop can be calculated based by noting that x
dV a dt = − , so D dV F m dt
= − . Integrating and solving for t gives the time to decelerate:
0 1 ( ) s F x t V V a = − (7-36)
Where V 0 is the initial speed and V F is the final speed of the vehicle. Letting
V F = 0 gives the time required to come to a full stop.
0 0 1 1 SD x x t V V a gA
=
=
(7-37)Likewise, the distance required to come to a full stop is given by:
(7-38)
Note that the stopping time is proportional to the initial speed, whereas the stopping distance is proportional to the speed squared. If the initial speed is dou- bled, the stopping time is doubled, but the distance required to stop is quadrupled. For stops made from high initial speeds, aerodynamic drag becomes signifi- cant. With the drag term included in Equation 7-35, the deceleration is not con- stant, even for constant F D. The stopping distance with drag included is given by
2 ln 1 2 D SD o D D m C A x V C A F
ρ
=
+
ρ
(7-39)Neglecting the drag term, as in Equation 7-38, always results in overestimating the stopping distance. If used in design, this results in an extra margin of safety.
The difference between Equation 7-38 and Equation 7-39 increases with initial speed—the faster the initial speed the more wind resistance will help to stop the vehicle. Likewise, a more streamlined the vehicle (low drag area C D A) is affected
less by drag and requires more distance to stop. At low initial speeds the differ- ence between the two equations is slight.
Design of Human-Powered Machines Design of Human-Powered Machines
Example 7-4 Example 7-4
A pedicab loaded with two passengers must make an emergency brake to stop from 8 m/s on a downhill grade of 3%. The total vehicle mass is 300 kg. The coefficient of rolling resistance Crr is .008 and the drag area C D A is 1.5
m2. The braking friction coefficient is 0.80. The pedicab has a wheelbase of
2.0 m, with a center of gravity located 0.40 meters forward of the rear axle and 1.3 m above the ground.
Determine the time and distance to stop under maximum braking. Solution:
First, the limiting deceleration must be determined. The pitchover limit is given by , 2.0 0.40 1.23 1.3 x pitchover L b A h − − = = =
The skid limit is similarly given by:
A x, skid =mm = 0.80
Since A x,skid< A x,pitchover , the limiting factor is a skid. Assume that the pedi-
cab driver is sufficiently skilled so as to maintain braking just prior to initiat- ing a skid. The deceleration is then 0.80 Gs.
Next, find the retarding forces due to the downhill gradient and rolling resistance. 300 9.81 0.008 23.5 N N 300 9.81 ( .03) 88.3 N N rr rr grade F mgC F mgG = = × × = = = × × − = −
Notice that the force due to the grade is negative because the pedicab is rolling downhill. Neglect aerodynamic drag for the time being. The total re- tarding force F D is then equal to ma x, as in Equation 7-35. The braking force
is thus given by 300 9.81 0.80 23.5 ( 88.3) 2419 N N B D rr grade x rr grade B F F F F mA g F F F = − − = − − = × × − − − =
Road Road LoadsLoads
This is a very large braking force. Pedicabs must be equipped with power- ful brakes in order to stop the large loads they must carry. The time and dis- tance to stop, neglecting air resistance, is given by Equations 7-37 and 7-38:
2 2 1 1 8 1.5 s 0.8 9.81 1 1 8 9.2 m 2 2 0.89 .81 s o x s o x t V A g x V A g
=
=
×
=
=
=
=
×
×
Compare the stopping distance above with that obtained by Equation 7-39, which takes into account aerodynamic drag. Assuming a standard at- mosphere withρ = 1.226, the stopping distance is:
2 2 ln 1 2 300 1.2 1.5 ln 1 8 8.9 m 1.2 1.5 2 2354 D s o D D s m C A x V C A F x
ρ
=
+
ρ
×
=
+
=
×
×
A loaded pedicab is a large, heavy, and has a large drag area. A vehicle such as this one would reach terminal velocity on a 3% grade at about 8.5 m/s, so the operator in this case would not need to work to obtain the 8 m/s speed. However, even though drag was significant and the initial speed relatively slow (for most HPVs), the difference between including and neglecting drag was only 30 cm over 9 m of stopping distance, or 2.8%.
CHAPTER
CHAPTER
8
PEED AND
PEED AND
POWER MODELS
POWER MODELS
C
ycling performance can be modeled and predicted fairly easily once the road forces are understood. Road loads such as aerodynamic drag, rolling resis- tance, and hill climbing resist the rider’s efforts. If the rider is generating more power at the cranks than the total resistive power, the vehicle will speed up. Con- versely, if she is generating less power, the vehicle will slow down. Performance models relate the vehicle speed and ride conditions to the power required of the rider. They can predict performance for competitive events and the potential for success in record-breaking efforts. They are also useful for determining if a ve- hicle is “fast” or “easy to pedal.” In this case, “fast” is taken to mean that under typical operating conditions and with an average rider generating average power, the vehicle is faster than an average or typical vehicle. “Easy to pedal” may mean the same thing, but often the implication is that at average speeds the power re- quired is less than that for an average vehicle.The external factors that resist motion include aerodynamic drag, rolling re- sistance, bearing and drive train friction, and changes in potential and kinetic en- ergy. Martin et al.1 conducted a study to determine if mathematical models using
these terms can accurately predict a cyclist’s power requirements during cycling. Their findings demonstrated that such models can do so with high accuracy. Of course, the accuracy of any model depends on the accuracy of the parameter val- ues used. An accurate estimate of drag power, for example, requires an accurate value of the drag coefficient. Parameter estimation in the design stage can be
difficult, and is discussed in other chapters.
The power required to ride a vehicle under a given set of conditions is the sum of the road forces acting on the vehicle, each multiplied by the road velocity of
1Martin, James C., Douglas L. Milliken, John E. Cobb, Kevin L. McFadden, and Andrew R.
Coggan, “Validation of a Mathematical Model for Road Cycling Power,” Journal of Applied
Mechanics, 14, 276–291, 1988.
S
S
Design of Human-Powered Machines Design of Human-Powered Machines
the vehicle. Drive train friction is not a road load, and is accounted for by divid- ing the required power by the drive train efficiency. The road power formula in a generic form is given in Equation 8-1.
1 { } TOTAL AERO RR WB PE P P P P P P ∆ΚΕ ∆ = + + + + η (8-1) where
Mechanical efficiency of the drive train Aerodynamic drag power
Rolling resistance power Wheel bearing friction power
Change in kinetic energy Change in potential energy
AERO RR WB KE PE P P P P P ∆ ∆ η = = = = = =
Many of these terms have been discussed in the previous sections. They will be reviewed here for clarity and consistency.
Drive Train Efficiency Drive Train Efficiency
All mechanical drive systems incur some power losses. This is due to friction in the chain, pedal and bottom bracket bearings, idlers, and gears, if so equipped. (Frame flexure and torsion due to the cyclical pedal loads can also contribute to power losses.) These loads are generally not evaluated individually, but are lumped together as system losses.
When actually measuring power output while riding, the instruments used may or may not include the drive train power losses. Some commercially available systems use an instrumented rear hub, while others use instrumented cranks. (Other methods, ranging from chain vibration to heart rate, have also been used to measure power, usually with less accuracy.) If power is measured at the crank, drive train power losses are included in the power measurement. Hub based sys- tems measure power downstream from the drive train and will not capture these losses. That is, the cyclist is actually generating somewhat more power than indi- cated by the power meter.
Drive train efficiency is defined as the ratio of the power available at the drive wheel divided by the power supplied by the crank.
Speed Speed and and Power Power ModelsModels Drive Wheel Crank P P η = (8-2)
Drive train efficiencies for typical bicycle chain drives in good condition and properly lubricated are quite good, often exceeding 95% and under ideal conditions exceeding 98%. The best hub gears approach these efficiencies, particularly at high power levels. Intermediate drives, chain guides and other components sometimes used on recumbent and front-wheel drive vehicles can generally be expected to re- duce efficiency. Drive train efficiencies are discussed in more depth in Chapter 12.
Aerodynamic Drag Aerodynamic Drag
Aerodynamic drag is the most significant factor in the power equation for mod- erate and high speed riding. Only for low speed vehicles or on steep grades does its significance diminish. The formula for computing aerodynamic drag presented in Chapter 7 is given as if the vehicle were placed in a wind tunnel, with the wind always coming from directly in front of the vehicle. On the road, this is seldom the case. The wind component produced by the motion of the bicycle (known as the ground speed,V G) must be added to the natural windV W to obtain the relative
wind velocityV R. The drag force depends on the component of the relative wind
velocity that is tangent to the direction of the vehicle’s motion. This is thehead-
wind component, designatedV H . The power, however, is the product of the drag
force (which depends on the head wind speedV H ) and the ground speedV G. Thus
both velocity terms appear in the drag power formula:
2 1 ( ) 2 AERO H D G P
=
V Cρ
A V
(8-3)Given the direction of the wind DW and the direction of the vehicle DV , the
headwind component can be found by
V H =V G +V W cos( DW – D B) (8-4)
WhereV G is the vehicle speed over the road,V W is the wind speed, and DW and D B are respectively the directions of the wind and the bicycle. The bicycle direc-
tion is the compass heading that the vehicle is going, while the wind direction is the compass direction from which the wind is blowing. The termV W cos( DW – D B)
Design of Human-Powered Machines Design of Human-Powered Machines
is the headwind component of the natural wind. The crosswind component, also known as the normal wind component, is given by:
V CW =V W sin( DW – D B) (8-5)
The relative wind strikes the vehicle at some angle to the direction of motion of the vehicle. This angle is known as the aerodynamic yaw angle, and is given by
1 ( 1 cos( ) tan tan sin( ) CW G W W B A H W W B V V V D D Y V V D D −
−+
−
=
=
−
(8-6)The aerodynamic yaw angle may be significant because drag area—the prod- uct of drag coefficient and area—will vary with wind yaw angle. Depending on the vehicle, the drag area may increase or decrease. The drag force and consequently the power to overcome drag usually increase, particularly at high yaw angles, even if the headwind component remains constant. This is why a cyclist riding with a direct crosswind will not ride as fast as on a calm day for the same power output. A noticeable tailwind component must be present to help the rider. Gen- erally, the wind direction must be more than 100º from the direction of the vehicle to be beneficial. Streamliners can be exceptions to this rule. It is possible for a streamliner to develop a thrust force in the presence of a crosswind. The thrust force can more than compensate for the increased drag, making even small yaw angles advantageous. However, even some unfaired road bikes may experience a