Yeah i gave up a few pages ago. Call me a simpleton but i think i'll just go for a ride and not sweat all the uber tech talk.DHS said:
Oh, and i like honda bike anyway.
HONDA... A picture says a thousand words
- Thread starter humprabbit
- Start date
Oh, and i like honda bike anyway.
That begs the question of whether the observed bobbing is due to the bike's reaction to acceleration or to the weight shift.ohio said:
I'm mainly concerned with the bobbing you can see in the rear on some bikes when you are smoothly pedaling (to the best of your ability) while sitting. If that bobbing is due to acceleration, the front end should show the bobbing as much, if not more, than the rear. The accelerational load shift affects both ends equally, and the front is normally sprung and damped more softly. I don't see any evidence of extension in the front to match (or exceed) the compression in the rear.
Uh, in your last post you specifically said you were interested in an increase in downward force. So which is it? Smoothly pedalling, or weighting and unweighting while pedalling?Steve from JH said:
But to address the above statement: yes, the fork extends under acceleration. depending on the frame it is as observable and usually moreso than any rear suspension action.
The parenthetical phrase "to the best of your ability" was meant to indicate that it is impossible to pedal without some weight shift.ohio said:
I don't observe fork motion unless out of the saddle. I can't tell whether any of that is acceleration induced because I'm leaning on the bars and moving them a lot from weight shift.
You guys are getting too tech... I'm going to ride the factory Jesus bike for 2006.. It will take care of everything if you just have faith in it... Ahhh the miracle in faith. Me=1 Satan/competitors=0
It is true that you can't pedal 100% smoothly, if you could then in theory you could also spin perfect circles and the rear end would find an equilibrium point at which it wouldn't bob anyway.Steve from JH said:
However, I definitely notice my Boxxer bobbing when I pedal. If I sit down and put in a hard crank, it extends noticeably. So I'd say any observations of this kind (without properly instrumenting it and quantitatively taking into account all factors) are pretty futile really.
BTW, the position of the IC (for a given axle tangent path) is unrelated to how the bike pedals. As I said before, when the only external forces acting on the swingarm are through the axle, you can only generate a point force, not a couple moment. If that were not true, then by pedalling you could theoretically induce bobbing WITHOUT accelerating the bike at all. But as you know, the only way that could ever happen is if you did that with the brake applied... which goes back to what I just said.
Best post in this thread yetsayndesyn said:
I'll leave the question of fork bobbing to someone with the proper instruments. I'll just say that I sometimes make my bike wheelie from a single pedal stroke when launching from a standstill on an uphill. The fork has obviously extended all the way before the wheel leaves the ground. But I was talking about a typical pedal stroke when cruising along, working hard but at a sustainable effort. And this would refer to bikes I've had in the past, low single pivots, that visibly bobbed in the rear no matter the gear or how I pedaled.thaflyinfatman said:
You might be interested in the analysis by Vittore Cossalter, head of the PhD program in motorcycle technology at the University of Padova, Italy. The principles governing the chain drive effects on the rear suspension should be the same for motorcycles and bicycles.
Leaving out the load transfer from acceleration, his formula for the combined chain and drive forces acting to create a moment around the swingarm pivot or IC is:
S[r+Lsin(d)] - T[c-Lsin(d-e)] where S = the drive force at the contact patch, T = the force in the chain, r = the wheel radius, c = the rear cog radius, L = the swingarm (or virtual swingarm) length, d = the angle from horizontal of the swingarm, and e = the angle from horizontal of the chain.
What this reduces down to in plain english is that the drive force moment equals the force at the ground times the height of the pivot. And the chain force moment equals the force in the chain times the perpendicular distance from the pivot to the chain line.
I highlighted the last sentence because it contradicts what you said. The torque around the pivot caused by the chain is the same as if the wheel were welded to the swingarm.
It also means that chain line parallel to swingarm produces a compressing moment equal to the radius of the cog times the chain force. And it means that the chain line passing through the pivot produces no moment.
Now it turns out that when you add in the drive force at the ground and consider the load transfer from the drive force, everything cancels out so that your statement about the location of the IC is true. All you need to know are the height of the CM, the length of the wheelbase, the angle of the swingarm, the angle of the chain, the size of the cog, and the size of the wheel.
But it's interesting that Cossalter starts with the assumption that there is a compressing moment if the chain passes above the pivot. And that becomes important in later analyses, such as what happens when the motorcycle suddenly loses rear traction.
Note where I said "COUPLE moment". You can generate a moment about the main pivot using a point force, but you can't generate a moment ABOUT THE AXLE with only a point force. And that is why the position of the IC is irrelevant. The compressive or extensive force acting on the axle (relative to the frame, and neglecting weight action/reaction forces or assuming them to be in equilibrium at all times) is the chain tension force T multiplied by the cosine of the angle from the tangent to the axle path (which is the same no matter how far away the IC is) to the chainline (call that angle alpha) plus the horizontal driving force D (worked out via basic gearing, relative to chain tension) multiplied by the cosine of the angle between it and the tangent to the axle path (call that angle beta). Pretty simple: Force acting to extend or compress the suspension, acting on the axle, relative to the front triangle, neglecting weight shift =Steve from JH said:
Tcos(alpha) + Dcos(beta).
Where you can most easily see that your understanding is flawed is that if you had an IC at infinity (parallel linkage), would the moment generated by the forces acting on the axle be infinite? Clearly not. The swingarm pivot in a singlepivot is not JUST the IC, it is also the CC. That considered, it is pretty obvious that the IC and a pivot DO NOT BEHAVE IN THE SAME MANNER.
Why his formula agrees with mine is that the swingarm length L and pivot height (relative to the axle) are directly related by the angle from swingarm to horizontal. That angle gives the axle path tangent. The axle path tangent is STILL THE CRITICAL FACTOR, like I've said the whole time.
As for the suspension behaving exactly the same as if the wheel is welded to the swingarm, that is outright wrong. All forces induced on the swingarm are, as I've said, COMPLETELY UNABLE TO GENERATE A COUPLE MOMENT (neglecting the action/reaction pairs of the point forces at the axle and the main pivot). You might be interested in reading this http://www.mtbcomprador.com/pa/english/chapter2_3.htm#CenterofMass if you still don't understand the difference between couple moments and point force reactions.
Finally, if/when you reply to this, please provide free body diagrams of the wheel and the swingarm. You will quickly see where you've gone wrong.
thaflyinfatman said:
You seem to have missed the part where I agreed that the exact position of the IC (or CC) doesn't matter when everything is considered. If you leave aside the load transfer from torque around the center of mass, all you need to know are the angles of the chain and swingarm and the relative sizes of the cog and wheel.
It's amusing that you refer to Ken Sasaki's Path Analysis. I'm not only familiar with it, to a great extent I am the person he was arguing with when he came up with it. So when he says something in the passive voice like "It has been argued that . . .", I'm often the one who did the arguing.
If you think I should be accepting Ken as an authority, who the last I heard was still trying to get a Masters in physics, instead of Cossalter, who is probably the world's leading authority on motorcycle dynamics, I don't think so.
Cossalter most definitely says that the chain passing above the pivot, which is the way motorcycles are usually set up, causes a compressive moment around the swingarm pivot.
And he definitely says that the drive force acts at the ground, not the axle. That has to be the case if you are going to base the load transfer on the height of the center of mass from the ground. If you base the load transfer on the height of the CM above the axle, then it would be possible to have no load transfer by getting the CM down to the axle level. Take a look at those exotic dragsters that have the big rear wheels, a very long low slung chassis, and little wheels on the front. They get the CM down at or below the axle height. And yet you can see front end lift when they accelerate violently. That's empirical proof that it's the height of the CM over the ground that matters. Your argument about point force vs. couple moment would seem to contradict that.
The problem must ultimately boil down to the fact that a rolling wheel under power can not be regarded as freely spinning about its axle.
I'll try to attach a diagram that is my summation of several of Cossalter's diagrams and arguments.
Attachments
Wait, who are you two guys? How do you have so much to say about this?
Nerds!
And proud of it.dropmachine.com said:Nerds!
Wait. How much is "that much"? Did we find out the price?jake133 said:
geez, it's still vapourware. I dunno what thread you guys are reading.downhillzeypher said:
Steve from JH said:
yea keep it up but dont forget to touch base with us mere mortals now and then!
Sasaki's work does not contradict Cossalter's. If you paid attention to the link I gave, you would have seen that the parallel force/acceleration theory does NOT directly take into account the reaction of a frictional (point) force at a point on the wheel/barrel/whatever - it is simply to demonstrate that any given net force acting anywhere on any body will generate both an angular acceleration (if the line of force does not pass through the CoM) and a linear acceleration THAT IS PARALLEL TO THE LINE OF FORCE. This is very well known and understood. Looking macroscopically at a bicycle (or motorbike, or car, or anything with wheels), if we assume all internal forces and reactions to make negligible difference to the geometry and external reactions, then the line of net force acting on the body is applied at the bottom of the wheel, which is always below the CoM because otherwise you'd have to have a car/bike/whatever that sat half below ground level (maybe a roller coaster could do this). So the line of force is not through the CoM, hence you get the linear acceleration (horizontally) and the angular acceleration (rotation backwards due to the inertial reaction acting through the CoM). I can see you understand this.Steve from JH said:
Now to examine the forces acting on the wheel, the swingarm and thus the body as a whole. If you can accurately argue against any of the following then I'll eat my words.
FBD 1: wheel. Assuming wheel to have negligible mass (no moment of inertia or weight)
Forces acting on the wheel are the chain tension (X and Y components, no doubt you're familiar with how to calculate those separately), the driving force at the contact point, the weight reaction force (of the body as a whole) and the axle reaction force (separated into X and Y components for simplicity). You can see that the wheel doesn't touch anything else, and that as such obviously no other forces are acting on it.
Equating moments about the axle (assuming constant forces/accelerations):
Chain tension x cog radius = driving force x wheel radius. Let's give some values here, say T = 1000N, c (cog radius) = 0.05m (5cm), wheel radius R = 0.325m (13" or so), and thus D = ~154N. Equating forces in the X direction:
D + T(x) - A(x) = 0, .'. A(x) = D + T(x) = 154 + 1000cos(alpha). Letting alpha = 15 degrees, A(x) = ~1120N.
Equating forces in the Y direction, letting N = 800N (just a random figure):
N + T(y) - A(y) = 0, .'. A(y) = 800 + 1000sin(15) = ~1059N
So we have A(x) = 1120N, A(y) = 1059N.
So now for a swingarm FBD, assuming it's a basic singlepivot for the time being. Note that I'm assuming in all this, that the position of the CoM is an arbitrary constant and I'm not bothering to calculate how much weight shift would occur, hence the arbitrary values for the normal reaction at the wheel. Also note that the only place the wheel is attached to the swingarm is at the axle, again, you can only transfer a point force through the axle (as you can see from the first FBD). The moment this point force can generate about the main pivot is a separate thing.
Anyway, equating moments about the main pivot taking clockwise as positive; S(x) is the force applied by the shock, assume it's horizontal only for ease of calculation:
A(y) x 0.45 - A(x) x 0.1 - S(x) x 0.15 = 0
.'. S(x) = 1059 x 0.45 - 1120 x 0.1 = 365N
From the above FBDs we can see a number of things:
- Only a point force is transferred to the axle via acceleration, this point force can generate a moment about the main pivot if the net force's line of action does not pass through the main pivot (which is obvious).
- The total point force acting on the axle (considering the axle as part of the swingarm) relative to the front triangle, is directly related to both the chainline and the chain tension magnitude, as well as the normal reaction force at the contact patch (which changes as weight shifts under acceleration - note that I have not calculated how much weight will shift, nor am I willing to at the moment) and the driving force at the contact patch. I have quite clearly shown how driving force, chain tension, cog size, wheel radius and the resultant force on the axle and thus the swingarm are related, and I'm pretty confident that this can be completely conciled with Cossalter's calculations. If you can see any errors in my FBDs, let me know.
- The aim of the game in this situation is to keep the swingarm's position with respect to the front triangle constant. Obviously this entails balancing weight shift with the moments about the swingarm main pivot, or generally speaking the forces acting to compress/extend the suspension. I'm not trying to or willing to calculate the total amounts of anti-squat as a percentage or anything like that.
- Assuming drive force to be completely horizontal, if the pivot was the same height as the axle, and the chainline was parallel to a line drawn through the pivot and swingarm, then there would be no forces acting on the axle that were not colinear with the pivot. As such, the only thing that would cause an extension or compression of the suspension would be the weight shift under acceleration. This is quite clear after looking at the FBDs I drew (yes I suck at Paint, you don't need to tell me).
- The relationship between the moments generated on the swingarm can be easily shown to be related to axle tangent path only, at any given point. Due to the lack of couple moment acting about the axle (I still don't see why you're being so stubborn about this, it's blindingly obvious that it simply cannot occur, once again the axle is the ONLY place the wheel attaches to the swingarm and thus cannot transfer a moment between them) on the swingarm, it is clear that the axle tangent path is the only thing - chain line/tension, weight shift, gearing etc being held constant - that will affect the bike's internal reactions to pedaling. If the axle tangent path (and not the rotation of the swingarm) is the only thing that matters, then the distance from the axle to the IC is irrelevant. All forces and calculations shown above CAN be expressed as functions of the axle tangent path only - not the pivot (or IC or even CC) position.
And this is what DW was saying all along. If you can find any error in the above, feel free to point it out.
BTW, I think there are some areas where we both agree but aren't getting the message across. Oh well.
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Themselves or "each other"? hahaha...ohio said:
My name is Steve and apparently so is his. We're both obviously bike nerds.James | Go-Ride said:
P.S. I don't actually ride
Sorry, I think it looks like something you would buy at target!
Originally Posted by thaflyinfatman
In Cossalter's method the height of the axle R is added to the height of the pivot above the axle--L*sin(d). d is the angle of the swingarm from the horizontal and is negative if below the pivot. In the special case we are considering L*sin(d) is zero. So we have an extending moment D*R. The compressing moment from the chain is T[c-L*sin(d-e)]. Since d = e (angle of chain from the horizontal) and both are zero, the compressing moment reduces to T*c. Now D = T*c/R. So D*R = T*c/R*R = T*c. The compressing moment is T*c and the extending moment is T*c. They cancel out. In your method there is nothing to cancel out. The chain is pulling the axle straight into the pivot and the drive force is doing the same and there are no moments.
To understand the compressive torque when the chain line runs above the pivot, consider that the chain tension works both ways. The same torque is being created at the pivot to rotate the main frame back around it as is rotating the wheel. And that reverse rotation of the frame is compressive. It depends on the perpendicular distance of the chain line above the pivot.
To understand the extending torque from the drive force at the ground, consider that the drive force produces a parallel force at the axle; the axle produces a parallel force at the pivot; the pivot produces a parallel force at the center of mass. You get the extending moment from the drive force by summing the ground axle moment and the axle pivot moment: D times the height of the axle and D times the height of the pivot above the axle.
What difference does it make if the suspension is neutral because of cancelling moments or because of no moments? In motorcycle racing it can show up when traction is suddenly lost. Now you have no extending torque. That happens instantaneously and the frame takes awhile to adjust its pitch attitude to the new situation. The compressing moment turns into rapid acceleration of the wheel, but that also takes time. Because of this time lag, the immediate effect is a violent compression of the suspension from the loss of extending torque. This manifests itself as an upward rotation of the swingarm that gets absorbed by the frame and sometimes pops the rider right off the bike. This has all been experimentally verified.
With your method of explanation, nothing violent would happen. The motorcycle would simply gradually return to its static sagged position until traction was regained. I probably haven't explained this very well, but I'm all typed out. No, I'm not going to go ride. I already rode for a couple of hours, like I do every day if I don't do more.
One other thing, in your post above where you used cos(alpha) and cos(beta)? Well my sin(d-e) is the same as cos(alpha), and my sin(d) is the same as cos(beta). Since the R term and the c term always cancel out in Cossalter's method, and since resultant forces (what you are using) involve dividing the moments by L and therefore L cancels out, the two methods are in fact pretty close. The main difference is that in Cossalter's method you must know the actual R and c measurements so that you can locate the ground contact point with respect to the intersection of the chain and swing lines.
Let's just look at this case for a minute. The chain line, swingarm line, and ground line are all parallel. Both methods would agree that the case is neutral. There is no net extension or compression around the pivot or up or down resultant force at the axle or contact patch except for the compression caused by the load shift to the rear. But the explanations for why that is true are different.
In Cossalter's method the height of the axle R is added to the height of the pivot above the axle--L*sin(d). d is the angle of the swingarm from the horizontal and is negative if below the pivot. In the special case we are considering L*sin(d) is zero. So we have an extending moment D*R. The compressing moment from the chain is T[c-L*sin(d-e)]. Since d = e (angle of chain from the horizontal) and both are zero, the compressing moment reduces to T*c. Now D = T*c/R. So D*R = T*c/R*R = T*c. The compressing moment is T*c and the extending moment is T*c. They cancel out. In your method there is nothing to cancel out. The chain is pulling the axle straight into the pivot and the drive force is doing the same and there are no moments.
To understand the compressive torque when the chain line runs above the pivot, consider that the chain tension works both ways. The same torque is being created at the pivot to rotate the main frame back around it as is rotating the wheel. And that reverse rotation of the frame is compressive. It depends on the perpendicular distance of the chain line above the pivot.
To understand the extending torque from the drive force at the ground, consider that the drive force produces a parallel force at the axle; the axle produces a parallel force at the pivot; the pivot produces a parallel force at the center of mass. You get the extending moment from the drive force by summing the ground axle moment and the axle pivot moment: D times the height of the axle and D times the height of the pivot above the axle.
What difference does it make if the suspension is neutral because of cancelling moments or because of no moments? In motorcycle racing it can show up when traction is suddenly lost. Now you have no extending torque. That happens instantaneously and the frame takes awhile to adjust its pitch attitude to the new situation. The compressing moment turns into rapid acceleration of the wheel, but that also takes time. Because of this time lag, the immediate effect is a violent compression of the suspension from the loss of extending torque. This manifests itself as an upward rotation of the swingarm that gets absorbed by the frame and sometimes pops the rider right off the bike. This has all been experimentally verified.
With your method of explanation, nothing violent would happen. The motorcycle would simply gradually return to its static sagged position until traction was regained. I probably haven't explained this very well, but I'm all typed out. No, I'm not going to go ride. I already rode for a couple of hours, like I do every day if I don't do more.
One other thing, in your post above where you used cos(alpha) and cos(beta)? Well my sin(d-e) is the same as cos(alpha), and my sin(d) is the same as cos(beta). Since the R term and the c term always cancel out in Cossalter's method, and since resultant forces (what you are using) involve dividing the moments by L and therefore L cancels out, the two methods are in fact pretty close. The main difference is that in Cossalter's method you must know the actual R and c measurements so that you can locate the ground contact point with respect to the intersection of the chain and swing lines.
Steve from JH said:
Steve from JH said:
I haven't read the whole of this conversation, but I caught this while skimming...Steve from JH said:
Can you describe more this "sudden loss of traction?" When in motorcycle racing is traction suddenly lost? Are you referring to a situation under forward acceleration or braking, or lateral (turning) acceleration. If you understand the dynamics of traction and slip, then you know that there is no such thing as a sudden loss of traction unless the vehicle leaves the ground. It could happen on the road if ones hits gravel, but you specifically said it shows up in racing (where tracks are swept clean).
Can you give an example?
WheelieMan said:
I shouldn't have said racing. Here's the first sentence of that section of his book:ohio said:
Later he says:
Later he discusses those systems like Bilever and Tracklever that keep the chain and swing lines parallel:
I think you may be misinterpreting Cossalter's theories somehow (don't take this as an insult or anything, I'm not trying to be nasty). I pretty clearly showed in my FBDs above that you can't take the rear wheel and the link to which it is attached as being fixed.Steve from JH said:
An interesting way of considering it however, that may reconcile our differences of opinion (which I have a suspicion are actually the same general understanding), is to consider a (thin, to the point where cross-sectional area is negligible) horizontal beam welded to whatever fixed support. At the end of the beam there is a pulley. A weight hangs from a cable, which runs up to the pulley, then parallel to the beam, where it is attached to the fixed support (so it turns 90 degrees basically). Obviously the tension in the cable (assuming negligible friction) is equal to the weight hanging on the cable. (If you find this hard to visualise, say so and I'll draw up a quick diagram of what I mean)
To find the net bending moment acting on the beam, you can do EITHER of the following:
1. multiply the length of the beam plus the radius of the pulley by the tension in the cable (for the horizontal moment arm) and subtract from that the tension in the cable multiplied by the radius of the pulley (which is the vertical moment arm).
2. Use resultant forces at the centre of the pulley: cable tension multiplied by the length of the beam (horizontal moment arm), then subtract cable tension multiplied by the vertical moment arm (which is zero).
Notably, both methods have exactly the same result, because the radius of the pulley in each case cancels out and the tension and tension lines become the critical factors. I think this is how you are misunderstanding my approach, and why I don't think it disagrees at all with Cossalter, as well as why I don't think Cossalter considers the wheel to be a fixed part of the swingarm.
Steve from JH said:That's the reason I was confused. The anti-squat diagram you posted over on MTBR is not anything like the diagram Ellsworth used for their patents. So I guess you agree partially with Ellsworth's ICT theory.WheelieMan said:
After viewing this experiment I don't know how I could agree with you.
Also worth checking this http://www.mtbcomprador.com/pa/english/chapter3_3.htm#ForcesBetweenLinearlyConstrainedParticles out. I realise you (Steve) may disagree with Sasaki's theories, but this part of it is definitely not arguable. Relatively basic physics will fully explain this, and it meshes perfectly with my resultant force explanation.
You don't need to repeat your arguments or quote K.S. I understood both long ago.thaflyinfatman said:
The question is am I misinterpreting Cossalter. Well here is a quote from the text at the beginning of the discussion of our topic:
Besides the static elastic moment exerted by the suspension spring and the moment generated by the static vertical load, which is dependant on the weight force, another four moments operate on the swinging arm: the moment generated by the load transfer N that compresses the suspension;
the moment generated by the driving force S that tends to extend the suspension;
the moment generated by the chain force T that compresses the suspension;
the additional elastic moment generated by the suspension M that can be positive or negative.
the moment generated by the driving force S that tends to extend the suspension;
the moment generated by the chain force T that compresses the suspension;
the additional elastic moment generated by the suspension M that can be positive or negative.
I emphasized the compressive moment because if we applied your method it would not be a compressive moment but an extending one.
The diagram that accompanies this text looks something like the top diagram in my attachment--typical of a motorcycle wheel, swingarm, and sprocket setup. The chain converges toward the swingarm to the front but passes above the pivot.
The lower diagram in my attachment represents a solid structure like a cog welded to a swingarm. Imagine a chain or cable wrapped around the round part so that it always stays tangent when pulled.
The moment calculation for the two is the same and is what Cossalter says applies to the motorcycle.
Attachments
Steve from JH said:
That makes much more sense, thanks.
This is under the assumption that speed is constant (steady state equal to the wind resistance at a given speed) and the suspension geometry creates squat under driving force? Given that, yes, this makes sense.
Realize that because your at a steady state speed (the driving force equals the wind resistance) there is NO acceleration, so no load transfer, so no additional compression/sag over a standstill situation (other than the fact that wind resistance actually creates a moment that can compress the rear end... but we'll ignore that). That may have been your point, was to illustrate a situation with driving force but no acceleration... I'm not sure. But that situation does not exist on a bicycle, unless you are climbing and have a perfect pedal stroke.
I have no idea how this related to the bike conversation I wasn't following... so back to your regularly scheduled arguments...
Do you want a real conspiracy theory? Honda is in cahoots with Shimano. Honda takes 1% (or less) of their R&D budget to build a robust gear box bike. That same amount would probably be a much more sizeable chunk out of Shimano's R&D budget. Honda sells to consumers for a short amount of time, regains a certain percentage of total money spent, and then sells to Shimano. Basically break even. Shimano can then drop their stupid dual control platform without loosing face, and then have a proven platform that will revolutionize the industry. This way they can remain in the MTB game while dealing with the upshot SRAM (and their products that seem to be gaining market share each year). A Japanease business keeps another viable. Plausible? Probably not, but fun to think about.neversummersnow said:
Right, I think I'm starting to get my head around how you're doing it now. The way I'm looking at it has to consider both chain tension and driving force acting on the axle at any time, your way however seems to use one without the other (and doing it that way it appears to make sense). Am I right in thinking that, when both driving force and chain tension moments are added, the results are the same? Since the driving force moment would appear to be substantially higher using Cossalter's method because he uses the height of the pivot above the ground (rather than above the axle), this (in my head at least) would cancel out the apparent moment arm of the cog (because the ratio of the radii of the wheel and the cog is the same as the ratio of driving force to chain tension)? If that is the case, yes I agree with what you're saying.Steve from JH said:
I do see what you're getting at with the driving force/slip thing now (if it is as Ohio explained it) - when the friction disappears and there is no pro-squat moment anymore, the bike will pitch back to its original (static) state.
Yes, you've got the cancellation thing down now. But the two ways of looking at it are still not the same as far as the actual physics is concerned. Two fairly large moments cancelling out is not the same as no moments. They can't both be right.thaflyinfatman said:
No, the drive force/slip thing is not as Ohio said. It's complicated and I'm not sure I really understand it. Basically Cossalter is saying that the sudden disappearance of the drive force leaves both the compression from the load shift and the compression from the chain force still active for a brief time. And in that time the suspension compresses so violently that it destabilizes the bike, which if it doesn't crash will oscillate for a while.
I've seen the phenomenon on TV in Formula motorcycle racing. It's quite amazing. The rider can be popped up like toast from a toaster.
The vehicle does not have to be accelerating for there to be load shift. I used to think that too. I came to realize it wasn't true just a little while before I got this Cossalter book.ohio said:
The load shift is dependent on a force couple between the thrusting force at the ground and an equal and opposite force acting at the CM. If we assume the air resistance acts at the CM, it doesn't matter whether the force is all reaction to acceleration (launching off from a standstill) or whether part of it is reaction to acceleration and part wind resistance, or whether it's all wind resistance (steady velocity on level ground).
An interesting corollary of this is that a motorcycle (or bicycle if the rider is Superman) has a theoretical maximum velocity that is reached when the load transfer is complete to the rear. The front wheel has zero load. Any more thrust and the bike will start to flip over backwards instead of accelerating further.
Another corollary of this is that it's easier to wheelie from acceleration on a motorcycle the faster you're going when you start the acceleration. It's hardest (takes the most acceleration) when you wheelie from a standstill.
A further interesting idea occurred to me this morning during my regular bathroom visit, which is when I do some of my deepest thinking. When the front wheel leaves the ground, you're now actually riding a unicycle. The way to go faster on a unicycle is to lean farther forward as you pedal harder. The maximum velocity on a bicycle being ridden as a unicycle would be right before the front wheel touched down. That's the point where the unicycle turns to bicycle and vice versa. And it's also the point where the bicycle hits its maximum velocity. Nice symmetry there.
all i know is i wish i had one