| Author | Message | |||
     Xl1200r |
Jdugger - yes, that is a correct statement. | |||
| Stirz007 |
XL - the quote sounds a lot like Lee Parks' analogy: The traction pie is only so big, you have to choose how many pieces you want to use for braking, steering, etc. Jim - I think the more "upright" the bike is, the more contact patch you have - ergo - more available for braking/acceleration. Obviously, dual or triple compound tires make that a little more complicated than a single compound tire. Not looking to argue here - I think we're saying the same thing, just different ways. | |||
| Xl1200r |
Traction and lean angle are independent, and in many cases, there's more traction at lean than upright with a typical sportbike tire contour. The lean angle of the bike has no input on the traction pie. I'm not looking to argue, either, but the same things aren't being said. | |||
| Stirz007 |
XL - I hope we are just saying the same thing two different ways, but: "traction and lean angle are independent" "there's more traction at lean than upright..." These are conflicting statements - only one can be true. I disagree with "the lean angle of the bike has no input on the traction pie" - Jim - bust my chops if you disagree. At a steeper lean angle, you are also diminishing the turn radius, right? So at a given speed, the tighter the radius, the more centrifugal force is generated. In order to reduce the tendency of centrepital forces acting on the CG to want to make the bike lean outward, we lean inward to compensate. So: more lean = smaller radii = more centrifugal force = less traction available for other inputs (unless my understanding of physics is flawed). | |||
| Xl1200r |
These are conflicting statements - only one can be true. I misspoke - what I mean was there's a larger contact patch, but for the sake of argument we have to assume the contact patch size doesn't change (a perfectly round tire profile). I disagree with "the lean angle of the bike has no input on the traction pie" - Jim - bust my chops if you disagree. So you're telling me that at a standstill, the tires are more likely to just slip out from under the bike when its leaned over vs. sitting upright? So: more lean = smaller radii = more centrifugal force = less traction available for other inputs (unless my understanding of physics is flawed). You're understanding is spot on and just proved my point. It's not the lean angle that diminishes traction, it's the lateral forces. Yes, those forces require lean angle, but altering the lean angle by moving the COG doesn't change the lateral forces acting on the tires. Can we agree that with a constant rubber compound, a given number of lateral g's are needed to break traction? If then, when following a given line through a turn, are you suggesting that moving the COG changes the number of g's acting on the tires? You can change the laws of physics by merely moving your ass, lol. | |||
| Drawkward |
Xl:
Moving the COG by body position allows for the bike to be stood up more through the same turn on the same line. A straighter bike gives more traction. So, by moving your ass you don't change physics, but you DO indeed change the amount of traction the bike has in turn changing your little equation. Less lean means more traction. (Message edited by drawkward on October 18, 2010) | |||
| Stirz007 |
XL - "So you're telling me that at a standstill, the tires are more likely to just slip out from under the bike when its leaned over vs. sitting upright?" Yes I am. We could get into moment coupling and all that, but the easiest way to explain it is - have you ever skated? Standing straight up - kinda no problem. Lean? End up on your ass. In an ideal physics world, if you had two frictionless surfaces (tires and ground) and the center of gravity was directly over (assuming gravity acts downward) the tire/ground connection, the bike would stay upright. Any deviation from perfectly vertical would result in the tires slipping sideways out from under the bike. In the real world, the tire friction keeps the bike from slipping sideways when you lean the bike (at a standstill). Try leaning a stand-still bike on an oily surface and see how much harder it is to keep it upright. If the center of gravity is not over the support (tires), then a moment (torque) results. The lateral restraining forces created by tire friction keep it in balance. JM | |||
| Xl1200r |
Moving the COG by body position allows for the bike to be stood up more through the same turn on the same line. A straighter bike gives more traction. So, by moving your ass you don't change physics, but you DO indeed change the amount of traction the bike has in turn changing your little equation. Your argument equates to, "It has more traction standing up because it does." I don't buy it. In an ideal physics world, if you had two frictionless surfaces (tires and ground) and the center of gravity was directly over (assuming gravity acts downward) the tire/ground connection, the bike would stay upright. Any deviation from perfectly vertical would result in the tires slipping sideways out from under the bike. How so? Gravity places a force directly downward. The only way for the tires to move out from under the bike is for there to be a lateral force. Leaning the bike doesn't create this. Leaning the bike, while in motion, is CAUSED by it. If you're imagining a string supporting the bike from above in your example, you have to move the string as the bike leans in order to keep the string perfectly vertical as this is how gravity works. If you keep the other end of the string directly above the tires, the string will be at an angel as the bike leans and therefor putting a lateral force on the tires. And even so, once the tires slipped, they would settle such that the string is again perfectly vertical. The tires slip in a turn because of lateral acceleration. This lateral acceleration is a function of the combined mass of the bike and rider, and the speed the bike is traveling through the turn - it's irrelevant where the center of gravity is. Moving the center of gravity affects neither the mass of the bike & rider nor the speed they're traveling through the turn. (Message edited by xl1200r on October 18, 2010) | |||
| Xl1200r |
BTW, your example also doesn't make much sense in this equation as moving the center of gravity wouldn't cause the tires to slip but rather the whole unit to just tip over. | |||
| Stirz007 |
"The only way for the tires to move out from under the bike is for there to be a lateral force." You are correct, grasshopper. There are two lateral forces, actually, offsetting the two vertical forces. Because when bike is not perfectly plumb, the vertical gravity force (down) is not over the tires where the direct (up) force is, a moment results Again, the question had to do with a bike standing still. See my cheesy diagram (Word doc, sorry) Gravity and the ground pushing up balance vertical forces, but create a clockwise moment (torque) going into the page RED ARROWS. Lateral tire friction and the pressure you apply sideways to hold the bike up balance horizontal forces and create a counterclockwise moment to balance the one created by gravity GREEN ARROWS. Take any of these forces away and the bike becomes unstable - it either falls over, or flies off into space. Let go of the bike, it falls, spray some Pam under the tires, it falls. Now, if the bike is on the sidestand, NO lateral forces are created because the CG falls between the tire and the sidestand. XL - Now you start adding velocity to the deal and "gravity" doesn't necessarily act downward any more. Depending upon the lateral G's resulting from a turn, the idea is to keep the resultant vector in line with the traction surface (contact patch) - That gets beyond todays lesson plan. There are entire volumes written on this - interesting stuff for us engineer-types. No more science lessons tonight, kiddies - I gotta cook dinner for the real kiddies.
| |||
| Jdugger |
I don't know the physics of it all exactly, but I do know that getting off the bike reduces the required lean angle for a given radius and speed, and that *is* better. I can go faster with equal safety margin off the bike than upright on it. Part of it may be the "recovery factor". When I'm completely committed, a low side is more likely than when I've got some lean angle left, mostly just because I know I can pick the bike up so the tires grab again in a slide. Rear wheels coming loose is a relatively common "problem" I deal with anymore, and it's sure easier to manage that when you can tuck your head into the turn and push the bike up and away from you to get the tires to grab than when on top of the bike. Plus, when it does snag traction again, you aren't likely to get flung off the other side... being deep into the corner off the bike, your body just sort of helps the bike regain grip. And, then I pit out, and promptly clean my leathers. | |||
| Xl1200r |
I can't see your diagram (I'm on a Mac) but will look tomorrow on my work computer. However... Lateral forces cannot offset a vertical force. When going through a turn, the bike is held up when leaned over by the lateral acceleration. Since this is not in place when the bike is stationary, you need have something in place to hold it at a constant angle. Use your sidestand example - should you put the sidestand and both wheels under zero-friction casters, in your example, the bike would move sideways because it's leaned over. BTW, gravity ALWAYS acts directly downward, regardless of velocity. | |||
| Jng1226 |
I've re-attached Stirz's diagram as a PDF. I just know how to ride, not the engineering details and find this interesting. Pass the popcorn and break out those ME degrees...
| |||
| Catalan42 |
You guys! You are missing the whole point! Because of the sidewall, tires have a maximum lean angle before you start to ride off the edge of the tread. Example: Michelin Pilot Power is 50.6 deg (see http://www.michelinmotorcycle.com/index.cfm?event= pilotpower). Other tires can be more or less max lean angle before hitting the sidewall depending on their design (Harley cruising tires <-> MotoGP slicks). For this example, there is no benefit to hanging off of the bike unless the combination of speed & curve radius requires more than 50 deg of lean angle. If the speed/radius combo requires >50 deg, you can keep the bike angle under 50 deg by hanging off. Say you need 55 deg of lean. You could probably get away with just riding a little off of the edge of your tires on a clean/dry surface. You would have a smaller contact patch, but it would probably hold for such a small error (not recommended, though!). A better solution is to hang your body off the side of the bike. Then the center-of-gravity (CG) of your body is greater than 55 deg (say, 65 deg). This allows the bike CG to lean at only 50 deg, say. The combined lean angle will still be 55 deg. Since a 150 lb body is about 1/3'rd the weight of a 450 lb bike, the CG of your body will move farther off of 55 deg (e.g. 65) than the CG of the bike moves (e.g. 50 deg). I will leave out the exact equations but that is the gist of it. Note that the above ignores the small effects of tire shape. Most tires are approx. round in profile. Some racing tires are advertised as being "pointier" (picture a rounded off triangle) for many reasons. This has several effects, but they are all separate from the big effect caused by the presence of the sidewall. It is the sidewall location & size that cause any particular tire design to have a "maximum lean angle". | |||
| Drawkward |
XL: I'm sorry you didn't understand what I'm talking about. This is what I meant...straight from the master himself... Many riders think there is less rubber on the ground when the bike is leaned over but it is the opposite, there is more. They think that because they can’t add gobs of throttle when it is leaned over. In actual fact, as we bring the bike up we can add more throttle because the tires do not have to deal with the leaned over side-loading from the cornering. When the bike is straight up it has the least rubber on the ground but no side loading to take away from the available traction.} http://forums.superbikeschool.com/index.php?s=a9bc d78cddbe2378e25981eafa813e7a&showtopic=877 The less the bike is leaned over in a turn, the less traction it has. By using body position to lower the CoG and lessen the lean angle of the bike, the greater speed one can carry through a turn because more traction is available. It's a really simple concept to grasp. You're missing the part where the CoG comes into play. Moving the center of gravity around for a single lean angle doesn't make a difference. But changing the CoG so that you can maintain the same speed with less lean angle (more available traction), or a greater speed with the same lean angle but more hanging off causing the center of gravity to lower (more available traction) is why moving the CoG is important to motorcycle racing. EDIT Catalan: All true, but you're missing a very important point. At max lean angle a 50deg tire may only be able to sustain speeds of 120mph. But if you move your CoG to allow you to use only 40deg of lean yet go 130mph, then you've widened your effective usage of the tire. (Message edited by drawkward on October 19, 2010) | |||
| Xl1200r |
Okay, I'm hoping this is my last post on this, hopefully it gets the point across. This statement: Many riders think there is less rubber on the ground when the bike is leaned over but it is the opposite, there is more. They think that because they can’t add gobs of throttle when it is leaned over. In actual fact, as we bring the bike up we can add more throttle because the tires do not have to deal with the leaned over side-loading from the cornering. When the bike is straight up it has the least rubber on the ground but no side loading to take away from the available traction. ...is being taken out of context for your argument (yes, I read the article). Nowhere in there is he speaking about altering the COG to alter the lean angle, he's merely speaking about lean angle, which increases with speed through a given turn increases, which increases side loading. He never said that hanging off the bike produces more grip or reduces those lateral forces. The less the bike is leaned over in a turn, the less traction it has. By using body position to lower the CoG and lessen the lean angle of the bike, the greater speed one can carry through a turn because more traction is available. Read what you wrote and tell me what you think. Moving the center of gravity around for a single lean angle doesn't make a difference. Correct, no argument from me and what I've been saying all along. But changing the CoG so that you can maintain the same speed with less lean angle (more available traction), or a greater speed with the same lean angle but more hanging off causing the center of gravity to lower (more available traction) is why moving the CoG is important to motorcycle racing. Yes, hanging off the bike equates to freeing up some lean angle, but again you contradict yourself. Leaning the bike further increases grip, but standing it up also increases grip? Part of your argument is essentially stating that a smaller contact patch means more grip. All I'm stating is that the part of the tire in contact with the road is level with the road - what does it matter what the angle of the bike is relative to it? It's arbitrary (unless we're riding off the sidewall as Catalan mentioned). You could apply the argument to cars - when in a hard turn, the inside tires aren't taking any of the side loading and the COG is in the center of the car, essentially creating a 'virtual lean angle' with the COG relative to the contact patch and the road surface. You're saying that if I put narrower tires on the car, it would carry a higher speed through the turn? See my cheesy diagram (Word doc, sorry) Okay, I've looked at the diagram. When the bike is stationary, the only forces acting upon it is gravity pulling it straight down. When leaned over, this doesn't change. You've added an arbitrary lateral force on the top of the bike to use the tire/road contact as a pivot point to push the bike back upright and counteract the force of gravity that would otherwise make it tip over. All you need to do is counteract the act of gravity on the bike by putting a force directly upward on the leaned over side. Now there is no lateral force anywhere and no reason for the bike to have a higher propensity to slide than if upright. Your diagram does hold true with a bike in motion going through a turn, but still does not illustrate that decreasing the lean angle by altering the COG would somehow lessen the lateral forces on the tires. Another argument: The COG, the tire/road contact patch, and (we'll assume) the level road surface all create an angle going through a given turn. Let's assume we're going fast enough to warrant 1g of lateral forces. This requires a lean angle of 45 degrees. If we stay in line with the bike vertically, the bike will lean at 45 degrees. If we hang off, the bike will lean less than 45, but our body will be leaning more - but the combined lean angle, by moving the center of gravity out from the bike remains 45. Let's just list what we know to be constant: The mass of the bike and rider The path of travel through the turn The speed we're carrying through the turn and, because lateral forces are a function of velocity and mass and not affected by the COG, the side loading on the tire remains the same. and, because the lateral forces are the same, the angle the COG makes with the contact patch and road surface remains the same. You're arguing that hanging off the bike to stand it up a little increases available grip. The only way to increase available grip are: -To increase the size of the contact patch: For argument's sake we've been keeping this constant, but in fact it gets larger as the bike is leaned over. -To reduce the side loading on the tire: In order to do this, you would have either make the bike/rider combination weigh less and/or reduce the speed the bike is traveling by hanging off the bike (we've already identified both items as constants, so you can't), because, again, where the COG is has no effect on the lateral forces acting on the tires (because, again, the angle of the COG with the contact patch vs. the road surface will remain the same at a given speed no matter the body position of the rider. | |||
| Skntpig |
My brother (newer rider)called and said he had an epiphany. He went around a low speed corner while hanging off and the bike was virtually upright. Now he gets it. You will always have more grip with the bike more upright. Don't believe me? Crack it open full throttle at full lean and tell me what happens. | |||
| Xl1200r |
That example has nothing to do with what's being discussed. If I hang off the bike going in a straight line, the bike is leaned over. There are still no lateral forces on the bike and the contact patch has gotten larger... but some will argue that I have less grip because the bike is leaned??? I'm still looking for someone to explain to me how a smaller contact patch = more grip, or why the contact patch, which is always level with the road surface, gives 2 sh!ts about the angle of the bike... | |||
| Drawkward |
I can't tell if you're being a hardhead or if you're actually genuinely not understanding the physics. I don't know of any other way to explain this to you, so I'm out. | |||
| Xl1200r |
I'm being hardhead? You posted an article out of context that proved nothing, and then contradicted yourself multiple times. I understand the physics perfectly. Please don't be offended when I pass you on my slippery, leaned-over tires. | |||
| Xbniner |
I know I'm showing up late to this, but I can see what you are both saying. The problem is you are equating lean angle with turning because, well- they happen at the same time. XL is mostly right, and the very helpful article that was posted actually proves it. Your tire has more grip leaned over than upright. Which is good, that happens to be when you need it the most. Not because of the lean angle though, but because of the actual turn. If you want anecdotal proof, then go thru a tight turn using just your bars to turn, and don't lean the bike or yourself. That force that tips you over and ruins your bike is the same force that limits your traction and force you to counteract it- whether you prefer to drag pegs or knees. I didn't look at that word document, but I assume it is a static free-body diagram. One point to consider is that you and the bike will share a single center of gravity. You can move said CG by moving mass further into the turn, whether that mass is meat or metal is irrelevant. On the other side of it, however, there are good reasons to use your body positioning to partially cancel out that lateral force instead of only using lean angle. Such as changing the radius of your turn to avoid an obstacle. | |||
| Xl1200r |
I didn't look at that word document, but I assume it is a static free-body diagram. One point to consider is that you and the bike will share a single center of gravity. You can move said CG by moving mass further into the turn, whether that mass is meat or metal is irrelevant. On the other side of it, however, there are good reasons to use your body positioning to partially cancel out that lateral force instead of only using lean angle. Such as changing the radius of your turn to avoid an obstacle. Yes, you can move the center of gravity, but only relative to the bike or yourself for a given speed going around a given radius. Hanging off the bike will move the COG inward and downward relative to the bike, but the angle it makes with the tires and road surface never change - again, when speed and radius are the same. And this is because the lateral forces are a function purely of the radius of the turn and the velocity of the object going through it. Hanging off can change neither of these things. However, hanging off cannot cancel out a lateral force, but I know what you mean. It will alter the lean angle of the bike only, which will allow you to lean the bike further before dragging hard parts. | |||
| Gbaz |
Read: Motorcycle Dynamics (Second Edition) by Vittore Cossalter (Paperback - Oct 2, 2006) | |||
| Xl1200r |
That's some pretty heavy stuff. I found the chapter which discusses the lateral grip of the tires, but I can't decipher anything that mentions the rider's position having an effect or not... | |||
| Drawkward |
Xl:
You've officially made an ass out of yourself. Congrats  | |||
| Drawkward |
Older but good read for those who want good discussion on lean angle. There are some great riders posting in this thread... and look who turns up later on... http://www.r1-forum.com/forums/showthread.php?t=22 7167 | |||
| Thruster |
Does the front suspension work better with less lean angle? I think yes, since surface irregularities would present a more vertically-oriented force to the fork. I'm supposing, then, that a better-behaving fork would provide more "effective traction", by keeping the tire in contact with the road more consistently. Does the rider apply more pressure to the handlebars in order to maintain the same line through a curve if leaned over more? My own experience is yes (though I'm no expert and have not been to a track yet). Surface irregularities, such as bumps, holes, and changes in traction (e.g., metal expansion joints with morning dew on them on an elevated ramp) can then cause a greater amount of unsettling of the bike's steering (because you'll push the steering more off-line when the tire briefly loses traction), for which you then need more traction in order to regain stability (and avoid headshake or the dreaded tankslapper). Also, I think that lowering your overall CG by hanging off can increase effective traction by bringing the outward lateral force closer to the opposing inward lateral resistive force afforded by the traction of the tire upon the pavement. Small amounts of skipping can be compensated for more quickly in that case. The effect is hard to describe, but if you hold a pen vertically by the tip and try to slide its opposite end along a carpet, it will skip, but if you hold it at the bottom end (just above the carpet), it will skip less. When riding "on the edge", the tires are constantly losing and regaining traction as they encounter variations in the pavement. With a lower CG, this effect is more controlled because you regain traction more quickly, before the bike gets a chance to unsettle further (for which you'd then need more traction to re-settle). Thus, although I agree in principle with some of the physics discussed above, where isolated effects under perfect conditions are assumed, I think things are more complex when considering everything as a whole. | |||
| Thruster |
Just another detail to add to my first point -- The fork is less compressed with less lean angle, so that also helps it maintain tire contact. If I'm not mistaken, I believe the notion of "effective traction" that I've described, however amateurishly, ought to figure in to Lee Parks' pie, one way or another. | |||
| Tmchcrk |
one word "Eigenvalues" | |||
| Thruster |
Perhaps, though I think there may not be sufficient eigenvectors to span the entire space, given its complexity. Furthermore, at the required number of dimensions, calculation of the eigenvalues could only be accomplished via tediously iterative numerical approximation. Better to just get out and ride! |
