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Blake
| Posted on Sunday, December 16, 2001 - 11:57 pm: |
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Greg: "Coefficient of traction" I like that. Sounds a lot like "coefficient of friction" though. I did mention "speed". It's up there right between "tire wear" and "lean angle". And no, I did not edit my post after the fact, promise. I was wondering about how one would even determine a friction coefficient for a rolling tire. I mean, if the tire is simply rolling like the front tire would on a straightaway, it's not sliding (kinetic), yet it's not static either. That's why I used the term "rolling coefficient of friction." That's another interesting point about the viscous properties of rubber. I can see how that, especially on a racing slick, would make a big diffrence in traction. I understand that there is much more to tire choice than traction. There's probably even more than stiffness, wear, and temperature performance to consider right? I don't think racers spend much if any time worrying about coefficients of friction/traction. My point was simply that, given a reasonably accurate coefficient of friction for the conditions of interest, it is possible to estimate the critical lean angle. I never meant to imply that such a simplistic calculation is important to racing teams. The point was related to the countersteering discussion and the forces involved. Here's a brain tickler for the non-engineers. Assuming our hypothetical bike/rider is not yet at the threshold of traction, let's say that we estimate that at 1.73 g's lateral force, the bike/rider still have about another 1 g of safe traction left (2.73 g's total). The bike is nearing the exit of the curve; time to get on the throttle baby! Here's the brain teaser... If the bike and rider weight totals 700 lb, how much thrust can the rear tire support at the current lean angle before risking a lowside?
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Blake
| Posted on Monday, December 17, 2001 - 12:58 am: |
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Hint: You may need to assume a certain weight distribution between front and rear axles. |
Blake
| Posted on Monday, December 17, 2001 - 07:15 am: |
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From Newfie Buell: I found this real interesting article on gyroscopic effects on motorcycles in a book called "The World of Motorcycles". From what I can determine the book set (about 22 in all) was published by Orbis Publishing in London in 1979 and distributed by Columbia House. Blake (For Newfie Buell) |
Bomber
| Posted on Monday, December 17, 2001 - 12:37 pm: |
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sheesh . . . .ok, fellahs . .. sorry I asked (Blake, my brain hurts!) also, anyone that's ridden with me for awhile KNOWS I don't tghink too much wow, that'll teach me not to log on over the weekend, yes? |
Newfie_Buell
| Posted on Monday, December 17, 2001 - 05:16 pm: |
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Thanks Blake, I never really thought of breaking it up. I guess it was too much thinking from all of the other reading. Bill |
Jima4media
| Posted on Tuesday, December 18, 2001 - 01:28 am: |
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Read Tony Foale's 1984 book "Motorcycle Chassis Design" It is out of print, but he is doing a new version with more content, pictures, and computer applications on a CD-ROM. Well worth the time to research him. Jim X-2.5 BTW, he designed a bike with a perimeter brake in 1987. |
Blake
| Posted on Tuesday, December 18, 2001 - 02:58 am: |
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Comments on the above article that Newfie Buell dug up... The first photograph of the gyroscope on a string is erroneously rototated 90 degrees counter-clockwise. "(Gyroscopic precession)is why a speeding motorcycle must have its handlebars turned towards the left when it is required to turn right." Wrong! See my explanation above posted on December 15th. Read on and you will understand that gyroscopic precession is actually a hindrance to turning a motorcycle. "...a force applied to deflect a spinning body in a plane perpendicular to it's axis of rotation will be resisted by an equal force tending to deflect the body in a plane perpendicular to that in which the original force acted and also perpendicular to the axis of rotation." Wrong. A moment (torque) applied in any plane parallel to the axis of rotation of a spinning body will result in an equal magnitude reactive moment lying in a plane also parallel to the axis of rotation but perpendcular to the plane of the applied moment. Got a picture. Wanna see it? Here it is... The blue arrows represent the moment applied to the front wheel through the forks and axle via the handlebars and ultimately your arms. The red arrows indicate the equal magnitude reactive moment that the wheel concurrently applies to the bike via axle and forks. "(The front wheel of a motorcycle) constitutes a pretty effective gyro, and plays a vital part in controlling the machine." Gyroscopic yes. Effective for controlling the motorcycle... NO WAY. Gyroscopic precession is actually a hindrance to turning a motorcycle. "High speed steering is accomplished by simply pressing the handlebars rather than actually turning them, which would be difficult (steering deflections on a motorcycle seldom exceed 1.5o at speeds over 20 mph) in the direction opposite to that in which which the rider wishes to turn, whereupon gyroscopic procession prompts the wheel to bank over in the desired direction, taking the bike with it." Huh? While turning/steering the front wheel becomes progressively more difficult at higher speeds due to it's gyroscopic properties, it most definitely does TURN! Put an angle sensor on the steering head and the output for negotiating a curve at speed will look something like the graph below. The article's author seems to imply that a steering angle of 1.5o is too small to navigate a curve in a conventional manner. There aren't many curves that a 1.5o steering angle wouldn't easily handle. A 1.5o degree steering angle on a 55" wheelbase amounts to no more than a 150 FT turning radius. An unbanked 150 FT radius curve would be posted at no more than 20 mph, meaning to imply that it is a relatively sharp curve. Gyroscopic precession is actually a hindrance to turning a motorcycle. Once leaned over, the front wheel of a motorcycle MUST turn into the curve at an angle consistent with the radius of the curve; that follows from simple geometry and common sense. Consider then that once the bike is leaned over, the front wheel must assume an attitude consistent with the radius of the curve being navigated. This means that if the wheel was turned 1.5o in countersteer, it would need to turn as much as 3o in the opposite direction to stabilize the bike while taking the curve. If gyroscopic precession of the front wheel is so powerful, when turning the front wheel 3o into the curve, the bike would flip back up and over to lean in the other direction. It is not precession but rather centrifugal and gravitational forces that cause our motorcycles to lean into a curve. If precession of the front wheel is so powerful, the heavier rear wheel along with the much higher rpm flywheel would resist ANY leaning far more effectively than the precession of the front wheel would instigate it. If precession of the front wheel is that powerful, then upon leaning, the resulting rear wheel precession would forcefully push the rear of the bike outwards causing a low side crash. If precession of the front wheel is indeed the mechanism responsible for turning a motorcycle at speed. The effort/force required to lean the bike at 100 mph would be no different from that required at 30 mph. Anyone who has ever ridden a motorcycle aggressively knows that it takes significantly more effort to get a bike turned at high speeds compared to low and moderate speeds. Why? Because at higher speeds, it becomes progressively more difficult to overcome the forces of gyroscopic precession of the front wheel, the engine, and the rear wheel. "It sounds improbable, but that is how it works; the precession of a wheel weighing a few pounds can move and steer a motorcycle laden to a quarter of a ton" Nope, sorry. That is simply totally incorrect. The primary mechanisms responsible for turning a motorcycle are centrifugal force and gravity as explained above. Blake PS: Arvel, you can still show the calculations backing this up if you want. |
Roc
| Posted on Tuesday, December 18, 2001 - 03:03 am: |
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Wolseley - If the Segway, another gyro-toy, looked that modern it would be a sure sell. |
Werewulf
| Posted on Tuesday, December 18, 2001 - 08:43 am: |
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check out the january issue of motorcycle consumer news in the letters to the editors section. it just came in the mail and its got a good argument on the countersteering! |
Blake
| Posted on Tuesday, December 18, 2001 - 11:11 am: |
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Can you summarize what they say? I don't have access to a January copy. |
Tripper
| Posted on Tuesday, December 18, 2001 - 01:31 pm: |
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Gyroscopic Precession turns me... ...and I can prove it. |
Jima4media
| Posted on Tuesday, December 18, 2001 - 02:05 pm: |
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This explains why the Firebolt turns as well as it does. The lighter weight wheel and brake assembly lessen the gyroscopic force, and the more vertical forks put more weight on the front tire, adding to over-all stability. Result - Ultra Flickability. Wait till you give it a try. Jim X-2.5 BTW, this is also the reason the V-ROD steering sucks big time.
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Jima4media
| Posted on Tuesday, December 18, 2001 - 02:15 pm: |
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This is what Buell should do to prove the superiority of the Firebolt to other motorcycles. Mount a front wheel and forks in a demo stand and spin the wheel with an electric motor to get it up to speed. Then let people flick the handlebars to feel the difference between a light wheel, and say - a V-Rod solid disc front wheel and mondo dual brakes. It would be as effective as the 45% tilt stand for the Firebolt that they are taking to shows now. Jim X-2.5 Of course, Willie G. would be none too pleased with the results.
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Sarodude
| Posted on Tuesday, December 18, 2001 - 03:42 pm: |
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Jim- They could just pick some kind of Brand-J front wheel assembly... Of course, I don't really know how much practical weight difference there is... -Saro |
Davegess
| Posted on Tuesday, December 18, 2001 - 05:57 pm: |
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sarodude, the weight difference is several pounds. I remember, probably incorrectlt 'cause it seems so large a number, that is is 7 pounds lighter than any other wheel tire brake combo on the market. What ever the number is, it is a spectacular difference and a huge accomplishment. Dave |
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