| Author | Message | ||
     Bomber |
er . . . .fellahs . . . . .after dipping back into my own archives (printed matter and gray matter-based), if I understand countersteering aright (and if I do, I''l be as surprised as the next guy!) . . . . isn't the front wheel "pointing" in a direction oposite of the direction of the turn? (even is only a little bit) .. if thtat's the case, the front wheel drive would certainly still help propel the scoot forward, but I can't help bu believe it's raise hell with the stability duiring the time it took to get through the corner . . . . and, as is obvious by know, even a lifetime of working with engineers does not make one such . . . . | ||
| Mikej |
Countersteering is used to initiate a turn, not maintain one. If you keep countersteering in the turn you'll fall over. Riding a motorcycle is very dynamic, adjustments always being made and so forth. Don't think too much, just ride. And as for 2wd bikes, I don't think they're the future, and I'd be very surprised if they do well in actual field conditions. Just my opinion. | ||
| Blake |
Bomber: What Mike said. The only time your front wheel points the opposite direction of a turn is when you first initiate the turn (the counter-steering part). Counter-steering is simply a way to quickly get the bike leaned over to take a curve. Once leaned over and in equilibrium, your front wheel will point left in a left-hander and right in a right-hander. Otherwise, you would meet pavement VERY quickly.  Simply put, countersteering is moving the wheels of your bike out from under the center of gravity (cg) so that the bike/rider will lean over. The gyroscopic precession of the front wheel can also help to get the bike leaned over, but at normal street speeds it is secondary to the centrifugal force. Arvel has offered to give us a physics lesson on that.  Why do we need to lean? So that the lateral acceleration (g-force or centrifugal force) created by changing trajectory is balanced/offset by gravity. Once leaned over in a left hand turn, the centrifugal force acts to tip you to the right while gravity acts to tip you to the left. It is simply a situation of two torques (properly called "moments" in this scenario) acting in opposition through the center of gravity of the rider/bike and about the line of contact between tires and road surface. If you can imagine for an instant the rider/bike leaned hard over in a flat left hander, now freeze that image. Imagine the center of gravity (cg) of the bike/rider; it will be a single point somewhere below your tank/seat and roughly midway between the axles. We like to imagine this point, because the laws of physics allow us to apply all external accelerations (gravity acts like an acceleration) acting on the rider/bike as if they were all acting only through the cg. There are two forces acting on the cg. Gravity, acting straight downwards towards the center of the Earth, and the centrifugal force acting laterally to the right (at 3 'o clock some might say), ninety degrees from the gravitational force. So we are imagining our rider/bike, frozen in time, leaned over in the midst of a flat left hand curve with two forces acting on (pushing on) the cg. One force is due to gravity. Suspend a plumb bob from the cg. That plumb bob, only being acted upon by gravity will point straight down to the asphalt. Now imagine the entire weight of rider/bike pulling down on that plum bob. That would tend to tip the bike over pretty darn quickly right? The tipping force is actually a torque or more properly, since nothing is being twisted or spun, a "moment". Now draw a chalkline from the front tire's contact patch to the rear tire's contact patch. Measure the shortest distance from the plumb bob (where it just touches the asphalt) to the chalk line. That's our moment arm. Multipy the weight of rider/bike (pounds) by that moment arm (feet) and you have just calculated the tipping moment (FT-LB or in metric N-m) acting on rider/bike. The centrifugal force Fn is can be easily calculated from the equation or in terms of weight (W) instead of mass (m) Fn=(W/g)V2/r or to get the force in terms of g's Fn=V2/r/g where "m" is the total mass of the rider/bike (slugs or Kg), "V" is the velocity or speed (FT/s or m/s), and "r" is the radius of the bike's curving path (FT or m). For the American system "m" is confusingly in units of "Slugs". To get your mass in slugs, divide your weight by 32.2, the gravitational acceleration constant. To calculate the force in g's, simply divide the weight (W) by Fn or use the third equation where "g" is 32.2 FT/S2 or 9.81 m/s2 for metric units as appropriate. Now, To find the moment arm about which the centrifugal force acts, measure the length of the plumb bob and it's line from cg to asphalt. Now multiply Fn times that length, and you have calculated the restoring moment. Once set into a curve, the tipping moment and restoring moment are equal and opposite. In order to maintain the turn your wheel must be pointed in the direction of the curve so that the radius is maintained and the restoring moment continues to oppose the gravity induced tipping moment. To get back on a straight path you need to move the tires under the cg to eliminate the tipping moment.; this is done by turning harder into the curve (effectively reducing the radius (r) so that the restoring moment briefly exceeds the tipping moment. When initiating a lean/turn we "counter-steer" or turn the front wheel in the direction opposite of the curve. This effectively creates what we previously called the restoring moment, but in the opposite direction, so it is actually an initiating tipping moment induced by centrifugal force. Does that make sense? Ain't it great that we don't actually have to understand physics to be able to ride a bike around a curve? Blake (geeshwhatanerd) | ||
| Jim_Sb |
Blake: If one assumes gravity acts straight downward (as in your plumb bob example) wouldn't centrifugal (or centripetal?) force act at a 90 deg. angle to the plumb bob line? Or horizontally? Then the two forces (or moments) would balance each other out (gravity wanting a leaned over bike to fall over on it's side, while centrifugal force wants to shove it over in a high side maneuver) resulting in a bike in a stabilized turn? Wouldn't we want the downforce to plant the bike into the tarmac in line with the bikes vertical axis? Or in this case, at a 60 deg. angle to both load the suspension and firmly plant the tires? Here's another question for you. Assume the bikes 2 wheels were fixed in a straight line. Could you still turn the bike? I think you could. Instead of countersteering you'd need to lean your weight excessively to one side to establish the lean angle. Once established, I believe the bike would turn. Further, what really causes the turn? Is the force of gravity slightly overpowering the centrifugal force therefore the bike is getting pulled ~slightly~ around the corner? Or (if it could be measured) is the front tire turned ever so slightly into the turn causing the bike to track left? You've got me wondering. I know how it works in airplanes, now I am curious as to how it works on a bike... Jim in Santa Barbara | ||
| Blake |
JimSB: "... wouldn't centrifugal force act at a 90 deg. angle to the plumb bob line? Or horizontally? Then the two forces (or moments) would balance each other out (gravity wanting a leaned over bike to fall over on it's side, while centrifugal force wants to shove it over in a high side maneuver) resulting in a bike in a stabilized turn?" Yes sir! Dang, ain't that what I said. Guess I was a little hard to follow. But, yeah, you are kerekt. "Wouldn't we want the downforce to plant the bike into the tarmac in line with the bikes vertical axis? Or in this case, at a 60 deg. angle to both load the suspension and firmly plant the tires? No. Actually, that would be of no benefit since the majority of that force would be pushing the bike to slide. We want more force straight downwards, with gravity, to push the tires into the asphalt without any additional slide instigating load (sideforce). Traction is equal to the tire/track friction coefficient times the weight on the rear wheel, with weight being defined as the force in-line with gravity, which without aerodynamic downforce can never exceed the weight of the bike/rider. At 30 degrees from horizontal the rider is pushing 2 g's total (along vertical axis of bike); the gravity force is 1g, and the sliding/centrifugal force is 1.73g's. If you recall your trigonorhea that is the dreaded 30/60/90 degree triangle with relative side lengths of 1/2/SQRT(3) (SQRT(3)=1.73). Forces are vectors (they have both magnitude and direction) and, when oriented tip to tail, follow the same rules for resolution of their resultant (summation) magnitude and direction as the sides of a triangle. "Here's another question for you. Assume the bikes 2 wheels were fixed in a straight line. Could you still turn the bike?" You may be thinking of how, when you release the handlebars and lean, the bike will indeed track nicely around a curve. It does that because the front wheel, due to rake and trail geometry, automatically turns into the curve. If the bars were fixed so the front wheel only stayed straight, the bike would be utterly uncontrollable. "Further, what really causes the turn? Is the force of gravity slightly overpowering the centrifugal force therefore the bike is getting pulled ~slightly~ around the corner? Or (if it could be measured) is the front tire turned ever so slightly into the turn causing the bike to track left?" That last thing you said. Think of a bike turning just like a car, except that instead of an extra set of wheels to counteract the centrifugal force (tipping moment), we use gravity and lean angle. Zat make sense now? I agree with Saro. I like seeing race bikes that are close to the versions I can put on the street. Or at least they resemble them. I wouldn't mind if GP went into some aerodynamic enhancements though. Otherwise, it seems that SBK and GB bikes are soon to be one in the same. I like the idea of GP bikes being the very hottest cutting edge of two wheeled racing technology like F1 is to car racing. Saro: VERY interesting comment concerning the Bernouli-Venturi effect between fairing and tarmac. Unfortunately the bike is already pretty much leaned to the extreme at that point isn't it? I was thinking of movable control surfaces so that only one side would deploy at a time, depending on the direction of the curve. Like flaps on the sides of the front forks/fender, and in the rear on the swingarm. They could be activated by any number of means, even automatically via sensors/processor control. Leaning over on a lefthander, the flaps on the right side would deploy gradually with increasing lean angle. The resulting force would be mostly downwards with some actually opposing the centrifugal force. BTW, I remembered. Centripetal force acts to oppose centrifugal force. So the centripetal force is the lateral force keeping the bike from sliding (the side load imparted to the tires by the track), while the centrifugal force is the one acting to cause the bike to slide. Geesh, whatever. Greg/S32002: Good call on the third gyro; ya beat me to it. And, I agree that countersteering mid turn can be used to tighten or relax your line. It works the same way when leaned over as it does when travelling upright and straight. It's all relative. That's an interesting point, that countersteering while leaned over would tend when tightening your line to reduce the distance between front/rear contact patches while the opposite would be true when relaxing your line. As long as you aren't pushing the limit of traction like Ben Bostrom, I'd guess the effect would be pretty negligible. Blake | ||
| Blake |
There are two technical goofs on this diagram. Anyone notice what they are?  Notice how all forces acting on bike/rider sum to zero and all moments also sum to zero.  Trigonorhea Is this fun or what? (just drawing, no math!) Blake | ||
| Jmartz |
So Blake: That is why a bike stays up in a curve! I finally understand it. Graphics are far better than ... well anything else Jose | ||
| Jb2 |
Blake: The first goof is the rider has no foot pegs and second he's looking straight ahead instead of around the corner... you really should look where you're going you know. :-) JB2 | ||
| Jb2 |
Blake: Three goofs... the brake light goes on the back of the bike. ;-) JB2 | ||
| Jim_Sb |
Blake: Your rider has only one leg... This is all very interesting. Thanks for the effort. Pullin' 2 g's in a 60 degree turn sounds like fun to me... Jim in Santa Barbara | ||
| Iggy |
where's the picture fer when the centrifugal force overcomes the frictive force from the tarmac. i was trying to imagine the original eq'ns with their vector components, the cartoons help. iG | ||
| S320002 |
Blake, As I begin this I can already hear the collective groan. There is more to the traction equation than contact patch and the coefficient of friction. Modern tire compounds add something that could be descibed as "gear effect". This is the result of the tire surface conforming to the iregularites in the track surface. The tiny bumps and knobs of the track surface act somewhat like the cleats or spikes on atheletic shoes. Softer compounds conform better but at some point become so soft that they smear or roll right across the surface irregularities resulting in reduced traction. This is why racers and tire companies are constantly playing with different compound for differnt tracks. Bottom line there are many more variables on the track than you will ever find in the lab. Greg | ||
| S320002 |
Blake, Me again, the blue vector in your vector diagram is pointed in the wrong direction. Combined forces should equal zero. Greg | ||
| Mikej |
Blake, The arrows for the tire/track interface are pointing in the wrong direction. The track just lays there, the tire does all the pushing. The track is static, therefore the arrows have to tie to the movable object. ie: the chair I'm sitting on is not "pushing" my rear, my rear is pushing down on the chair. Also, there is no axle bolt holding the forks to the tire/wheel assembly, therefore the rider is about to crash (and I'm amazed he was even able to get riding in the first place). Must be all that positive mental/kinetic power stuff. Oh well, time to go someplace. | ||
| Raymaines |
Would someone please expalin how cntripetal force and cntrifugal force effect the size and shape of the rear tire contact patch. . . . . . . Oh, never mind |
