(i) Introduction ^^^^^^^^^^^^ This is an essay about braking on a bicycle. Its conclusions must not be transferred to tandems, recumbents, three-wheels 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 maths-physics 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. 111-113). 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 counter-clockwise. 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 action-reaction 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 non-inertial (that is, the bike doesn't move evenly), we have to define a so-called inertial force. It is -m*a. _________________________________ | | | (3a) mf*a*r = (Bf-Ff)r | | (3b) mr*a*r = (Br-Fr)r | |_________________________________| The torques (turning moments of forces) applied to the wheels are non-zero because they don't rotate evenly. The scheme is: "Moment of inertia * Angular acceleration = Sum(Force*Moment arm)" "Moment of inertia" is a mass-like 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. Counter-clockwise 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(h-r) + ((Nr-mr*g) - (Nf-mf*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 h-r 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(r-h) = (Nf-Nr)L/2 + (mr-mf)g*L/2 - (Bf+Br)r + m*g*d Substituting "a" from (6) and using M'= m + 2(mf+mr): (Bf+Br)m(r-h) + (Nf-Nr)M'*L/2 + (mr-mf)M'*g*L/2 + m*g*d*M'- (Bf+Br)M'*r = 0 Hence (Nf-Nr)M'*L/2 = (Bf+Br)(2r(mf+mr) + m*h) - m*g*d*M' - (mr-mf)M'*g*L/2 So / h m 2(mf+mr) 2r\ Nf - Nr = (Bf+Br)|2 - * - + -------- * -- | + (mf-mr)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 mr-mf 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 well-known 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(L-C) 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 \ C-L | | (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*(B-L)/H). Br in- ^ / creas- | / ing | / / | u | / | | / / | / | / | / | / / | / / --0------/-------X------------/---> Bf | / / / | B / | / / M*g* - / | / / / /H | / / / |// / / B-L |/ / / 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 u-adhesive 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 non-flipping 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 coaster-braked 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 (quasi-horizontal) and front (quasi-vertical). 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| -0-------------------------X-----> 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 made-adhesive 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 constant-line 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\ _\ constant-line |OOOOOOOO\\ |OOOOOOOOO/ \ |OOOOOOOO/ | |OOOOOOO/ front skidline | --0--------------------------X-----> 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 \ B-L (14) Br (>)= Bf*| --- -1| + M*g* --- \u*H / H Equating them: 1 B / L \ B-L -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 ] B-L / 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)(L-u*H) u*M*g*B L-u*H B-L (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 | _- - _ | |_- - _ | -0-------------------------X-----> Bf | The optimal curve goes through the flipping-point 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 life-saving, 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 endo-point, 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 well-known 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 well-known 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 re-transform 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 fellow-cyclists 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,beginplace-40)-(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,beginplace-size*y/ratio) LINE (1,beginplace-y1*size/ratio)-(size*x,beginplace-size*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".