Why do bikes hardly ever skid while braking with the front wheel?
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When abruptly braking with the rear wheel on a bike, it tends to skid pretty easily. Doing the same with the front wheel is a very different experience. Instead of skidding, the bike lifts the rear wheel. I've never seen the front wheel of a bike skid.
According to this answer the torque generated on a bike during breaking is set around the front wheel contact point. While this explains why the rear wheel is lifted, it doesn't really explain why the wheel almost never slips.
What causes the front wheel to behave like it had more friction/grip with the ground?
newtonian-mechanics rotational-dynamics friction everyday-life free-body-diagram
asked Nov 15 '18 at 3:16
Tiago MarinhoTiago Marinho
307127
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When abruptly braking with the rear wheel on a bike, it tends to skid pretty easily. Doing the same with the front wheel is a very different experience. Instead of skidding, the bike lifts the rear wheel. I've never seen the front wheel of a bike skid.
According to this answer the torque generated on a bike during breaking is set around the front wheel contact point. While this explains why the rear wheel is lifted, it doesn't really explain why the wheel almost never slips.
What causes the front wheel to behave like it had more friction/grip with the ground?
newtonian-mechanics rotational-dynamics friction everyday-life free-body-diagram
asked Nov 15 '18 at 3:16
Tiago MarinhoTiago Marinho
307127
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Do you mean drift (as in skid outwards on a bend "11. a controlled four-wheel skid, used by racing drivers to take bends at high speed " (Collins)) or simply skid in forward motion?
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– Chris H
Nov 15 '18 at 13:05
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I have managed to skid my front wheel (without crashing) but only in a dead-straight line, on wet leaves on tarmac. Any form of side-slipping the front wheel (while braking) is likely to be painful, and bikes never stay straight for long (there's another question in that). I've also skidded the front wheel on bends (with or without braking) and invariably hit the road rather hard.
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– Chris H
Nov 15 '18 at 13:08
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Skidding the front wheel occurs quite easily in deep, soft snow. It's also easier to handle than on wet road, because even though the wheel stops spinning it still works as a ski, retaining some control :)
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– jpa
Nov 15 '18 at 13:37
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I've never seen the front wheel of a bike skid. grab a good fist full of front brake while you're riding on gravel, ice, wet metal or any other surface with diminished friction, and your story will abruptly change.. I recommend not to ride around utilizing the front brake as though it's impossible to lock the front wheel, if you wish to avoid injury..
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– Caius Jard
Nov 15 '18 at 20:40
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Also note that it's not all that hard to skid with both wheels. (Not the question, but still interesting.)
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– Jasper
Nov 16 '18 at 14:09
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show 6 more comments
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When abruptly braking with the rear wheel on a bike, it tends to skid pretty easily. Doing the same with the front wheel is a very different experience. Instead of skidding, the bike lifts the rear wheel. I've never seen the front wheel of a bike skid.
According to this answer the torque generated on a bike during breaking is set around the front wheel contact point. While this explains why the rear wheel is lifted, it doesn't really explain why the wheel almost never slips.
What causes the front wheel to behave like it had more friction/grip with the ground?
newtonian-mechanics rotational-dynamics friction everyday-life free-body-diagram
asked Nov 15 '18 at 3:16
Tiago MarinhoTiago Marinho
307127
$endgroup$
When abruptly braking with the rear wheel on a bike, it tends to skid pretty easily. Doing the same with the front wheel is a very different experience. Instead of skidding, the bike lifts the rear wheel. I've never seen the front wheel of a bike skid.
According to this answer the torque generated on a bike during breaking is set around the front wheel contact point. While this explains why the rear wheel is lifted, it doesn't really explain why the wheel almost never slips.
What causes the front wheel to behave like it had more friction/grip with the ground?
newtonian-mechanics rotational-dynamics friction everyday-life free-body-diagram
newtonian-mechanics rotational-dynamics friction everyday-life free-body-diagram
asked Nov 15 '18 at 3:16
Tiago MarinhoTiago Marinho
307127
asked Nov 15 '18 at 3:16
Tiago MarinhoTiago Marinho
307127
edited Nov 17 '18 at 11:53
Tiago Marinho
asked Nov 15 '18 at 3:16
Tiago MarinhoTiago Marinho
307127
asked Nov 15 '18 at 3:16
Tiago MarinhoTiago Marinho
307127
asked Nov 15 '18 at 3:16
Tiago MarinhoTiago Marinho
307127
307127
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Do you mean drift (as in skid outwards on a bend "11. a controlled four-wheel skid, used by racing drivers to take bends at high speed " (Collins)) or simply skid in forward motion?
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– Chris H
Nov 15 '18 at 13:05
7
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I have managed to skid my front wheel (without crashing) but only in a dead-straight line, on wet leaves on tarmac. Any form of side-slipping the front wheel (while braking) is likely to be painful, and bikes never stay straight for long (there's another question in that). I've also skidded the front wheel on bends (with or without braking) and invariably hit the road rather hard.
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– Chris H
Nov 15 '18 at 13:08
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Skidding the front wheel occurs quite easily in deep, soft snow. It's also easier to handle than on wet road, because even though the wheel stops spinning it still works as a ski, retaining some control :)
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– jpa
Nov 15 '18 at 13:37
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I've never seen the front wheel of a bike skid. grab a good fist full of front brake while you're riding on gravel, ice, wet metal or any other surface with diminished friction, and your story will abruptly change.. I recommend not to ride around utilizing the front brake as though it's impossible to lock the front wheel, if you wish to avoid injury..
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– Caius Jard
Nov 15 '18 at 20:40
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Also note that it's not all that hard to skid with both wheels. (Not the question, but still interesting.)
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– Jasper
Nov 16 '18 at 14:09
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show 6 more comments
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Do you mean drift (as in skid outwards on a bend "11. a controlled four-wheel skid, used by racing drivers to take bends at high speed " (Collins)) or simply skid in forward motion?
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– Chris H
Nov 15 '18 at 13:05
7
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I have managed to skid my front wheel (without crashing) but only in a dead-straight line, on wet leaves on tarmac. Any form of side-slipping the front wheel (while braking) is likely to be painful, and bikes never stay straight for long (there's another question in that). I've also skidded the front wheel on bends (with or without braking) and invariably hit the road rather hard.
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– Chris H
Nov 15 '18 at 13:08
2
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Skidding the front wheel occurs quite easily in deep, soft snow. It's also easier to handle than on wet road, because even though the wheel stops spinning it still works as a ski, retaining some control :)
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– jpa
Nov 15 '18 at 13:37
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I've never seen the front wheel of a bike skid. grab a good fist full of front brake while you're riding on gravel, ice, wet metal or any other surface with diminished friction, and your story will abruptly change.. I recommend not to ride around utilizing the front brake as though it's impossible to lock the front wheel, if you wish to avoid injury..
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– Caius Jard
Nov 15 '18 at 20:40
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Also note that it's not all that hard to skid with both wheels. (Not the question, but still interesting.)
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– Jasper
Nov 16 '18 at 14:09
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2
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Do you mean drift (as in skid outwards on a bend "11. a controlled four-wheel skid, used by racing drivers to take bends at high speed " (Collins)) or simply skid in forward motion?
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– Chris H
Nov 15 '18 at 13:05
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Do you mean drift (as in skid outwards on a bend "11. a controlled four-wheel skid, used by racing drivers to take bends at high speed " (Collins)) or simply skid in forward motion?
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– Chris H
Nov 15 '18 at 13:05
7
7
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I have managed to skid my front wheel (without crashing) but only in a dead-straight line, on wet leaves on tarmac. Any form of side-slipping the front wheel (while braking) is likely to be painful, and bikes never stay straight for long (there's another question in that). I've also skidded the front wheel on bends (with or without braking) and invariably hit the road rather hard.
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– Chris H
Nov 15 '18 at 13:08
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I have managed to skid my front wheel (without crashing) but only in a dead-straight line, on wet leaves on tarmac. Any form of side-slipping the front wheel (while braking) is likely to be painful, and bikes never stay straight for long (there's another question in that). I've also skidded the front wheel on bends (with or without braking) and invariably hit the road rather hard.
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– Chris H
Nov 15 '18 at 13:08
2
2
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Skidding the front wheel occurs quite easily in deep, soft snow. It's also easier to handle than on wet road, because even though the wheel stops spinning it still works as a ski, retaining some control :)
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– jpa
Nov 15 '18 at 13:37
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Skidding the front wheel occurs quite easily in deep, soft snow. It's also easier to handle than on wet road, because even though the wheel stops spinning it still works as a ski, retaining some control :)
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– jpa
Nov 15 '18 at 13:37
2
2
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I've never seen the front wheel of a bike skid. grab a good fist full of front brake while you're riding on gravel, ice, wet metal or any other surface with diminished friction, and your story will abruptly change.. I recommend not to ride around utilizing the front brake as though it's impossible to lock the front wheel, if you wish to avoid injury..
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– Caius Jard
Nov 15 '18 at 20:40
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I've never seen the front wheel of a bike skid. grab a good fist full of front brake while you're riding on gravel, ice, wet metal or any other surface with diminished friction, and your story will abruptly change.. I recommend not to ride around utilizing the front brake as though it's impossible to lock the front wheel, if you wish to avoid injury..
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– Caius Jard
Nov 15 '18 at 20:40
2
2
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Also note that it's not all that hard to skid with both wheels. (Not the question, but still interesting.)
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– Jasper
Nov 16 '18 at 14:09
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Also note that it's not all that hard to skid with both wheels. (Not the question, but still interesting.)
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– Jasper
Nov 16 '18 at 14:09
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9 Answers
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Because, when you brake, your weight is being shifted towards the front wheel. The inertia coupled with gravity puts your weight and that of the bike onto the front wheel. More weight ⇒ more pressure ⇒ more friction/grip with the ground.
answered Nov 15 '18 at 3:43
Alex DoeAlex Doe
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@henning But when you brake then the moment of inertia wants to rotate the bike (and the rider) such that there is more "weight" on the front wheel and less on the back.
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– phuzi
Nov 15 '18 at 9:48
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To be specific - if the rear wheel lifts then indeed all weight is on the front wheel, as explained by phuzi due to the forces affecting the bike. If you brake slightly lighter then some weight still remains on the rear wheel but still the force/weight on the front wheel is significantly larger.
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– Ister
Nov 15 '18 at 10:10
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and if you brake just a touch more than the back wheel lifting, you go over the bars and land on your head; maximal braking is close to instability. (@henning)
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– Chris H
Nov 15 '18 at 13:09
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@henning Incidentally, that's why you should almost never brake using the rear brake alone. As your weight shifts off the rear wheel while braking, the maximum frictional force of the rear wheel decreases. Using the front brake alone lets you stop in about half the distance as using the rear brake alone, and using both is even better.
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– Nuclear Wang
Nov 15 '18 at 14:01
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@NuclearWang if you have perfect (constant) grip and a smooth road, the back brake contributes nothing to stopping in minimal distance. On real roads you're right, as you don't want to assume quite such ideal conditions
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– Chris H
Nov 15 '18 at 15:22
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In addition to Alex Doe's answer: front wheel drift does occur on slippery surfaces, Unfortunately a bike is really unstable in this mode so you'll fall over very soon (whereas a rear wheel drift is pretty stable and controllable).
answered Nov 15 '18 at 8:48
HobbesHobbes
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Right. Correctly dosing an “anti-lock front brake maneuver” is an essentiall skill both for MTB and for survival in rainy cities without very well-designed cycle roads.
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– leftaroundabout
Nov 15 '18 at 12:16
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in a sharp turn, a front-wheel skid causes something that bikers call a high-side crash. they are spectacularly destructive and youtube is full of videos of them. search under "highside mulholland" to find some.
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– niels nielsen
Nov 15 '18 at 18:18
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CORRECTION! front-wheel drift causes a LOW-SIDE CRASH! My bad- NN
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– niels nielsen
Nov 15 '18 at 21:37
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@nielsnielsen I've managed to achieve both high and low-side crashes - depends if the tyre regains grip at any point after the loss of traction, and whether the weight is in a suitable place at that instant.
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– Criggie
Nov 15 '18 at 21:41
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A number of years ago now whilst on holiday in Corfu, I dropped a rental scooter in precisely this way. I braked too sharply with the front brake just as I hit a patch of gravel across the road. The front wheel locked, I couldn't control the skidding wheel, and everything went a bit pear-shaped. I was wearing jeans so my legs were only bruised, but I had a decent road rash on my left shoulder and arm.
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– Graham
Nov 16 '18 at 1:20
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Why rear wheel easily drifts?
Consider the torque around contact of front tire from the frame of the cycle.
The pseudo force on the rider and the cycle is in forward direction
Thus, the weight and thus the normal force on rear tire tends to decrease.
Thus smaller friction force and bicycle tends to drift.
Why front wheel rarely drifts?
- Considering the pseudo force and the tendency to rotate, the normal force on front tire is much larger (due to forward pseudo force on the centre of gravity). Thus, the grip is very tight.
answered Nov 15 '18 at 4:00
m__m__
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it doesn't really explains why the wheel never slips
Well, that's just not true. Front wheels can, and do slip. However, normally, the contact between front wheel and road is just too good for the slipping to set in before the wheel blocks and you go over the bars.
It's actually all down to two factors:
the braking force that your front wheel can transmit to the road without slipping
the angle between the center of gravity, the wheel-road contact point, and the road
The later factor is only a question of frame geometry. When your front wheel is about to block, the effective force on your center of gravity is exactly at that angle. Because gravity is fixed, and that angle is fixed, so is the braking force that is applied in this situation.
Note that the entire weight of the bike and rider is on the front wheel when it's about to block, so the vertical force on the wheel-road contact point is fixed too.
Now we can compare two forces:
The max braking force that your wheel material can transmit to the road in its current condition, given that the wheel is pressed to the road with the entire weight.
The braking force that is required to get you to the tipping point.
If the first is the greater, your front wheel will block, and you will go over the bars. If the later is the greater, your front wheel will start slipping.
Now, why is the first force the greater for almost all bikes?
Well, a good tire allows about as much horizontal force to be transmitted to a dry road as the vertical force that is applied to press it onto the road. I.e., the angle between center of gravity, contact point, and road can get as low as 45°, and the front wheel is still able to block. Yet, most bikes are built in such a way that this angle is significantly greater than 45°.
So, because of the geometry of most bikes, you need quite slippery road conditions in order for the front wheel to be able to slip. The longer the frame and the further back the center of gravity is behind the front wheel, the harder it becomes to block the front wheel and to go over the bars; up to the point that going over the bars becomes impossible.
answered Nov 15 '18 at 11:07
cmastercmaster
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I had 2-3 occasions where both my wheels locked. The occasion I remember well I was in high school. I was riding like 5 or 6 km/h going around people on a beach alley. Unexpectedly an old woman stepped into in front of me. I immediately squeezed the brakes hard. Due to inclination and sand on road, both wheel locked. I slid maybe 1m and realized that I'm not decelerating at all, and I'm going to hit the lady. So I released the bakes and managed to change my direction to miss her by just a little. Felt very happy I avoided that incident. The sand on road really surprised me.
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– akostadinov
Nov 16 '18 at 11:46
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Rumors have it that most people who go over the bars do so before (i.e. without) the front wheel blocking: they go over because of bending their elbow i.e. braking harder than expeced and unable to catch the additional weight on the arms fast enough. As a plausible guesstimate, I'd suggest that pretty much everyone who goes over the bars and after "landing" has the bike behind them with the rear wheel not in front of the front wheel did not have the bike flipping over with them but was instead "jumping" off to the front/did not decelerate as fast as their bike.
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– cbeleites
Nov 16 '18 at 16:29
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@cbeleites Fortunately, I cannot say whether you are right since I never went over my bars. My frame is long enough to make that impossible. But I did skid my front wheel several times. By the geometry of most frames, and by the quality of the tires that I know, it's very possible to go over the bars by actually blocking the front wheel. My frame is quite a strong exception in this. And I'm very happy that it makes this kind of accident impossible: Skidding your front wheel is bad enough as it is, going head-over is nothing I like thinking about...
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– cmaster
Nov 16 '18 at 17:21
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"contact between front wheel and road is just too good" I'm not sure I agree with that. I think if someone is prepared for it and shifts their weight back as far possible then the front wheel will skid on most surfaces+bikes. Wouldn't want to try it out though. :)
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– Nobody
Nov 18 '18 at 22:15
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@Nobody That also depends on the tire you are riding. The ones I ride definitely have a skidding angle below 45° on a dry road (it has happened that I've touched the road with my pedals when doing sharp and fast corners). I've ridden a tire once that I could never have dreamed of riding in such extreme corners, I felt it crawling away already in normally tight corners. Needless to say that I've never purchased a second tire of that brand. Point is, bike tire behave very differently, and the better ones are definitely impossible to skid on a dry road with most bikes, especially racing bikes.
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– cmaster
Nov 18 '18 at 23:07
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The brief answer to this is 'because the centre of mass of a bicycle is high up', and especially it is high up compared to a car, where often the front wheels can slip.
To see how this works, consider a vastly oversimplified model of a bike: assume that the structure of the bike is light compared to its rider, and represent its rider as a point mass (see below for why this works even when that's not true). And we're going to be interested in the moment when the rear wheel lifts, so we can completely ignore the rear wheel, and concentrate just on the front wheel, and specifically on the point at which the front wheel touches the road. So the system looks something like this:
So, here, $c$ is the point at which the front tyre is touching the road, $m$ is the rider, and the horizontal & vertical distances between $c$ and $m$ are $l$ & $h$ respectively. And the bike is decelerating at $a$. And I've drawn the forces exerted by the front wheel on the road at $c$ (remember the back wheel is, by assumption, just lifting, so can be ignored: it's not exerting any forces on anything).
For now assume that the coefficient of friction between the front wheel and the road is sufficiently high that the wheel does not slip and lets work out the point at which the bike just begins to tumble over the front wheel: this will tell us the maximum possible value of $a$, however sticky the front wheel is.
It's pretty easy to see that the force on $m$ has two components: a vertical component which is $-mg$, where $g$ is acceleration due to gravity, and a horizontal component which is $ma$, where $a$ is the horizontal acceleration. And the bike will tumble when this vector points above the front wheel. Well, just by drawing the appropriate components you can see that this is true when
$$fracmamg gt fraclh$$
or in other words, for the bike not to tumble
$$a le fraclgh$$
or
$$a_textmax = fraclgh tag1$$
You can convince yourself this is right: a very tall bike ($h gg l$) will tumble really easily, and a completely flat bike ($h ll l$) will almost never tumble. And a bike in very low gravity will tumble more easily than one in high gravity. So (1) tells us how big $a$ can be, however sticky the front wheel is.
Now consider the coefficient of friction at the front wheel. The coefficient of friction, $mu$ is defined as the force with which the wheel is trying to slide along the road and the force which it is being pressed down onto the road, at the point where the wheel just slips. So it's obvious that,
$$mu = fracma_textslipmg = fraca_textslipg$$
where $a_textslip$ is the point at which the wheel slips. In other words
$$a_textslip = mu g tag2$$
And now we can use (1) & (2) to give us the answer we're looking for: the bike will tumble before it drifts if $a_textslip gt a_textmax$, in other words if
$$mu gt fraclh$$
And you can now see the problem here: bikes are rather short and rather tall, so $l/h$ tends to be rather small, which means that the bike will tumble over the front wheel with a lower critical value of $mu$. And modern tyres on dry roads have values of $mu$ which can be fairly close to $1$ (I think $0.8$ to $0.9$ is plausible), while $l/h$ is generally significantly less than $1$.
This is why bikes tumble before they drift.
This approximation can be used even if the bike (or other vehicle) isn't light compared to the rider: you just need to work out where the centre of mass of the vehicle is and use that. For vehicles with suspension (some bikes have this of course, and even for bikes that don't the forks deflect under braking) you have to take into account the change in geometry under braking as well.
answered Nov 15 '18 at 15:14
tfbtfb
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I can't answer why but I can tell you that it is a property of multi-wheeled vehicles. Take a toy car and block the rear wheels so they don't move and push it on a smooth surface and you will see the locked rear wheels will always swap to the front. This is why a "bootleg turn" is possible by using the rear emergency brake to lock them down and the car will swap directions without changing the original vector by much. This is why the front brakes on a car does 90% of the braking and why the brakes on the front wheel of a motorcycle can help maintain control far more than the rear.
answered Nov 15 '18 at 4:28
Kathmandu GilmanKathmandu Gilman
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It's because the force of friction is approximately propertional to two things:
- The "coefficient of friction" between the two surfaces (i.e. how sticky they are)
- The "weight" or normal force (perpendicular force) between the two surfaces.
The latter means, for example, that if you double the weight on a sled then you more or less double the friction between the sled and the ground.
If you move all (or most of) your weight off the back wheel and onto the front wheel of a bike, then:
- The force due to friction on the front wheel more or less doubles (i.e. you'd need twice as much sideways or backward force to make it skid).
- The force due to friction on the back wheel goes towards zero (i.e. it skids more easily as it begins t lift off the ground)
answered Nov 15 '18 at 12:58
ChrisWChrisW
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In the front wheel you have a much stronger vertical component when you use the breaks because of the momentum caused by inertia. With the distribution of forces and weight of a normal person, the back wheel will tend to lift adding more friction I'm the front wheel and stopping you harder and harder. To drift you need to have inertia in the back wheel but when you stop like this all the energy is released stopping you and lifting your back wheel
When you use the break of the back wheel only, the momentum caused by inertia tends to lift the back wheel, such as in the case above. That causes friction to be lower I'm the back wheel and thus energy is not released suddenly, allowing you to drift.
I believe that if you used a wheel with bad grip in the front wheel you should be able to drift
answered Nov 15 '18 at 12:11
F. S.F. S.
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I think it's because you can turn the handle bars into the best angle to maintain your center of gravity near the body.
While the back wheel can't turn, so it loses grip, because the center of gravity goes outside your body during a drift.
edited Nov 16 '18 at 7:56
Peter Mortensen
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answered Nov 15 '18 at 4:09
eromoderomod
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How would turning the handlebars affect the centre of gravity? One arm moves further out, the other comes closer so there will be little, if any, change. Your second sentence makes no sense.
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– Transistor
Nov 16 '18 at 19:17
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Why would it matter whether the centre of gravity of me and my bike is within my body?
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– David Richerby
Nov 17 '18 at 11:47
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9 Answers
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Because, when you brake, your weight is being shifted towards the front wheel. The inertia coupled with gravity puts your weight and that of the bike onto the front wheel. More weight ⇒ more pressure ⇒ more friction/grip with the ground.
answered Nov 15 '18 at 3:43
Alex DoeAlex Doe
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@henning But when you brake then the moment of inertia wants to rotate the bike (and the rider) such that there is more "weight" on the front wheel and less on the back.
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– phuzi
Nov 15 '18 at 9:48
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To be specific - if the rear wheel lifts then indeed all weight is on the front wheel, as explained by phuzi due to the forces affecting the bike. If you brake slightly lighter then some weight still remains on the rear wheel but still the force/weight on the front wheel is significantly larger.
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– Ister
Nov 15 '18 at 10:10
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and if you brake just a touch more than the back wheel lifting, you go over the bars and land on your head; maximal braking is close to instability. (@henning)
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– Chris H
Nov 15 '18 at 13:09
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@henning Incidentally, that's why you should almost never brake using the rear brake alone. As your weight shifts off the rear wheel while braking, the maximum frictional force of the rear wheel decreases. Using the front brake alone lets you stop in about half the distance as using the rear brake alone, and using both is even better.
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– Nuclear Wang
Nov 15 '18 at 14:01
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@NuclearWang if you have perfect (constant) grip and a smooth road, the back brake contributes nothing to stopping in minimal distance. On real roads you're right, as you don't want to assume quite such ideal conditions
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– Chris H
Nov 15 '18 at 15:22
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Because, when you brake, your weight is being shifted towards the front wheel. The inertia coupled with gravity puts your weight and that of the bike onto the front wheel. More weight ⇒ more pressure ⇒ more friction/grip with the ground.
answered Nov 15 '18 at 3:43
Alex DoeAlex Doe
843312
$endgroup$
5
$begingroup$
@henning But when you brake then the moment of inertia wants to rotate the bike (and the rider) such that there is more "weight" on the front wheel and less on the back.
$endgroup$
– phuzi
Nov 15 '18 at 9:48
22
$begingroup$
To be specific - if the rear wheel lifts then indeed all weight is on the front wheel, as explained by phuzi due to the forces affecting the bike. If you brake slightly lighter then some weight still remains on the rear wheel but still the force/weight on the front wheel is significantly larger.
$endgroup$
– Ister
Nov 15 '18 at 10:10
12
$begingroup$
and if you brake just a touch more than the back wheel lifting, you go over the bars and land on your head; maximal braking is close to instability. (@henning)
$endgroup$
– Chris H
Nov 15 '18 at 13:09
7
$begingroup$
@henning Incidentally, that's why you should almost never brake using the rear brake alone. As your weight shifts off the rear wheel while braking, the maximum frictional force of the rear wheel decreases. Using the front brake alone lets you stop in about half the distance as using the rear brake alone, and using both is even better.
$endgroup$
– Nuclear Wang
Nov 15 '18 at 14:01
5
$begingroup$
@NuclearWang if you have perfect (constant) grip and a smooth road, the back brake contributes nothing to stopping in minimal distance. On real roads you're right, as you don't want to assume quite such ideal conditions
$endgroup$
– Chris H
Nov 15 '18 at 15:22
|
show 10 more comments
$begingroup$
Because, when you brake, your weight is being shifted towards the front wheel. The inertia coupled with gravity puts your weight and that of the bike onto the front wheel. More weight ⇒ more pressure ⇒ more friction/grip with the ground.
answered Nov 15 '18 at 3:43
Alex DoeAlex Doe
843312
$endgroup$
Because, when you brake, your weight is being shifted towards the front wheel. The inertia coupled with gravity puts your weight and that of the bike onto the front wheel. More weight ⇒ more pressure ⇒ more friction/grip with the ground.
answered Nov 15 '18 at 3:43
Alex DoeAlex Doe
843312
edited Nov 16 '18 at 15:35
answered Nov 15 '18 at 3:43
Alex DoeAlex Doe
843312
answered Nov 15 '18 at 3:43
Alex DoeAlex Doe
843312
answered Nov 15 '18 at 3:43
Alex DoeAlex Doe
843312
843312
5
$begingroup$
@henning But when you brake then the moment of inertia wants to rotate the bike (and the rider) such that there is more "weight" on the front wheel and less on the back.
$endgroup$
– phuzi
Nov 15 '18 at 9:48
22
$begingroup$
To be specific - if the rear wheel lifts then indeed all weight is on the front wheel, as explained by phuzi due to the forces affecting the bike. If you brake slightly lighter then some weight still remains on the rear wheel but still the force/weight on the front wheel is significantly larger.
$endgroup$
– Ister
Nov 15 '18 at 10:10
12
$begingroup$
and if you brake just a touch more than the back wheel lifting, you go over the bars and land on your head; maximal braking is close to instability. (@henning)
$endgroup$
– Chris H
Nov 15 '18 at 13:09
7
$begingroup$
@henning Incidentally, that's why you should almost never brake using the rear brake alone. As your weight shifts off the rear wheel while braking, the maximum frictional force of the rear wheel decreases. Using the front brake alone lets you stop in about half the distance as using the rear brake alone, and using both is even better.
$endgroup$
– Nuclear Wang
Nov 15 '18 at 14:01
5
$begingroup$
@NuclearWang if you have perfect (constant) grip and a smooth road, the back brake contributes nothing to stopping in minimal distance. On real roads you're right, as you don't want to assume quite such ideal conditions
$endgroup$
– Chris H
Nov 15 '18 at 15:22
|
show 10 more comments
5
$begingroup$
@henning But when you brake then the moment of inertia wants to rotate the bike (and the rider) such that there is more "weight" on the front wheel and less on the back.
$endgroup$
– phuzi
Nov 15 '18 at 9:48
22
$begingroup$
To be specific - if the rear wheel lifts then indeed all weight is on the front wheel, as explained by phuzi due to the forces affecting the bike. If you brake slightly lighter then some weight still remains on the rear wheel but still the force/weight on the front wheel is significantly larger.
$endgroup$
– Ister
Nov 15 '18 at 10:10
12
$begingroup$
and if you brake just a touch more than the back wheel lifting, you go over the bars and land on your head; maximal braking is close to instability. (@henning)
$endgroup$
– Chris H
Nov 15 '18 at 13:09
7
$begingroup$
@henning Incidentally, that's why you should almost never brake using the rear brake alone. As your weight shifts off the rear wheel while braking, the maximum frictional force of the rear wheel decreases. Using the front brake alone lets you stop in about half the distance as using the rear brake alone, and using both is even better.
$endgroup$
– Nuclear Wang
Nov 15 '18 at 14:01
5
$begingroup$
@NuclearWang if you have perfect (constant) grip and a smooth road, the back brake contributes nothing to stopping in minimal distance. On real roads you're right, as you don't want to assume quite such ideal conditions
$endgroup$
– Chris H
Nov 15 '18 at 15:22
5
5
$begingroup$
@henning But when you brake then the moment of inertia wants to rotate the bike (and the rider) such that there is more "weight" on the front wheel and less on the back.
$endgroup$
– phuzi
Nov 15 '18 at 9:48
$begingroup$
@henning But when you brake then the moment of inertia wants to rotate the bike (and the rider) such that there is more "weight" on the front wheel and less on the back.
$endgroup$
– phuzi
Nov 15 '18 at 9:48
22
22
$begingroup$
To be specific - if the rear wheel lifts then indeed all weight is on the front wheel, as explained by phuzi due to the forces affecting the bike. If you brake slightly lighter then some weight still remains on the rear wheel but still the force/weight on the front wheel is significantly larger.
$endgroup$
– Ister
Nov 15 '18 at 10:10
$begingroup$
To be specific - if the rear wheel lifts then indeed all weight is on the front wheel, as explained by phuzi due to the forces affecting the bike. If you brake slightly lighter then some weight still remains on the rear wheel but still the force/weight on the front wheel is significantly larger.
$endgroup$
– Ister
Nov 15 '18 at 10:10
12
12
$begingroup$
and if you brake just a touch more than the back wheel lifting, you go over the bars and land on your head; maximal braking is close to instability. (@henning)
$endgroup$
– Chris H
Nov 15 '18 at 13:09
$begingroup$
and if you brake just a touch more than the back wheel lifting, you go over the bars and land on your head; maximal braking is close to instability. (@henning)
$endgroup$
– Chris H
Nov 15 '18 at 13:09
7
7
$begingroup$
@henning Incidentally, that's why you should almost never brake using the rear brake alone. As your weight shifts off the rear wheel while braking, the maximum frictional force of the rear wheel decreases. Using the front brake alone lets you stop in about half the distance as using the rear brake alone, and using both is even better.
$endgroup$
– Nuclear Wang
Nov 15 '18 at 14:01
$begingroup$
@henning Incidentally, that's why you should almost never brake using the rear brake alone. As your weight shifts off the rear wheel while braking, the maximum frictional force of the rear wheel decreases. Using the front brake alone lets you stop in about half the distance as using the rear brake alone, and using both is even better.
$endgroup$
– Nuclear Wang
Nov 15 '18 at 14:01
5
5
$begingroup$
@NuclearWang if you have perfect (constant) grip and a smooth road, the back brake contributes nothing to stopping in minimal distance. On real roads you're right, as you don't want to assume quite such ideal conditions
$endgroup$
– Chris H
Nov 15 '18 at 15:22
$begingroup$
@NuclearWang if you have perfect (constant) grip and a smooth road, the back brake contributes nothing to stopping in minimal distance. On real roads you're right, as you don't want to assume quite such ideal conditions
$endgroup$
– Chris H
Nov 15 '18 at 15:22
|
show 10 more comments
$begingroup$
In addition to Alex Doe's answer: front wheel drift does occur on slippery surfaces, Unfortunately a bike is really unstable in this mode so you'll fall over very soon (whereas a rear wheel drift is pretty stable and controllable).
answered Nov 15 '18 at 8:48
HobbesHobbes
34113
$endgroup$
6
$begingroup$
Right. Correctly dosing an “anti-lock front brake maneuver” is an essentiall skill both for MTB and for survival in rainy cities without very well-designed cycle roads.
$endgroup$
– leftaroundabout
Nov 15 '18 at 12:16
$begingroup$
in a sharp turn, a front-wheel skid causes something that bikers call a high-side crash. they are spectacularly destructive and youtube is full of videos of them. search under "highside mulholland" to find some.
$endgroup$
– niels nielsen
Nov 15 '18 at 18:18
$begingroup$
CORRECTION! front-wheel drift causes a LOW-SIDE CRASH! My bad- NN
$endgroup$
– niels nielsen
Nov 15 '18 at 21:37
1
$begingroup$
@nielsnielsen I've managed to achieve both high and low-side crashes - depends if the tyre regains grip at any point after the loss of traction, and whether the weight is in a suitable place at that instant.
$endgroup$
– Criggie
Nov 15 '18 at 21:41
2
$begingroup$
A number of years ago now whilst on holiday in Corfu, I dropped a rental scooter in precisely this way. I braked too sharply with the front brake just as I hit a patch of gravel across the road. The front wheel locked, I couldn't control the skidding wheel, and everything went a bit pear-shaped. I was wearing jeans so my legs were only bruised, but I had a decent road rash on my left shoulder and arm.
$endgroup$
– Graham
Nov 16 '18 at 1:20
|
show 1 more comment
$begingroup$
In addition to Alex Doe's answer: front wheel drift does occur on slippery surfaces, Unfortunately a bike is really unstable in this mode so you'll fall over very soon (whereas a rear wheel drift is pretty stable and controllable).
answered Nov 15 '18 at 8:48
HobbesHobbes
34113
$endgroup$
6
$begingroup$
Right. Correctly dosing an “anti-lock front brake maneuver” is an essentiall skill both for MTB and for survival in rainy cities without very well-designed cycle roads.
$endgroup$
– leftaroundabout
Nov 15 '18 at 12:16
$begingroup$
in a sharp turn, a front-wheel skid causes something that bikers call a high-side crash. they are spectacularly destructive and youtube is full of videos of them. search under "highside mulholland" to find some.
$endgroup$
– niels nielsen
Nov 15 '18 at 18:18
$begingroup$
CORRECTION! front-wheel drift causes a LOW-SIDE CRASH! My bad- NN
$endgroup$
– niels nielsen
Nov 15 '18 at 21:37
1
$begingroup$
@nielsnielsen I've managed to achieve both high and low-side crashes - depends if the tyre regains grip at any point after the loss of traction, and whether the weight is in a suitable place at that instant.
$endgroup$
– Criggie
Nov 15 '18 at 21:41
2
$begingroup$
A number of years ago now whilst on holiday in Corfu, I dropped a rental scooter in precisely this way. I braked too sharply with the front brake just as I hit a patch of gravel across the road. The front wheel locked, I couldn't control the skidding wheel, and everything went a bit pear-shaped. I was wearing jeans so my legs were only bruised, but I had a decent road rash on my left shoulder and arm.
$endgroup$
– Graham
Nov 16 '18 at 1:20
|
show 1 more comment
$begingroup$
In addition to Alex Doe's answer: front wheel drift does occur on slippery surfaces, Unfortunately a bike is really unstable in this mode so you'll fall over very soon (whereas a rear wheel drift is pretty stable and controllable).
answered Nov 15 '18 at 8:48
HobbesHobbes
34113
$endgroup$
In addition to Alex Doe's answer: front wheel drift does occur on slippery surfaces, Unfortunately a bike is really unstable in this mode so you'll fall over very soon (whereas a rear wheel drift is pretty stable and controllable).
answered Nov 15 '18 at 8:48
HobbesHobbes
34113
answered Nov 15 '18 at 8:48
HobbesHobbes
34113
answered Nov 15 '18 at 8:48
HobbesHobbes
34113
answered Nov 15 '18 at 8:48
HobbesHobbes
34113
34113
6
$begingroup$
Right. Correctly dosing an “anti-lock front brake maneuver” is an essentiall skill both for MTB and for survival in rainy cities without very well-designed cycle roads.
$endgroup$
– leftaroundabout
Nov 15 '18 at 12:16
$begingroup$
in a sharp turn, a front-wheel skid causes something that bikers call a high-side crash. they are spectacularly destructive and youtube is full of videos of them. search under "highside mulholland" to find some.
$endgroup$
– niels nielsen
Nov 15 '18 at 18:18
$begingroup$
CORRECTION! front-wheel drift causes a LOW-SIDE CRASH! My bad- NN
$endgroup$
– niels nielsen
Nov 15 '18 at 21:37
1
$begingroup$
@nielsnielsen I've managed to achieve both high and low-side crashes - depends if the tyre regains grip at any point after the loss of traction, and whether the weight is in a suitable place at that instant.
$endgroup$
– Criggie
Nov 15 '18 at 21:41
2
$begingroup$
A number of years ago now whilst on holiday in Corfu, I dropped a rental scooter in precisely this way. I braked too sharply with the front brake just as I hit a patch of gravel across the road. The front wheel locked, I couldn't control the skidding wheel, and everything went a bit pear-shaped. I was wearing jeans so my legs were only bruised, but I had a decent road rash on my left shoulder and arm.
$endgroup$
– Graham
Nov 16 '18 at 1:20
|
show 1 more comment
6
$begingroup$
Right. Correctly dosing an “anti-lock front brake maneuver” is an essentiall skill both for MTB and for survival in rainy cities without very well-designed cycle roads.
$endgroup$
– leftaroundabout
Nov 15 '18 at 12:16
$begingroup$
in a sharp turn, a front-wheel skid causes something that bikers call a high-side crash. they are spectacularly destructive and youtube is full of videos of them. search under "highside mulholland" to find some.
$endgroup$
– niels nielsen
Nov 15 '18 at 18:18
$begingroup$
CORRECTION! front-wheel drift causes a LOW-SIDE CRASH! My bad- NN
$endgroup$
– niels nielsen
Nov 15 '18 at 21:37
1
$begingroup$
@nielsnielsen I've managed to achieve both high and low-side crashes - depends if the tyre regains grip at any point after the loss of traction, and whether the weight is in a suitable place at that instant.
$endgroup$
– Criggie
Nov 15 '18 at 21:41
2
$begingroup$
A number of years ago now whilst on holiday in Corfu, I dropped a rental scooter in precisely this way. I braked too sharply with the front brake just as I hit a patch of gravel across the road. The front wheel locked, I couldn't control the skidding wheel, and everything went a bit pear-shaped. I was wearing jeans so my legs were only bruised, but I had a decent road rash on my left shoulder and arm.
$endgroup$
– Graham
Nov 16 '18 at 1:20
6
6
$begingroup$
Right. Correctly dosing an “anti-lock front brake maneuver” is an essentiall skill both for MTB and for survival in rainy cities without very well-designed cycle roads.
$endgroup$
– leftaroundabout
Nov 15 '18 at 12:16
$begingroup$
Right. Correctly dosing an “anti-lock front brake maneuver” is an essentiall skill both for MTB and for survival in rainy cities without very well-designed cycle roads.
$endgroup$
– leftaroundabout
Nov 15 '18 at 12:16
$begingroup$
in a sharp turn, a front-wheel skid causes something that bikers call a high-side crash. they are spectacularly destructive and youtube is full of videos of them. search under "highside mulholland" to find some.
$endgroup$
– niels nielsen
Nov 15 '18 at 18:18
$begingroup$
in a sharp turn, a front-wheel skid causes something that bikers call a high-side crash. they are spectacularly destructive and youtube is full of videos of them. search under "highside mulholland" to find some.
$endgroup$
– niels nielsen
Nov 15 '18 at 18:18
$begingroup$
CORRECTION! front-wheel drift causes a LOW-SIDE CRASH! My bad- NN
$endgroup$
– niels nielsen
Nov 15 '18 at 21:37
$begingroup$
CORRECTION! front-wheel drift causes a LOW-SIDE CRASH! My bad- NN
$endgroup$
– niels nielsen
Nov 15 '18 at 21:37
1
1
$begingroup$
@nielsnielsen I've managed to achieve both high and low-side crashes - depends if the tyre regains grip at any point after the loss of traction, and whether the weight is in a suitable place at that instant.
$endgroup$
– Criggie
Nov 15 '18 at 21:41
$begingroup$
@nielsnielsen I've managed to achieve both high and low-side crashes - depends if the tyre regains grip at any point after the loss of traction, and whether the weight is in a suitable place at that instant.
$endgroup$
– Criggie
Nov 15 '18 at 21:41
2
2
$begingroup$
A number of years ago now whilst on holiday in Corfu, I dropped a rental scooter in precisely this way. I braked too sharply with the front brake just as I hit a patch of gravel across the road. The front wheel locked, I couldn't control the skidding wheel, and everything went a bit pear-shaped. I was wearing jeans so my legs were only bruised, but I had a decent road rash on my left shoulder and arm.
$endgroup$
– Graham
Nov 16 '18 at 1:20
$begingroup$
A number of years ago now whilst on holiday in Corfu, I dropped a rental scooter in precisely this way. I braked too sharply with the front brake just as I hit a patch of gravel across the road. The front wheel locked, I couldn't control the skidding wheel, and everything went a bit pear-shaped. I was wearing jeans so my legs were only bruised, but I had a decent road rash on my left shoulder and arm.
$endgroup$
– Graham
Nov 16 '18 at 1:20
|
show 1 more comment
$begingroup$
Why rear wheel easily drifts?
Consider the torque around contact of front tire from the frame of the cycle.
The pseudo force on the rider and the cycle is in forward direction
Thus, the weight and thus the normal force on rear tire tends to decrease.
Thus smaller friction force and bicycle tends to drift.
Why front wheel rarely drifts?
- Considering the pseudo force and the tendency to rotate, the normal force on front tire is much larger (due to forward pseudo force on the centre of gravity). Thus, the grip is very tight.
answered Nov 15 '18 at 4:00
m__m__
38113
$endgroup$
add a comment |
$begingroup$
Why rear wheel easily drifts?
Consider the torque around contact of front tire from the frame of the cycle.
The pseudo force on the rider and the cycle is in forward direction
Thus, the weight and thus the normal force on rear tire tends to decrease.
Thus smaller friction force and bicycle tends to drift.
Why front wheel rarely drifts?
- Considering the pseudo force and the tendency to rotate, the normal force on front tire is much larger (due to forward pseudo force on the centre of gravity). Thus, the grip is very tight.
answered Nov 15 '18 at 4:00
m__m__
38113
$endgroup$
add a comment |
$begingroup$
Why rear wheel easily drifts?
Consider the torque around contact of front tire from the frame of the cycle.
The pseudo force on the rider and the cycle is in forward direction
Thus, the weight and thus the normal force on rear tire tends to decrease.
Thus smaller friction force and bicycle tends to drift.
Why front wheel rarely drifts?
- Considering the pseudo force and the tendency to rotate, the normal force on front tire is much larger (due to forward pseudo force on the centre of gravity). Thus, the grip is very tight.
answered Nov 15 '18 at 4:00
m__m__
38113
$endgroup$
Why rear wheel easily drifts?
Consider the torque around contact of front tire from the frame of the cycle.
The pseudo force on the rider and the cycle is in forward direction
Thus, the weight and thus the normal force on rear tire tends to decrease.
Thus smaller friction force and bicycle tends to drift.
Why front wheel rarely drifts?
- Considering the pseudo force and the tendency to rotate, the normal force on front tire is much larger (due to forward pseudo force on the centre of gravity). Thus, the grip is very tight.
answered Nov 15 '18 at 4:00
m__m__
38113
answered Nov 15 '18 at 4:00
m__m__
38113
answered Nov 15 '18 at 4:00
m__m__
38113
answered Nov 15 '18 at 4:00
m__m__
38113
38113
add a comment |
add a comment |
$begingroup$
it doesn't really explains why the wheel never slips
Well, that's just not true. Front wheels can, and do slip. However, normally, the contact between front wheel and road is just too good for the slipping to set in before the wheel blocks and you go over the bars.
It's actually all down to two factors:
the braking force that your front wheel can transmit to the road without slipping
the angle between the center of gravity, the wheel-road contact point, and the road
The later factor is only a question of frame geometry. When your front wheel is about to block, the effective force on your center of gravity is exactly at that angle. Because gravity is fixed, and that angle is fixed, so is the braking force that is applied in this situation.
Note that the entire weight of the bike and rider is on the front wheel when it's about to block, so the vertical force on the wheel-road contact point is fixed too.
Now we can compare two forces:
The max braking force that your wheel material can transmit to the road in its current condition, given that the wheel is pressed to the road with the entire weight.
The braking force that is required to get you to the tipping point.
If the first is the greater, your front wheel will block, and you will go over the bars. If the later is the greater, your front wheel will start slipping.
Now, why is the first force the greater for almost all bikes?
Well, a good tire allows about as much horizontal force to be transmitted to a dry road as the vertical force that is applied to press it onto the road. I.e., the angle between center of gravity, contact point, and road can get as low as 45°, and the front wheel is still able to block. Yet, most bikes are built in such a way that this angle is significantly greater than 45°.
So, because of the geometry of most bikes, you need quite slippery road conditions in order for the front wheel to be able to slip. The longer the frame and the further back the center of gravity is behind the front wheel, the harder it becomes to block the front wheel and to go over the bars; up to the point that going over the bars becomes impossible.
answered Nov 15 '18 at 11:07
cmastercmaster
5709
$endgroup$
1
$begingroup$
I had 2-3 occasions where both my wheels locked. The occasion I remember well I was in high school. I was riding like 5 or 6 km/h going around people on a beach alley. Unexpectedly an old woman stepped into in front of me. I immediately squeezed the brakes hard. Due to inclination and sand on road, both wheel locked. I slid maybe 1m and realized that I'm not decelerating at all, and I'm going to hit the lady. So I released the bakes and managed to change my direction to miss her by just a little. Felt very happy I avoided that incident. The sand on road really surprised me.
$endgroup$
– akostadinov
Nov 16 '18 at 11:46
1
$begingroup$
Rumors have it that most people who go over the bars do so before (i.e. without) the front wheel blocking: they go over because of bending their elbow i.e. braking harder than expeced and unable to catch the additional weight on the arms fast enough. As a plausible guesstimate, I'd suggest that pretty much everyone who goes over the bars and after "landing" has the bike behind them with the rear wheel not in front of the front wheel did not have the bike flipping over with them but was instead "jumping" off to the front/did not decelerate as fast as their bike.
$endgroup$
– cbeleites
Nov 16 '18 at 16:29
$begingroup$
@cbeleites Fortunately, I cannot say whether you are right since I never went over my bars. My frame is long enough to make that impossible. But I did skid my front wheel several times. By the geometry of most frames, and by the quality of the tires that I know, it's very possible to go over the bars by actually blocking the front wheel. My frame is quite a strong exception in this. And I'm very happy that it makes this kind of accident impossible: Skidding your front wheel is bad enough as it is, going head-over is nothing I like thinking about...
$endgroup$
– cmaster
Nov 16 '18 at 17:21
$begingroup$
"contact between front wheel and road is just too good" I'm not sure I agree with that. I think if someone is prepared for it and shifts their weight back as far possible then the front wheel will skid on most surfaces+bikes. Wouldn't want to try it out though. :)
$endgroup$
– Nobody
Nov 18 '18 at 22:15
$begingroup$
@Nobody That also depends on the tire you are riding. The ones I ride definitely have a skidding angle below 45° on a dry road (it has happened that I've touched the road with my pedals when doing sharp and fast corners). I've ridden a tire once that I could never have dreamed of riding in such extreme corners, I felt it crawling away already in normally tight corners. Needless to say that I've never purchased a second tire of that brand. Point is, bike tire behave very differently, and the better ones are definitely impossible to skid on a dry road with most bikes, especially racing bikes.
$endgroup$
– cmaster
Nov 18 '18 at 23:07
add a comment |
$begingroup$
it doesn't really explains why the wheel never slips
Well, that's just not true. Front wheels can, and do slip. However, normally, the contact between front wheel and road is just too good for the slipping to set in before the wheel blocks and you go over the bars.
It's actually all down to two factors:
the braking force that your front wheel can transmit to the road without slipping
the angle between the center of gravity, the wheel-road contact point, and the road
The later factor is only a question of frame geometry. When your front wheel is about to block, the effective force on your center of gravity is exactly at that angle. Because gravity is fixed, and that angle is fixed, so is the braking force that is applied in this situation.
Note that the entire weight of the bike and rider is on the front wheel when it's about to block, so the vertical force on the wheel-road contact point is fixed too.
Now we can compare two forces:
The max braking force that your wheel material can transmit to the road in its current condition, given that the wheel is pressed to the road with the entire weight.
The braking force that is required to get you to the tipping point.
If the first is the greater, your front wheel will block, and you will go over the bars. If the later is the greater, your front wheel will start slipping.
Now, why is the first force the greater for almost all bikes?
Well, a good tire allows about as much horizontal force to be transmitted to a dry road as the vertical force that is applied to press it onto the road. I.e., the angle between center of gravity, contact point, and road can get as low as 45°, and the front wheel is still able to block. Yet, most bikes are built in such a way that this angle is significantly greater than 45°.
So, because of the geometry of most bikes, you need quite slippery road conditions in order for the front wheel to be able to slip. The longer the frame and the further back the center of gravity is behind the front wheel, the harder it becomes to block the front wheel and to go over the bars; up to the point that going over the bars becomes impossible.
answered Nov 15 '18 at 11:07
cmastercmaster
5709
$endgroup$
1
$begingroup$
I had 2-3 occasions where both my wheels locked. The occasion I remember well I was in high school. I was riding like 5 or 6 km/h going around people on a beach alley. Unexpectedly an old woman stepped into in front of me. I immediately squeezed the brakes hard. Due to inclination and sand on road, both wheel locked. I slid maybe 1m and realized that I'm not decelerating at all, and I'm going to hit the lady. So I released the bakes and managed to change my direction to miss her by just a little. Felt very happy I avoided that incident. The sand on road really surprised me.
$endgroup$
– akostadinov
Nov 16 '18 at 11:46
1
$begingroup$
Rumors have it that most people who go over the bars do so before (i.e. without) the front wheel blocking: they go over because of bending their elbow i.e. braking harder than expeced and unable to catch the additional weight on the arms fast enough. As a plausible guesstimate, I'd suggest that pretty much everyone who goes over the bars and after "landing" has the bike behind them with the rear wheel not in front of the front wheel did not have the bike flipping over with them but was instead "jumping" off to the front/did not decelerate as fast as their bike.
$endgroup$
– cbeleites
Nov 16 '18 at 16:29
$begingroup$
@cbeleites Fortunately, I cannot say whether you are right since I never went over my bars. My frame is long enough to make that impossible. But I did skid my front wheel several times. By the geometry of most frames, and by the quality of the tires that I know, it's very possible to go over the bars by actually blocking the front wheel. My frame is quite a strong exception in this. And I'm very happy that it makes this kind of accident impossible: Skidding your front wheel is bad enough as it is, going head-over is nothing I like thinking about...
$endgroup$
– cmaster
Nov 16 '18 at 17:21
$begingroup$
"contact between front wheel and road is just too good" I'm not sure I agree with that. I think if someone is prepared for it and shifts their weight back as far possible then the front wheel will skid on most surfaces+bikes. Wouldn't want to try it out though. :)
$endgroup$
– Nobody
Nov 18 '18 at 22:15
$begingroup$
@Nobody That also depends on the tire you are riding. The ones I ride definitely have a skidding angle below 45° on a dry road (it has happened that I've touched the road with my pedals when doing sharp and fast corners). I've ridden a tire once that I could never have dreamed of riding in such extreme corners, I felt it crawling away already in normally tight corners. Needless to say that I've never purchased a second tire of that brand. Point is, bike tire behave very differently, and the better ones are definitely impossible to skid on a dry road with most bikes, especially racing bikes.
$endgroup$
– cmaster
Nov 18 '18 at 23:07
add a comment |
$begingroup$
it doesn't really explains why the wheel never slips
Well, that's just not true. Front wheels can, and do slip. However, normally, the contact between front wheel and road is just too good for the slipping to set in before the wheel blocks and you go over the bars.
It's actually all down to two factors:
the braking force that your front wheel can transmit to the road without slipping
the angle between the center of gravity, the wheel-road contact point, and the road
The later factor is only a question of frame geometry. When your front wheel is about to block, the effective force on your center of gravity is exactly at that angle. Because gravity is fixed, and that angle is fixed, so is the braking force that is applied in this situation.
Note that the entire weight of the bike and rider is on the front wheel when it's about to block, so the vertical force on the wheel-road contact point is fixed too.
Now we can compare two forces:
The max braking force that your wheel material can transmit to the road in its current condition, given that the wheel is pressed to the road with the entire weight.
The braking force that is required to get you to the tipping point.
If the first is the greater, your front wheel will block, and you will go over the bars. If the later is the greater, your front wheel will start slipping.
Now, why is the first force the greater for almost all bikes?
Well, a good tire allows about as much horizontal force to be transmitted to a dry road as the vertical force that is applied to press it onto the road. I.e., the angle between center of gravity, contact point, and road can get as low as 45°, and the front wheel is still able to block. Yet, most bikes are built in such a way that this angle is significantly greater than 45°.
So, because of the geometry of most bikes, you need quite slippery road conditions in order for the front wheel to be able to slip. The longer the frame and the further back the center of gravity is behind the front wheel, the harder it becomes to block the front wheel and to go over the bars; up to the point that going over the bars becomes impossible.
answered Nov 15 '18 at 11:07
cmastercmaster
5709
$endgroup$
it doesn't really explains why the wheel never slips
Well, that's just not true. Front wheels can, and do slip. However, normally, the contact between front wheel and road is just too good for the slipping to set in before the wheel blocks and you go over the bars.
It's actually all down to two factors:
the braking force that your front wheel can transmit to the road without slipping
the angle between the center of gravity, the wheel-road contact point, and the road
The later factor is only a question of frame geometry. When your front wheel is about to block, the effective force on your center of gravity is exactly at that angle. Because gravity is fixed, and that angle is fixed, so is the braking force that is applied in this situation.
Note that the entire weight of the bike and rider is on the front wheel when it's about to block, so the vertical force on the wheel-road contact point is fixed too.
Now we can compare two forces:
The max braking force that your wheel material can transmit to the road in its current condition, given that the wheel is pressed to the road with the entire weight.
The braking force that is required to get you to the tipping point.
If the first is the greater, your front wheel will block, and you will go over the bars. If the later is the greater, your front wheel will start slipping.
Now, why is the first force the greater for almost all bikes?
Well, a good tire allows about as much horizontal force to be transmitted to a dry road as the vertical force that is applied to press it onto the road. I.e., the angle between center of gravity, contact point, and road can get as low as 45°, and the front wheel is still able to block. Yet, most bikes are built in such a way that this angle is significantly greater than 45°.
So, because of the geometry of most bikes, you need quite slippery road conditions in order for the front wheel to be able to slip. The longer the frame and the further back the center of gravity is behind the front wheel, the harder it becomes to block the front wheel and to go over the bars; up to the point that going over the bars becomes impossible.
answered Nov 15 '18 at 11:07
cmastercmaster
5709
answered Nov 15 '18 at 11:07
cmastercmaster
5709
answered Nov 15 '18 at 11:07
cmastercmaster
5709
answered Nov 15 '18 at 11:07
cmastercmaster
5709
5709
1
$begingroup$
I had 2-3 occasions where both my wheels locked. The occasion I remember well I was in high school. I was riding like 5 or 6 km/h going around people on a beach alley. Unexpectedly an old woman stepped into in front of me. I immediately squeezed the brakes hard. Due to inclination and sand on road, both wheel locked. I slid maybe 1m and realized that I'm not decelerating at all, and I'm going to hit the lady. So I released the bakes and managed to change my direction to miss her by just a little. Felt very happy I avoided that incident. The sand on road really surprised me.
$endgroup$
– akostadinov
Nov 16 '18 at 11:46
1
$begingroup$
Rumors have it that most people who go over the bars do so before (i.e. without) the front wheel blocking: they go over because of bending their elbow i.e. braking harder than expeced and unable to catch the additional weight on the arms fast enough. As a plausible guesstimate, I'd suggest that pretty much everyone who goes over the bars and after "landing" has the bike behind them with the rear wheel not in front of the front wheel did not have the bike flipping over with them but was instead "jumping" off to the front/did not decelerate as fast as their bike.
$endgroup$
– cbeleites
Nov 16 '18 at 16:29
$begingroup$
@cbeleites Fortunately, I cannot say whether you are right since I never went over my bars. My frame is long enough to make that impossible. But I did skid my front wheel several times. By the geometry of most frames, and by the quality of the tires that I know, it's very possible to go over the bars by actually blocking the front wheel. My frame is quite a strong exception in this. And I'm very happy that it makes this kind of accident impossible: Skidding your front wheel is bad enough as it is, going head-over is nothing I like thinking about...
$endgroup$
– cmaster
Nov 16 '18 at 17:21
$begingroup$
"contact between front wheel and road is just too good" I'm not sure I agree with that. I think if someone is prepared for it and shifts their weight back as far possible then the front wheel will skid on most surfaces+bikes. Wouldn't want to try it out though. :)
$endgroup$
– Nobody
Nov 18 '18 at 22:15
$begingroup$
@Nobody That also depends on the tire you are riding. The ones I ride definitely have a skidding angle below 45° on a dry road (it has happened that I've touched the road with my pedals when doing sharp and fast corners). I've ridden a tire once that I could never have dreamed of riding in such extreme corners, I felt it crawling away already in normally tight corners. Needless to say that I've never purchased a second tire of that brand. Point is, bike tire behave very differently, and the better ones are definitely impossible to skid on a dry road with most bikes, especially racing bikes.
$endgroup$
– cmaster
Nov 18 '18 at 23:07
add a comment |
1
$begingroup$
I had 2-3 occasions where both my wheels locked. The occasion I remember well I was in high school. I was riding like 5 or 6 km/h going around people on a beach alley. Unexpectedly an old woman stepped into in front of me. I immediately squeezed the brakes hard. Due to inclination and sand on road, both wheel locked. I slid maybe 1m and realized that I'm not decelerating at all, and I'm going to hit the lady. So I released the bakes and managed to change my direction to miss her by just a little. Felt very happy I avoided that incident. The sand on road really surprised me.
$endgroup$
– akostadinov
Nov 16 '18 at 11:46
1
$begingroup$
Rumors have it that most people who go over the bars do so before (i.e. without) the front wheel blocking: they go over because of bending their elbow i.e. braking harder than expeced and unable to catch the additional weight on the arms fast enough. As a plausible guesstimate, I'd suggest that pretty much everyone who goes over the bars and after "landing" has the bike behind them with the rear wheel not in front of the front wheel did not have the bike flipping over with them but was instead "jumping" off to the front/did not decelerate as fast as their bike.
$endgroup$
– cbeleites
Nov 16 '18 at 16:29
$begingroup$
@cbeleites Fortunately, I cannot say whether you are right since I never went over my bars. My frame is long enough to make that impossible. But I did skid my front wheel several times. By the geometry of most frames, and by the quality of the tires that I know, it's very possible to go over the bars by actually blocking the front wheel. My frame is quite a strong exception in this. And I'm very happy that it makes this kind of accident impossible: Skidding your front wheel is bad enough as it is, going head-over is nothing I like thinking about...
$endgroup$
– cmaster
Nov 16 '18 at 17:21
$begingroup$
"contact between front wheel and road is just too good" I'm not sure I agree with that. I think if someone is prepared for it and shifts their weight back as far possible then the front wheel will skid on most surfaces+bikes. Wouldn't want to try it out though. :)
$endgroup$
– Nobody
Nov 18 '18 at 22:15
$begingroup$
@Nobody That also depends on the tire you are riding. The ones I ride definitely have a skidding angle below 45° on a dry road (it has happened that I've touched the road with my pedals when doing sharp and fast corners). I've ridden a tire once that I could never have dreamed of riding in such extreme corners, I felt it crawling away already in normally tight corners. Needless to say that I've never purchased a second tire of that brand. Point is, bike tire behave very differently, and the better ones are definitely impossible to skid on a dry road with most bikes, especially racing bikes.
$endgroup$
– cmaster
Nov 18 '18 at 23:07
1
1
$begingroup$
I had 2-3 occasions where both my wheels locked. The occasion I remember well I was in high school. I was riding like 5 or 6 km/h going around people on a beach alley. Unexpectedly an old woman stepped into in front of me. I immediately squeezed the brakes hard. Due to inclination and sand on road, both wheel locked. I slid maybe 1m and realized that I'm not decelerating at all, and I'm going to hit the lady. So I released the bakes and managed to change my direction to miss her by just a little. Felt very happy I avoided that incident. The sand on road really surprised me.
$endgroup$
– akostadinov
Nov 16 '18 at 11:46
$begingroup$
I had 2-3 occasions where both my wheels locked. The occasion I remember well I was in high school. I was riding like 5 or 6 km/h going around people on a beach alley. Unexpectedly an old woman stepped into in front of me. I immediately squeezed the brakes hard. Due to inclination and sand on road, both wheel locked. I slid maybe 1m and realized that I'm not decelerating at all, and I'm going to hit the lady. So I released the bakes and managed to change my direction to miss her by just a little. Felt very happy I avoided that incident. The sand on road really surprised me.
$endgroup$
– akostadinov
Nov 16 '18 at 11:46
1
1
$begingroup$
Rumors have it that most people who go over the bars do so before (i.e. without) the front wheel blocking: they go over because of bending their elbow i.e. braking harder than expeced and unable to catch the additional weight on the arms fast enough. As a plausible guesstimate, I'd suggest that pretty much everyone who goes over the bars and after "landing" has the bike behind them with the rear wheel not in front of the front wheel did not have the bike flipping over with them but was instead "jumping" off to the front/did not decelerate as fast as their bike.
$endgroup$
– cbeleites
Nov 16 '18 at 16:29
$begingroup$
Rumors have it that most people who go over the bars do so before (i.e. without) the front wheel blocking: they go over because of bending their elbow i.e. braking harder than expeced and unable to catch the additional weight on the arms fast enough. As a plausible guesstimate, I'd suggest that pretty much everyone who goes over the bars and after "landing" has the bike behind them with the rear wheel not in front of the front wheel did not have the bike flipping over with them but was instead "jumping" off to the front/did not decelerate as fast as their bike.
$endgroup$
– cbeleites
Nov 16 '18 at 16:29
$begingroup$
@cbeleites Fortunately, I cannot say whether you are right since I never went over my bars. My frame is long enough to make that impossible. But I did skid my front wheel several times. By the geometry of most frames, and by the quality of the tires that I know, it's very possible to go over the bars by actually blocking the front wheel. My frame is quite a strong exception in this. And I'm very happy that it makes this kind of accident impossible: Skidding your front wheel is bad enough as it is, going head-over is nothing I like thinking about...
$endgroup$
– cmaster
Nov 16 '18 at 17:21
$begingroup$
@cbeleites Fortunately, I cannot say whether you are right since I never went over my bars. My frame is long enough to make that impossible. But I did skid my front wheel several times. By the geometry of most frames, and by the quality of the tires that I know, it's very possible to go over the bars by actually blocking the front wheel. My frame is quite a strong exception in this. And I'm very happy that it makes this kind of accident impossible: Skidding your front wheel is bad enough as it is, going head-over is nothing I like thinking about...
$endgroup$
– cmaster
Nov 16 '18 at 17:21
$begingroup$
"contact between front wheel and road is just too good" I'm not sure I agree with that. I think if someone is prepared for it and shifts their weight back as far possible then the front wheel will skid on most surfaces+bikes. Wouldn't want to try it out though. :)
$endgroup$
– Nobody
Nov 18 '18 at 22:15
$begingroup$
"contact between front wheel and road is just too good" I'm not sure I agree with that. I think if someone is prepared for it and shifts their weight back as far possible then the front wheel will skid on most surfaces+bikes. Wouldn't want to try it out though. :)
$endgroup$
– Nobody
Nov 18 '18 at 22:15
$begingroup$
@Nobody That also depends on the tire you are riding. The ones I ride definitely have a skidding angle below 45° on a dry road (it has happened that I've touched the road with my pedals when doing sharp and fast corners). I've ridden a tire once that I could never have dreamed of riding in such extreme corners, I felt it crawling away already in normally tight corners. Needless to say that I've never purchased a second tire of that brand. Point is, bike tire behave very differently, and the better ones are definitely impossible to skid on a dry road with most bikes, especially racing bikes.
$endgroup$
– cmaster
Nov 18 '18 at 23:07
$begingroup$
@Nobody That also depends on the tire you are riding. The ones I ride definitely have a skidding angle below 45° on a dry road (it has happened that I've touched the road with my pedals when doing sharp and fast corners). I've ridden a tire once that I could never have dreamed of riding in such extreme corners, I felt it crawling away already in normally tight corners. Needless to say that I've never purchased a second tire of that brand. Point is, bike tire behave very differently, and the better ones are definitely impossible to skid on a dry road with most bikes, especially racing bikes.
$endgroup$
– cmaster
Nov 18 '18 at 23:07
add a comment |
$begingroup$
The brief answer to this is 'because the centre of mass of a bicycle is high up', and especially it is high up compared to a car, where often the front wheels can slip.
To see how this works, consider a vastly oversimplified model of a bike: assume that the structure of the bike is light compared to its rider, and represent its rider as a point mass (see below for why this works even when that's not true). And we're going to be interested in the moment when the rear wheel lifts, so we can completely ignore the rear wheel, and concentrate just on the front wheel, and specifically on the point at which the front wheel touches the road. So the system looks something like this:
So, here, $c$ is the point at which the front tyre is touching the road, $m$ is the rider, and the horizontal & vertical distances between $c$ and $m$ are $l$ & $h$ respectively. And the bike is decelerating at $a$. And I've drawn the forces exerted by the front wheel on the road at $c$ (remember the back wheel is, by assumption, just lifting, so can be ignored: it's not exerting any forces on anything).
For now assume that the coefficient of friction between the front wheel and the road is sufficiently high that the wheel does not slip and lets work out the point at which the bike just begins to tumble over the front wheel: this will tell us the maximum possible value of $a$, however sticky the front wheel is.
It's pretty easy to see that the force on $m$ has two components: a vertical component which is $-mg$, where $g$ is acceleration due to gravity, and a horizontal component which is $ma$, where $a$ is the horizontal acceleration. And the bike will tumble when this vector points above the front wheel. Well, just by drawing the appropriate components you can see that this is true when
$$fracmamg gt fraclh$$
or in other words, for the bike not to tumble
$$a le fraclgh$$
or
$$a_textmax = fraclgh tag1$$
You can convince yourself this is right: a very tall bike ($h gg l$) will tumble really easily, and a completely flat bike ($h ll l$) will almost never tumble. And a bike in very low gravity will tumble more easily than one in high gravity. So (1) tells us how big $a$ can be, however sticky the front wheel is.
Now consider the coefficient of friction at the front wheel. The coefficient of friction, $mu$ is defined as the force with which the wheel is trying to slide along the road and the force which it is being pressed down onto the road, at the point where the wheel just slips. So it's obvious that,
$$mu = fracma_textslipmg = fraca_textslipg$$
where $a_textslip$ is the point at which the wheel slips. In other words
$$a_textslip = mu g tag2$$
And now we can use (1) & (2) to give us the answer we're looking for: the bike will tumble before it drifts if $a_textslip gt a_textmax$, in other words if
$$mu gt fraclh$$
And you can now see the problem here: bikes are rather short and rather tall, so $l/h$ tends to be rather small, which means that the bike will tumble over the front wheel with a lower critical value of $mu$. And modern tyres on dry roads have values of $mu$ which can be fairly close to $1$ (I think $0.8$ to $0.9$ is plausible), while $l/h$ is generally significantly less than $1$.
This is why bikes tumble before they drift.
This approximation can be used even if the bike (or other vehicle) isn't light compared to the rider: you just need to work out where the centre of mass of the vehicle is and use that. For vehicles with suspension (some bikes have this of course, and even for bikes that don't the forks deflect under braking) you have to take into account the change in geometry under braking as well.
answered Nov 15 '18 at 15:14
tfbtfb
15.4k43251
$endgroup$
add a comment |
$begingroup$
The brief answer to this is 'because the centre of mass of a bicycle is high up', and especially it is high up compared to a car, where often the front wheels can slip.
To see how this works, consider a vastly oversimplified model of a bike: assume that the structure of the bike is light compared to its rider, and represent its rider as a point mass (see below for why this works even when that's not true). And we're going to be interested in the moment when the rear wheel lifts, so we can completely ignore the rear wheel, and concentrate just on the front wheel, and specifically on the point at which the front wheel touches the road. So the system looks something like this:
So, here, $c$ is the point at which the front tyre is touching the road, $m$ is the rider, and the horizontal & vertical distances between $c$ and $m$ are $l$ & $h$ respectively. And the bike is decelerating at $a$. And I've drawn the forces exerted by the front wheel on the road at $c$ (remember the back wheel is, by assumption, just lifting, so can be ignored: it's not exerting any forces on anything).
For now assume that the coefficient of friction between the front wheel and the road is sufficiently high that the wheel does not slip and lets work out the point at which the bike just begins to tumble over the front wheel: this will tell us the maximum possible value of $a$, however sticky the front wheel is.
It's pretty easy to see that the force on $m$ has two components: a vertical component which is $-mg$, where $g$ is acceleration due to gravity, and a horizontal component which is $ma$, where $a$ is the horizontal acceleration. And the bike will tumble when this vector points above the front wheel. Well, just by drawing the appropriate components you can see that this is true when
$$fracmamg gt fraclh$$
or in other words, for the bike not to tumble
$$a le fraclgh$$
or
$$a_textmax = fraclgh tag1$$
You can convince yourself this is right: a very tall bike ($h gg l$) will tumble really easily, and a completely flat bike ($h ll l$) will almost never tumble. And a bike in very low gravity will tumble more easily than one in high gravity. So (1) tells us how big $a$ can be, however sticky the front wheel is.
Now consider the coefficient of friction at the front wheel. The coefficient of friction, $mu$ is defined as the force with which the wheel is trying to slide along the road and the force which it is being pressed down onto the road, at the point where the wheel just slips. So it's obvious that,
$$mu = fracma_textslipmg = fraca_textslipg$$
where $a_textslip$ is the point at which the wheel slips. In other words
$$a_textslip = mu g tag2$$
And now we can use (1) & (2) to give us the answer we're looking for: the bike will tumble before it drifts if $a_textslip gt a_textmax$, in other words if
$$mu gt fraclh$$
And you can now see the problem here: bikes are rather short and rather tall, so $l/h$ tends to be rather small, which means that the bike will tumble over the front wheel with a lower critical value of $mu$. And modern tyres on dry roads have values of $mu$ which can be fairly close to $1$ (I think $0.8$ to $0.9$ is plausible), while $l/h$ is generally significantly less than $1$.
This is why bikes tumble before they drift.
This approximation can be used even if the bike (or other vehicle) isn't light compared to the rider: you just need to work out where the centre of mass of the vehicle is and use that. For vehicles with suspension (some bikes have this of course, and even for bikes that don't the forks deflect under braking) you have to take into account the change in geometry under braking as well.
answered Nov 15 '18 at 15:14
tfbtfb
15.4k43251
$endgroup$
add a comment |
$begingroup$
The brief answer to this is 'because the centre of mass of a bicycle is high up', and especially it is high up compared to a car, where often the front wheels can slip.
To see how this works, consider a vastly oversimplified model of a bike: assume that the structure of the bike is light compared to its rider, and represent its rider as a point mass (see below for why this works even when that's not true). And we're going to be interested in the moment when the rear wheel lifts, so we can completely ignore the rear wheel, and concentrate just on the front wheel, and specifically on the point at which the front wheel touches the road. So the system looks something like this:
So, here, $c$ is the point at which the front tyre is touching the road, $m$ is the rider, and the horizontal & vertical distances between $c$ and $m$ are $l$ & $h$ respectively. And the bike is decelerating at $a$. And I've drawn the forces exerted by the front wheel on the road at $c$ (remember the back wheel is, by assumption, just lifting, so can be ignored: it's not exerting any forces on anything).
For now assume that the coefficient of friction between the front wheel and the road is sufficiently high that the wheel does not slip and lets work out the point at which the bike just begins to tumble over the front wheel: this will tell us the maximum possible value of $a$, however sticky the front wheel is.
It's pretty easy to see that the force on $m$ has two components: a vertical component which is $-mg$, where $g$ is acceleration due to gravity, and a horizontal component which is $ma$, where $a$ is the horizontal acceleration. And the bike will tumble when this vector points above the front wheel. Well, just by drawing the appropriate components you can see that this is true when
$$fracmamg gt fraclh$$
or in other words, for the bike not to tumble
$$a le fraclgh$$
or
$$a_textmax = fraclgh tag1$$
You can convince yourself this is right: a very tall bike ($h gg l$) will tumble really easily, and a completely flat bike ($h ll l$) will almost never tumble. And a bike in very low gravity will tumble more easily than one in high gravity. So (1) tells us how big $a$ can be, however sticky the front wheel is.
Now consider the coefficient of friction at the front wheel. The coefficient of friction, $mu$ is defined as the force with which the wheel is trying to slide along the road and the force which it is being pressed down onto the road, at the point where the wheel just slips. So it's obvious that,
$$mu = fracma_textslipmg = fraca_textslipg$$
where $a_textslip$ is the point at which the wheel slips. In other words
$$a_textslip = mu g tag2$$
And now we can use (1) & (2) to give us the answer we're looking for: the bike will tumble before it drifts if $a_textslip gt a_textmax$, in other words if
$$mu gt fraclh$$
And you can now see the problem here: bikes are rather short and rather tall, so $l/h$ tends to be rather small, which means that the bike will tumble over the front wheel with a lower critical value of $mu$. And modern tyres on dry roads have values of $mu$ which can be fairly close to $1$ (I think $0.8$ to $0.9$ is plausible), while $l/h$ is generally significantly less than $1$.
This is why bikes tumble before they drift.
This approximation can be used even if the bike (or other vehicle) isn't light compared to the rider: you just need to work out where the centre of mass of the vehicle is and use that. For vehicles with suspension (some bikes have this of course, and even for bikes that don't the forks deflect under braking) you have to take into account the change in geometry under braking as well.
answered Nov 15 '18 at 15:14
tfbtfb
15.4k43251
$endgroup$
The brief answer to this is 'because the centre of mass of a bicycle is high up', and especially it is high up compared to a car, where often the front wheels can slip.
To see how this works, consider a vastly oversimplified model of a bike: assume that the structure of the bike is light compared to its rider, and represent its rider as a point mass (see below for why this works even when that's not true). And we're going to be interested in the moment when the rear wheel lifts, so we can completely ignore the rear wheel, and concentrate just on the front wheel, and specifically on the point at which the front wheel touches the road. So the system looks something like this:
So, here, $c$ is the point at which the front tyre is touching the road, $m$ is the rider, and the horizontal & vertical distances between $c$ and $m$ are $l$ & $h$ respectively. And the bike is decelerating at $a$. And I've drawn the forces exerted by the front wheel on the road at $c$ (remember the back wheel is, by assumption, just lifting, so can be ignored: it's not exerting any forces on anything).
For now assume that the coefficient of friction between the front wheel and the road is sufficiently high that the wheel does not slip and lets work out the point at which the bike just begins to tumble over the front wheel: this will tell us the maximum possible value of $a$, however sticky the front wheel is.
It's pretty easy to see that the force on $m$ has two components: a vertical component which is $-mg$, where $g$ is acceleration due to gravity, and a horizontal component which is $ma$, where $a$ is the horizontal acceleration. And the bike will tumble when this vector points above the front wheel. Well, just by drawing the appropriate components you can see that this is true when
$$fracmamg gt fraclh$$
or in other words, for the bike not to tumble
$$a le fraclgh$$
or
$$a_textmax = fraclgh tag1$$
You can convince yourself this is right: a very tall bike ($h gg l$) will tumble really easily, and a completely flat bike ($h ll l$) will almost never tumble. And a bike in very low gravity will tumble more easily than one in high gravity. So (1) tells us how big $a$ can be, however sticky the front wheel is.
Now consider the coefficient of friction at the front wheel. The coefficient of friction, $mu$ is defined as the force with which the wheel is trying to slide along the road and the force which it is being pressed down onto the road, at the point where the wheel just slips. So it's obvious that,
$$mu = fracma_textslipmg = fraca_textslipg$$
where $a_textslip$ is the point at which the wheel slips. In other words
$$a_textslip = mu g tag2$$
And now we can use (1) & (2) to give us the answer we're looking for: the bike will tumble before it drifts if $a_textslip gt a_textmax$, in other words if
$$mu gt fraclh$$
And you can now see the problem here: bikes are rather short and rather tall, so $l/h$ tends to be rather small, which means that the bike will tumble over the front wheel with a lower critical value of $mu$. And modern tyres on dry roads have values of $mu$ which can be fairly close to $1$ (I think $0.8$ to $0.9$ is plausible), while $l/h$ is generally significantly less than $1$.
This is why bikes tumble before they drift.
This approximation can be used even if the bike (or other vehicle) isn't light compared to the rider: you just need to work out where the centre of mass of the vehicle is and use that. For vehicles with suspension (some bikes have this of course, and even for bikes that don't the forks deflect under braking) you have to take into account the change in geometry under braking as well.
answered Nov 15 '18 at 15:14
tfbtfb
15.4k43251
edited Nov 15 '18 at 15:20
answered Nov 15 '18 at 15:14
tfbtfb
15.4k43251
answered Nov 15 '18 at 15:14
tfbtfb
15.4k43251
answered Nov 15 '18 at 15:14
tfbtfb
15.4k43251
15.4k43251
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add a comment |
$begingroup$
I can't answer why but I can tell you that it is a property of multi-wheeled vehicles. Take a toy car and block the rear wheels so they don't move and push it on a smooth surface and you will see the locked rear wheels will always swap to the front. This is why a "bootleg turn" is possible by using the rear emergency brake to lock them down and the car will swap directions without changing the original vector by much. This is why the front brakes on a car does 90% of the braking and why the brakes on the front wheel of a motorcycle can help maintain control far more than the rear.
answered Nov 15 '18 at 4:28
Kathmandu GilmanKathmandu Gilman
162
$endgroup$
add a comment |
$begingroup$
I can't answer why but I can tell you that it is a property of multi-wheeled vehicles. Take a toy car and block the rear wheels so they don't move and push it on a smooth surface and you will see the locked rear wheels will always swap to the front. This is why a "bootleg turn" is possible by using the rear emergency brake to lock them down and the car will swap directions without changing the original vector by much. This is why the front brakes on a car does 90% of the braking and why the brakes on the front wheel of a motorcycle can help maintain control far more than the rear.
answered Nov 15 '18 at 4:28
Kathmandu GilmanKathmandu Gilman
162
$endgroup$
add a comment |
$begingroup$
I can't answer why but I can tell you that it is a property of multi-wheeled vehicles. Take a toy car and block the rear wheels so they don't move and push it on a smooth surface and you will see the locked rear wheels will always swap to the front. This is why a "bootleg turn" is possible by using the rear emergency brake to lock them down and the car will swap directions without changing the original vector by much. This is why the front brakes on a car does 90% of the braking and why the brakes on the front wheel of a motorcycle can help maintain control far more than the rear.
answered Nov 15 '18 at 4:28
Kathmandu GilmanKathmandu Gilman
162
$endgroup$
I can't answer why but I can tell you that it is a property of multi-wheeled vehicles. Take a toy car and block the rear wheels so they don't move and push it on a smooth surface and you will see the locked rear wheels will always swap to the front. This is why a "bootleg turn" is possible by using the rear emergency brake to lock them down and the car will swap directions without changing the original vector by much. This is why the front brakes on a car does 90% of the braking and why the brakes on the front wheel of a motorcycle can help maintain control far more than the rear.
answered Nov 15 '18 at 4:28
Kathmandu GilmanKathmandu Gilman
162
answered Nov 15 '18 at 4:28
Kathmandu GilmanKathmandu Gilman
162
answered Nov 15 '18 at 4:28
Kathmandu GilmanKathmandu Gilman
162
answered Nov 15 '18 at 4:28
Kathmandu GilmanKathmandu Gilman
162
162
add a comment |
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$begingroup$
It's because the force of friction is approximately propertional to two things:
- The "coefficient of friction" between the two surfaces (i.e. how sticky they are)
- The "weight" or normal force (perpendicular force) between the two surfaces.
The latter means, for example, that if you double the weight on a sled then you more or less double the friction between the sled and the ground.
If you move all (or most of) your weight off the back wheel and onto the front wheel of a bike, then:
- The force due to friction on the front wheel more or less doubles (i.e. you'd need twice as much sideways or backward force to make it skid).
- The force due to friction on the back wheel goes towards zero (i.e. it skids more easily as it begins t lift off the ground)
answered Nov 15 '18 at 12:58
ChrisWChrisW
35518
$endgroup$
add a comment |
$begingroup$
It's because the force of friction is approximately propertional to two things:
- The "coefficient of friction" between the two surfaces (i.e. how sticky they are)
- The "weight" or normal force (perpendicular force) between the two surfaces.
The latter means, for example, that if you double the weight on a sled then you more or less double the friction between the sled and the ground.
If you move all (or most of) your weight off the back wheel and onto the front wheel of a bike, then:
- The force due to friction on the front wheel more or less doubles (i.e. you'd need twice as much sideways or backward force to make it skid).
- The force due to friction on the back wheel goes towards zero (i.e. it skids more easily as it begins t lift off the ground)
answered Nov 15 '18 at 12:58
ChrisWChrisW
35518
$endgroup$
add a comment |
$begingroup$
It's because the force of friction is approximately propertional to two things:
- The "coefficient of friction" between the two surfaces (i.e. how sticky they are)
- The "weight" or normal force (perpendicular force) between the two surfaces.
The latter means, for example, that if you double the weight on a sled then you more or less double the friction between the sled and the ground.
If you move all (or most of) your weight off the back wheel and onto the front wheel of a bike, then:
- The force due to friction on the front wheel more or less doubles (i.e. you'd need twice as much sideways or backward force to make it skid).
- The force due to friction on the back wheel goes towards zero (i.e. it skids more easily as it begins t lift off the ground)
answered Nov 15 '18 at 12:58
ChrisWChrisW
35518
$endgroup$
It's because the force of friction is approximately propertional to two things:
- The "coefficient of friction" between the two surfaces (i.e. how sticky they are)
- The "weight" or normal force (perpendicular force) between the two surfaces.
The latter means, for example, that if you double the weight on a sled then you more or less double the friction between the sled and the ground.
If you move all (or most of) your weight off the back wheel and onto the front wheel of a bike, then:
- The force due to friction on the front wheel more or less doubles (i.e. you'd need twice as much sideways or backward force to make it skid).
- The force due to friction on the back wheel goes towards zero (i.e. it skids more easily as it begins t lift off the ground)
answered Nov 15 '18 at 12:58
ChrisWChrisW
35518
answered Nov 15 '18 at 12:58
ChrisWChrisW
35518
answered Nov 15 '18 at 12:58
ChrisWChrisW
35518
answered Nov 15 '18 at 12:58
ChrisWChrisW
35518
35518
add a comment |
add a comment |
$begingroup$
In the front wheel you have a much stronger vertical component when you use the breaks because of the momentum caused by inertia. With the distribution of forces and weight of a normal person, the back wheel will tend to lift adding more friction I'm the front wheel and stopping you harder and harder. To drift you need to have inertia in the back wheel but when you stop like this all the energy is released stopping you and lifting your back wheel
When you use the break of the back wheel only, the momentum caused by inertia tends to lift the back wheel, such as in the case above. That causes friction to be lower I'm the back wheel and thus energy is not released suddenly, allowing you to drift.
I believe that if you used a wheel with bad grip in the front wheel you should be able to drift
answered Nov 15 '18 at 12:11
F. S.F. S.
1
$endgroup$
add a comment |
$begingroup$
In the front wheel you have a much stronger vertical component when you use the breaks because of the momentum caused by inertia. With the distribution of forces and weight of a normal person, the back wheel will tend to lift adding more friction I'm the front wheel and stopping you harder and harder. To drift you need to have inertia in the back wheel but when you stop like this all the energy is released stopping you and lifting your back wheel
When you use the break of the back wheel only, the momentum caused by inertia tends to lift the back wheel, such as in the case above. That causes friction to be lower I'm the back wheel and thus energy is not released suddenly, allowing you to drift.
I believe that if you used a wheel with bad grip in the front wheel you should be able to drift
answered Nov 15 '18 at 12:11
F. S.F. S.
1
$endgroup$
add a comment |
$begingroup$
In the front wheel you have a much stronger vertical component when you use the breaks because of the momentum caused by inertia. With the distribution of forces and weight of a normal person, the back wheel will tend to lift adding more friction I'm the front wheel and stopping you harder and harder. To drift you need to have inertia in the back wheel but when you stop like this all the energy is released stopping you and lifting your back wheel
When you use the break of the back wheel only, the momentum caused by inertia tends to lift the back wheel, such as in the case above. That causes friction to be lower I'm the back wheel and thus energy is not released suddenly, allowing you to drift.
I believe that if you used a wheel with bad grip in the front wheel you should be able to drift
answered Nov 15 '18 at 12:11
F. S.F. S.
1
$endgroup$
In the front wheel you have a much stronger vertical component when you use the breaks because of the momentum caused by inertia. With the distribution of forces and weight of a normal person, the back wheel will tend to lift adding more friction I'm the front wheel and stopping you harder and harder. To drift you need to have inertia in the back wheel but when you stop like this all the energy is released stopping you and lifting your back wheel
When you use the break of the back wheel only, the momentum caused by inertia tends to lift the back wheel, such as in the case above. That causes friction to be lower I'm the back wheel and thus energy is not released suddenly, allowing you to drift.
I believe that if you used a wheel with bad grip in the front wheel you should be able to drift
answered Nov 15 '18 at 12:11
F. S.F. S.
1
answered Nov 15 '18 at 12:11
F. S.F. S.
1
answered Nov 15 '18 at 12:11
F. S.F. S.
1
answered Nov 15 '18 at 12:11
F. S.F. S.
1
1
add a comment |
add a comment |
$begingroup$
I think it's because you can turn the handle bars into the best angle to maintain your center of gravity near the body.
While the back wheel can't turn, so it loses grip, because the center of gravity goes outside your body during a drift.
edited Nov 16 '18 at 7:56
Peter Mortensen
1,94411323
answered Nov 15 '18 at 4:09
eromoderomod
856
$endgroup$
2
$begingroup$
How would turning the handlebars affect the centre of gravity? One arm moves further out, the other comes closer so there will be little, if any, change. Your second sentence makes no sense.
$endgroup$
– Transistor
Nov 16 '18 at 19:17
1
$begingroup$
Why would it matter whether the centre of gravity of me and my bike is within my body?
$endgroup$
– David Richerby
Nov 17 '18 at 11:47
add a comment |
$begingroup$
I think it's because you can turn the handle bars into the best angle to maintain your center of gravity near the body.
While the back wheel can't turn, so it loses grip, because the center of gravity goes outside your body during a drift.
edited Nov 16 '18 at 7:56
Peter Mortensen
1,94411323
answered Nov 15 '18 at 4:09
eromoderomod
856
$endgroup$
2
$begingroup$
How would turning the handlebars affect the centre of gravity? One arm moves further out, the other comes closer so there will be little, if any, change. Your second sentence makes no sense.
$endgroup$
– Transistor
Nov 16 '18 at 19:17
1
$begingroup$
Why would it matter whether the centre of gravity of me and my bike is within my body?
$endgroup$
– David Richerby
Nov 17 '18 at 11:47
add a comment |
$begingroup$
I think it's because you can turn the handle bars into the best angle to maintain your center of gravity near the body.
While the back wheel can't turn, so it loses grip, because the center of gravity goes outside your body during a drift.
edited Nov 16 '18 at 7:56
Peter Mortensen
1,94411323
answered Nov 15 '18 at 4:09
eromoderomod
856
$endgroup$
I think it's because you can turn the handle bars into the best angle to maintain your center of gravity near the body.
While the back wheel can't turn, so it loses grip, because the center of gravity goes outside your body during a drift.
edited Nov 16 '18 at 7:56
Peter Mortensen
1,94411323
answered Nov 15 '18 at 4:09
eromoderomod
856
edited Nov 16 '18 at 7:56
Peter Mortensen
1,94411323
edited Nov 16 '18 at 7:56
Peter Mortensen
1,94411323
edited Nov 16 '18 at 7:56
Peter Mortensen
1,94411323
1,94411323
answered Nov 15 '18 at 4:09
eromoderomod
856
answered Nov 15 '18 at 4:09
eromoderomod
856
answered Nov 15 '18 at 4:09
eromoderomod
856
856
2
$begingroup$
How would turning the handlebars affect the centre of gravity? One arm moves further out, the other comes closer so there will be little, if any, change. Your second sentence makes no sense.
$endgroup$
– Transistor
Nov 16 '18 at 19:17
1
$begingroup$
Why would it matter whether the centre of gravity of me and my bike is within my body?
$endgroup$
– David Richerby
Nov 17 '18 at 11:47
add a comment |
2
$begingroup$
How would turning the handlebars affect the centre of gravity? One arm moves further out, the other comes closer so there will be little, if any, change. Your second sentence makes no sense.
$endgroup$
– Transistor
Nov 16 '18 at 19:17
1
$begingroup$
Why would it matter whether the centre of gravity of me and my bike is within my body?
$endgroup$
– David Richerby
Nov 17 '18 at 11:47
2
2
$begingroup$
How would turning the handlebars affect the centre of gravity? One arm moves further out, the other comes closer so there will be little, if any, change. Your second sentence makes no sense.
$endgroup$
– Transistor
Nov 16 '18 at 19:17
$begingroup$
How would turning the handlebars affect the centre of gravity? One arm moves further out, the other comes closer so there will be little, if any, change. Your second sentence makes no sense.
$endgroup$
– Transistor
Nov 16 '18 at 19:17
1
1
$begingroup$
Why would it matter whether the centre of gravity of me and my bike is within my body?
$endgroup$
– David Richerby
Nov 17 '18 at 11:47
$begingroup$
Why would it matter whether the centre of gravity of me and my bike is within my body?
$endgroup$
– David Richerby
Nov 17 '18 at 11:47
add a comment |
protected by ACuriousMind♦ Nov 15 '18 at 18:35
Thank you for your interest in this question.
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2
$begingroup$
Do you mean drift (as in skid outwards on a bend "11. a controlled four-wheel skid, used by racing drivers to take bends at high speed " (Collins)) or simply skid in forward motion?
$endgroup$
– Chris H
Nov 15 '18 at 13:05
7
$begingroup$
I have managed to skid my front wheel (without crashing) but only in a dead-straight line, on wet leaves on tarmac. Any form of side-slipping the front wheel (while braking) is likely to be painful, and bikes never stay straight for long (there's another question in that). I've also skidded the front wheel on bends (with or without braking) and invariably hit the road rather hard.
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– Chris H
Nov 15 '18 at 13:08
2
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Skidding the front wheel occurs quite easily in deep, soft snow. It's also easier to handle than on wet road, because even though the wheel stops spinning it still works as a ski, retaining some control :)
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– jpa
Nov 15 '18 at 13:37
2
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I've never seen the front wheel of a bike skid. grab a good fist full of front brake while you're riding on gravel, ice, wet metal or any other surface with diminished friction, and your story will abruptly change.. I recommend not to ride around utilizing the front brake as though it's impossible to lock the front wheel, if you wish to avoid injury..
$endgroup$
– Caius Jard
Nov 15 '18 at 20:40
2
$begingroup$
Also note that it's not all that hard to skid with both wheels. (Not the question, but still interesting.)
$endgroup$
– Jasper
Nov 16 '18 at 14:09