osco53

It's funny how our rear brakes almost always wear out first,,, Yah !

Yellowbeard

Ok. This is probably gonna be the first of a two or three posts, so hold your breath. All will not be covered right away, but I said I'd draw it, so I'm gonna draw it but It'll probably look like crap because I'm basically trying to pull an infographic out of my ass. Hopefully it'll make sense, at least. *Edit: Images are now embedded for easier viewing and non-members.

And yes, I may make errors. I'm doing this in my head as I go along. Please point them out if you see any.

Starting with the basics; this is just a setup so that the stuff we're actually arguing about will be clearer, when I get around to it, that is.
Diagrams are not to scale.

Ok, let's assume that:
-air resistance and rolling resistance are negligible
-the coefficients of friction for rubber on dry asphalt are about 0.9 (static), 0.65 (sliding).
-our rider and his/her bike combined weigh 85 kg (187.4 lbs). Let's assume a wheelbase of 1 metre (39.4 inches)
-the center of gravity of the rider & bike is 60 cm (23.6 inches) behind the front wheel and 1.2 metres (49.2 inches) high.

Ok, Figure 1: All we're doing here is going through the equilibrium equations for coasting on level ground (given our assumptions). According to Newton's laws, any action has an equal and opposite reaction, and acceleration is equal to net force divided by mass. If all forces in a certain direction cancel out then there is no net force, and therefore no acceleration in that direction. The weight of the bike pushing down on the ground causes a reaction force that pushes up from the ground.

The only forces here are gravity (pushing down) and the reactions (pushing up). For the purposes of friction, the reactions are the normal forces, but we'll get to that.

The sum of all forces must equal zero. If not, then the rider is accelerating in the direction of the net force. Here the sum IS zero.



Figure 2: Rotation works the same way as translation (linear motion). Any two forces that do not act along the same line cause a rotational force (moment, or torque). If the sum of all moments around any single point does not equal zero, then the whole bike and rider is gaining or losing rotational speed around that point. Here we assume that the rider is not somersaulting so that we can calculate how much weight is on each wheel.



With me so far?

Figure 3: Now we throw light braking into the mix. At such low intensity, front vs. rear braking have the same mathematical effect. Note that MAX friction has no effect on our braking force here. The friction equation only tells us the MAXIMUM POSSIBLE frictional force for a given amount of weight on the wheel. Also note that we don't know what max friction is until we calculate the weight shift.



So about that weight shift. Braking produces a new moment.

**IMPORTANT** because the rider is slowing down, we are not in equilibrium anymore. We can't just sum all the moments around the front wheel and get the same answer as before (forces through the point of rotation produce zero moment) because we would have to make up an imaginary force of inertia at the center of gravity to account for the deceleration. I'd rather consider real forces than inertial ones, so let's take the moments around the CG this time. **

Figure 4: Here we use the sum of moments (we're not in equilibrium, but we can still use equilibrium equations to tell us by how much) around the center of gravity to calculate how much weight shifts from the rear wheel to the front wheel under light (0.12g) braking to compensate for the traction force and keep the bike level. I haven't bothered showing my algebra, but it's pretty trivial (two equations, two unknowns, substitution).



What the math tells us is that a higher center of gravity gives a larger moment due to braking, that means more weight shift for the same deceleration. That's why we want to crouch low when we brake hard. It also tells us that if our center of gravity is further back we start with more weight on the rear wheel available to shift forward.

Ok, that's all I've got drawn so far. Stay tuned for Episode 2: Attack of the Figures 5-and-so-on, where we go to the limits of weight shift and beyond.

imi

Wow! Yellowbeard! You stepped up. Thank you. I will be just able to get my head around this, learning lots
Keep it coming!

Megalith

How fast would you need to be going and how hard would you have to brake to get thrown over the bike? I mean, it's practically impossible unless you're speeding downhill, right?

I only use a front brake and appreciate how stopping throws my body forward just enough for a convenient dismount on top of my bike tube.

Yellowbeard

We can calculate it. I'll post an example when I get around to it.

Obviously in real life you're not going to be applying math (or know where your center of gravity is) but it's definitely possible. I almost did it when I was a kid and got my first bike with V-brakes. Rolling slowly down the sidewalk I thought I'd squeeze the brakes as hard as I could to see how well they stopped. I nearly flipped (luckily wasn't moving fast enough) and came back down on my saddle hard enough to tilt it down about 18 notches.

cale

I was wondering where you got your screen name...

79pmooney

If you ride long enough and in enough different situations, you will find a situation that you can get out of still upright with both brakes, but not just a front. Might take 100,000 miles to find it.

I'll keep both. And roll to my next 100,000 next year perhaps one or two spills fewer. Good enough for me.

Favorite bike: A fix gear with two excellent brakes. And good gripping tires.

Ben

79pmooney

I did yellowbeard's calcs a while back. As I recall the max decelleration is around 1/3 g, quite low because our CG is so high. But yellowbeard assumes we can hit 0.9 g with our tires. If that is true, we have well over twice the stopping power required to flip us over the bars if our brakes are up to it. And as I have known the nature of these numbers for decades, I have never been a fan of powerful brakes except rim brakes in the wet. I like to leave end-over-end to the gymnasts.

Freedom2015

Does getting your bag of aluminum cans sucked in the front tire on a bike with no brakes count? Trust me, this story didnt end well. One word "SUPERMAN"!!!!!!!!!

Yellowbeard

Whoa there. I think Cyco said something about 0.9 Gs but what I assumed was a 0.9 coefficient of friction. That's a real, if approximate, value that you can look up. Deceleration in Gs has to be derived.

**Edit: wait, no, I'm half-wrong. While the source of the 0.9 value is the coefficient of static friction for rubber on dry asphalt, it does actually mean that we can theoretically decelerate at a maximum of 0.9 Gs before skidding (ignoring the bar-flipping).

That's because the coefficient of friction is the ratio between the normal force (in this case, weight of rider) and the frictional limit. So a coefficient of 0.9 means that for every lb you weight you get 0.9 lbs of horizontal force before skidding.

cyccommute 

If, as Wilfred Laurier says, if nothing is done by the rider they will eventually go over the handlebars. The rate of deceleration will remain the same until the rider reaches that point where deceleration changes sign and the rider accelerates as they fall to the ground. But that is a different problem.


First, the only point where you couldn't stop a bike on a hill is if the hill is steep enough to put the CG in front of the contact patch. Decelerating on a steep slope is still possible although the possibility of reaching the point where the rider can be thrown over the bars is greater and requires more skill.

That's not what Wilfred Laurier was saying. He said that the rate of deceleration decreases as the rear wheel rises (assuming level ground) or, more to the point, as the CG shifts further forward. The implications his idea is that you will eventually reach a point where you can't stop the bike. That point would be where the CG is directly over the contact patch.

Think of it this way: Application of the brake causes the rear of any vehicle to lift. If deceleration decreased as the rear of the vehicle lifted, there would be no driving force to lift the rear of the vehicle. One way to get the vehicle to settle back on to the rear wheels is to release the brakes...i.e. decrease the deceleration. If the deceleration decrease naturally due to the rear wheel lift, then the rear would never lift.

Look at where the center of gravity on a bicycle rider system is. It's in the middle of the rider because the rider is the most massive part of the system. Raising the rear wheel forces the CG forward but not up...at least not significantly.

No, it is not. Since the CG is behind the contact patch, there is a vector component that points rearward towards the CG and gravity is pulling down on the rear of the bike. Simple releasing the front brake allows the bike to settle back onto the rear wheel. There is a component of gravitational force acting to pull the CG back down to the ground. If the CG is directly over the contact patch, there is nothing to pull the bike back down to the ground. We can't hold at that point all that well due to minor weight shifts ...and, often, lack of nerve.

Let's take out the wheel and the problems that rotation put into the problem. Assume a seat on a 4x4 post (or any size for that matter). You can balance a mass on that post forever as long as nothing moves. But move the GC of the mass you've balanced in any direction and the post falls over because gravity is pulling on the CG of the mass off the axis of the system.

Okay, wrong word then. "Gravity" is still pulling done on the GC because the CG isn't directly over the contact patch. Still works the same. Release the brake and the bike settles back onto the rear wheel.

What happens when you release the brake and decrease the deceleration? The normal force isn't the same on the front wheel because the rear wheel will settle back to the ground unless the CG is directly over the contact patch.

If you are assuming that the deceleration reaches zero when the CG is directly over the contact patch, you are assuming that the rider is actively reacting to the rear wheel lift. Wilfred Laurier said above that you don't have to be at zero speed when in a nose wheelie. I agree. But, if you assume that deceleration is zero at that point, how are you going to stop the bicycle in a nose wheelie? If you apply the brakes and there is zero deceleration when the CG is over the contact patch, you couldn't stop the bicycle and would remain at that speed forever.

Then you aren't being clear.


Then what are we disagreeing about?

cyccommute 

As Yellowbeard says, it can be calculated. Wilson's book (see post 92 for a reference) has the formula for calculating the deceleration. The example I've cited is for a 170lb rider on a bike with a 42" wheelbase is 0.56g (5.5m/s^2) assuming a "normal" crouched position. More weight will change that and, more importantly, a change in the center of gravity has a huge impact. Moving the center of gravity forward decreases the deceleration you can achieve before you go over the bars. Moving the CG rearward and down increases it significantly. I have the equation at home and will post it later.

Your use of only the front brake and your experienced with being thrown on to the top tube demonstrates why you should use both brakes. If you use only the front brake to avoid skidding the rear wheel, you are forgoing an important indicator of whether you are approaching rear wheel lift and, ultimately, pitch over. If you use the rear brake and can feel the tire start to slide, you can release the front brake not end up on the top tube. Yes, you are going to sacrifice a small amount of deceleration but you gain more control and, more importantly, don't hit the dangly bits on hard stuff.

cyccommute 

A coefficient of 0.9 is high for tires from what I've seen. Between 0.7 and 0.8 is what I commonly see for rubber on asphalt.

The 0.9 g (8.8m/s^2) value that I have calculated for stopping is from Wilson's formulas based on moving the CG back about 4" and down about 2".

Yellowbeard

Yes.

That's exactly what I said in the paragraph that you quoted.

That's not what Wilfred Laurier said at all. He said the same thing that I'm saying. That while actual deceleration depends only on traction and braking force, the maximum braking force you can apply to the front wheel without flipping over (which starts with the rear wheel lifting) decreases as the somersault continues.

No one in this thread is claiming the opposite. You're confusing the maximum potential deceleration with actual instantaneous deceleration.



In my illustrated example above I assume 60 centimeters (0.6 of the wheelbase) back, and 120 cm up from the front wheel patch to the CG. Surely not dead on, but not wildly unrealistic. That CG is going to rise 1 cm for every 2 that it moves forward at the instant the rear wheel lifts off (though of course it will level off eventually). One-third of the initial decrease in maximum braking will be due to the vertical rise. Not insignificant. And as it stops rising, it'll be moving forward quicker. Both have the same effect, and the same importance, mathematically.

Nothing you say here is particularly wrong, but it doesn't refute what I said.

Thought experiment: if the rear wheel is half a millimeter above the ground, how much weight is it supporting? Zero. Now subtract that from your total weight and you've got the weight on the front wheel. 100%-0 = 100%. 100% of your weight is already on the front wheel (because of the deceleration moment) and putting your CG further forward doesn't make you heavier.

No one is assuming that, or arguing that. We are arguing that the maximum deceleration that can be achieved without further lifting the rear wheel (and ultimately somersaulting over the bars) decreases in that case. NOT that the actual, effective force of braking at the front tire or the center of mass of the rider decreases fundamentally.

I'm drawing more pictures.

Your apparent confusion of the actual braking force with the mathematically defined maximum braking force.

cyccommute 

It's not "exactly" what you said. My point is that if you are on a hill steep enough to get in front of the contact patch, the hill is no longer a hill but a cliff. You would be falling down the cliff under the influence of gravity. Up to that point, the deceleration you could develop would be the same as on level ground.

No. That isn't what he said. Here's what he said:

The deceleration doesn't decrease as the rear wheel raises. There is no change in the force being applied to the wheel and the only thing that is changing is the location of the CG. What mechanism is changing the rate? And, as I said above, a way to decrease the deceleration is to release the brake. The bike will settle back down on the rear wheel provided the CG isn't forward of the contact patch. If the deceleration decreases like Laurier says and reaches zero deceleration when the CG is over the front wheel, then the bike should never lift off the ground at all. Zero deceleration isn't reached until the rider transitions from decelerating to falling off the bike. That doesn't occur when the CG is directly over the front wheel but when the CG has move forward of the contact patch.

I will grant you that the closer you get to the pitch-over point, the less wiggle room you have before you are thrown over the bars which is what makes doing a nose wheelie difficult but that doesn't mean that deceleration decreases.



Laurier is claiming exactly that. He's saying that the deceleration decreases as the rear wheel lifts and reaches zero when the CG is over the contact patch. If the deceleration were zero when the CG is over the contact patch and the bike still had speed...which is possible in a nose wheelie...you couldn't stop the bicycle.


You can lift the rear wheel off the ground and still be below the maximum possible deceleration without going over the bars.. Laurier says that the "pitch-over" point is when the rear wheel first leaves the ground. That's not a definition of pitch-over that I've ever heard anyone else use.

What you, and many who say to use only the front brake, are missing...and what I've been saying all along...is that there is a maximum possible deceleration that can be reached. I have said here and in many other places that it's only a mathematical oddity and not a practical limit. If you are going to use only the front brake, you are wasting a significant amount of braking (from 10% to 20%) by not using the rear brake until such time as the rear wheel loses contact with the ground. And once the rear wheel does leave the ground, you are losing a significant amount of control.

Not confused at all. I know that the maximum possible deceleration is a mathematical artifact and not really a practical way to ride a bike. I've said this many, many times and been met with these same arguments.

Let's consider this from a more practical point of view. Laurier says that once the rear wheel lifts of the ground you've reached the pitch-over point. Using the example that Wilson lays out in his book, a 89 kg rider on a 1067mm wheelbase bike braking at 4.9 m/s^2 has only nebulous contact of the rear wheel with the ground. Based on Wilson's calculations, the maximum possible deceleration the rider can develop is 5.5 m/s^2. If the rider were to apply 5.0 m/s^2, the rear wheel is not going to be weighted any more. The rider isn't in danger of being pitched over the bars at this point but also isn't at maximum possible deceleration. The rider could apply more deceleration and lift the wheel higher up to the point where the rider is finally at the pitch-over point. That extra deceleration could come about with the rear wheel off the ground and the deceleration wouldn't decrease.

Another conclusion to draw from this example is that could exceeded the maximum possible deceleration but you wouldn't know until such time as you are either near the pitch-over point (not at maximum deceleration) or you are flying over the bars (exceeded maximum possible deceleration). I'm not sure about you but I'm not a real good judge of whether I'm at 5.4m/s^2 or 5.6 m/s^2 (assuming that I weigh 89 kg...which I don't). I doubt that I could tell you if I'm decelerating at 4 m/s^2 or 6 m/s^2 and I also doubt that other's could as well. Like I've said many times, it's a theoretical limit, not a real practical one.

I use both brakes with regularity because I understand that both brakes contribute to deceleration until the point of rear wheel lift. I also let off the front brake when the rear starts to lift or even skid to get the rear wheel back on the ground. I'm sacrificing a little braking power...i.e. I'm not reaching maximum possible deceleration...for control.

People who use the front brake only are sacrificing more deceleration at the start of the braking curve...where it is more important...because they think they can reach that theoretical maximum or, more likely, because they are afraid they are going to slide the rear tire and are frightened of that possibility. In my opinion, people who are afraid to use the rear brake because the rear wheel could skid are worse than newbies who are afraid to use the front brake. They should know better.

Yellowbeard

No it wouldn't. I'll include this case in the force diagrams to prove it.

I count the word "maximum" three times in that paragraph and zero times in yours because, again, you wrongly conflated the maximum deceleration with the actual deceleration. You're not arguing against what is actually being said.

See, now you're assuming that the rider releases the brake during the wheelie. I never did and I never saw Laurier assume it, either. If you don't, then you're sure as hell going over the bars by that point (and even if you do, really, because your flip will have momentum).

Couldn't stop the bicycle without flipping over the bars.

But you're not just saying (or at least your words aren't) that it isn't practical, you're saying that you can stop faster with both brakes than just the front, and that you can stop equally fast in a partial nose-wheelie to having both wheels on the ground. Both are mathematically impossible.

Not having all the parameters of Wilson's example, I'm not going to analyze it. I'm don't doubt that it's correct because it's very, very simple math. And yes, technically, you can brake a little harder for a certain distance at the expense of raising your wheel a certain distance, but you're not gaining anything because to even to stop and HOLD your nose-wheelie (not reverse it) you'll have to return to a lesser degree of braking than you had before you're rear wheel popped up because (like I said before) your center of gravity is higher and further forward and both of those things alter the moments to your detriment.

Also, your argument from practicality doesn't hold water, because surely you're not arguing that it's practical to use a nose-wheelie to maximize braking? Even if it had that effect (which it doesn't).

How 'bout this: which way do you shift your weight for better braking? Lower and further back, right? So what effect does shifting it up and further forward have? Which of those two things does a nose-wheelie do?

PepeM

Wow, this is still going.

Yellowbeard

Ok, so continuing from Post #127 Figure 5: From the situation in Figure 4 all we've done is increase the braking force until all the rider's weight has shifted onto the front wheel. The rear wheel hasn't risen, but it's completely unweighted. This is what we are talking about when we say "maximum braking:" the maximum braking force (and therefore deceleration) without lifting the rear wheel. Once the rear wheel rises, maximum braking is reduced (see Figure 7 for why).




Note that while I said "only two forces left" even though gravity makes a third, what I mean is that there are only two forces causing moments around the CG. Gravity can't, because it acts through the CG.

Also note that these values are only true for my assumed geometry. Vary the geometry and the forces and limits will vary as well.


Figure 6: This is the instant that braking force exceeds the maximum. Again, "maximum" is referring to the maximum braking force/deceleration possible without causing the rear wheel to lift. Here, that is exceeded by an arbitrary amount. Even though there is no more weight available to put on the front wheel, the braking force still has a 50% margin before skidding occurs (see bottom left calculation).





In the very next instant, the rear wheel will leave the ground and the rider will accelerate (rotationally) around the front wheel contact patch

Figure 7: the rear wheel has lifted an arbitrary amount (10 cm, or 5.71 degrees of tilt) due to the harder braking in the previous instant. The rotation of the rider has moved the CG forward and up, changing the moment arms. You can calculate the shift with trigonometry and/or sketch it out to scale. The effect of gravity pulling the rear wheel down (mathematically represented by the front wheel normal force) is weakened by the horizontal shift, and the opposite effect of braking is magnified by the vertical shift.



So the maximum possible braking without the rear wheel lifting further is less. In fact, it's only 76% of maximum braking with the wheels level (in this example). Easing off on the brakes this much will only let you maintain your wheelie; to get your rear wheel back on the ground you'll actually have to brake even less than that.

cyccommute 

We'll see.

I do not see how you reach that conclusion and it is not what I am assuming. The only way for a rider to be pitched over the bars while braking is to continue to apply the brakes to the wheel until such time as the rider is forward of the contact patch. Releasing the brake during or approaching a nose wheelie before the CG is over the contact patch will result in the rear wheel falling back to the ground. To balance in the nose wheelie,the rider has to have delicate control over the brakes when approaching the pitch-over point and do subtle shifts in the CG to reach the balance point. It's not an easy trick to pull off on a bicycle.

And yet, if you have speed while in a nose wheelie, you can roll to a stop without falling. It's a delicate process but it can be done. Therefore you can stop the bike with the brakes so the deceleration isn't zero.

That's not what I'm saying at all. I'm saying that while you can reach the maximum possible deceleration with only the front brake, the point where you can reach that maximum possible deceleration is at the point where you are going to be pitched over the bars. It's not a practical point to try and reach in any situation. You've quoted me out of context. Further down I said that I'm willing to sacrifice some braking power for control and, frankly, going over the bars hurts too much to do it with any regularity.

Nowhere have I said that anything about being able to stop equally fast with both wheels on the ground nor that I can stop faster with both wheels then just the front. It is safer to use both brakes and avoid rear wheel lift because a freewheeling unicycle...i.e. balanced on the front wheel...isn't a stable vehicle.



You have all the parameters of Wilson's example. I gave them to you. I'll also repeat the quote that I gave to Laurier:

Bicycling Science, 3rd edition, David Gordon Wilson, 2004, p245.

Please note that Wilson literally wrote the book on the subject. He's no fly-by-night internet yahoo, either. I trust his calculations more than I do yours or mine.

Are you actually reading what I write? What I have been saying since page 3, post 52 is that the maximum possible braking isn't a practical way to slow down a bicycle. It may be possible and it may give you the greatest stopping power but it just isn't practical. Even just lifting the rear wheel isn't a practical method of attaining the best braking. Keeping both wheels on the ground, rolling and under control may not lead to the maximum possible deceleration but it's the best practical method of deceleration. The possibility of disaster is much lower.

Yellowbeard

Are we boring you?

cale

I'm bored. But it's okay I'm goi...
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Yellowbeard

Here ya go

Figure 8: braking on a downward slope.



Figure 9: why you can't decelerate as fast on a downward slope without lifting the rear wheel (and therefore reducing your braking further and risking a somersault) even when the CG is behind the front wheel.

jfowler85

Rule in favor of Yellowbeard. The court finds Cycco guilty of pretending he is an actual scientist...again.