General Cycling Discussion - Front brake - Flying over the handlebars

It is known that maximum braking occurs when the front brake is applied so hard that the rear wheel is just about to lift off. Skilled cyclists use the front brake alone probably 95% of the time.

However, improperly modulating the force applied on the front brake will bring the bike to a sudden stop, rider and bicycle's weight shifts to the front thus generating a momentum (force) that will send the rider "sailing" forward over the handlebars and eventually the bike will flip.

Q1): Does anyone know what is the name in physics of this generated force.

Q2): Based on speed, bike/rider mass-height how can this force be calculated.

Any links?

Thank you anticipated

Why do you want this, when you don't know much about physics to start with?

You can do it with a free body diagram with inputs of mass and position of the center of gravity relative to the front wheel. Even then, you'll only find the maximum braking force; a force significantly greater than this is required for flight.

Moment of (polar) inertia may be the term you're looking for.

It is known that maximum braking occurs when the front brake is applied so hard that the rear wheel is just about to lift off. Skilled cyclists use the front brake alone probably 95% of the time.

This is incorrect.




Q1): Does anyone know what is the name in physics of this generated force.

Which force in particular are you talking about, as there are a few at work?

Biology major here but IIRC wouldn't the basic mass in motion tends to stay in motion theory and rules apply:rolleyes: And FWIW I have yet to do an endo using just the front brake if it is used correctly, and on at least one of my bikes that is the only fully functioning brake:eek:

Aaron:)

It is known that maximum braking occurs when the front brake is applied so hard that the rear wheel is just about to lift off. Skilled cyclists use the front brake alone probably 95% of the time.
Actually, maximum braking occurs when you spot the sign that says "free beer" with an arrow pointing down a driveway you just passed..... or when your brakes fail and you stop by hitting the side of a building...... and when your parachute doesn't open (but that generally will only ever happen to you once).


However, improperly modulating the force applied on the front brake will bring the bike to a sudden stop, rider and bicycle's weight shifts to the front thus generating a momentum (force) that will send the rider "sailing" forward over the handlebars and eventually the bike will flip.

Q1): Does anyone know what is the name in physics of this generated force.

Q2): Based on speed, bike/rider mass-height how can this force be calculated.

Any links?

Thank you anticipated

Q1): That would be the "OhSh!tI'mAboutToDie" force. It's one of the most powerfull forces in the universe. It's so powerfull that survivors often report a slowing down of time and everything happening in slow motion while under its influence.

Q2): You can't actually calculate it based on just your speed and bike/rider mass-height as Newtonian physics assumes time to be a constant, which clearly it is not when under the influence of the OhSh!tI'mAboutToDie force.

:roflmao: :roflmao: :roflmao: :roflmao: :roflmao:

"Skilled cyclists use the front brake alone probably 95% of the time." Did not know that. Then again, I don't have a front brake.

" thus generating a momentum (force) that will send the rider "sailing" forward"
There is not a force sending the rider forward. There is a force slowing the rider down. But it is not applied at the center of mass of the man/machine, and therefore also tends to rotate the assembly.

Q1): Does anyone know what is the name in physics of this generated force. The force is "friction", slowing the brake down. The force is applied eccentrically, causing a "moment" or "torque". The polar moment of inertia doesn't enter into it until you start calculating how fast the bike rotates over.

Additionally, some of the rotational momentum of the wheels would be transferred to the bicycle, but it's probably a minor issue. But if you had a bike floating in space with the wheels turning, and then locked the brakes, the whole bike would then rotate but much more slowly. "Conservation of Angular Momentum" at work there.

Q2): Based on speed, bike/rider mass-height how can this force be calculated.
Not really. I mean, you can calculate it, but not with any accuracy because you don't know the data you need. What's the coefficient of friction between tires and ground in a particular case? Where's the center of gravity of a person and of a bicycle? The fact that a person/bicycle combo is not a rigid assembly also complicates things enormously if you are trying to get an accurate result.

In theory if you knew all variables you could calculate the forces required to flip you over the bars, but there are so many variables: the force of braking, the coefficient of friction on the road (which prevents the front wheel from skidding) the location of your centre of gravity (which will move forward as you brake), the angle between your centre of gravity and front tire contact patch, the angle of your arms to the handlebars, the wheel length of the bike since a long bike is less likely to endo, any load you may be carrying on the rear rack, whether your feet are clipped in or not, etc., etc.

But has anybody really BEEN over the bars on braking force alone? I'm sure you can manage a front endo (in fact I've been there), but in braking the rider will shift his weight to the rear, and if the back wheel lifts, he'll ease up on the lever a little.

But has anybody really BEEN over the bars on braking force alone? I'm sure you can manage a front endo (in fact I've been there), but in braking the rider will shift his weight to the rear, and if the back wheel lifts, he'll ease up on the lever a little.

Yes, and it has resulted in fatalities. Link to follow.

Darwinism at its finest.

Riding a bike? Hold on.

Duh!

It is known that maximum braking occurs when the front brake is applied so hard that the rear wheel is just about to lift off. Skilled cyclists use the front brake alone probably 95% of the time. This is incorrect.

However, improperly modulating the force applied on the front brake will bring the bike to a sudden stop, rider and bicycle's weight shifts to the front thus generating a momentum (force) that will send the rider "sailing" forward over the handlebars and eventually the bike will flip. All braking will transfer weight/force towards the front. It's true of bikes, cars, trains, everything. At some point you may lift the rear wheel and continue all the way over.

Q1): Does anyone know what is the name in physics of this generated force.

Q2): Based on speed, bike/rider mass-height how can this force be calculated.

Any links?

Thank you anticipated

What. The. Hell?

If you're riding a normal road bike, the force required is proportionate to the materials being used in construction of your rig: carbon will immediately flip, owing to its immaculate loftiness. Aluminum will flip you out of spite, as it wanted to be born as a mountain bike. Steel will base its punishment on the amount of lycra you're wearing, and if you're wearing a bright neon jersey, steel will wait patiently until you need to brake suddenly when surrounded by broken glass.

Some over the handlebar experiences:

http://www.cyclingforums.com/t217749-ac-joint-separation---recovery-time.html

"Managed to avoid being killed by a blind motorist this morning, but my effective panic-braking technique saw me go flying over the handlebars and landing on my shoulder and head."

http://query.nytimes.com/gst/fullpage.html?res=9C0CE1D8123AF934A35753C1A966958260&sec=&spon=&pagewanted=3

" few years ago, a bee flew down Sheri Taylor's shirt while she was cycling. She slammed on the front brake, which locked the front wheel, and she flew over the handlebars. "

There is a thread on BF which I will find when the search function is restored, about a guy who died due to panic braking and endoing when cut off by a truck.

"Skilled cyclists use the front brake alone probably 95% of the time. This is incorrect."

I've read that something like 80% of the braking capacity is in the front tire- as the rear lifts & skids under heavy braking.

But then I'll see statements that 80% of the braking that is done is with the front tire. I think that is likely an incorrect interpretation of the capacity statement. Unless people just choose to brake mostly with the front, or unless they routinely brake so hard they are forever skidding the rear tire.

On something like a long slope, where you need prolonged braking but not hard braking, the capacity of the two brakes should be about equal, based on heating, rather than skidding.

The process is simply too dynamic to assign one static value. I adjust my braking accordingly.

Modulation is the key.

But has anybody really BEEN over the bars on braking force alone? I'm sure you can manage a front endo (in fact I've been there), but in braking the rider will shift his weight to the rear, and if the back wheel lifts, he'll ease up on the lever a little.

I've done it while drunk. I guess the alcohol caused me to pull the lever stupidly hard. Then my face hit the ground before my impaired reactions had a chance to kick in.

I think if you have good brakes and really yank the lever you don't have a lot of time to react. The key is to know the limit so if you do have to brake in a panic you are less likely to exceed it.

"Skilled cyclists use the front brake alone probably 95% of the time. This is incorrect."

I've read that something like 80% of the braking capacity is in the front tire- as the rear lifts & skids under heavy braking.

But then I'll see statements that 80% of the braking that is done is with the front tire. I think that is likely an incorrect interpretation of the capacity statement. Unless people just choose to brake mostly with the front, or unless they routinely brake so hard they are forever skidding the rear tire.

On something like a long slope, where you need prolonged braking but not hard braking, the capacity of the two brakes should be about equal, based on heating, rather than skidding.

Both of you are off somewhat. The fact is that up to 100% of your braking capability comes from your front brake. When you are decelerating as fast as possible, the rear wheel contributes 0% of the braking force, since it will have virtually no traction under maximum braking (if your rear wheel has enough traction to make the rear brake useful, you're not all that close to maximum braking). Under dry conditions, there is no need to bother with the rear brake, and no sense in doing so. When the roads are wet or conditions otherwise make for the possibility of a front skid preceding the rear wheel lifting off, I use both brakes.

Emergency stopping "techniques" that suggest the coordination of front and rear brakes on dry roads aren't much good, because (again), if your rear brake is useful, you could be decelerating harder.

Not braking too hard and going over the bars is a matter of practicing a skill that requires the development of some degree of muscle memory, not a matter of braking hard with the front brake being inherently unsafe. If you aren't braking so hard in an emergency that your rear brake is useless, it's also possible that you could end up under the wheels of a truck. It is undoubtedly true that many people have been killed because they weren't able to slow their bicycles fast enough, likely more than those that died as the result of an endo. Not to resort to hyperbole, or anything, but being able to stop in the shortest distance possible is an important safety skill, and the only way to do it is to master the front brake.

^Only if your weight is shifted forward, at no time should you be using the traction of only one tire.

While it is right that you get a huge amount of braking, you'll never get 100% unless you're riding your front wheel like a unicycle. As long as that wheel is on the ground you have some sort of friction and braking power available to you.

However, the need for a rear brake isn't exactly huge, being as you do a lot of stopping with the front.


Generally I use my rear for slowing descents slightly. I'll mix with front more and more depending on how much more braking I need. You could easily do it with a front brake only. However I have this idea that if I lose traction due to bad road conditions (dirt, water, etc), having the back slide is going to be much easier to control than having the front slide.

Then again, in theory while that may be nice, in reality it would be quite rare.


Anyway, I've gone over the bars once, panic situation that if I had more experience I would have avoided, gripped the front brakes hard as hell, considered it a lesson learned. I was also down low and quite forward on the bike, if I was sitting in a more normal cruising position I would have never gone over the bars. Whats really fun is in this position you're going FACE FIRST with your arms under the bars.

Well, on a serious note instead of inputting more silliness, like CastIron mentions, this is one of those things that simply has far too many variables to take into account to really be easily calculated.

The one endo i've done as a result of braking was as a kid, on my mountain bike with moderate quality indirect pull cantilever brakes (not exactly the pinnacle of stopping power). Normally, I wouldn't even consider these brakes potent enough to do such a thing, but they were apparently :)

People go over the bars when they dont brace their elbows for braking. They apply the front brake, tha bike stops and they fly forward because their arms are not holding them back with the bike. Their body then hits the bars and their momentum carries them over.

I still believe that it has to be a formula that this force (Angular Momentum I believe) can be calculated taking in consideration the speed, height and weight of the bike, rider's weight and gravity center.

Its like this; A bike with a mass of 15Kg, having a rider with a mass of 80 kg hits a parking concrete barrier at 5 km/h, where hitting the barrier equals with brake force.

In this example, at this speed, chances are the rider will not fly over the handlebar.

Yet in the same equation, if we change the variable "speed" with 15 km/h, chances are the rider will start sailing thru the air.

Now the case scenario, only that the rider would have 120 kg. Most likely the bike wouldn't tip over. (due to rider's mass)

Now if we take this last rider's mass of 120kg but we increase the variable "speed at 40 km/h, chances are the bike will tip over

The question is how do I calculate the force of the impact occurred at 5km/h and at 15 km/h knowing the speed, and the mass of both rider and bike.

Thank you!

This is incorrect.


My dear friend, dismissing something without contributing with a replacement theory, thus substituting the previous one, does not make you more knowledgeable, nor demonstrate that I am wrong and your are right (or me incorrect and you correct), and certainly you are not entertaining a constructive discussion.

Your answer would have been more regarded if you would have say: "This is incorrect. This is what I believe to be correct.. .. etc, etc... based on ... . etc, etc"


But thank you for your insight anyway.

By the way, do you have any "correct" theory that you'd like to share with us?

Actually, maximum braking occurs when you spot the sign that says "free beer" with an arrow pointing down a driveway you just passed..... or when your brakes fail and you stop by hitting the side of a building...... and when your parachute doesn't open (but that generally will only ever happen to you once).



Q1): That would be the "OhSh!tI'mAboutToDie" force. It's one of the most powerfull forces in the universe. It's so powerfull that survivors often report a slowing down of time and everything happening in slow motion while under its influence.


My dear friend, I don't know if your answer is more funny than pathetic or more pathetic than funny.

So... . by the way.. .. what's your point?

There is no "force" pushing the rider over the bars. The only force is the bike pushing backwards due to the deceleration from the friction of the pads on rim and tyre on ground. You can calculate this force indirectly by measuring braking-distance from a given-speed at maximum-deceleration. Plug into these two physics equations to arrive at deceleration-G:

V=at
D=1/2at^2 ; substitute for t in 2nd-equation gives

a=1/2(V^2/D) = deceleration rate in G



The actual weight-transfer amount to the front-tyre can be calculated based upon deceleration-G, wheelbase and COG-height:

http://www.gamedev.net/reference/articles/phor/_7138_tex2html_wrap64.gif
http://www.gamedev.net/reference/articles/phor/_7138_tex2html_wrap65.gif where:

Lf = load on front-wheel
Lr = load on rear-wheel
d = static weight-distribution on front wheel (fraction or percentage)
G = static total weight
B = braking-force in G
h = center-of-gravity height
w = wheelbase


And at some point when braking-force is increased more and more, the rear-wheel WILL lift off the ground and you'll get 100% of the braking-force from the front-tyre. Skilled riders are able to brake with the rear-tyre skipping up and down slightly. Adding more lever-force at this point will not result in any faster deceleration, but it will raise the rear-tyre ever higher and higher...

http://i42.photobucket.com/albums/e346/DannoXYZ/Motorcycles/ElenaMyers2.jpg

I still believe that it has to be a formula that this force (Angular Momentum I believe) can be calculated taking in consideration the speed, height and weight of the bike, rider's weight and gravity center.

Its like this; A bike with a mass of 15Kg, having a rider with a mass of 80 kg hits a parking concrete barrier at 5 km/h, where hitting the barrier equals with brake force.

In this example, at this speed, chances are the rider will not fly over the handlebar.

Yet in the same equation, if we change the variable "speed" with 15 km/h, chances are the rider will start sailing thru the air.

Now the case scenario, only that the rider would have 120 kg. Most likely the bike wouldn't tip over. (due to rider's mass)

Now if we take this last rider's mass of 120kg but we increase the variable "speed at 40 km/h, chances are the bike will tip over

The question is how do I calculate the force of the impact occurred at 5km/h and at 15 km/h knowing the speed, and the mass of both rider and bike.

Thank you!

My dear friend,
I realize that winter is upon us, and we are often shut up within our respective laboratories, but seriously, put your slide rule away and go ride your bike. Clearly, you are smart as a whip (whatever that is supposed to mean), and will figure out this mystery before anyone else. Or, are you just looking for an argument to pass the winter solstice, hmm?

I went over my bike's bars on braking alone -- the trick is to be going downhill.

My dislocated shoulder took about 10 months to fully recover.

I seriously wonder how much saddle time some of you folks have.

Danno, you really should drop by more often. As always, your posts elaborate far better then I ever could.

There is no "force" pushing the rider over the bars. The only force is the bike pushing backwards due to the deceleration from the friction of the pads on rim and tyre on ground. You can calculate this force indirectly by measuring braking-distance from a given-speed at maximum-deceleration. Plug into these two physics equations to arrive at deceleration-G:

V=at
D=1/2at^2 ; substitute for t in 2nd-equation gives

a=1/2(V^2/D) = deceleration rate in G



The actual weight-transfer amount to the front-tyre can be calculated based upon deceleration-G, wheelbase and COG-height:

http://www.gamedev.net/reference/articles/phor/_7138_tex2html_wrap64.gif
http://www.gamedev.net/reference/articles/phor/_7138_tex2html_wrap65.gif where:

Lf = load on front-wheel
Lr = load on rear-wheel
d = static weight-distribution on front wheel (fraction or percentage)
G = static total weight
B = braking-force in G
h = center-of-gravity height
w = wheelbase


And at some point when braking-force is increased more and more, the rear-wheel WILL lift off the ground and you'll get 100% of the braking-force from the front-tyre. Skilled riders are able to brake with the rear-tyre skipping up and down slightly. Adding more lever-force at this point will not result in any faster deceleration, but it will raise the rear-tyre ever higher and higher...

Thank you, thank you , thank you!!!
Trying to analyze your theory and calculations I realized that I was looking in the wrong place for answer. Thus in order to come up with the right answer I have to look at the longitudinal stability because we have to consider in the first place mechanical forces generated by the a bike.

Where the wheelbase (L) and a center of mass halfway between the wheels and at height (h), with both wheels locked, reveals that the normal (vertical) forces at the wheels are

There is no "force" pushing the rider over the bars. The only force is the bike pushing backwards due to the deceleration from the friction of the pads on rim and tyre on ground. You can calculate this force indirectly by measuring braking-distance from a given-speed at maximum-deceleration. Plug into these two physics equations to arrive at deceleration-G:

V=at
D=1/2at^2 ; substitute for t in 2nd-equation gives

a=1/2(V^2/D) = deceleration rate in G



The actual weight-transfer amount to the front-tyre can be calculated based upon deceleration-G, wheelbase and COG-height:

http://www.gamedev.net/reference/articles/phor/_7138_tex2html_wrap64.gif
http://www.gamedev.net/reference/articles/phor/_7138_tex2html_wrap65.gif where:

Lf = load on front-wheel
Lr = load on rear-wheel
d = static weight-distribution on front wheel (fraction or percentage)
G = static total weight
B = braking-force in G
h = center-of-gravity height
w = wheelbase


And at some point when braking-force is increased more and more, the rear-wheel WILL lift off the ground and you'll get 100% of the braking-force from the front-tyre. Skilled riders are able to brake with the rear-tyre skipping up and down slightly. Adding more lever-force at this point will not result in any faster deceleration, but it will raise the rear-tyre ever higher and higher...

Thank you, thank you , thank you!!!
Trying to analyze your theory and calculations I realized that I was looking in the wrong place for answer. Thus in order to come up with the right answer I have to look at the longitudinal stability because we have to consider in the first place mechanical forces generated by the a bike.

Where the wheelbase (L) with a center of mass halfway between the wheels at height (h) and where both wheels are locked, will make that the normal (vertical) forces at the wheels for the rear wheel http://mcthongs.com/images/calc1.gif and http://mcthongs.com/images/calc2.gif

Now, the frictional (horizontal) force is Fr = μNr (rear wheel) and Ff = μNf (front wheel), where (μ) is the friction coefficient, (m) is the mass and (g) is the acceleration of gravity.

Based on this theorem we have http://mcthongs.com/images/calc3.gif

At his point the normal force of the rear wheel will be negative and the bike will flip over.

Please, please... .. no pictures, no autographs.. . (crowd cheering in awe)

My dear friend,
I realize that winter is upon us, and we are often shut up within our respective laboratories, but seriously, put your slide rule away and go ride your bike. Clearly, you are smart as a whip (whatever that is supposed to mean), and will figure out this mystery before anyone else. Or, are you just looking for an argument to pass the winter solstice, hmm?

My dear friend,

I wish was that easy. If I would have choose any pass time subject that would have been girls, soccer or 9/11 conspiracy theory.

"smart as a whip".. .. is that a good or a bad thing?!?

http://www.youtube.com/watch?v=efD-tiNmgR4

Do a search on youtube for "bike" and "endo"- there's a lot on there, with bicyles and motorcycles.

Note that the intentional endos involve a good bit of body movement relative to the bike, and are nothing like a free-body problem.

My dear friend,

I wish was that easy. If I would have choose any pass time subject that would have been girls, soccer or 9/11 conspiracy theory.

"smart as a whip".. .. is that a good or a bad thing?!?

:D Well, I don't think there would be much argument about girls, maybe some concerning soccer - but 9/11 conspiracy theories will soon out number those regarding the Kennedy thing by some quantum factor.

I think "smart as a whip" is a play on words from the 19th century, and may have little meaning in our highly technical existance: a drover who was good at driving a team of oxen or horses knew exactly how to just barely touch the animal to get it's attention. He would be considered "a smart whip." A stupid drover just beat the animals because it got them going, but it was considered poor form.

I think "smart as a whip" is a play on words from the 19th century, and may have little meaning in our highly technical existance: a drover who was good at driving a team of oxen or horses knew exactly how to just barely touch the animal to get it's attention. He would be considered "a smart whip." A stupid drover just beat the animals because it got them going, but it was considered poor form.

Super nice word disection. Nice to know. Thanks!

Based on this theorem we have http://mcthongs.com/images/calc3.gif

At his point the normal force of the rear wheel will be negative and the bike will flip over.Nice work! We still need to turn this into a differential-equation with respect to time as several of those variables (h & u) change during the braking process. As many people mentioned, you don't want to simply grab the brakes all-out, but rather over the course of about 0.1s, start with a 50/50 split initially then gradually ease up on the rear-brakes as you increase the front one. By the time the rear starts to skip & slide, you want to modulate the front at the limit to prevent going over.

And at the same time, you want to slide your butt back and down so your tummy is on the seat. This changes COG-height (h) dynamically. Also the tyre's friction-coefficient (u) is not constant and varies with vertical-loading. So while total total load (N) remains constant, the change from 50/50 weight-distribution to 100/0% on the front (Nf), doubling the front-tyre's normal-force won't result in double the traction. Depending upon how you juggle the braking-modulation and lowering height h, you may actually slide the front-tyre before developing enough deceleration-force to go over the bars:

http://i42.photobucket.com/albums/e346/DannoXYZ/AutoTech/TireEfficiency.jpg

My dear friend, dismissing something without contributing with a replacement theory, thus substituting the previous one, does not make you more knowledgeable, nor demonstrate that I am wrong and your are right (or me incorrect and you correct), and certainly you are not entertaining a constructive discussion.

Your answer would have been more regarded if you would have say: "This is incorrect. This is what I believe to be correct.. .. etc, etc... based on ... . etc, etc"


But thank you for your insight anyway.

By the way, do you have any "correct" theory that you'd like to share with us?

Sure sure, I just couldn't be bothered listing problems with what you originally wrote at the time. So I might go ahead now since you felt so...hmmm....disgruntled? Of course there was no need to show my whole post where I actually asked you to clarify information that would help formulate a better discussion. So feel free to take what other people have said out of context and attempt to slam them for it, because I am sure, you wouldn't mind it happening to you at all.


It is known that maximum braking occurs when the front brake is applied so hard that the rear wheel is just about to lift off.
How do you know this? I believe that if you knew this was true, you have not have to ask this question. Maximum breaking occurs when a number of factors are in place, including some that people have said here, including rider weight position, condition of the road and brakes and use of the rear brake.


Skilled cyclists use the front brake alone probably 95% of the time.

Now, let's all have fun with generalisations. Even though I do not have reliable statistics to back myself up (but neither do you!) I disagree with this figure. And it doesn't exactly make you sound credible when you use the word 'probably'. I would instead hedge my bets that a skilled cyclist knows the appropriate time to use both front and rear brakes and both of them together. And of course you wouldn't be including mountaing bikers in particular who may use the rear brake more often due to the inherent nature of the activity. And of course you wouldn't be talking about fixed gear riders who can manage quite a bit of braking power using just the pedals. These people are skilled riders of course, and I wouldn't want you to discount them.

And while I have the chance, thanks to the people who have posted the humorous replies, you've made my night.

I've braked with rear wheel liftoff successfully a few times but on one occasion I had almost slowed to a stop when my wheel hit a small obstacle and up I went. I hung there for a bit, still on the bike and looking straight ahead at the ground below me, and then my front wheel folded. I would have been better off if I released the brake and rolled over the obstacle but it was a small bike on its side and and the twelve year old owner was still on it.

Sure sure, I just couldn't be bothered listing problems with what you originally wrote at the time.

My dear friend no one twisted your arm to look at this post, nor is compulsory to read it. As a good friend of you that I am, I will give you a practical advice. Really, no need to waist your precious time on everything bothers you. No good for your heart... .. I have enough things to worry about, other than you health condition.



How do you know this?

Common sense my dear friend.. .. common sense.. .





Now, let's all have fun with generalisations. Even though I do not have reliable statistics to back myself up (but neither do you!)

Ohh but there is my friend.. .. just do a brief Google search and you'll find tons of statistics.


"Maximum braking occurs when the front brake is applied so hard that the rear wheel is just about to lift off. At that point, the slightest amount of rear brake will cause the rear wheel to skid" (http://www.sheldonbrown.com/brakturn.html)
"the most effective technique for powerful stopping is to use the front brake almost exclusively" (http://www.sheldonbrown.com/brakturn.html)
"the most effective technique for powerful stopping is to use the front brake almost exclusively" (http://en.wikipedia.org/wiki/Bicycle_brake_systems)
"When you’re stopping -- in a car, on a bike or on foot -- your weight shifts to the front." (http://www.dot.state.pa.us/Internet/Bureaus/pdBikePed.nsf/infoChapterSix?OpenForm)
"The front brake is actually 70-90% of your braking power" (http://www.vulturesknob.com/Mountain-Bike/4175.htm)




... just to name a few


I disagree with this figure. And it doesn't exactly make you sound credible when you use the word 'probably'. I would instead hedge my bets that a skilled cyclist knows the appropriate time to use both front and rear brakes and both of them together.
Based on... .. what? I came with the supporting data for my argument.. .. Can you?




And while I have the chance, thanks to the people who have posted the humorous replies, you've made my night.
Same here!!!


Any way.. .. take good care of your heart my dear friend.. .. you got me worried a tad knowing that you get bothered so easily.. ..



Best regards,


SR

I don't know what you call it but when I went over my bars a couple weeks ago I called it "Oh sh_t.

Just a couple of observations and these have been extreme cases that have happened to me. First of all on a Tandem. Lots of weight to hold a bike down and top quality disc brakes. Tested the brakes to see how effective they are and jammed on the brakes on a dry road at 40mph. Rear wheel lifted but there is no way a Tandem is going to endo. Offroad on an mtb and just about to tackle a technical bit of track downhill at slow speed. Too much front brake and straight over into face plant.

Now on the road- And take it that the road is perfectly dry and that sufficient speed is there to be able to lift the rear wheel under braking. It would take a very effective brake to keep the amount of braking force on the front wheel to enable it to stop the front wheel and if there was enough force to do that- The Sninny tyres would skid in preference to keep the wheel braking to enable the bike to endo. I have had this happen once and I did not endo. Just lost the front wheel grip and fell over sideways. What would normally happedn is that under hard braking the rear wheel would have the weight lifted off it and the rear tyre lose grip and skid. At that point the natural rider reaction would be to release the brakes a bit or put more weight over the rear wheel to regain braking control.

Then again- Endo's are possible on all bikes but this is normally caused by the front wheel hitting an imovable object like a Dog- Car or tree. Not the usual situation on a normal bike ride.

Or Do you Know Better?

...Now on the road- And take it that the road is perfectly dry and that sufficient speed is there to be able to lift the rear wheel under braking. It would take a very effective brake to keep the amount of braking force on the front wheel to enable it to stop the front wheel and if there was enough force to do that....
Normally I would agree that a traditional road bike does not quite have the braking power to do a clean front-flip--but now we see road bikes coming out with disk brake mounts. Hydraulic disk brakes (especially with larger rotors) give a LOT more braking force than any rim brakes I know of. Of course then the question becomes if the road-bike forks can withstand that? I haven't looked, but so far I have not seen anyplace making carbon forks with disk mounts, and I can guess the reason why.


...Then again- Endo's are possible on all bikes but this is normally caused by the front wheel hitting an imovable object like a Dog- Car or tree. Not the usual situation on a normal bike ride.
Or Do you Know Better?
Many recumbent bikes can't be flipped at all from front braking, even with fat slick tires on clean pavement. The wheelbases are long enough and the CG low enough and rearward enough (sometimes 66% or more on the rear) to prevent it. The braking force you can generate is the primary limit, with the fork and wheel strength beyond that.

And some are less likely to flip forward in a crash than others, as well: witness the Ryan/FOMAC/Avatar-2000 (http://images.google.com/images?svnum=30&um=1&hl=en&c2coff=1&safe=off&q=ryan+fomac+avatar-2000&btnG=Search+Images) types, one current example from another company would be the Longbikes Slipstream (http://www.longbikes.com/Slipstream.html). With this style of bike, if you crash, you always do so feet-first, and there's no parts of the bicycle to get entangled in as you are making your "unscheduled dismount". The riding position is a matter of taste however, and the Slipstream is a nice but expensive piece of work.
~

"unscheduled dismount".

LOL!!! That was funny! I'll remember this one





With this style of bike, if you crash, you always do so feet-first,

I you ride at a regular speed 20-25 km/h yes. However, if the bike would be tested in the crash lab environment, and the speed will be increased at 120km/h, and the front brakes will be instantly deployed at 100% braking power, chances are, Slipstream too will flip over.

I do understand that some bike are more stable than others, and in the ideal riding conditions the wheel base parameter will make all the difference in an event of a frontal (and perfectly perpendicular) crash... ...

As well I know that the rider's skill makes all the difference in the emergency braking situation as well as the amount of force the riders applies on the front brake (well.. .. technically on the brake levers on the handle bar) and all that, but that was not my initial query "if someone can flip over or not"

The question was, as far as the physics is concerned, what are called the forces pushing the rider and how can one calculate them. (mass, distance and speed in meters per second, all the good stuff)

My topic does not refer to this... the beautiful blue bicycle
http://www.mcthongs.com/images/BlueBikeMed.jpg


but to this...
http://www.mcthongs.com/images/CrashBike2Med.jpg

and this...
http://www.mcthongs.com/images/indexu1.jpg (http://www.youtube.com/watch?v=NrIN0qVBpZ4)
(see video on Youtube)

...being able to stop in the shortest distance possible is an important safety skill, and the only way to do it is to master the front brake.

Yeah, what Grolby said right there. :D

Which leads back to the age old question of brake lever/caliper orientation.
I'm right handed and have the front brake lever on the RIGHT side of the handlebars.
Motorcycle-esque.
That way my skilled hand is controling the important front brake for better modulation.
And when braking hard, I shift my weight towards the back.

That's what Miguel Duhammel does when he hits 180 on the straightaway and still enters the next corner at 70. That's mastering the front brake.

Like Grolby said. ;)

47% of all statistics are made up on the spot...or maybe it was 48%.

-Barry-

[



Ohh but there is my friend.. .. just do a brief Google search and you'll find tons of statistics.


"Maximum braking occurs when the front brake is applied so hard that the rear wheel is just about to lift off. At that point, the slightest amount of rear brake will cause the rear wheel to skid" (http://www.sheldonbrown.com/brakturn.html)
"the most effective technique for powerful stopping is to use the front brake almost exclusively" (http://www.sheldonbrown.com/brakturn.html)
"the most effective technique for powerful stopping is to use the front brake almost exclusively" (http://en.wikipedia.org/wiki/Bicycle_brake_systems)
"When you’re stopping -- in a car, on a bike or on foot -- your weight shifts to the front." (http://www.dot.state.pa.us/Internet/Bureaus/pdBikePed.nsf/infoChapterSix?OpenForm)
"The front brake is actually 70-90% of your braking power" (http://www.vulturesknob.com/Mountain-Bike/4175.htm)




... just to name a few


Thanks for the links. I had an "over the bar experience" with my first 10 speed when I was 15 and I used the front brake only. Sheldon is right when he says applying the right pressure because if you "lock em up", and you going at least 16 mph, you will go over the bars.

[
... just to name a few


Thanks for the links. I had an "over the bar experience" with my first 10 speed when I was 15 and I used the front brake only. Sheldon is right when he says applying the right pressure because if you "lock em up", and you going at least 16 mph, you will go over the bars.

Hmmmm... .. Sheldon is right (only because is Sheldon). I said the very same thing but apparently I wasn't right

Hmmmm... .. Sheldon is right (only because is Sheldon). I said the very same thing but apparently I wasn't right

No, you are right also. I was just reading from the link. Since we have this issue taken care of, we can now move on to the real issue. Let's design some "Anti-Lock" front brakes and market them.

Hmmmm... .. Sheldon is right (only because is Sheldon). I said the very same thing but apparently I wasn't right

Actually, with the "skilled cyclists use only the front brake probably 95% of the time" thing, I think Sheldon is wrong too.

I agree with his theory in general, but his "dry pavement" approximation just doesn't fit many of the road surfaces I ride on throughout the year. And of course, road cycling is only one branch of the sport.