A post about brakes
02-19-07, 02:35 AM
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i ride a bicycle
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A post about brakes
A Preface:
This is NOT intended to claim anyone is wrong for choosing to ride with or without brakes.
^^ Read that again, I mean it.
I wrote this to give everyone an idea of the dynamics of what exactly happens when you have to decelerate your bike. If you are curious, read on.
A quick description of how a bike brakes on a level road, with as little “engineer speak” as possible. A couple assumptions: 1)The bike is “traction limited” (the rider can lock the wheel, through brakes or skidding) and 2)Aero drag is neglected, as it is relatively small compared to braking force and it would be the same for the brake or brakeless rider.
Here is a diagram of the relevant forces acting on a bike during braking.
W is the combined weight of the rider and bike acting at the center of mass, positioned approximately halfway along the wheelbase of the bike, at height ‘h’.
Wf and Wr are the weights at each wheel, before braking. During braking, weight will shift to the front wheel and we will call these new weights Wfb and Wrb.
Ff and Fr are the friction forces between the tires and road that are actually slowing the bike.
The friction force at the wheel can be approximated by multiplying a friction coefficient, ‘u’, by the weight on the wheel. So during braking,
(eqn1) Ff=u*Wfb
at the front wheel. In other words, more weight on the wheel means more friction force.
I’ll spare you the DiAllembert and summation of moments stuff. The weight on the front wheel when the rider brakes is described as:
(eqn2) Wfb = Wf + W*(a/g)*(h/L)
And similarly the rear wheel:
(eqn3) Wrb = Wr - W*(a/g)*(h/L)
The easy way to describe this equation is the weight on the wheel during braking equals the ‘original’ weight plus or minus (in respect to front or rear wheel) the weight transferred from the effect of braking.
In the case of front wheel only braking, the summation of forces equation is:
Ff = W(a/g)
Substitute eqn1:
u*Wfb = W(a/g)
Substitute eqn2:
u*( Wf + W*(a/g)*(h/L)) = W(a/g)
Solve this equation for (a/g) and the answer is your deceleration in g’s:
(a/g) = (u*(Wf/W)) / (1-u*(h/L))
For the case of rear wheel braking only, the deceleration is:
(a/g) = (u*(Wr/W)) / (1+u*(h/L))
Make up some weights and dimensions:
W = 200 lbs (rider and bike)
Wf = Wr = 100 lbs (using the assumption the center of mass is at the center of the wheelbase)
L = 40 inches
h = 40 inches
u = 0.7
These values show the front brake rider could decelerate at 1.17 g’s and the brakeless rider could only manage 0.206 g’s. Quite a difference.
The brake-equipped bike can’t really decelerate quite that fast. Substitute this deceleration, a/g = 1.17, into eqn3 and you will see that the weight on the rear tire is negative – huh? In other words, the rear wheel has come off the ground and the rider has (possibly) gone over the bars.
Yes, there are TONS of variables I didn’t include in this explanation: condition of the brakes, the skidder shifting his weight, etc. Still, it should give you an idea of just why a brake-equipped bike can stop quicker than its brakeless equivalent. If a brakeless rider tells you he can stop faster than a bike with brakes, recite this explanation to him, and ask him why Mr. Newton’s laws don’t happen to apply to him.
I am open to any comments, questions, corrections, flames, smart-ass comments that you all have for me. Again, please believe me when I say I did not write this to upset anyone. I truly believe everyone out there should ride the bike that makes them happy, brakes or not. I’m just trying to give a little something back (in the form of engineering-nerd knowledge) to a forum I really enjoy reading every day.
Mac
This is NOT intended to claim anyone is wrong for choosing to ride with or without brakes.
^^ Read that again, I mean it.
I wrote this to give everyone an idea of the dynamics of what exactly happens when you have to decelerate your bike. If you are curious, read on.
A quick description of how a bike brakes on a level road, with as little “engineer speak” as possible. A couple assumptions: 1)The bike is “traction limited” (the rider can lock the wheel, through brakes or skidding) and 2)Aero drag is neglected, as it is relatively small compared to braking force and it would be the same for the brake or brakeless rider.
Here is a diagram of the relevant forces acting on a bike during braking.
W is the combined weight of the rider and bike acting at the center of mass, positioned approximately halfway along the wheelbase of the bike, at height ‘h’.
Wf and Wr are the weights at each wheel, before braking. During braking, weight will shift to the front wheel and we will call these new weights Wfb and Wrb.
Ff and Fr are the friction forces between the tires and road that are actually slowing the bike.
The friction force at the wheel can be approximated by multiplying a friction coefficient, ‘u’, by the weight on the wheel. So during braking,
(eqn1) Ff=u*Wfb
at the front wheel. In other words, more weight on the wheel means more friction force.
I’ll spare you the DiAllembert and summation of moments stuff. The weight on the front wheel when the rider brakes is described as:
(eqn2) Wfb = Wf + W*(a/g)*(h/L)
And similarly the rear wheel:
(eqn3) Wrb = Wr - W*(a/g)*(h/L)
The easy way to describe this equation is the weight on the wheel during braking equals the ‘original’ weight plus or minus (in respect to front or rear wheel) the weight transferred from the effect of braking.
In the case of front wheel only braking, the summation of forces equation is:
Ff = W(a/g)
Substitute eqn1:
u*Wfb = W(a/g)
Substitute eqn2:
u*( Wf + W*(a/g)*(h/L)) = W(a/g)
Solve this equation for (a/g) and the answer is your deceleration in g’s:
(a/g) = (u*(Wf/W)) / (1-u*(h/L))
For the case of rear wheel braking only, the deceleration is:
(a/g) = (u*(Wr/W)) / (1+u*(h/L))
Make up some weights and dimensions:
W = 200 lbs (rider and bike)
Wf = Wr = 100 lbs (using the assumption the center of mass is at the center of the wheelbase)
L = 40 inches
h = 40 inches
u = 0.7
These values show the front brake rider could decelerate at 1.17 g’s and the brakeless rider could only manage 0.206 g’s. Quite a difference.
The brake-equipped bike can’t really decelerate quite that fast. Substitute this deceleration, a/g = 1.17, into eqn3 and you will see that the weight on the rear tire is negative – huh? In other words, the rear wheel has come off the ground and the rider has (possibly) gone over the bars.
Yes, there are TONS of variables I didn’t include in this explanation: condition of the brakes, the skidder shifting his weight, etc. Still, it should give you an idea of just why a brake-equipped bike can stop quicker than its brakeless equivalent. If a brakeless rider tells you he can stop faster than a bike with brakes, recite this explanation to him, and ask him why Mr. Newton’s laws don’t happen to apply to him.
I am open to any comments, questions, corrections, flames, smart-ass comments that you all have for me. Again, please believe me when I say I did not write this to upset anyone. I truly believe everyone out there should ride the bike that makes them happy, brakes or not. I’m just trying to give a little something back (in the form of engineering-nerd knowledge) to a forum I really enjoy reading every day.
Mac
02-19-07, 02:40 AM
#2
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thanks, i haven't had a physics lesson since HS.
but seriously, this makes sense with or without equations added to it. honestly, if you get a brake rider and a brakeless rider and have a stop-off on a level stretch of road, i'd put my money on the brake rider. i don't really know anyone who wouldn't.
but seriously, this makes sense with or without equations added to it. honestly, if you get a brake rider and a brakeless rider and have a stop-off on a level stretch of road, i'd put my money on the brake rider. i don't really know anyone who wouldn't.
02-19-07, 09:08 AM
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if you think riding bikes is something that can be reduced to mute equations then you are missing many many whole points
02-19-07, 09:12 AM
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Originally Posted by sac02
A Preface:
This is NOT intended to claim anyone is wrong for choosing to ride with or without brakes.
^^ Read that again, I mean it.
I wrote this to give everyone an idea of the dynamics of what exactly happens when you have to decelerate your bike. If you are curious, read on.
A quick description of how a bike brakes on a level road, with as little “engineer speak” as possible. A couple assumptions: 1)The bike is “traction limited” (the rider can lock the wheel, through brakes or skidding) and 2)Aero drag is neglected, as it is relatively small compared to braking force and it would be the same for the brake or brakeless rider.
Here is a diagram of the relevant forces acting on a bike during braking.
W is the combined weight of the rider and bike acting at the center of mass, positioned approximately halfway along the wheelbase of the bike, at height ‘h’.
Wf and Wr are the weights at each wheel, before braking. During braking, weight will shift to the front wheel and we will call these new weights Wfb and Wrb.
Ff and Fr are the friction forces between the tires and road that are actually slowing the bike.
The friction force at the wheel can be approximated by multiplying a friction coefficient, ‘u’, by the weight on the wheel. So during braking,
(eqn1) Ff=u*Wfb
at the front wheel. In other words, more weight on the wheel means more friction force.
I’ll spare you the DiAllembert and summation of moments stuff. The weight on the front wheel when the rider brakes is described as:
(eqn2) Wfb = Wf + W*(a/g)*(h/L)
And similarly the rear wheel:
(eqn3) Wrb = Wr - W*(a/g)*(h/L)
The easy way to describe this equation is the weight on the wheel during braking equals the ‘original’ weight plus or minus (in respect to front or rear wheel) the weight transferred from the effect of braking.
In the case of front wheel only braking, the summation of forces equation is:
Ff = W(a/g)
Substitute eqn1:
u*Wfb = W(a/g)
Substitute eqn2:
u*( Wf + W*(a/g)*(h/L)) = W(a/g)
Solve this equation for (a/g) and the answer is your deceleration in g’s:
(a/g) = (u*(Wf/W)) / (1-u*(h/L))
For the case of rear wheel braking only, the deceleration is:
(a/g) = (u*(Wr/W)) / (1+u*(h/L))
Make up some weights and dimensions:
W = 200 lbs (rider and bike)
Wf = Wr = 100 lbs (using the assumption the center of mass is at the center of the wheelbase)
L = 40 inches
h = 40 inches
u = 0.7
These values show the front brake rider could decelerate at 1.17 g’s and the brakeless rider could only manage 0.206 g’s. Quite a difference.
The brake-equipped bike can’t really decelerate quite that fast. Substitute this deceleration, a/g = 1.17, into eqn3 and you will see that the weight on the rear tire is negative – huh? In other words, the rear wheel has come off the ground and the rider has (possibly) gone over the bars.
Yes, there are TONS of variables I didn’t include in this explanation: condition of the brakes, the skidder shifting his weight, etc. Still, it should give you an idea of just why a brake-equipped bike can stop quicker than its brakeless equivalent. If a brakeless rider tells you he can stop faster than a bike with brakes, recite this explanation to him, and ask him why Mr. Newton’s laws don’t happen to apply to him.
I am open to any comments, questions, corrections, flames, smart-ass comments that you all have for me. Again, please believe me when I say I did not write this to upset anyone. I truly believe everyone out there should ride the bike that makes them happy, brakes or not. I’m just trying to give a little something back (in the form of engineering-nerd knowledge) to a forum I really enjoy reading every day.
Mac
This is NOT intended to claim anyone is wrong for choosing to ride with or without brakes.
^^ Read that again, I mean it.
I wrote this to give everyone an idea of the dynamics of what exactly happens when you have to decelerate your bike. If you are curious, read on.
A quick description of how a bike brakes on a level road, with as little “engineer speak” as possible. A couple assumptions: 1)The bike is “traction limited” (the rider can lock the wheel, through brakes or skidding) and 2)Aero drag is neglected, as it is relatively small compared to braking force and it would be the same for the brake or brakeless rider.
Here is a diagram of the relevant forces acting on a bike during braking.
W is the combined weight of the rider and bike acting at the center of mass, positioned approximately halfway along the wheelbase of the bike, at height ‘h’.
Wf and Wr are the weights at each wheel, before braking. During braking, weight will shift to the front wheel and we will call these new weights Wfb and Wrb.
Ff and Fr are the friction forces between the tires and road that are actually slowing the bike.
The friction force at the wheel can be approximated by multiplying a friction coefficient, ‘u’, by the weight on the wheel. So during braking,
(eqn1) Ff=u*Wfb
at the front wheel. In other words, more weight on the wheel means more friction force.
I’ll spare you the DiAllembert and summation of moments stuff. The weight on the front wheel when the rider brakes is described as:
(eqn2) Wfb = Wf + W*(a/g)*(h/L)
And similarly the rear wheel:
(eqn3) Wrb = Wr - W*(a/g)*(h/L)
The easy way to describe this equation is the weight on the wheel during braking equals the ‘original’ weight plus or minus (in respect to front or rear wheel) the weight transferred from the effect of braking.
In the case of front wheel only braking, the summation of forces equation is:
Ff = W(a/g)
Substitute eqn1:
u*Wfb = W(a/g)
Substitute eqn2:
u*( Wf + W*(a/g)*(h/L)) = W(a/g)
Solve this equation for (a/g) and the answer is your deceleration in g’s:
(a/g) = (u*(Wf/W)) / (1-u*(h/L))
For the case of rear wheel braking only, the deceleration is:
(a/g) = (u*(Wr/W)) / (1+u*(h/L))
Make up some weights and dimensions:
W = 200 lbs (rider and bike)
Wf = Wr = 100 lbs (using the assumption the center of mass is at the center of the wheelbase)
L = 40 inches
h = 40 inches
u = 0.7
These values show the front brake rider could decelerate at 1.17 g’s and the brakeless rider could only manage 0.206 g’s. Quite a difference.
The brake-equipped bike can’t really decelerate quite that fast. Substitute this deceleration, a/g = 1.17, into eqn3 and you will see that the weight on the rear tire is negative – huh? In other words, the rear wheel has come off the ground and the rider has (possibly) gone over the bars.
Yes, there are TONS of variables I didn’t include in this explanation: condition of the brakes, the skidder shifting his weight, etc. Still, it should give you an idea of just why a brake-equipped bike can stop quicker than its brakeless equivalent. If a brakeless rider tells you he can stop faster than a bike with brakes, recite this explanation to him, and ask him why Mr. Newton’s laws don’t happen to apply to him.
I am open to any comments, questions, corrections, flames, smart-ass comments that you all have for me. Again, please believe me when I say I did not write this to upset anyone. I truly believe everyone out there should ride the bike that makes them happy, brakes or not. I’m just trying to give a little something back (in the form of engineering-nerd knowledge) to a forum I really enjoy reading every day.
Mac
I think all brakeless riders know that. They just dont care. Im a brake lover but ive been riding brakeless. Ive had alot more fun.
02-19-07, 09:13 AM
#6
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Originally Posted by eddiebrannan
if you think riding bikes is something that can be reduced to mute equations then you are missing many many whole points
02-19-07, 10:36 AM
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Originally Posted by sac02
Yes, there are TONS of variables I didn’t include in this explanation: condition of the brakes, the skidder shifting his weight, etc.
I'm not saying your physics isn't good and stuff, I'm just saying your quantifications might be a little off. I'd bet the factor of difference is more like 1.5 or 2 and not 6. But that's just an armchair physicists view.
02-19-07, 11:17 AM
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Originally Posted by eddiebrannan
if you think riding bikes is something that can be reduced to mute equations then you are missing many many whole points
+1. stopping quickly, with or without brakes is as much a matter of skill as mechanics. even a bike with handbrakes requires skills such as knowing when and to apply pressure and how much to apply, weight distribution, etc. the biggest variable is the rider.
02-19-07, 11:19 AM
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Originally Posted by Aeroplane
Weight shifting is the biggest thing that isn't taken into account. Since the primary cause of the changes between your two equations is the rider's weight being moved forward, shifting the weight back would cause a huge difference in the two. The main thing that made me realize this was your free body diagrams. When braking (hand brake or drivetrain), my arms are pushing the bars away.
I'm not saying your physics isn't good and stuff, I'm just saying your quantifications might be a little off. I'd bet the factor of difference is more like 1.5 or 2 and not 6. But that's just an armchair physicists view.
I'm not saying your physics isn't good and stuff, I'm just saying your quantifications might be a little off. I'd bet the factor of difference is more like 1.5 or 2 and not 6. But that's just an armchair physicists view.
true. also, a good skidder will know how to skid while basically seated.
02-19-07, 11:21 AM
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Originally Posted by humancongereel
+1. stopping quickly, with or without brakes is as much a matter of skill as mechanics. even a bike with handbrakes requires skills such as knowing when and to apply pressure and how much to apply, weight distribution, etc. the biggest variable is the rider.
except that anyone who is a half decent rider can stop faster with brakes then the best brakeless rider can so it really isn't unless you are talking about rank noobs and kids who just lock up the rear wheel in which case they are stopping about as fast as the best brakeless rider anyway.
02-19-07, 11:22 AM
#12
hello
I already knew that, even without all the formulas......but thanks...
Last edited by roadfix; 02-19-07 at 03:52 PM.
02-19-07, 11:25 AM
#13
aka mattio
Originally Posted by dutret
except that anyone who is a half decent rider can stop faster with brakes then the best brakeless rider can so it really isn't unless you are talking about rank noobs and kids who just lock up the rear wheel in which case they are stopping about as fast as the best brakeless rider anyway.
02-19-07, 11:27 AM
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Originally Posted by humancongereel
true. also, a good skidder will know how to skid while basically seated.
02-19-07, 11:38 AM
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No matter what your stopping method is, the COM is going to move forward when stopping from a decent speed. Your body isn't completely locked into the same frame of reference as the bike and carries a lot more inertia.
All that really matters is this:
1. You can stop your bike.
2. You are comfortable stopping your bike at any given moment.
Avoidance is best, but not always an option.
All that really matters is this:
1. You can stop your bike.
2. You are comfortable stopping your bike at any given moment.
Avoidance is best, but not always an option.
02-19-07, 11:43 AM
#16
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Originally Posted by mihlbach
I don't think the OP was making that claim, but obviously you can't cheat the laws of physics no matter how much of a zen ninja you may think you are. At any rate, crashing into a bus without brakes looks way cooler than flipping over your bars because your front braked stopped you too fast.
Too bad that isn't the point for (most) brakeless riders. If you want to follow a bus ridiculously close and feel safe with your brakes that is fine. The true spirit of brakeless I think doesn't relate to riding the same way with brakes, but compensating and problem solving. There are more ways to avoid danger than franticly mashing at your brakes. Just learn to ride, DEWWWWWD!
02-19-07, 12:34 PM
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I'm constantly amazed by how many skilled riders seem totally unable to operate handbrakes or wrap handlebars.
02-19-07, 02:40 PM
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Originally Posted by dutret
except that anyone who is a half decent rider can stop faster with brakes then the best brakeless rider can so it really isn't unless you are talking about rank noobs and kids who just lock up the rear wheel in which case they are stopping about as fast as the best brakeless rider anyway.
but the rider is such a big variable, you can't quantify that. how are you going to prove that statement? you can't. let's say you have a braked bike and a brakeless bike compete in a quick stop. the braked bike wins. that could mean the braked bike is better. it could also mean the brakeless rider is teh sux00rz. and if the brakeless bike wins, the opposite would be true. if the same rider tries both methods and is better at one than the other, that proves only that they can use one better than the other, not that one is inherently better than the other. the formula may prove efficiency, but not necessarily an accurate analysis of the advantages/disadvantages of the two methods in ever day riding. that's about the rider, not the machine.
i know people who suck at riding road bikes. they could suck at track bikes, too. or they could rule on them. i know people who suck at track bikes. same goes for road bikes with them. it has as much to do with avoidance and technique as it does with any mechanical advantage. yes, there is that mechanical edge as the formula above proves. whether or not it will be of any practical value in every day situations is a matter that is dependent wholly upon the rider.
02-19-07, 02:45 PM
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Most skilled riders will scrub speed and carve around or avoid the obstacle compleletly, braked or brakeless.
02-19-07, 02:46 PM
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Originally Posted by humancongereel
i know people who suck at riding road bikes. they could suck at track bikes, too. or they could rule on them. i know people who suck at track bikes. same goes for road bikes with them. it has as much to do with avoidance and technique as it does with any mechanical advantage. yes, there is that mechanical edge as the formula above proves. whether or not it will be of any practical value in every day situations is a matter that is dependent wholly upon the rider.
with very little effort nearly anyone could learn to use a handbrake to stop faster and with more control than even a good brakeless rider
someone who "sucks at riding road bikes" should try a little harder learn a little more or smoke a little less
dont tell me your friends dont have hands
Last edited by doofo; 02-19-07 at 11:13 PM.
02-19-07, 02:53 PM
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Originally Posted by thurstonboise
Your body isn't completely locked into the same frame of reference as the bike and carries a lot more inertia.
02-19-07, 02:55 PM
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Originally Posted by humancongereel
but the rider is such a big variable, you can't quantify that. how are you going to prove that statement? you can't. let's say you have a braked bike and a brakeless bike compete in a quick stop. the braked bike wins. that could mean the braked bike is better. it could also mean the brakeless rider is teh sux00rz. and if the brakeless bike wins, the opposite would be true. if the same rider tries both methods and is better at one than the other, that proves only that they can use one better than the other, not that one is inherently better than the other. the formula may prove efficiency, but not necessarily an accurate analysis of the advantages/disadvantages of the two methods in ever day riding. that's about the rider, not the machine.
i know people who suck at riding road bikes. they could suck at track bikes, too. or they could rule on them. i know people who suck at track bikes. same goes for road bikes with them. it has as much to do with avoidance and technique as it does with any mechanical advantage. yes, there is that mechanical edge as the formula above proves. whether or not it will be of any practical value in every day situations is a matter that is dependent wholly upon the rider.
i know people who suck at riding road bikes. they could suck at track bikes, too. or they could rule on them. i know people who suck at track bikes. same goes for road bikes with them. it has as much to do with avoidance and technique as it does with any mechanical advantage. yes, there is that mechanical edge as the formula above proves. whether or not it will be of any practical value in every day situations is a matter that is dependent wholly upon the rider.
What you seem to miss is that the BEST ANY brakeless rider can do is shift their weight as little forward or up as possible while locking their rear wheel. This is roughly equivalent to the worst any braked rider can do. Watch a novice cyclist make a panic stop sometime... They do exactly what you claim the best brakeless riders do: "keep their weight close to where it would be when they are in the saddle."
Avoidance and technique are fine but there are plenty of situations where stopping fast is all that matters. Your previous claim had even less to do with that however since you claimed that "stopping quickly with or without brakes is as much a matter of skill as mechanics." That simply isn't true outside the fantasy land brakeless rider use to justify their choice.
02-19-07, 02:57 PM
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Originally Posted by dutret
Retem
02-19-07, 02:59 PM
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Originally Posted by Aeroplane
It really should be (or damn near it) unless you are trying to endo.
After I posted that I thought:
**** I posted in a brakeless thread. After that I thought, it's probably not enough to really matter. I was also thinking about panic stops from 15-20MPH.
02-19-07, 03:14 PM
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Originally Posted by humancongereel
whether or not it will be of any practical value in every day situations is a matter that is dependent wholly upon the rider.
You can't concede the physics (which you DID do) and then argue that it's all about the rider.
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Just say you don't care if you can't stop as well. Everyone's fine with that. Pretending that you can is a lie, to both yourself and the forum.