(i) Introduction
^^^^^^^^^^^^
This is an essay about braking on a bicycle. Its conclusions
must not be transferred to tandems, recumbents, threewheels or
cars without pondering the possible consequences. Its pieces of
advice are not compulsory to anyone. All remarks are welcome.
I am Hungarian, male, aged 22; my profession will be mathsphysics
teacher. I don't take the responsibility for the English of this
essay. I allow everyone to correct my grammar mistakes.
Almost everything written below is the product of my brain. The only
reference I used is a Hungarian book on cycling (Dr. Sandor NAGY:
Cyclists' Book. Sport, Budapest, 1988). I verified two of my formulae
from there: (11) and (13). The author didn't give the deductions but
only the results (pp. 111113).
I also read the article of John Forester (jforester@cup.portal.com)
found in rec.bicycles FAQ 9.17. It is maintained by Mike Iglesias
(iglesias@draco.acs.uci.edu). This article didn't contain formulae
but rather summed up the gist of the problem in a nutshell.
Here I authorize everyone to read, copy, distribute etc. this essay
in whatever form, by all media, in its entirety or excerpts from it.
Indication of my name is desirable. However, there are some parts of
it (namely vi/b) which can't stand on their own. They are integrated
into the course of the whole essay for pedagogical considerations,
and they are further elaborated right after their occurrence.
I hope that having completed secondary school and general sensitivity
towards everyday's physics will be sufficient to understand the essay.
,
Written by Ferenc NEMETH, Budapest, 1996
in the honour of the 1100 years of Hungary.
Program
^^^^^^^
We are going to discuss the problems of braking a bicycle.
Basically with centerpull/sidepull brakes, but our observations
can be extended to coaster brakes, too. First we realize what
kind of forces are applied to the various parts of the bike,
then set up our model: rigid body with two wheels. We write up
the determining equations, solve them and arrive at interesting
conclusions. Air resistance and other dissipative effects are
neglected. There are some simplifying assumptions about the
wheels, too. The results apply to flat terrain.
We shan't deal with situations involving continuous skidding;
when skidding starts then our research ends.
(ii) Notation
^^^^^^^^
LET THE SYSTEM OF REFERENCE MOVE TOGETHER WITH THE BIKE.
Let the cyclist move to the left.
Let the positive x axis point to the right.
Let the positive y axis point upwards.
Let us put the origin into the point which is halfway between the axles.
Let us appoint the positive sense of rotation counterclockwise.
a = acceleration
g = gravitational acceleration (=9.81 m/s^2)
u = coefficient of adhesive friction (0...1)
*f = front *
*r = rear *
m = mass of the frame and the cyclist together
mf, mr = mass of the wheels
L = the distance between the axles
r = radius of wheels
h = height of the mass centre of the "bike"
d = position of the mass centre of the "bike" behind the point
which is halfway between the axles
B = braking force
A = force at the axle (horizontal)
F = frictional force at the ground
N = normal force at the ground
All the forces are of positive values. Directions are represented by +/.
(iii/a) Forces applied to the wheel
^^^^^^^^^^^^^^^^^^^^^^^^^^^
Braking force
______>
/ \
/  \
/ Forces  exerted \ Rotation:
/ by the v fork \ ___
/ < \ / \
 <  v O ^
\ Inertial Weight / \ ___ /
\ force v /
\ ^ Normal /
\  force /
\ _____ /
_________________>_____________________________ground_______
Adhesive friction
(iii/b) Forces applied to the frame and the cyclist
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
Inertial force <X

====================
Braking force  Weight /\
</ \  <\ Braking force
/ \ v / \
^ / \ / \ ^
 / \ / \ 
/ \ / \ 
o> \ /================= o>
Forces exerted O Forces exerted
by the axle by the axle
With a little imagination everyone can understand why the forces at the
axle and at the brakes are indicated twice: they constitute actionreaction
pairs. Thus their magnitude is equal and their direction is opposite.
The weight of the cyclist actually presses the saddle, the pedals, and
the handlebars, yet it is clear that these forces can be substituted by
one single force drawn from the mass centre of his body. Namely, because
the substitutory downward force exerted by the cyclist on the bike equals
the upward force with which the bike upholds him. This force, in turn,
is equal to the weight of the cyclist, in order that the resultant of
the forces he experiences vertically, could be zero.
For these considerations we can depict the cyclist and the bike (without
the wheels) as a united rigid body. We shall call it "bike" from now on.
(iv) Equations
^^^^^^^^^
The following equations describe the motion of the bicycle:
_____________________________________
 
 (1) (m+mf+mr)g + Nf + Nr = 0 
_____________________________________
The vertical forces between the forks and the axles, as internal ones,
are dropped out by adding the equations of the vertical equilibrium of
the wheels and that of the "bike". (1) is the result.
_______________________________________
 
 (2) m*a + Af + Ar  Bf  Br = 0 
_______________________________________
It shows the horizontal equilibrium.
As the motion is noninertial (that is, the bike doesn't move evenly),
we have to define a socalled inertial force. It is m*a.
_________________________________
 
 (3a) mf*a*r = (BfFf)r 
 (3b) mr*a*r = (BrFr)r 
_________________________________
The torques (turning moments of forces) applied to the wheels are
nonzero because they don't rotate evenly. The scheme is:
"Moment of inertia * Angular acceleration = Sum(Force*Moment arm)"
"Moment of inertia" is a masslike property characteristic of the
wheel. The greater it is, the more difficult it is to make the (still,
removed) wheel rotate. It is proportional to m and r^2. I used the
approximation m*r^2 because the heavy parts are the rim and the
tyre. Actually it might be 0.8 m*r^2.
Angular acceleration is a(tangential)/r, where a(tangential) is of
the same magnitude as the deceleration of the bike. It is true only
in the absence of skidding.
"Moment arm" is the distance of the force vector from the axis of
rotation. Counterclockwise torques are positive in our system.
_____________________________________
 
 (4a) mr*a + Fr + Br  Ar = 0 
 (4b) mf*a + Ff + Bf  Af = 0 
_____________________________________
They refer to the horizontal equilibrium the wheels.
Of course, I could have united them with (2) but I didn't
do so because of pedagogical reasons. The reader must see
this method in operation, and if possible, in the presence
of more variables.
______________________________________________________________________
 
 (5) m*a(hr) + ((Nrmr*g)  (Nfmf*g))L/2 + (Bf+Br)r  m*g*d = 0 
______________________________________________________________________
This one is the most sophisticated of all. Its meaning is that the
resultant torque applied to the "bike"is zero.
m*a is the inertial force (pointing forward, as the inertial force
has to be the opposite of m*a observed from still system) applied
to the mass centre. Its height is hr above the origin, which is
the axis of the possible rotation of the bike. It doesn't mean
that the bike can dig itself into the ground. It is known from
physics that the rotational equilibrium of a rigid body can be
described with whatever possible axis I want. If it doesn't rotate
about our chosen axis then it won't do it about other axes either,
provided the two observers placed on the axes don't sense each
other rotate.
The following part is the torque caused by the vertical forces at
the axles. Their arms are L/2. The braking forces also have turning
effect on the "bike", with arms of r length. At last, due to the
biker's posture, his weight turns the "bike" clockwise, that is, in
negative sense. Other forces (those horizontal ones at the axles)
don't have turning moments because they have zero moment arms, the
direction of their vectors passing through the axis of the (non)
rotation.
(v) Solution
^^^^^^^^
The goal we pursue is to find a relation between the braking forces,
normal forces and the coefficient of friction, in order to predict
what happens. Therefore, from (4ab) we get
Af = Ff  Bf + mf*a and Ar = Fr  Br + mr*a.
Write it into (2):
m*a = Ff + Fr  a(mf+mr)
__________________________
Using (3ab), we get  
 Bf + Br 
m*a = Bf + Br  2a(mf+mr), so  (6) a =  
 m + 2(mf+mr) 
__________________________
We can now turn to the most awesome task: first multiply (5) by 1:
a*m(rh) = (NfNr)L/2 + (mrmf)g*L/2  (Bf+Br)r + m*g*d
Substituting "a" from (6) and using M'= m + 2(mf+mr):
(Bf+Br)m(rh) + (NfNr)M'*L/2 + (mrmf)M'*g*L/2 + m*g*d*M' (Bf+Br)M'*r = 0
Hence
(NfNr)M'*L/2 = (Bf+Br)(2r(mf+mr) + m*h)  m*g*d*M'  (mrmf)M'*g*L/2
So
/ h m 2(mf+mr) 2r\
Nf  Nr = (Bf+Br)2  *  +  *   + (mfmr)g  m*g*d
\ L M' M' L/
Comparing it with (1) (m+mf+mr)g = Nf+Nr:
/h m 2(mf+mr) r\ m*g / 2d\
Nf = (Bf+Br)  *  +  *   + 1    + mf*g
\L M' M' L/ 2 \ L/
and
/h m 2(mf+mr) r\ m*g / 2d\
Nr = (Bf+Br)  *  +  *   + 1 +   + mr*g
\L M' M' L/ 2 \ L/
We can introduce new properties, namely
m 2(mf+mr)
H := h *  + r * 
M' M'
and M and D that satisfy the following equations:
M*g / 2D\ m*g / 2d\
a) 1    = 1    + mf*g
2 \ L/ 2 \ L/
M*g / 2D\ m*g / 2d\
b) 1 +   = 1 +   + mr*g
2 \ L/ 2 \ L/
Adding a) to b), we obtain:
M = m + mf + mr
and subtracting one from the other:
m mrmf
D = d *  + L * 
M 2M
Thus we can express the normal forces by means of the braking forces:
______________________________________
 
 H M*g / 2D\ 
 (7) Nf = (Bf+Br)  + 1    
 L 2 \ L/ 
 
 H M*g / 2D\ 
 (8) Nr = (Bf+Br)  + 1 +   
 L 2 \ L/ 
______________________________________
Before proceeding further, let's make some remarks.
(a)
From (7) and (8) we can see that both the front and the rear normal
forces are dependent on the sum of the two braking forces, and not
on them individually.
(b)
The push of the front wheel on the road increases with braking,
and that on the rear one decreases. It is invaluable to know this.
(c)
We could get rid of the radius of the wheels with a formula in which
the coefficient of r is negligible. That is, r causes a small correction
in the height. Likewise, in the definition of D the correction caused by
the mass of the wheels is also negligible.
Thus we reduced the problem to the case of radiusless and weightless
wheels. They just act as transferring agents of force. They are, however,
of great importance because they transform the rubbing effect of the
brake pads into adhesive friction at the ground. The lighter the wheel,
the greater the efficiency, because the mass of the wheel has rotational
inertia: it also requires force to slow it down.
(d)
It is relieving to see that if the rider sits back (so he increases D)
then the rear wheel will be burdened heavier, and the front one lighter.
This remark, however trivial it may be, isn't intended to consider the
reader a fool. In solving such problems, one has to constantly verify
if he didn't make an arithmetical mistake somewhere, and also that the
physical model, with all the simplifying assumptions, describes the
problem properly. And if the results agree with such trivial facts as
the above one then we can still hope that we are on the right path.
From now on I'll consider the mass of the wheels zero, so the efficiency
is 100%. This means that the braking forces are equal to the frictional
forces at the ground (Bf=Ff and Br=Fr). Of course, I could have declared
it at the beginning, too, but then no one would believe me that I have
the right to do so. But it's obvious that I have it: it causes only some
percents' deviation from the truth.
(vi) Simple applications
^^^^^^^^^^^^^^^^^^^
All we have now is two equations (7,8) including four unknowns:
Bf, Br, Nf, Nr. That said, we are pressed for new equations.
It is a wellknown empirical fact that the maximum dragging force
which a body lying on the ground can experience without skidding,
is proportional to the force that pushes it downwards:
_________________________
 
 (9) Bf <= u * Nf 
 (10) Br <= u * Nr 
_________________________
where u is the coefficient of adhesive friction.
To be able to compute things, we'll treat these inequations as
equations, and pose the question "when will the wheels skid?"
It's useful now to introduce new parameters: B and C.
B:= L/2 + D
C:= L/2  D
Origin Mass centre
 
<L/2><L/2>
<D>
X====================O====================X
Front  Rear
axle  axle
<><>
B C
With these, (7) and (8) get the following form:
___________________________________
 
 H C 
 (7') Nf = (Bf+Br)  + M*g*  
 L L 
 
 H B 
 (8') Nr = (Bf+Br)  + M*g*  
 L L 
___________________________________
We'll use H=1.2 m
L=1.5 m
D=0.05 m
B=0.8 m
C=0.7 m
u=0.9 as standard parameters.
(vi/a) Skidding the rear wheel with the rear brake only
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
Two equations refer to this:
B H
(10) Nr=Br/u and (8') Nr = M*g*   Br* 
L L
Solving them, we learn that
B
Br = M*g 
L/u + H
For the frictional force is the only external force which can
slow down the bike, M*a = Br. So the result is
_________________________
 
 B 
 (11) a <= g *  
 L/u + H 
_________________________
Substituting the standard parameters (good road, typical bike) into it,
we conclude that the rear brake cannot cause more deceleration than
0.3 g. Not because all rear brakes are inherently bad, but due to the
geometry of the bike: at 0.3 g so little weight is left on the rear
wheel that it skids even at a minor increase of breaking force.
(vi/b) Skidding the front wheel with the front brake only
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
Two equations refer to this:
C H
(9) Nf=Bf/u and (7') Nf = M*g*  + Bf* 
L L
Solving them, we get
C
Bf = M*g* 
L/u  H
With the same train of thought, M*a = Bf. Thus the result is
__________________________
 
 C 
 (12) a <= g *  
 L/u  H 
__________________________
The divisor is much less than in the previous case, so it can be
expected that the maximum achievable deceleration with the front brake
is much greater than with the rear one. So far it would be totally
in accord with everyday experience. Well, if we substitute the
standard parameters into this formula then we get a(max) = 1.5 g.
It's a very surprising result because from the theory of the simpler
models (eg. a mass point on the floor) we know that a body skids at
once when the horizontal force applied to it exceeds u*M*g. On the
grounds of analogy, one would expect of the bike that it also obeyed
this more general rule.
Where is the error hidden?
I wilfully concealed the details of the rear wheel. If we examine the
(8') formula of the rear normal force with the value Bf = 1.5 M*g, it
turns out to be negative! So the rear wheel "sticks" to the ground!
What an arrant nonsense! Of course it will leave the ground and over
throw the careless rider. Consequently, the solution of the dilemma
posed above is that at a certain level of front braking force,
flipping will take place instead of skidding. But exactly what
is the necessary frictional coefficient and the maximum
applicable braking force?
(vi/c) Flipping the bike with front brake
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
Suppose that the bike just manages not to flip the rider, so Nr = 0.
Other relevant equations: Nf = M*g and (9) Bf <= u*Nf.
Using the simplified (7'):
H C
Nf = Bf*  + M*g* , we get
L L
H C
Mg = Bf*  + M*g* 
L L
Hence
M*g(LC) M*g*B
Bf =  = ; cf. with (9) Bf <= u*Nf = u*M*g:
H H
_________________
 
 (13) u >= B/H 
_________________
It is required so that flipping could take place.
With the standard parameters: u >= 0.66
To such u's the maximum applicable braking force is M*g*B/H.
Otherwise the rear wheel won't remain on the ground.
(vii) Using both brakes
^^^^^^^^^^^^^^^^^
We have still not dealt with the simultaneous brake usage, which is
infected with many indelible superstitions, as John Forester writes.
For example, that using the rear brake can prevent flipping, no matter
how hard I apply the front one. To do away with them, again much
mathematics is needed.
(vii/a) Skidding of the front wheel and flipping, using both brakes
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
We have to use
H M*g C
(7') Nf = (Bf+Br)  +  * 
L 2 L
and (9) Nf (>)= Bf/u.
Equating these, we get
___________________________________________
 
 / L \ CL 
 (14) Br (>)= Bf*  1 + M*g*  
 \u*H / H 
___________________________________________
Let's draw the graph of Br against Bf! Depending on the value of u,
these are straightlines, all going through the point (0;M*g*(BL)/H).
Br in
^ / creas
 / ing
 / /  u
 / 
 / /  /
 /  /
 / /  / /
0/X/> Bf
 / / /  B /
 / / M*g*  /
 / / / /H
 / / /
// / /
BL / / /
M*g*  _//
H 
Take a particular u and the straightline belonging to it. The
area which is on its right hand side denotes the combinations
of (Bf;Br) to which the front wheel will skid on a uadhesive road.
A very important observation: if flipping is possible then
the weight on the rear wheel tends to zero with increasing force
on the front brake lever, so the application of the rear brake
will just skid the rear wheel but won't help with avoiding the
"endo" because it can't increase the rear normal force.
So a vertical line can be drawn at Bf = M*g*B/H, the critical
front brake force, which is alone responsible for flipping.
The critical u, to which the moving straightline intersects
the magic flipline (that is, in the "valid" quarter of the plane,
where Bf>=0, Br>=0) is B/H, as we deduced formerly in (vi/b).
Then we substituted Bf = M*g*B/H and Br = 0. Now we see it
in a more general picture: instead of moving along a straightline,
we move on the whole plane.
On the other hand, while the rear braking is effective (no skidding
of the rear wheel) it further increases deceleration, putting even
more weight on the front wheel, thus allowing a bit stronger front
grip to the cyclist. See the graph: if your brake combination is on
the nonflipping side of a straightline, and you increase the front
grip, then your combination goes to the right and you may skid.
But if you increase both grips simultaneously then you probably still
remain safe because your combination is more likely not to go over
to the right hand side of the skidline.
Alas, this technique can hardly be tested in practice. As the
measurable graph will show, the front skidlines are either very
close to the vertical (on slippery road), or they meet the rear
skidlines discussed below so soon (on sticky road) that in order
to remain on the safe area, you have to handle the rear brake
very finely. But who dares to cope with such subtleties when
he is just about to skid the front wheel, which is much more
dangerous than doing this with the other one?
(vii/b) Skidding of the rear wheel, using both brakes
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
We have to use
H M*g B
(8') Nr = (Bf+Br)  +  * 
L 2 L
and (10) Nr (>)= Br/u.
Equating these, we get
____________________________________________________
 
 1 B 
 (15) Br (<)= Bf*  + M*g*  
 L/(u*H) + 1 L/u + H 
____________________________________________________
Again, drawing the graph of Br against Bf we get a range of straightlines,
depending on the value of u. It's very convincing that they all go
through the point (M*g*B/H;0), that is, the magic border of flipping.
Just substitute Bf = Mg*B/H into the above equation, you will get zero.
Br
^
 u
\ing \
creas \ \
In \ \ \
 \ _ \ \
 \ _ \ \ 
 \ _ \\
0\_> Bf
  B
 M*g* 
  H
As u tends to infinity, the grades of the straightlines approach 1.
In extreme case (infinitely great frictional coefficient) the
triangle which is cut from the valid area, will be isosceles.
Thus the interception on the vertical axis converges to M*g*B/H.
Due to this circumstance, the lamentable fact is that on a very
adhesive surface, greater deceleration than g*B/H cannot be achieved
with one brake. The front brake will flip you, the rear will skid
the wheel.
Despite this "convergence" bluff, one has to bear in mind that the
interceptions on the vertical axis (denoting the maximum applicable
rear braking force when using it exclusively) are about thrice less
than those to the front wheel, when the frictional coefficient has
reasonable values (0.1...1). So if you are allowed to put only one
more brake on your coasterbraked bike, let it be a front one.
(vii/c) Optimal braking
^^^^^^^^^^^^^^^
Let's sum up the facts about the skidlines. They are straightlines
in the graph of (Bf;Br); their points denote certain combinations of
braking forces. There are two kinds of them: rear (quasihorizontal)
and front (quasivertical). To each frictional coefficient one rear
and one front skidline belongs. You have to remain under the rear one
to avoid rear skidding. You have to remain to the left from the front
one in order not to skid the front wheel. In addition, you have to
be to the left from the magic vertical line of flipping, positioned
at Bf = M*g*B/H. And of course, you are supposed to apply positive
or zero braking forces, negative never (because the brakes are not
engines). So the valid combinations fall into the positive quarter.
From the previous diagrams it is clear that the allowed combinations
constitute a quadrilateral, and to bigger frictional coefficients
a triangle, on the diagram.
Br
^

:uuu
O::::uuu
++OOO::::uuu
o+++++OOO::::uuu 
ooooo++++OOOO::::uuu  Flipping
oooo++++OOOOOOOOO::::uuu 
oooo++++OOOOOOOO::::::::u
0X> Bf

ooo are the combinations you can use on icy road;
ooo and +++ are those on damp road;
ooo, +++ and OOO are those on sandy road;
ooo, +++, OOO and ::: are those on asphalt road;
ooo, +++, OOO, ::: and uuu are those on madeadhesive road.
On slippery roads no endo is possible because the front wheel
will skid instead. (The interception of the skidlines, which is the
rightermost point of the allowed area, lies left from the flipline.)
As we know, to combinations near flipping, almost no weight remains
on the rear wheel so it skids to the finest rear braking. I've never
made experiments to verify this, but it could be a safe method
of testing whether this greater danger is ahead. One must apply
the rear brake gently, increase the front grip gradually, and
if the rear wheel skids then slacken it a bit.
Optimal braking is achieved by the greatest allowed sum of braking
forces because they are equal to the resultant of the external
(frictional) forces which are alone able to slow down the bike.
Looking at the diagram it's easy to realize that the optimal combination
(to which Bf+Br is the maximum) is the interception of the skidlines
belonging to the same u. One has to draw straightlines with the grade
1 through the graph; along these Bf+Br is constant. The farther they
are from the origin, the bigger this constant is. We have to find that
particular constantline which meets the allowed area in one point.
It goes through the mentioned interception because the grades of all
rear skidlines are greater than 1, and those of the front ones are
positive.
Br
^
rear  skidline
\ _ \
OOOO\ _\ constantline
OOOOOOOO\\
OOOOOOOOO/ \
OOOOOOOO/ 
OOOOOOO/ front skidline 
0X> Bf

Let's compute the value of the optimal sum of braking forces!
To this, we use the equations defining the skidlines.
The rear one:
1 B
(15) Br (<)= Bf*  + M*g* 
L/(u*H) + 1 L/u + H
The front one:
/ L \ BL
(14) Br (>)= Bf*  1 + M*g* 
\u*H / H
Equating them:
1 B / L \ BL
Bf*  + M*g*  = Bf*  1 + M*g* 
L/(u*H) + 1 L/u + H \u*H / H
Multiplying both sides by L/(u*H) + 1:
[ / L \2 ] BL / L \ B
Bf*[   1] + M*g*   + 1 = Bf + M*g* 
[ \u*H/ ] H \u*H / H
So
/ L \2 B / L B\ / L \
Bf*   = M*g*  + M*g*   M*g*    + 1 =
\u*H/ H \ H H/ \u*H /
M*g*L^2 M*g*L M*g*B*L M*g*L u*H + L  B
=  +    =  * 
u*H^2 H u*H^2 H u*H
____________________________________
 
 u*M*g 
Finally,  (16) Bf =  * (u*H + L  B). 
 L 
____________________________________
Such is the front component of the optimal braking pair.
To determine the rear one, we substitute our result into
the equation of either skidline, let it be the front:
u*M*g (L+u*H)(Lu*H) u*M*g*B Lu*H BL
(14) Br =  *    *  + M*g* 
L u*H L u*H H
Dismissing the tedious computation (verify it yourself):
__________________________________
 
 u*M*g 
 (17) Br =  * (B  u*H) 
 L 
__________________________________
Here is the punchline! If you add Bf and Br, you'll get u*M*g,
greater friction than which is theoretically impossible!
So the bike as a rigid body can live up to our expectations
and it is able to exploit the sources of friction fully!
However...
If we have a look at the following sketch showing the route of
the optimal point as the function of u, a shocking thing is revealed:
Br
^


 
 _  _  Flipping
 _  _ 
_  _ 
0X> Bf

The optimal curve goes through the flippingpoint X, ... and what
happens if u >= B/H? Does it sink under zero?
From the mathematical point of view, we have never made use of the
fact that Bf and Br are positive or zero. So to remain on the optimal
curve, one has to apply negative braking force, in order to achieve
negative normal force under the rear wheel, according to the relation
between friction and normal force: Br <= u*Nr.
Mathematically it's possible, but it's an obvious stupidity from a
physical point of view. The cyclist cannot produce negative push under
his wheels, even by pedalling (which would be "negative braking").
Our equation should be abs(Br) <= u*Nr. Omitting "abs", we actually
excluded pedalling from the scope of our research.
So on adhesive road we have to put up with a deceleration g*B/H,
even if the theoretical u*g is greater than that. The reason is
that the bike is not a mass point, to which u*g applies. It is
a body with macroscopic size: it can be tilted and overthrown.
Let me explain it on the simplest model of rigid body. The bike
is imagined as a brick which we pull to the left. (The pull
equals the m*a inertial force which the bike experiences due
to its deceleration.)
L
<>
_______________________
  ^
  
 Pull (m*a)  
 <O   H
  Weight  
 v (m*g)  
o_______________________ v
/^\
Let's suppose that the friction is great enough to prevent the brick
from skidding. (We can put an axle fixed to the ground across its
lower left edge.) The condition of its not revolving about this axle
is that the resultant torque of the indicated two forces about the
axis be clockwise or zero. The moment arms are H/2 and L/2.
Thus, it is required that M*a*H/2  M*g*L/2 <= 0,
L B
or a <= g* . It resembles our result to the bike: (11) a <= g* .
H H
All what you can do in order to enlarge your bike's capacity to slow
down, is to move your bottom backwards and down. Ten centimetres can
be lifesaving, however, this gain is counteracted by the evident in
stability of the strange posture. And you have to plan it beforehand!
(viii) Closing remarks
^^^^^^^^^^^^^^^
Well, we are past the simplest part of the problem. We know how to
decelerate the bike with one brake. We have learnt the difference
between front and rear brake efficiency, and its reason. We know
about the danger of endo and we have a method of testing its proximity.
We can also use both brakes simultaneously and we are skilful in
finding the optimal combination of the front and rear grip.
Yet there are some tasks ahead.
(a)
We don't know the exact nature of the optimal curve. With dull
mathematical methods one could prove that it is a parabola whose axis
is tilted in 135 degrees (points to 4:30 direction). The parabola
goes through the origin in all cases but its axis  only if D = 0.
It could be shown that the maximum of the curve (which requires
the greatest rear braking force among all optimal combinations
belonging to various u's) has the frictional coefficient B/(2*H),
that is, the half of the critical u of flipping. The height of
the maximum is M*g*B^2/(4*H*L), which is approximately 0.09*M*g.
If you are braking optimally you'll never have to apply the
rear brake harder!
The maths part is called "parametred curve". It can be compared
to a tourist who climbs a hill between two villages. The hill
plays the role of our optimal curve. He climbs up slowly and
walks down quickly, so it is possible that although the peak is
not exactly in the middle between the villages, climbing takes
the same time as descending.
In this simile "time" stands for the "parameter" of the curve, that
is, u. After u hours of elapsed time the tourist is on the optimal
point belonging to the frictional coefficient u. After B/(2*H) hours
he is on the peak. At B/H hours he reaches the endopoint, viz. the
second village.
While walking, he writes down his height above sea level, and also
his horizontal distance from the initial village (eg. in every five
minutes) and finally gives us his notebook. We have to reconstruct
the form and the height of the hill using this book.
(b)
It can be proven without much labour that the position of the
brakes relative to the frame doesn't alter the formulae (7) and
(8), thus no one can make bikes that can't be overthrown by
merely placing the brakes elsewhere.
Speculatively, it's indeed impossible to change the interaction
between the bike and the road by effecting internal changes on the
bike. All bikes can be compared to bricks, and all bricks can
skid, be flipped, etc.
(c)
Another interesting problem is that of the braking on a slope.
If the depression angle is y then we could set the axes of the
coordinate system parallel to the ground and perpendicular to it;
include m*g*sin(y) between the horizontal forces and change all
m*g's to m*g*cos(y). After tremendous efforts, the following
general equations would be the result:
___________________________________________________________________
 
 H / C \ 
 (7'') Nf = (Bf+Br)  + m*g*  + mf*g*cos(y) + Q*m*g*sin(y) 
 L \ L / 
 
 H / B \ 
 (8'') Nr = (Bf+Br)  + m*g*  + mr*g*cos(y)  Q*m*g*sin(y) 
 L \ L / 
___________________________________________________________________
where Q is about 0.01. So the part containing sin(y) can be neglected
to all angles and braking forces. After introducing G:=g*cos(y) and
A:=a+g*sin(y) our equations get the same structure as those describing
the motion on flat surface. The results will be of the same form,
with G instead of g, and A instead of a. At given u, according to our
results, A<=u*G. You can maintain your speed on a long slope (a=0)
if g*sin(y)<=u*g*cos(y) or
_____________________
 
 (18) u >= tan(y) 
_____________________
This is a wellknown criterion of a mass point's not sliding down
a slope, so our model goes over into a more general one again.
However, there is a snag here. This formula allows the cyclist
to remain in still state on all slopes if u is great enough.
Whereas it's nonsense: he will be flipped on a "wall".
Of course, the cause again is that the bike is not a mass point.
According to the general solution, flipping will take place if
u>=B/H and A>G*B/H. Substituting a=0, we get g*sin(y)>g*cos(y)*B/H,
or ____________________
 
 B 
 (19) tan(y) >  
 H 
____________________
Summing up the dangers on slopes: if u<=B/H then you won't be able
to travel at the same speed on slopes exceeding arctan(u) because
you will skid; and in case of u>B/H you mustn't risk life and limb
on slopes steeper than arctan(B/H) (67%=33.7 degrees) because you'll
be flipped.
(d)
Even air resistance can be squeezed into the course of our essay.
There is a wellknown formula for it: F(air)=k*v^2, where k is a
coefficient characteristic of the bike, the rider, and the posture,
and v is the velocity relative to the air. We have to simplify the
case: let's suppose that the force of the air resistance can be drawn
as a horizontal vector pointing backwards at the height of the mass
centre. It's not a necessary fact because air resistance doesn't
depend on the vertical weight distribution but on how the cross
sectional area of the bike and the rider is arranged vertically.
With this assumption, let's realize that the (unbraked) bike is
pushed forward with a force equalling m*g*sin(y)  k*v^2.
If we can introduce new properties or modify the old ones so that
the expressions of the forces could be of the same form as in the
beginning of the previous section (g and y), we are ready.
Let's introduce g' and y' so that
g'*cos(y') = g*cos(y) and g'*sin(y') = g*sin(y)  k*v^2/m.
Dividing the respective sides of the equations, we get
______________________________________________
 
 (20) tan(y') = tan(y)  k*v^2/(m*g*cos(y) 
______________________________________________
From here we can determine y'. Substituting it into the first equation:
g' = g*cos(y)/cos(y').
These manoeuvres can be compared to placing the cyclist on another
planet (because of the new g), and on another slope (due to the new y).
The second consideration is very plausible because we all know that
no one can be accelerated on a slope over a certain speed: the air
resistance will make him approach a maximum speed and not exceed it.
When will it take place? When the resultant "horizontal" force will
be zero. That is, with no braking, k*v^2 = m*g*sin(y). Looking at
(20), we learn that tan(y') is zero. It corresponds flat road.
More generally, a cyclist rushing down a slope in the presence of air
resistance can instantaneously be pictured as if he were travelling on
a less steep slope (or even climb) on another planet without atmosphere.
As we saw, a simple slope can further be turned into flat terrain,
by introducing G and A. Thus we managed to give our equations the
same form as the initial (1)(5). However, when applying the results
of (c) and (d), one has to bear in mind that they are written in the
language of G. So he has to retransform them to the normal g using
the equations which define the new properties.
(e)
Our research could be extended to swerving, too. However, I am going
to write another essay which will basically deal with continuous motion.
(As everyone could realize, so far I shied away from hinting at
the description of the motion as a process.) I put the issue of
swerving into this new essay in order to excite your interest.
(ix) Appendix
^^^^^^^^
For those fellowcyclists who are bored of my vain sophistry, and
demand tangible data, I offer the list of a Quickbasic program which
draws a measurable graph of the most important diagram. It has to be
used with SCREEN 12 in order to be enjoyable, and with SCREEN 2 to
be printable. (Run GRAPHICS.COM and then in QB hit PrintScreen.)
SCREEN 2 requires some correction of the coordinates to be measurable.
First I drew a CIRCLE(50,50),50 and a box with LINE(1,1)(100,100),,B
then experimented with the box sizes until it looked like a square.
The "ratio" in the program is x(box)/y(box).
The list of the program is the following:

L = 1.5 :REM These parameters can be altered.
H = 1.2 :REM You can build various "bikes"
B = .8 :REM However, it is required that B <= L.
size = 930 :REM It is the size of the screen.
ratio = 2.536 } { ratio = 1
beginplace = 173 } or { beginplace = 330
SCREEN 2 } { SCREEN 12
LINE (1,beginplace)(630,beginplace)
LINE (1,1)(1,beginplace)
LINE (size*B/H,beginplace40)(size*B/H,beginplace)
FOR u = 0.04 TO 0.64 STEP 0.04 :REM the step can be altered.
x = u * (u * H + L  B) / L
y = u * (B  u * H) / L
x1 = (L  B) / (L / u  H)
y1 = B / (H + L / u)
LINE (size*x1,beginplace)(size*x,beginplacesize*y/ratio)
LINE (1,beginplacey1*size/ratio)(size*x,beginplacesize*y/ratio)
NEXT u
LOCATE 1,2: PRINT "Rear"
LOCATE 2,2: PRINT "braking"
LOCATE 3,2: PRINT "force"
LOCATE 2,30: PRINT "Measurable graph"
LOCATE 17,70: PRINT "Flipline"
LOCATE 18,70: PRINT "(M*g*B/H)"
LOCATE 23,55: PRINT "Front braking force"
t: t$=inkey$: if t$ = "" THEN GOTO t

Note: As I said, all comments are welcome. (About grammar, physics,
biking, typos, unclear why's, etc.) If you have some, write
to MX%"nemo@ludens.elte.hu".