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Improving road bike braking power

Old 02-01-10, 04:26 PM
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daven1986
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Improving road bike braking power

Hi all,

I had briefly considered changing my fork so I could put a front disc brake on my road bike, however this, I decided, would be too much hassle. So in light of the fact that I have salmon kool stop pads and Shimano BR600 brakes, what can I do to improve braking power? Are there brakes out there that would give me much better braking power (without breaking the bank!)?

Thanks

Daven
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Old 02-01-10, 04:45 PM
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Not this again.....
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Old 02-01-10, 04:48 PM
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True your wheels, clean the sidewalls of your rims (no petroleum based solvents), and tighten your brake cables.

Also, your front brake does 80+% of the braking, so don't be afraid of using that one. If you keep your arms straight, and exercise common sense, you won't go over the bars.

Originally Posted by daven1986 View Post
Hi all,

I had briefly considered changing my fork so I could put a front disc brake on my road bike, however this, I decided, would be too much hassle. So in light of the fact that I have salmon kool stop pads and Shimano BR600 brakes, what can I do to improve braking power? Are there brakes out there that would give me much better braking power (without breaking the bank!)?

Thanks

Daven
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Old 02-01-10, 05:03 PM
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Originally Posted by daven1986 View Post
Hi all,

I had briefly considered changing my fork so I could put a front disc brake on my road bike, however this, I decided, would be too much hassle. So in light of the fact that I have salmon kool stop pads and Shimano BR600 brakes, what can I do to improve braking power? Are there brakes out there that would give me much better braking power (without breaking the bank!)?

Thanks

Daven

Bicycles can only generate about 0.5g of deceleration at the best of conditions. Most brakes...at least modern ones...are capable of providing that kind of deceleration. Discs don't make your bike stop any faster, it just take less pressure to get the same deceleration. At this point, changing equipment probably isn't going to give you any better results. Changing technique, however, will.

Many people...Sheldon Brown included...have a mistaken idea that you should only use the front brake on a bicycle to obtain maximum deceleration. You can determine the maximum deceleration of a bicycle when doing the calculations by setting the rear wheel's contribution to zero. And, if you are willing to lift the rear wheel off the ground each time you stop, you can get maximum deceleration but that is not the only way to reach the 0.5g maximum. It's only one way to do it. And lifting the rear wheel is required to reach that maximum. If the rear wheel is still in contact with the ground, you haven't reached maximum deceleration.

Using both brakes and using your body, will squeeze more deceleration out of your bike without having to do this all the time


Basically, you need to move your weight back and down as you brake to keep the rear wheel in contact with the ground while you brake, like this


As you brake push back off the saddle, this will extend your arms and naturally cause your body (the biggest contributor to the center of gravity of the system) to go lower on the bike. This keeps the rear of the bike on the ground and keeps providing deceleration from the rear wheel longer.

Learning how to mountain bike does wonders for road bike handling

If you want to plow through it, I've done some calculations for the weight distribution of a bike braking while the rider sits static and while he shifts his center of gravity back and down.

I used the estimate of weight distribution of 45% front/55% rear in static conditions, i.e. no acceleration. I used the weight distribution equations from here. It is for a car but the principles are the same. I assumed maximum deceleration of 0.5g (which comes from Bicycling Science). I also assumed a center of gravity of 39" (1m) and a CG shift of half that by moving back and down while braking. I also assumed a weight distribution change to 30/70 front to rear. The wheel base is a very generous 40" (1.01M)

Some of these are wild guesses, particularly the CG. However, don't look at them as real world numbers but numbers used for comparing staying in one spot while braking and moving rearward to keep the rear wheel from lifting.

Based on these numbers, a 200 lb system (890N) on level ground with no braking has a distribution for the front to rear of 90 lb (400N)/140 lb (490N). If the rider stays seated and brakes at 0.5g, the weight distribution changes to 200 lb front and 0 lb rear. You've moved 110lb to the front wheel and the rear wheel lifts.

If the rider moves the center of gravity down by half and moves 15% more of the weight rearwards (a 30/70 split), the weight transfer is 170 lb (750N) front and 30 lb (140N) rear. Braking moves 80 lbs to the front wheel. But the rear wheel stays in contact with the ground.
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Old 02-01-10, 05:18 PM
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And, if you are willing to lift the rear wheel off the ground each time you stop, you can get maximum deceleration but that is not the only way to reach the 0.5g maximum.
Lifting the rear wheel lifts the CG and moves it forward, making the system more prone to flip forward at even less deceleration. Hence, you see trick riders (like the pic you posted) riding on the front wheel at the balance point where they are on barely feathering the front brake. NOT maxmimally decelerating.

Maximum braking is achieved when the rear wheel has no weight on it, but does NOT lift off the ground. Since at this maximum deceleration limit there is no weight on the rear wheel, there is also no braking that can be done by the rear wheel. So maximum deceleration also implies 100% front brake usage.

I assumed maximum deceleration of 0.5g (which comes from Bicycling Science).
On dry pavement maximum deceleration is strictly limited by the CG location (which you should remember from that very same passage of Bicycling Science, since their 0.5g figure is derived from a typical CG location with respect to the front wheel contact patch in the first place); if you are looking at the effect of CG shifting on max. deceleration you should calculate max. deceleration from CG location and not "assume" max deceleration.

Which is to say, shifting weight down and back INCREASES the 0.5g max.decel. figure .... but that maximum deceleration for any given CG location is still only achieved when using 100% front brake.

The only exception is if the surface you are riding has limited traction, in wet or off road conditions. (very specifically: if the coefficient of friction between tire and pavement is less than the ratio of CG setback from front wheel divided by CG height off ground). In that case if you are not shifting weight you will do best by mixing front and rear brake. but if you do the physics you will still find something surprising: that even more stopping power can be had by shifting weight FORWARD (so that CG setback/ CG height = coefficient of friction) and front braking only.

Last edited by zzyzx_xyzzy; 02-01-10 at 06:21 PM.
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Old 02-01-10, 05:20 PM
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Thanks for the info. I use my front brake mainly and will try what cyccommute said with adjusting my body position more as I brake.
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Old 02-01-10, 06:34 PM
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Originally Posted by zzyzx_xyzzy View Post
Lifting the rear wheel lifts the CG and moves it forward, making the system more prone to flip forward at even less deceleration. Hence, you see trick riders (like the pic you posted) riding on the front wheel at the balance point where they are on barely feathering the front brake. NOT maxmimally decelerating.

Maximum braking is achieved when the rear wheel has no weight on it, but does NOT lift off the ground. Since at this maximum deceleration limit there is no weight on the rear wheel, there is also no braking that can be done by the rear wheel. So maximum deceleration also implies 100% front brake usage.
The nose wheelie picture I showed was to illustrate an extreme of what no weight on the rear wheel looks like. In practice, the rear wheel would be lifted only slightly off the pavement.

Maximum braking can be calculated by assuming no weight on the rear wheel. It is not the only way that maximum braking can be achieved. The maximum deceleration is still only 0.5g for a wheel on dry payment. If you load the rear wheel and use the rear brake, you can get up to that maximum out of both brakes. The rear wheel only contributes about 0.1g of deceleration but that is still 20% of the total.

Originally Posted by zzyzx_xyzzy View Post
On dry pavement maximum deceleration is strictly limited by the CG location (which you should remember from that very same passage of Bicycling Science, since their 0.5g figure is derived from a typical CG location with respect to the front wheel contact patch in the first place); if you are looking at the effect of CG shifting on max. deceleration you should calculate max. deceleration from CG location and not "assume" max deceleration.

Which is to say, shifting weight down and back INCREASES the 0.5g max.decel. figure .... but that maximum deceleration for any given CG location is still only achieved when using 100% front brake.

The only exception is if the surface you are riding has limited traction, in wet or off road conditions. (very specifically: if the coefficient of friction between tire and pavement is less than the ratio of CG setback from front wheel divided by CG height off ground). In that case if you are not shifting weight you will do best by mixing front and rear brake. but if you do the physics you will still find something surprising: that even more stopping power can be had by shifting weight FORWARD (so that CG setback/ CG height = coefficient of friction) and front braking only.
It's been a little while since I've looked at Bicycle Science but I'm fairly certain that the value of maximum deceleration he uses is 0.5g even with body shift. The body shift only loads the rear wheel so you can squeeze some braking power out of the rear wheel.

As for a shift of weight forward giving you better stopping power, this is just wrong. Slide forward on any bike while braking and you'll skid the rear tire. If you want long skids, you do exactly that. Any 10 year old knows this.
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Old 02-01-10, 07:00 PM
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That .5G figure isn't cast in stone. It'll depend on lot on how well the rider can shift the CG rearwards and down like the shot of that mountain bike fellow going down the rocky slope. For someone just sitting upright in the saddle and taking no special weight shifting movements it may well be less than .5G. For a rider aggresively hanging off the back and doing all they can to lower their torso the max possible would be well over .5G.

Daven, getting back to your original question. Is it pure stopping distance that you're looking to decrease or the lever effort needed to get what you're getting now? Your question isn't all that clear about this. It hasn't stopped the rest of us from running off on the stopping force issue though....
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Old 02-01-10, 07:22 PM
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I just went out and did 5 all-out maximum-braking tests from 20mph. I got the following results:

17.5ft
18.5ft
18.5ft
17ft
18ft

Someone can calculate the braking-Gs. I can say they're all definitely more than 0.5g. The calculations indicating maximum of 0.5g braking-force are erroneous because they don't account for weight-transfer under braking. And, there's more to grip than just pure friction between rubber and asphalt (f=mu).
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Old 02-01-10, 07:34 PM
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Not discussed yet, but you can get even better braking if you push your bike into the ground at the same time you grab the front brake.

Todo this raise your torso up, drop down onto the handlebars and just as you reach the lowest point start pushing your torso back up and very slightly rearward with your arms like doing a push-up while hitting the front brake. Easy to beat get .5g this way i'd bet. Just don't get mad if you endo, it takes some practice.

Disc brakes are superior for modulation, power(hydraulic) and all-weather performance... basically the latter makes the upgrade worthwhile IMHO.

Also a wider tire up front with proper psi will help

Last edited by electrik; 02-01-10 at 07:56 PM.
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Old 02-01-10, 07:45 PM
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Danno's numbers give approximately .75 g's of deceleration.
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Old 02-01-10, 07:51 PM
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Can your brakes lock up your wheels?? As long as they can then the coefficient of friction of the tires is the limiting factor. Try better tires.

What is the chance we are addressing the wrong problem?? Could the problem be that your brakes don't inspire confidence?? If e.g. your brake cable housing ends weren't cut/filed flat then there could be some compression and spongy feeling brakes. Could this be the kind of problem we should address?? If your road bike has cantilever brakes the leverage due to the straddle wire length could be an issue.
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Old 02-01-10, 08:02 PM
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Originally Posted by DannoXYZ View Post
I just went out and did 5 all-out maximum-braking tests from 20mph. I got the following results:

17.5ft
18.5ft
18.5ft
17ft
18ft

Someone can calculate the braking-Gs. I can say they're all definitely more than 0.5g. The calculations indicating maximum of 0.5g braking-force are erroneous because they don't account for weight-transfer under braking. And, there's more to grip than just pure friction between rubber and asphalt (f=mu).
Remembering my high school and college physics,

V=(2aS)^1/2 where V=velocity, S= distance and a = acceleration, all in consistant units.

Solving for a:

a = V^2/(2*S)

So for this case:

V= 20 mph = 29 ft/sec
S = 17 to 18.5 ft

a= 29^2/(2*17) = 24.7 ft/sec2.
a= 29^2/(2*18.5) = 22.7 ft/sec2

One "G" = 32.16 ft/sec2

So the best braking deceleration was 24.7/32.16 = 0.77G and the worst was 22.7/32.16 = 0.71G

Your rearward weight transfer must have been very good.
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Old 02-01-10, 08:55 PM
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What you say is, in my most charitable reading, "if the maximum braking is X with only the front brake, then you can shift weight back and get X braking with a combination of front and rear."

What I'm saying is, "If you are shifting weight back, you might as well continue to use front brake only and get MORE THAN X deceleration."

Originally Posted by cyccommute View Post
It's been a little while since I've looked at Bicycle Science but I'm fairly certain that the value of maximum deceleration he uses is 0.5g even with body shift. The body shift only loads the rear wheel so you can squeeze some braking power out of the rear wheel.
I've got a copy right in front of me. 0.5g is only given as an example of what kind of braking you might desire in an emergency--it is never called a "maximum". Later on they derive 0.56g maximum from an "average" bicycle geometry without considering any change of rider position. Then they point out that bicycles where the CG is further back, like tandems or recumbents, can get up to 0.8g (limit of coefficient of friction). So 0.56g is not a maximum, either, but increases as you shift the CG back.

Originally Posted by cyccommute View Post
As for a shift of weight forward giving you better stopping power, this is just wrong. Slide forward on any bike while braking and you'll skid the rear tire. If you want long skids, you do exactly that. Any 10 year old knows this.
Reread what I wrote:

[in low traction situations] more stopping power can be had by shifting weight FORWARD (so that CG setback/ CG height = coefficient of friction) and front braking only.
How are you going to skid the rear tire if you aren't touching the rear brake?

-----

We will address this and the other misconceptions with physics. Here is a free body diagram of a bicycle.

braking.png

Labeled quantites:
H: height of the center of mass of bicycle + rider above the ground
C: is the setback of CM from the front wheel contact patch
W: wheelbase
Wf: weight loaded on front wheel
Wr: weight loaded on rear wheel
Bf: Braking force applied by front wheel
Br: Braking force applied by rear wheel
Wg: weight of bicycle + rider

Now, the bicycle remains on the ground, so

Wf + Wr = Wg

and the bicycle does stays level, hence no net torque around its center of mass, so

(Bf + Br)H = Wf*C + Wr*(C-W)

Additionally we have constraints:

Wf ≥ 0, and
Wr ≥ 0, because tires will lift off if unloaded.

If you want to consider braking in limited traction situations, then there are additional constraints:

|Bf/Wf| ≤ , and
|Br/Wr| ≤ , where is the coefficient of friction of tire on ground.

-----------

I think we will both agree that the goal is to maximize deceleration, i.e. find the maximum B = (Bf + Br) that satisfies the constraints.

You can derive from the two equations above that

Wf = Wg * (1-C/W) + BH/W
Wr = Wg * (C/W) - BH/W

where B = Br + Bf is total braking. This should agree with your formulas for weight shift under braking.

Now, your challenge is to pick ANY values you choose for H, C, W, and Wg, and choose a front/rear braking split (Bf, Br) that satisfies the above constraints.

I will then show you that for the same geometry and body position, you can get even more more deceleration by using the front brake only and no rear.

I don't even need to look at your numbers, all I will do is set:

Br = 0,
Bf = Wg * C/H

THEOREM: You will not be able to find any combination of Br and Bf that meets or exceeds this amount of deceleration that does not have Br = 0.

PROOF. If you apply even a little bit of rear brake -- say, Br = x, then you must satisfy x ≤ *Wr, or else you skid the rear wheel. But then using the formula for Wr,

x/ ≤ Wg* (C/W) - BH/W, then

B ≤ Wg * C/H - W*x/

which is less braking than just applying Wg * C/H to the front wheel alone. So, if you apply even a little bit of rear brake, there is less maximum braking available. QED.

-----

If you want to consider limited traction, and shifts of the rider's center of mass, the optimum behavior is to shift your center of mass such that:

C/H = (which for low is a shift forward)
then apply

Bf = *Wg, and
Br = 0.

You will not be able to find any combination of weight shift and braking that exceeds this, and if your proposal includes any nonzero rear brake while staying within constraints, you will have less deceleration than this. (Do the math.)

-----

If you have limited traction, and you can't or won't shift weight far enough to satisfy C/H = , then (and only then) can you get more deceleration by adding rear brake. Exercise to the reader to find the optimum.

-----

All of this ignores that skidding the front wheel usually has much more dire consequences than skidding the rear. One uses more rear (i.e. less than optimal deceleration) off road or in the wet, for this reason.

Last edited by zzyzx_xyzzy; 02-01-10 at 09:07 PM.
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Old 02-01-10, 11:21 PM
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After rereading and doing some calculation, I did misread what Wilson presented.

His example is for a bicycle with a 1067mm (42") wheel base, a center of gravity height of 1067mm (45") that is 432mm (17") in front of the rear wheel. For that configuration, which is pretty high and far forward, the maximum deceleration is 0.58g.

Moving the center of gravity back and down does increase the deceleration. A center of gravity on the same bike of 216mm in front of the wheel and of 1016mm height, gives a deceleration of 0.84g.

The force on the wheels in the first case is 821N front and 52N rear. In the latter case, the force on the wheels is 592N front and 281N rear. In the former case, "90% of the braking force is on the front wheel and the rear wheel is only in tenuous contact with the ground. The rear wheel would skid with little effort."

Moving the center of gravity rearward, gives 68% of the braking force on the front wheel and 32% on the rear wheel.

Originally Posted by zzyzx_xyzzy View Post
What you say is, in my most charitable reading, "if the maximum braking is X with only the front brake, then you can shift weight back and get X braking with a combination of front and rear."

What I'm saying is, "If you are shifting weight back, you might as well continue to use front brake only and get MORE THAN X deceleration."
How can you get more deceleration when moving rearward by neglecting the rear brake especially when the rear wheel is now responsible for 32% of the braking force? I'll have to read and digest what you say below but, from an operational standpoint, no one who rides a mountain bike would rely on their front brake alone. The possibility of pitch over is far too great in those situations.


Originally Posted by zzyzx_xyzzy View Post
I've got a copy right in front of me. 0.5g is only given as an example of what kind of braking you might desire in an emergency--it is never called a "maximum". Later on they derive 0.56g maximum from an "average" bicycle geometry without considering any change of rider position. Then they point out that bicycles where the CG is further back, like tandems or recumbents, can get up to 0.8g (limit of coefficient of friction). So 0.56g is not a maximum, either, but increases as you shift the CG back.
As I stated earlier, I was in error.

Originally Posted by zzyzx_xyzzy View Post
Reread what I wrote:



How are you going to skid the rear tire if you aren't touching the rear brake?
I can see what you are saying but I disagree. Moving the center of gravity forward would put you in a situation where you are more likely to pitch over. You said in a previous post, that a nose wheelie requires less pressure on the brakes as you reach the balance point to keep the rider upright. Moving the CG forward while braking hard would cause the rider to reach, or go past, that point. If the wheel were to hit something that ******** its motion, the rider would go over the bars.

As an example, look at low traction situations you might encounter in mountain biking. Moving the weight forward in nearly any situation that you might encounter in mountain biking will result in lifting the rear wheel and pivoting the rider around the front wheel.

Originally Posted by zzyzx_xyzzy View Post
snip of material that I'll have to review when I have time.
Edit: The goofy filter has removed the word I used. The word I used was for a slowing down or stopping of the wheel, not for someone who is mentally handicapped.
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Old 02-01-10, 11:24 PM
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Old 02-02-10, 12:20 AM
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Cyccommute, in the case of the rider moving his weight back the CG goes back so the point where the front lifts the the back end off the ground occurs at a higher braking G force. This assumes that the front tire is not in any danger of skidding and that the only issue to be avoided is pitching over. However if the rider should manage to move his weight far enough back to where the front can be felt to be on the virge of skidding THEN applying the rear to take advantage of what weight that remains on the rear would help.

While I agree with your take on not relying soley on the front for mountain biking that's sort of a different issue. In those cases the rider is trying to make the best of the situation without actually locking the front and ending up washing out the front. To do that while riding over the constantly changing conditions on most trails this means keeping a little in reserve. In this case using both is certainly the right thing to do. As a bonus what the rear is doing gives the rider a guide as to what they can get away with on the front.

The bit zzyzx_xyzz describes about moving forward was about riding in excessively slippery conditions. The idea in that case being to put more weight on the front to try to achieve at least SOME traction for braking. Under such conditions pitching isn't an issue. Locking up and dumping it would be though. zzyzx_xyzz, putting more load onto the front may indeed allow the wheel to stop the bike faster in such conditions but MAN! if the traction does let go it's face plant time with a vengeance....
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Old 02-02-10, 12:44 AM
  #18  
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Originally Posted by cyccommute View Post
Moving the center of gravity rearward, gives 68% of the braking force on the front wheel and 32% on the rear wheel.
Don't get cause & effect mixed up. The weight-transfer is the result of deceleration forces pushing back on the bike (while the rider continues forward). The amount of weight-transfer then determines the front/rear split in total-traction. With zero-deceleration and zero weight-transfer, the front/rear split is roughly 40/60 F/R. The actual split varies based up on deceleration-G. The higher the braking-force, the more the weight-transfers and the greater the different in F/R traction. It's not a fixed quantity, but varies with deceleration-rate.

Here's how you actually calculate the F/R split in weight-distribution and resultant traction-limits of each tyre under braking:

Lf = dG+Bh/w
Lr = (1-d)G-Bh/w ; where

d = initial static weight-distribution at front wheel
w = wheelbase
G = static mass of bike + rider system
h = height of center of gravity of bike + rider system
B = braking-force deceleration in G

As you can see, the Lf versus Lr ratio will change depending upon how hard you are braking. The higher the deceleration-G, the more weight transfers to the front-tyre, the higher the load on the front tyre and the higher the front tyre's grip. At some point, as others calculated, about 0.80-0.85g, ALL of the traction is on the front tyre. If you reduce braking-force by letting up on the front-lever to get the rear-tyre down, you'd only have 0.70-0.75g and not stop as fast.
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Old 02-02-10, 12:52 AM
  #19  
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Originally Posted by BCRider View Post
The bit zzyzx_xyzz describes about moving forward was about riding in excessively slippery conditions. The idea in that case being to put more weight on the front to try to achieve at least SOME traction for braking. Under such conditions pitching isn't an issue. Locking up and dumping it would be though. zzyzx_xyzz, putting more load onto the front may indeed allow the wheel to stop the bike faster in such conditions but MAN! if the traction does let go it's face plant time with a vengeance....
This reminds of the time in HS where I tried to go off-roading in my FWD auto after a decent rain. Had some fun down by a creek and tried to get out. No way with FWD. So I had my friends climb out and sit on the bonnet to add more weight to the front-tyre. It's amazing what moving a couple hundred pounds from the rear seat to the front-tyres will do for improving traction!

But yeah, be careful of the easier pitch-over on a bike.
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Old 02-02-10, 12:53 AM
  #20  
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I'll agree that keeping safety factors against pitchover and front wheel lockup are practical concerns for which you wouldn't really want to go right to the edge of maximum braking.

It would be interesting to extend the analysis to ask, what would you do if you wanted to give both wheels an appropriate amount of "reserve" traction. I haven't worked it out, but I'm not sure you would do much with the rear brake in that case either. One case where you need traction above and beyond what you need for braking is in a turn -- and here, Sheldon and Jobst think it's fine to use the front brake in a downhill hairpin, but a lot of motorcycle instructors say to never front brake in a turn, and yet pro motorcycle racers are hard on the brake right to the apex of a turn and accelerate right back out. So, hmm, might be worth trying to work that one out from the physics standpoint.

WHen i took dirtbike lessons one of the early lessons was to learn to love the front brake, and demonstrate to everyone that it stops better, especially while descending, as long as you are smooth about it. (When you have suspension, there's another complication, as you can't apply as much braking to the front until the forks fully compress.)

I also have the experience that a snow/ice bike with studded tires seems to ride more securely and stop better if I carry my load up on a front basket rather than on the back.
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Old 02-02-10, 01:07 AM
  #21  
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Originally Posted by zzyzx_xyzzy View Post
I also have the experience that a snow/ice bike with studded tires seems to ride more securely and stop better if I carry my load up on a front basket rather than on the back.
That makes sense. The more heavily loaded the front is, the less likely that front will lock up.
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Old 02-02-10, 04:33 AM
  #22  
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Interesting thread so far, especially the contributions from zzyzx_xyzzy.

One thing I haven't seen discussed so far is the fact that the coefficient of friction of a soft material (like rubber) on a hard surface is pressure dependent - as the pressure increases the coefficient of friction decreases.

This is because the usual assumption that true contact area is proportional to normal force breaks down if one material is soft enough to elastically stretch around the asperities of the other. A good practical illustration of this is race car tyres - the width and low pressure (just over 1 bar) give the highest grip possible.

In theory, therefore, lower front tyre pressure will give you better braking. Does the theory hold up in practice? I've not done the experiments but I'll give it a go tomorrow.
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Old 02-02-10, 04:46 AM
  #23  
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Wow, thanks for the replies. In answer to a question back there, I guess I am looking for "pressure on lever" improvement, at the moment it seems I have to use a lot of pressure on the lever for not much braking power. I will go out today and try it out so I can put into words exactly what I am looking for.

Thanks
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Old 02-02-10, 08:51 AM
  #24  
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Originally Posted by DannoXYZ View Post
Don't get cause & effect mixed up. The weight-transfer is the result of deceleration forces pushing back on the bike (while the rider continues forward). The amount of weight-transfer then determines the front/rear split in total-traction. With zero-deceleration and zero weight-transfer, the front/rear split is roughly 40/60 F/R. The actual split varies based up on deceleration-G. The higher the braking-force, the more the weight-transfers and the greater the different in F/R traction. It's not a fixed quantity, but varies with deceleration-rate.

Here's how you actually calculate the F/R split in weight-distribution and resultant traction-limits of each tyre under braking:

Lf = dG+Bh/w
Lr = (1-d)G-Bh/w ; where

d = initial static weight-distribution at front wheel
w = wheelbase
G = static mass of bike + rider system
h = height of center of gravity of bike + rider system
B = braking-force deceleration in G

As you can see, the Lf versus Lr ratio will change depending upon how hard you are braking. The higher the deceleration-G, the more weight transfers to the front-tyre, the higher the load on the front tyre and the higher the front tyre's grip. At some point, as others calculated, about 0.80-0.85g, ALL of the traction is on the front tyre. If you reduce braking-force by letting up on the front-lever to get the rear-tyre down, you'd only have 0.70-0.75g and not stop as fast.
The calculations I did aren't for the mass transfer but for the force at the tire patch. Wilson uses the same calculation to describe the amount of braking that the front and rear wheel contribute in Bicycle Science. I just recalculated with a different. I just realized I quoted verbatim from Wilson and I've edited the post to reflect that.

I haven't done the calculations yet but I doubt that the difference in deceleration is as high as you think. Wilson shows in his calculations...the first example I used...that the deceleration is 0.5g (where I got my erroneous notion from) but the maximum deceleration obtainable is 0.57g. Not a huge difference.
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Old 02-02-10, 09:08 AM
  #25  
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My stock Campy brakes can lock up either wheel it the levers are gripped hard enough. The rear brake is one of the few on the market that intentionally has far less braking power - like it should. One of the big mistakes is making front and rear brakes capable of applying the same braking force. That makes it very easy to lockup a rear wheel and ruin a tire in a matter of seconds. You never see a car or motorcyle with brakes like that; the front is always far more powerful.
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