Thursday 22 July 1993
We always used to think that a bent-back front fork
meant an impact with a solid object. After all, we had seen plenty of those and
we had also seen bicycles pitch over without damaging the front fork when the
front brake had been applied too hard. Logically, it appeared that since the
front fork was sufficiently strong to create pitchover, then a bent-back front
fork could only be caused by a stronger force, such as could only be created by
impact with a solid object. That is not quite true.
A mother and child were injured by falling from a bicycle. The child was being carried on a child's seat that was mounted on the top tube, between the mother's arms. The manufacturer of the seat had provided foot pegs that mounted on each side of the down tube, but had not provided a guard to prevent the child's feet from getting into the spokes of the front wheel. Witnesses said that the vehicle was crossing an intersection with no traffic in sight and no objects or potholes on the road surface. The vehicle pitched over, injuring the mother. After the accident the child's left shoe was off the child's foot. The fork blades were bent backwards several inches and three adjacent spokes on the left side of the front wheel were bent inwards, towards the centerplane of the bicycle, in gentle curves.
How did this accident happen? These analyses show that another mechanism, the sudden jamming of the front wheel, can develop sufficient force to bend the front fork blades backwards. There are two analyses, one by linear momentum and conservation of energy, the other by moment of momentum. The method of linear momentum I developed first. Then Jim Papadopoulos (PhD in mechanics from MIT, with a professional interest in bicycle mechanics) suggested the method of moments of momentum, and I worked the problem according to his suggestion.
The maximum possible braking deceleration is controlled
by the slope of the line joining the center of mass of the bicycle-and-rider
combination with the center of the front tire's contact patch. The maximum
possible braking deceleration, that at which pitchover occurs, is, in g-units,
the tangent of the angle between that line and the vertical. For typical
bicycles this is about 0.67g. Since the full weight of the bicycle and rider are
carried by the front wheel immediately before pitchover and the coefficient of
friction between tire and road usually exceeds 0.67, pitchover occurs rather
than skidding of the front wheel. At this point the decelerating force applied
to the front fork equals 0.67 times the weight of the bicycle and rider.
Furthermore, the force against the front fork appears to be limited by frictional effects to only a little more than that required for pitchover. Since at pitchover the front tire carries the entire weight, and the observed coefficient of friction has always been less than 1.0, the maximum decelerating force available was 1.0 times the weight, a value that forks withstand. If some process developed greater force, say by an inertial mechanism, then that extra force would be absorbed by skidding of the front wheel without increasing the decelerating force against the front fork.
When a bicycle hits a wall or similar object, it is obvious that the forces are not limited by the friction that is available between the front tire and the road. Backwardly-bent front forks and buckled frame tubes are then expected and are frequently found. From this analysis, it appeared that backwardly- bent front forks implied a frontal impact.
However, this conclusion is not accurate. There is another accident mechanism that will bend the front fork without frontal impact. When the front wheel jams suddenly, from something caught in the front spokes or by something jamming between wheel or tire and front fork blades or by extremely sudden application of the front brake to a far excessive value (not possible with the usual brake), then the front fork blades may be bent rearward. The critical factor is the speed with which the wheel- jamming force develops.
If the force gradually increases through the range that causes the rear wheel to lift, the bicycle and rider start to rotate at a slow acceleration and the final wheel locking doesn't take place until the bicycle and rider are rotating at a rate that is determined by the energy of the original forward motion. The sum of the energies of the remaining forward motion, the energy of the rotation, and the energy lost in the brake must equal the energy of the original forward motion. Since the front fork blades are sufficiently strong to lift the bicycle and rider through a couple applied to the front axle and the brake mounting bolt, no damage is done to the bicycle by this action.
However, if the wheel jamming action occurs extremely rapidly, the downward force on the front tire can become so large that the available frictional force exceeds the strength of the front fork blades, in which case the front fork blades can bend backwards instead of the front tire skidding, if some other mechanism produces a backward force. This downward force can be produced as the reaction against the effort of the bicycle to lift the cyclist into the air as the bicycle starts to rotate around the front-tire contact point.
If some action suddenly causes the front wheel to jam, the front forks attempt to rapidly accelerate the bicycle and rider to rotate around the contact point between the tire and the road. That rotation rate is determined by the division of the original energy between the remaining energy of forward motion and the energy of rotation. The following analysis assumes that the wheel is stopped so suddenly that no energy is lost through the friction of the jamming process and that the pitchover proceeds as the front tire rolls along the road surface without skidding. Therefore, the tangential speed of the cyclist about the front tire contact point is equal to the remaining forward velocity times the ratio between the wheel radius and the distance between the cyclist's center of mass and the front-tire contact point. The energy relationships are shown by Eq. 1 which reduces, in this case, to Eq. 2.
M*V*V/2 = M*v*v/2 + (M*v*v*/2)*(R*R/r*r) Eq 1
V*V = v*v + v*v*(R/r)*(R/r) Eq 2
Where M = mass of rider and bicycle
V = original speed
v = final speed
R = radius of gyration of bicycle and rider around tire contact point
r = radius of wheel
For a typical cyclist, both the saddle and the cyclist's center of mass are about 3.3 feet from the front-tire contact point, although not on the same radial line. A typical wheel radius is 1.1 feet.
With r = 1.1 foot and R = 3.3 feet, V*V = 10*v*v and the final forward velocity is 0.32 times the original velocity and the cyclist is rotating about the front-tire contact point at 0.95 times the original speed at an angle about 40 degrees above the horizontal.
The resulting rotational speed about the tire contact point is calculated by the ratio of v to R, adjusted by the factor 2 pi to convert from radians per second to revolutions per second, according to Eq 3.
Rev/sec = (v/R)/(2*pi) Eq 3
If the structure remained intact, in the typical case of 22 fps (15 mph) original speed with r = 1.1 ft and R = 3.3 ft, the resulting forward speed would be 6.8 fps (4.6 mph) and the rotational speed about 2.27 radians per sec, which equals 0.36 revs/sec. The cyclist's center of mass would be moving at 20.9 fps relative to the front-tire contact point (which is also moving) and inclined at an angle about 40 degrees above the horizontal. The upward component of his motion would be 13.4 fps.
The time in which the forward motion converts to the combined forward and rotary motion is quite short. It can be estimated from the time in which the bicycle would travel the distance that the front fork blades are bent backwards. A bend distance of 5 inches at 15 mph equals 0.02 secs. A shorter time would imply larger forces.
The acceleration to achieve the cyclist's upward velocity of 13.4 fps in 0.02 secs is 20.8 g. Even assuming that because of the flexibility of the cyclist's body only about 20 pounds of his body is actually moving at the end of the 0.02 secs, this requires a force of over 400 pounds. The reaction of the upward force from the bicycle's saddle to the cyclist's body is an equal downward force between the front wheel and the road. Thus the downward force at the front tire contact point increases from 150 pounds to over 550 pounds. This allows a retarding force of about 440 pounds, depending on the actual coefficient of friction, which is sufficient to bend the front fork.
This upward force is produced at a distance of about 1.8 feet behind the front tire contact point, so it requires a torque input through the front forks of about 730 pounds feet. Since the distance between the front brake blocks and the front axle is about 1 foot, this requires a force against the front axle of 730 pounds. This exceeds the maximum force available from friction between the front tire and the road, so that the actual backward force delivered to the front fork is that available from friction, which has been shown to be capable of bending the front fork.
There is another method of showing the large force produced against the front forks. In this same 0.02 second the bicycle has lost forward velocity, from 22 fps to 6.8 fps. This requires a deceleration of 23.6 g. The only force that could retard the bicycle is the rearward force at the front-tire contact point that is developed by the resistance of the front wheel to rotation. The mass of the cyclist should not be counted here, because he probably has decelerated only a small amount, being free to slip forward off the saddle and his arms not yet stiffened to resist that movement. However, the bicycle alone weighs about 30 pounds, so that over 700 pounds of force are required to produce the achieved deceleration.
This analysis shows that the impulse necessary to produce the probable changes in velocity in the probable time requires more force than the front fork blades can withstand. Therefore, they bend backwards and the motions that actually occur are modified accordingly.
This analysis is not complete. The rotational inertia of the bicycle itself is ignored, although it could be measured and used. The rotational inertia of the cyclist's body is ignored and only a small portion of its translational inertia is assumed to be active during the change in velocity. Experimental determination of these under sudden impulse conditions would be difficult. As a result of these and other omissions, the analysis is not complete. It does not accurately predict the final velocities, either translational or rotational, of the bicycle and cyclist, nor the cyclist's posture. It does not predict the magnitude of the wheel-jamming force that is necessary to bend the front fork. It does not analyze the actual movements while the front fork blades are collapsing, nor predict the amount of collapse. However, the analysis appears to be generally conservative in that the matters not treated increase the forces produced.
Therefore it is reasonable to conclude that the accident occurred as the child's left foot caught in the spokes of the front wheel. The motion of the spokes fortunately removed the child's shoe from his foot, thus saving his foot. That motion then carried the shoe into the gap between the spokes and the fork blade, which was too narrow for it. In a very few inches of forward motion the shoe bent three spokes toward the centerplane of the bicycle, producing in those few inches of forward motion, and hence in a very short time, a very high frictional grip between wheel and fork. Both pitchover and bending back of the front forks resulted from the sudden application of this grip.
Tuesday, 17 August, 1993
I enclose the printouts of the analysis of forkbending by linear momentum and by moments of momentum. As I have said before, I have learned from you. One thing that I learned is that the analysis according to angular momentum can be used even when there is no initial rotation, provided only that there are motions that would exert a torque if interfered with. The lm and the mm values are quite close, and also quite close, for the speed of 15 mph, to the value that I calculated by hand for that speed in the article. The force at the fork tips in the mm analysis should be corrected by a factor of about 0.9 to allow for the slope of the line between fork tip and fork crown. Then the mm values are about 0.85 of the lm values, for ostensibly the same bike and conditions. Whichever way one looks at this problem, there is obviously plenty of force to bend forks at any normal cycling speed. Naturally, I hope that you will review this work for accuracy.
I make the following replies to your comments. and questions.
Does a front-brake induced pitchover lock up the front wheel? I used to say that it did, that that was what caused the pitchover, particularly since static friction is greater than sliding friction. I said so only last week, in fact, under oath (but I don't think that the question or answer are significant to that case). If the brake did lock up the front wheel, then, by the logic of the problems that I have just analyzed, that would bend the front fork. Since such braking causes pitchover instead of bent forks, the logical conclusion is that the front brake merely lifts the cyclist while the front wheel is still sliding through the brake. Of course, once the cyclist is moving upward and the weight on the front wheel reduces, then, if the cyclist is still holding the brake levers tightly, the brake will lock the wheel. The critical factor appears to be the time between the initiation of pitchover and the lockup. If it is short, the forks will bend; if it is long, the force will be less than the forks can withstand. Suppose the lockup occurred after 45 degrees of rotation. That would be about 1 foot of forward motion, so that the forces would be about only 1/4 of those calculated for the forkbend case. Looks like forks can stand that until the speed gets high.
Is the horizontal component of force the proper measure of the tendency to bend the fork? I think it is. Consider a cantilever beam held horizontally with a weight hanging from its tip. Suppose that the weight was heavy enough to cause the beam to yield. Would applying another force at the tip of the beam directed horizontally toward the support make the beam stronger? I think that it would weaken the beam because of the additional buckling factor that would be introduced. In other words, the lower edge of the beam would be more inclined to buckle than before. The same effect would be seen with fork blades. Are there experimental data on this?
Saddle distance, more nearly 2.8 than 1.8 feet. My mistake.
While the tire contact patch cannot apply torque it can apply, or have applied to it, forces that produce or result from torques. If the cyclist's butt is accelerated upward by some force that is the result of a torque, the opposite side of the couple that developed the torque must appear somewhere. Consider the frame of the bicycle as a type of weightless beam rotating under a torque. It will lift the cyclist's butt with the same vertical force that it applies to the ground contact point. Consider one of those brake diagrams that you sent me, with a brakeblock on an arm that leads its pivot. The normal force on the brakeblock is equal to and opposite in direction to the parallel component of the force in the pivot.
Does rolling along the periphery of a jammed wheel imply translational movement as well as rotational movement. Maybe it depends on how you analyze the problem; one route needs that movement, another doesn't. I am still dubious about the arguments on both sides.
R is equal to the radius to the center of mass because the cyclist's butt is assumed to be a point mass, or at any rate one that doesn't rotate about itself to any appreciable extent during the short time in which the fork blades are being bent.
The radian rotation rate is defined as v/R because the radian rotation rate is determined by the energy in the rotating masses, whose radius is R, not in the wheel, whose radius is r.
I fully recognized that in using energy conservation I was assuming that the jamming effect occurred very rapidly and consumed negligible energy. Frankly, I used that approach because I was more sure of my knowledge about that than about momentum; the concept of this analysis came to me first. Only afterwards did I realize that I could have done it by momentum, and your urging and my curiosity led me to do so. Remember, I did not recognize that conservation of moment of momentum could be used to analyze this situation; you told me that.
If I read you correctly, you say that while the limit of front-wheel braking is pitchover (on normal surfaces), sliding might occur if the cyclist becomes entirely disconnected from the bicycle, because the c.m. of the bicycle would then be low enough to allow it. There is a simple test. Put a bicycle on a table that can be tilted, and see whether it slides or pitches. Naturally, the material of the table's surface must match typical road surfaces in coefficient of friction. I just did this, on a rough-sawn redwood plank, and the bike slid before tipping. However, you must also remember that once the cyclist becomes disconnected, his legs probably drag over the handlebars, and this might provide sufficient additional moment to cause pitchover.
I don't propose the above as the words for your letter to CS; my words may make some changes in your thoughts. If you want to send me another proposed letter, we can work it out and send it jointly. I think that it would be useful to say that we now have tried three different methods of computation, and they produce roughly equal results.
The next thing that I will do for CS is to mull over your words about analyzing self-energizing brakes and see what changes I should make in the paper. You have remarked that I approach these problems in unusual ways that often turn out to be accurate. That's in part because of modern computational aids coupled with an inadequate grounding in classical techniques followed by a lot of forgetting. I understood the physical process of the self- energizing brake (I have been a car enthusiast also, where these problems are more relevant). I understood that it was a cyclic feedback process with, normally, diminishing results with each cycle. I could set up the first cycle on my HP programmable, and then repeat the cycle many times by merely pushing the key until I saw no change to the number of significant digits I cared about. For me, that was faster and more reliable than trying to work out the proper mathematical expression for the sum of the series. With MathCad I can do it more reliably. This approach does limit the understanding that a more formal approach provides. When the terms in a denominator go to zero, there will be trouble, but if you never see the denominator in that form you won't pick up that understanding until you try different input values and find that some don't work.
With best regards,
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