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Old 12-12-07, 04:55 AM   #26
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Originally Posted by Cyclaholic View Post
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?
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Old 12-12-07, 07:21 AM   #27
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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:


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...


Last edited by DannoXYZ; 12-12-07 at 07:37 AM.
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Old 12-12-07, 11:59 AM   #28
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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?
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Old 12-12-07, 01:26 PM   #29
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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.
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Old 12-12-07, 03:21 PM   #30
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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.
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Old 12-12-07, 10:14 PM   #31
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Originally Posted by DannoXYZ View Post
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:


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
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Old 12-12-07, 11:11 PM   #32
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Originally Posted by DannoXYZ View Post
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:


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 and

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

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)
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Old 12-12-07, 11:16 PM   #33
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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?!?
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Old 12-12-07, 11:53 PM   #34
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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.
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Old 12-13-07, 03:57 AM   #35
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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?!?
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.
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Old 12-13-07, 05:17 AM   #36
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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!
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Old 12-13-07, 04:22 PM   #37
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Based on this theorem we have

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:

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Old 12-14-07, 07:18 AM   #38
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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.

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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.

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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.
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Old 12-14-07, 11:38 AM   #39
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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.
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Old 12-14-07, 05:19 PM   #40
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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.


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How do you know this?
Common sense my dear friend.. .. common sense.. .




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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.



... just to name a few

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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?



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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
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Old 12-15-07, 10:51 PM   #41
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physics?

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.
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Old 12-16-07, 03:24 AM   #42
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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?
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Old 12-16-07, 08:32 AM   #43
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...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.

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...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 types, one current example from another company would be the Longbikes Slipstream. 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.
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Old 12-16-07, 12:58 PM   #44
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"unscheduled dismount".
LOL!!! That was funny! I'll remember this one



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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



but to this...


and this...

(see video on Youtube)
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Old 12-16-07, 10:07 PM   #45
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...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.

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.
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Old 12-17-07, 02:29 AM   #46
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47% of all statistics are made up on the spot...or maybe it was 48%.

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Old 12-17-07, 03:53 AM   #47
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[



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



... 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.
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Old 12-17-07, 11:08 AM   #48
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[
... 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
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Old 12-17-07, 02:25 PM   #49
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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.
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Old 12-17-07, 03:12 PM   #50
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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.
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