racer Posted October 22, 2006 Report Share Posted October 22, 2006 Hey Tim, Only when the chain pulls at an angle to the swingarm. Respectfully, I think that rotation of the rear wheel and horizontal motion or force are irrelevant to the swingarm going down under acceleration. It is in fact the chain pulling down on the swingarm and shortening the wheel base that allows the rear wheel to claw under the bike, not the other way around. This is clearly demonstrated by orienting the chain pull parallel to the swingarm and observing no rise. Under acceleration, the friction or resistance (which could be considered horizontal) at the rear contact patch allows the chain to pull down on the swingarm. The wall is a convenient way of isolating the relevant action with "infinite" resistance or "infinite" acceleration. But I think it might create the impression of that horizontal force being the actor. Think of the bike rollin along at even throttle at about 30 mph. Now add some gas. Does it still feel like horizontal force is pushing the swingarm down? Hypothetically speaking, if we removed the engine and chain-drive entirely and installed a super-conducting electro-magnetic drive system inside the hub of the rear wheel, thereby eliminating all chain forces, perhaps a down angled swingarm would be affected by "horizontal force" at the rear wheel. But I suspect it would be more proper to say that rotational forces at the hub attempt to rotate the arm up and around the axle and are resisted by the mass of all the batteries replacing the engine...and, again, it would seem that the arm rotation allows the wheel to move horizontally inward, not vice-versa. Damn. I tried. Cheers, Racer Quote Link to comment Share on other sites More sharing options...
oldfrt Posted October 22, 2006 Report Share Posted October 22, 2006 I don't disagree with the chain force tending to rotate the swingarm. It may be more of a factor in moving the swingarm than the external forces I referred to. A detailed analysis could be made if the geometry, weights, and forces were known. If you say that the horizontal force at the rear wheel contact patch is irrelevent to the discussion, then I don't agree with that. In fact a simple static test could be done quite easily. Put the bike in gear (motor off) and push the front wheel toward the rear allowing the rear wheel to resist the movement. The pushing should be by a flat vertical surface pushing horizontally rearward. No front brake, obviously. Quote Link to comment Share on other sites More sharing options...
racer Posted October 22, 2006 Report Share Posted October 22, 2006 oops, hey tim, you were posting while i was editing. i need to go take care of something. i'll try to be back shortly. in the meantime, check out the edited post. tell me what you think of the new examples. cheers Quote Link to comment Share on other sites More sharing options...
racer Posted October 22, 2006 Report Share Posted October 22, 2006 hey tim, i think your static test is still demonstrating chain pull. Quote Link to comment Share on other sites More sharing options...
racer Posted October 22, 2006 Report Share Posted October 22, 2006 Hey Tim, I believe your static test is still using the chain to act on the swingarm. I think it is essentially the same example as the prior wall test except you apply force with the wall instead of the engine. At the moment, the only way I can think of to actually remove the chain from the equation while maintaining the component of acceleration is to make the chain parallel to the swingarm, which has been done and tested. Earlier I referred to a post by Keith Code in which he discusses these tests conducted with World Champion Eddie Lawson riding the bike. The relevant post is the last one in the thread titled "Weight transfers...Does the back really go up?" I originally found it by using the search function for the word "squat". In any case folks, Like I said in my first reply to MotoGB, the significant fact for a rider is that the swingarm does go down and the suspension does get "harder" under acceleration. Cheers all, Racer Quote Link to comment Share on other sites More sharing options...
oldfrt Posted October 22, 2006 Report Share Posted October 22, 2006 hey tim, i think your static test is still demonstrating chain pull. I give up Quote Link to comment Share on other sites More sharing options...
oldfrt Posted October 22, 2006 Report Share Posted October 22, 2006 My static test was not a good example because there is no difference from it to the engine force pushing against the wall. Instead of giving up I'm going for one more shot. Frequently in engineering we isolate the forces we want to analyse and ignore the others. Lets just look at the forces we add if we push back on the bike from the wall. The horizontal wall force is resisted by an equal and opposite force on the rear wheel acting at the surface of the pavement. Just taking this force acting parallel to the surface of pavement at the contact patch, it appears obvious that this force is tending to rotate the swing arm downard. This will happen regardless of angle of the chain to the swingarm in my opinion. OK, now I give up. PS I left out another force required to balance the system, that is an additional force upward on the rear wheel and downward on the front wheel to balance the overturning moments caused by the horizontal forces. I don't think this affects the gist of the discussion and only adds more confusing complications. Therefore ignore it. Quote Link to comment Share on other sites More sharing options...
racer Posted October 23, 2006 Report Share Posted October 23, 2006 Hey Tim, I'm not a mechanical engineer and I am simply trying to work through this complex problem in my head from memory without so much as a motorcycle sitting in front of me. As I alluded to in my original reply, I hadn't really finished thinking through the whole thing yet. And as much as I enjoy the game of trying to do that without referring to any other materials or asking for help, it does lead me to some half baked ideas based on faulty assumptions from time to time. And being relatively certain about most of what I talk about, it is also easy to fall into a 'habit' of speaking from a position of certainty. I try to remember to stay on top of that. That said, it was not really my intention to end our discussion, I simplynot sure I have really worked through it all yet and don't want to misspeak myself or confuse other people or devolve into an argument without basis as it were. I'm happy to be wrong. I just like to see if I can figure stuff out. And, in this case, it may be too much for someone lacking a complete education on the matter, no matter how sharp I am. And what I wanted to work out was the counter-steering thing I have been trying to work out on and off for about 8-9 months here. Oh well. I guess it's gonna have to wait a little longer. lol. That said, I was in the same place you are about this rear geometry thing half way through my original reply yesterday and did some ear steaming brain effort on it. And went around in several circles from several angles. And mostly decided to stick with what I knew for sure, which is the rotational force applied to the wheel. Which obviously tries to rotate the arm up. And the thought that it is this that tries to lift the arm as opposed to pull it down. And that the lever for that is quite small, namely the radius of the rear sprocket. (This is all with a parallel chain) And also that chain tension is going to fight the rotation at the pivot point against the mass of the bike pushing down. Even in and perhaps especially in a wheelie. And I had the thought that in a wheelie, varying the acceleration and tension might reduce that effect at the pivot. With chain force applied completely to the sprocket, the chain won't provide any linear force to the swingarm. I don't know if that means there is none, I simply worked from the premise that, at the very least, there is much less as was demonstrated from Keith's experiment with Eddie and tried to develop a theory to fit a possibly incomplete data set. That said, I believe that a horizontal component of force must probably be considered with some upward angle of force. And I am still working out how to sort that out. I have a gut feeling that a composite force vector might be correctly drawn from the contact patch to the CG or there abouts. And perhaps this is separate from the rotational force at the wheel which might be why the bike can wheelie and not slow down? The truth is gut insitnct and intuition commonly don't line up with the reality of complex physics like this...so, I'm gonna leave it at that. Tell me what you think, and I'm going to read up some material about wheels and angular momentum and such and how that getsw reduced to a singel force vector in this scenario...eh? Some if not much of what I have found on so called "motorcycle" sites is nearly laughable. And I feel like I'm cheating if I peek. My personal game. And perhpas inappropriate for this site? I don't know. But there it is. I'm off to bed. Good night. Bill Quote Link to comment Share on other sites More sharing options...
racer Posted October 23, 2006 Report Share Posted October 23, 2006 Ok...my last post was late at night after a long day and I was not clear in my descriptions. For the sake of clarity... In the third paragraph when I said, "rotate the arm up", I meant pivoting up around the rear axle. When I spoke about the chain tension fighting rotation at the pivot point, I meant the swingarm pivot point assuming a roller topped by the chain in the "parallel" configuration. This roller position has no basis in reality, just what my mind pictured. The ATK system that Keith mentioned would not seem to be quite like this. I can't find a picture of this system, but, from Keith's description, it sounds like the rollers would need to be above and below. Without seeing the set up I can only imagine. But Keith said that the torque force was greatly reduced... not eliminated. If the swingarm pivot is changed to be neutral at sag with rider, there would seem to be a progressive increase of torquing chain force directly proportional to the amount of swingarm deflection (in either direction) under acceleration. So, being that upward torque is not desirable, some position locating that neutral position should be near (or in the extreme above) the top of suspension travel (in case of hitting a big bump under accel), thereby making the amount of downward torque somewhat adustable. Anyway, the super-conducting electro-magnetic rear hub drive would seem to be the only absolute model for elimination of any chain force torquing the swingarm and isolating the rotational force. I'm not sure how that relates to "horzontal force" at the moment. So, wheel radius/circumference (and perhaps sprocket) radius with applied chain force assuming level ground. Must be a formula for that. And the amount of "horizontal force" is going to be limited by the mass of the motorcycle and the lever it has for resisting the rotational force at the rear wheel. Hence the idea that keeping the front wheel on the ground is faster. This would also seem painfully obvious. I have no illusions that I am discovering anything here (except for myself). That's all the time I have. Thanks. Bill Quote Link to comment Share on other sites More sharing options...
oldfrt Posted October 25, 2006 Report Share Posted October 25, 2006 I was trying to clarify and simplify the problem by only looking at the external forces on the bike and those initiated by the bike. Getting into the sproket, swing arm, and chain are "internal" , or component, forces and can be ignored if one is only trying to balance the external forces. The external forces are the weight of the bike acting vertically and momentum or inertia acting horizontally, and the forces resisting the inertia and weight which is the upward vertical force at each wheel passing through the axles and the tractive force at the rear wheel which is horizontal. My point is that the horizontal traction at the rear wheel is a force that passes under the swing arm pivot, and is therefore trying to rotate the swing arm downward. I don't think torsional inertia of the wheels and rotating mass is a large consideration in our discussion. I've given some thought to the quick turn and countersteering and as I've said before, it's my belief that countersteering causes a turn to be initiated because it makes the wheel LEAN not turn in the desired direction. I think the small degree of turn in the handlebar actually follows the lean. If you think about a turn where the bike is leaned over hard, there is only a small degree of actual turn in the handlebar, the bike is actually tracking as if it were inside of a conical structure. The people that ride in the spherical cages are actually not "turning" at all when they get to the point of doing vertical loops or going almost horizontal around the circumference. As for the quick turn, there is no limit to how fast the turn can be initiated except how quickly the bike can be leaned to a stable turning lean where the resultant of the gravity and the centrifugal accelleration passes through the c.g. of the bike and thru a line between the front and rear contact patches. Lucky for us, countersteering actually is an aid in leaning the bike. I've been away from the forum for a couple of days myself and probably will be off for another couple. I hope we are in agreement on these issues. If not, I'm sure you will tell me. It's an interesting and complex subject. I went to the Level 1 course back in May and learned a lot. I try to practice those lessons each time I ride now, and have become noticeably faster (to myself), but more importantly, more comfortable and competant in my riding. That translates also to safer, I hope. Quote Link to comment Share on other sites More sharing options...
racer Posted October 26, 2006 Report Share Posted October 26, 2006 Hey Tim, The downward force of the mass of the bike caused by gravity is balanced by the ground pushing back up. Inertia or forward momentum of the whole bike is essentially irrelevant to the effects of acceleration on the swingarm. We can have accelerataion from a dead stop or from a hundred miles an hour. In either case it is the acceleration that is applying force to the swingarm. The traction force at the contact patch pushes back against the tire with exactly the same force as the tire pushes against the road surface from acceleration. It is balanced. It is NOT an external horizontal force pushing the motorcycle forward. There is NO external horizontal force pushing on the back of the wheel or the axle. If we backed the bike up against a wall, that would be an external horizontal force being applied to the back of the wheel or axle. But that isn't what is happening. The only unbalanced force being applied to the wheel is rotational acceleration. If we prevent the rear wheel from rotating, it will rotate the swingarm up. If we allow the rear wheel to rotate, the wheel will roll forward, and acceleration will rotate the swingarm up. This is true no matter what angle the swingarm is to the frame. Hence the acceleration force applied to the frame of the bike is not horizontal. The acceleration force applied to the frame of the bike is at an upward angle. Hence the bike rises under acceleration thereby shortening the wheel base. Any chain pull offset below the angle of the swingarm will increase this effect. Cheers, Racer Quote Link to comment Share on other sites More sharing options...
oldfrt Posted October 26, 2006 Report Share Posted October 26, 2006 I shouldn't have referred to gravity, accelleration, and wheel reactions as external forces. I was only trying to simplify the problem. Here are the forces. The bike is accellerating forward and is resisted/balanced by the horzontal traction force at the rear wheel. The bike has weight which is resisted by the upward pressure on the wheel and these forces are vertical and pass through the axles. The forward accelleration force is through the c.g. of the bike and creates an overturning moment which is balanced by an additional load vertically downard at the rear wheel and upward at the front wheel. The rear wheel traction force has to get from the rear wheel/swingarm/suspension to the c.g. of the bike allowing that it is also part of the mass of the bike. All I am saying is that the resultant of the additional upward component due to balancing the overturning moment and the horizontal traction force balancing the accelleration at the rear wheel passes below the swingarm pivot and causes the rear to rise. I am discounting the rotational inertia of the rear wheel which I say is not significant to the discussion in my opinion. Visualize the bike rolling backward and applying the rear brake. This creates the same forces and may make it easier to visualize what I am trying to convey. Quote Link to comment Share on other sites More sharing options...
racer Posted October 26, 2006 Report Share Posted October 26, 2006 Hi Tim, I don't think that a line drawn from the contact patch to the CG will pass below the swingarm pivot. If I shift the swingarm forward such that the pivot is located forward of the CG which is now located above the center of the swingarm, the swingarm will still rotate around the axle under acceleration. The bike will still rise. Forward acceleration is resisted by the mass of the bike, aerodynamic drag, and friction in the wheel bearings and drivetrain. It is not resisted by a horizontal force at the contact patch. Thanks, Racer Quote Link to comment Share on other sites More sharing options...
oldfrt Posted October 27, 2006 Report Share Posted October 27, 2006 Hi Tim, I don't think that a line drawn from the contact patch to the CG will pass below the swingarm pivot. If I shift the swingarm forward such that the pivot is located forward of the CG which is now located above the center of the swingarm, the swingarm will still rotate around the axle under acceleration. The bike will still rise. Forward acceleration is resisted by the mass of the bike, aerodynamic drag, and friction in the wheel bearings and drivetrain. It is not resisted by a horizontal force at the contact patch. Thanks, Racer I didn't say a line from the contact patch to the c.g. , I said that the resultant of the additional forces at the contact patch would pass under the swingarm pivot. The additional forces at the contact patch are the balancing moment couple force acting vertical, and the horizontal friction force driving the bike and overcoming the inertia of the bike. The balancing moment couple force is required to balance the overturning (think wheelie) moment required to keep the bike stable. The overturning moment is the accelleration force acting at the c.g. times the height of the c.g. to the ground. The initial weight component force at the rear contact patch doesn't need to be considered because it is already balanced by the suspension, so we only need to consider the additional forces. The driving force for the bike is the horizontal force at the contact patch produced by the motor and drive chain. Otherwise the bike would only spin the rear wheel. All of the forces making the bike move forward are transmitted to the ground through the contact patch. Quote Link to comment Share on other sites More sharing options...
racer Posted October 27, 2006 Report Share Posted October 27, 2006 The bike is accellerating forward and is resisted/balanced by the horzontal traction force at the rear wheel. The forward accelleration force is through the c.g. of the bike and creates an overturning moment which is balanced by an additional load vertically downard at the rear wheel and upward at the front wheel. I guess it was these two sentences that confused me the most. Quote Link to comment Share on other sites More sharing options...
oldfrt Posted October 27, 2006 Report Share Posted October 27, 2006 It's hard to describe in words what is so simple if a sketch could be drawn. Quote Link to comment Share on other sites More sharing options...
racer Posted October 27, 2006 Report Share Posted October 27, 2006 Tim, I agree. About the sketch. I think I also know what it is you mean with your words, and I think I might even agree at this point, but, I'm just not quite clear. We may very well be thinking the same thing and coming at it from opposite directions. Anyway, as soon as I have a bit of time, I will get back ot offering a direct opinion rather than picking nits with yours. Cheers, Racer Quote Link to comment Share on other sites More sharing options...
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