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Yes - its a separate issue from the effects of pivot position
It happens even with a concentric pivot.
Pivot position is important but not in isolation
A carefully chosen pivot will allow the force I describe above and the effects from the non concentric pivot to balance ( in some gears)
MidlandTrailquestsGraham - MemberIt could stay still, rise or fall depending.
But it doesn't, because the force is so small it has no effect.
Nope. It all depends on the balance of forces - that vid shows clear squat and rise on a very short travel system
Some motorbikes don't squat or rise much - good sus design balancing the forces out.
The simple force created by the chain being not parallel to teh swingarm is easily big enough to cause squat or rise. However you can balance this by clever placement of the swingarm pivot.
It needs paper and pencil to describe all this stuff adequatly.
I shall give up.
TJ is your point above based on separating the again tension into its components along and perpendicular to the axle-bb line?
if so then it seems you are with Ade re. chain tension and pivot placement?
cynic-al - MemberTJ is your point above based on separating the again tension into its components along and perpendicular to the axle-bb line?
Yes
Dunno what ADE has said.
There are two different effects being confused here.
Pivot placement has an effect that can create a vertical force on the wheel. the angle of the chain produces a different vertical force. These two are completely separate.
Thse two can both be in either direction or zero depending on the variables.
Clever positioning can make them equal and opposite in some gear combos
With a simple swinging arm, the only variables are the position of the swinging arm pivot, the point where the chain meets the chain ring and the point where the chain meets the rear sprocket.
These three points form a triangle.
The chain forms one side of the triangle.
By putting tension on the chain, it tries to shorten that side of the triangle.
With the swinging arm pivot below the top run of the chain, it will try to raise the swinging arm, or squat the suspension.
However, the angle between the chain and the swinging arm is so small that the force required to move the swinging arm is too great and chain tension has no effect on suspension movement.
Back to this picture.
[img] 
[/img]
Imagine the jib almost horizontal so that the top rope is nearly parallel with it.
It would take a lot of effort raise a load on the hook by pulling on the top rope.
Now imagine the jib nearly vertical.
It takes very little effort to raise the jib all the way to vertical.
Jib = Swinging arm
Rope = Chain
However, the angle between the chain and the swinging arm is so small that the force required to move the swinging arm is too great and chain tension has no effect on suspension movement.
This is simply not shown by what you say.
You are still confusing the two different effects.
On the motorcycle clip you linked to it was moving 1/3 of the travel easily.
With respect sir, that did not move 1/3rd of the travel.
Anyone else care to estimate how much the suspension on this bike moves when the power is applied ?
Incidentally, I'm trying to work my way round to explaining why the Santa Cruz VPP [i]does[/i] use chain tension as part of the suspension system, but after 5 pages we're still not clear on what's happening with a single pivot, so I don't know if we'll ever get there.
The harley on the dyno I meant You can see it rise and fall. A harly sportster only has about 4 inch of rear travel.
A burnout is no test at all - not much power used at all.
ah ok. I don't believe analysing the tension in the chain that way is correct as I've said. happy to be proven wrong.
Al - its basic mechanics. vector addition - triangle law
I have attempted to explain it to you twice without success. if you have a force along the chain that is resisted by the swingarm / chainstay then if these two are not parallel you get a resultant vertical force.
http://hyperphysics.phy-astr.gsu.edu/hbase/vect.html
Ah, OK, wrong video.
The thing is, even with the swinging arm pivot below the top chain line, the back of the bike rises.
How could chain tension cause that ?
What I think is happening is the triangle thing again.
The three points are the rear tyre's contact with the roller, the swinging arm pivot and the front tyre's contact with the wall.
As the rear wheel pushes forward it tries to shorten the bottom line of the triangle by pushing the swinging arm upwards.
I think you're misleading yourself with the parallel thing as well.
Two sides of a triangle can't be parallel.
MTQG
You have got two different things confused. Its really impossible to explain without diagrammes.
This shows the effects of the chain and swingarm being non parallel. Both are force triangles.
[img] 
[/img]
This shows the effects of the swingarm pivot not being concentric to the bb
[img] 
[/img]
Both those diagrammes are horribly oversiplified as well
TJ I get vectors, I just disagree that looking at the position and angle of the chain is the correct analysis.
wish I could fine that Spanish website, it was linked to loads here back in the day and I thought this point was well established.
Al - there is a basic misunderstandfing / lack of comnprehension
If the chain and swingarm are not parallel then there will be a vertical force produced irregardless of the position of the swingam pivot. this is simple basic mechanics.
Have a look at the diagrammes
Al http://www.sportrider.com/tech/146_0404_motorcycle_traction_geometry/photo_02.html
Remember that on a bicycle you generally have a larger front sprocket so the vertical force is opposite that on a motorcycle - reffered to as "chain pull" here it is a squatting force on a bicyle in a typiocal gear.
The three forces all act on the swingarm at the rear axle, and their moments about the swingarm pivot can be calculated. As the rear suspension compresses, the geometry of the force diagram changes--sometimes significantly. Note that while the chain pull always creates an antisquat tendency, the driving force torque changes from antisquat to pro-squat as the suspension compresses and the swingarm passes horizontal.
hmmm. I'm still not sure it's ap relevant on I bike given the disparity in power and torque.
oh and perhaps it's fairer to say we disagree than I misunderstand?
it is fascinatng reading about suspension forces as written by a solicitor, a mental health professional, and a bus mechanic.
what will the three wise men do next
Nope - I think there may be confused concepts but you cannot state that this clearly demonstrated effect does not happen. Sorry.
A cyclist produces as much torque as most motorcycles
Cycle springing is much lighter.
MTQG
I'll try and do a picture later
But I've just been out on my bike. A single pivot (Fuel ex 6.5)
Before I set off I sat on the bike and leant on the shed. Locked the front fork and pulled on the front brake
In the middle ring waiting the leading pedal contracted the suspension visibley. I assume this was due to chain tension but obviuosly as you wait a pedal other thing happens like the reaction force on your bum decreases. The pivot is below the middle ring. The effect seemed less in the the big ring.
Then I got bored and went for a ride
ch well in that case TJ you misunderstand too 😀
LOLLERCOPTERS at the tag.
oh and spunkyard I'll have you know I studied Physics at university 😀
I think the whole motor bike thing is taking us no where. Obviously no one would design a motor bike so that chain tension had a huge effect. It seems obvious a motor bike designer has lots of options to reduce or control the effect of chain tension. Its not obvious to me that the tension in the chain on a motor bike is more than on a push bike. A motor bike has greater power but this could be achieved through greater chain velocity not greater chain tension
read these, they may not settle your arguments but if you guys are genuinely interested in these things then they can be an enlightening read.
[url= http://raystrax.com/PathAnalysis/index.htm ]Path analysis 1[/url]
[url= http://www.rdrop.com/~/twest/mtb/pathAnalysis/ ]Path analysis 2[/url]
what always fascinates me is how hung up on axle path some people get when in reality all the major players are so close that it's not really worth even dsicussing. Even the tiny spend VPP paths are so close to a simple arc that it is of almost no relevance.
pivot placements, spring rates (both shock and system) and the rider have a far bigger impact thatn anything else.
Over the years I've used a lot of different designs, SP both linkage driven and directly actuated , 4bar, VPP, and various softails which blur the lines and I know what I prefer and it genrally depends on what kind of riding I do. I still own and ride a single pivot, and two different 4 bar designs on a regular basis but believe it or not the bike I get on best with is a 2001 schwinn straight 6 which is a lawill design trail bike. It has its quirks, it's a bit heavy and has a horribly tendancy to extend under heavy breaking but good use of compression damping at the front end and running a smaller than normal rear rotor tames that to the point where I can cope with it, and it's bump performance is probably the best of any bike I've ever owned and ridden.
what you all seem to be doing is focusing on the back end, dont forget that on trail performance is a harmonic blned of system, shock, setup and the front end, a poorly performing or set up front end can accentuate the good and the bad of any rear end, remember to treat the bike as a whole and not just think of the back and front as seperate.
above all suspension feel is such a personal thing I dont think you can ever say one type is 'better' or 'worse' they all have their ups and downs even the same sytem in different guises so just ride and see what you get on with best 🙂
[img][URL= http://inlinethumb36.webshots.com/6115/2136887990036898833S600x600Q85.jpg [/IMG][/URL][/img][img][URL= http://rides.webshots.com/photo/2006672680036898833Xezzvp ][IMG] http://inlinethumb28.webshots.com/3611/2006672680036898833S600x600Q85.jpg [/IMG][/URL][/img][img][URL= http://rides.webshots.com/photo/2629527740036898833uDdUZk ][IMG] http://inlinethumb29.webshots.com/21468/2629527740036898833S600x600Q85.jp g" target="_blank">
ok we seem to have opened a huge can of worms here,, lets ignore the motorbikes for a while but bimota used to put a lot of effort to making the swingingarm pivot concentric to the engine sprocket,,
sorry side tracked there,,
i seemed to have massivly oversimplified the whole thing here,, there are lots of forces at play and all combined ,,
there are three screendumps from linkage one with pivot obove chainring one below and one to give 0 anti squat,,
the anti squat graph on the left is the one we are looking at the blue graph give you the amount of antisquat a +ve number means it's not squating and a -ve number means it is squating
it auto scales so take note of the vertical scale
i was surprised how low the pivot position needed to be to get zero antisquat ,, so there is someother force acting here
when pivot is placed on the bb it gives about -200% anti squat
what is not taken into account is the change in wheel rate (spring) due to the change in lever length of me moving the pivot
it goes from 1.9 pivot at bb to 3.5 pivot obove the chain,,
in conclusion we are all right ,, the height of the bb does alter the antisquat but there are other forces involved and it is a combination of these which determins the reaction of the swingingarm to pedaling forces and these cannot be taken in isolation
so there is someother force acting here
Thrust force from wheel acting along line from hub to pivot point?
my mistake al what a true polymath you are
TJ still knows noting though can we at least agree on that 😉
1) - chain growth
2) - when the chain is not parallel with teh line between the wheel and BB centres ( viewed from teh side) You get squat or rise depending on the angle
colinear works for me TJ but not simply parallel. parallel [but offset] will cause movement of the rear axle. take moments about the swingarm pivot - there is an unresolved moment that must be resisted by the suspension.
"in conclusion we are all right"
I KNEW I WAS RIGHT 😀
parallel [but offset] will cause movement of the rear axle. take moments about the swingarm pivot - there is an unresolved moment that must be resisted by the suspension.
This is where I get stuck. The chain tension doesn't act on the swing arm it acts on the sprocket/wheel, which are attached to the swing arm but via an axle which can't transmit any torque. If the chain tension is parallel to the swing arm this pushes the axle straight towards the pivot. So the force of the axle on the swing arm has no moment about the pivot
I'm not saying I'm correct I'm just saying I'm confused
the freehub can transmit torque to the wheel only, you're right. that's how they work. pawls etc. but the wheel will then exert a force perpendicular to the swingarm if it's [the wheel's] rotation is resisted by something e.g. contact with the ground, or the brake being applied for example as this locks the cassette in rotational position relative to the swingarm.
Forget about parallel. it's not a helpful term. think force = magnitude and direction instead. there are no shortcuts to understanding this, except computer software. basically you have to get into forces, moments and free body diagrams for various components of the assembly, from the rider to the wheel, the swingarm etc etc. This needs A level physics as a minimum I would say, not to discourage anyone or put anyone down in the slightest.
but via an axle which can't transmit any torque
there is a torque otherwise there would be no tension in the chain, and the chain line is offset from the line linking the axle and pivot as the chain leaves the rear sprocket at a distance from the axle even if the pivot is on the point where the chain meets the front ring
I think part of the issue is inexact use of words to describe concepts and thus misunderstandings over who means what.
Teh other issue is its hard to explain without drawings
ok back at linkage software this time std marin single pivot
only change is 36 rear sprocket and change to 11 tooth rear sprocket
36 gives 90% anti squat
11 gives 42% anti squat
well I think its interesting
I think you need to explain, carefully, and slowly 😉 what "90% anti-squat" means in the context of the software.
you don't get any chain tension effects with a BB-concentric-single-pivot.
i don't care how many long words, diagrams, or *****ng screen grabs from linkage we can link to.
you just don't.
so there.
X
think you need to tighten up on your terms a little there, ho ho ho 😀 no pedal feedback from suspension movement perhaps...
I think you need to explain, carefully, and slowly what "90% anti-squat" means in the context of the software.
i think i have to admit I am unsure as there is no units
what I do know is a positive number means the the swinging arm is being forced open trying to lift the back of the bike
a negative number is the the opposite back end being compressed
so in my last example 11 to 36 tooth give roughly twice as much anti squat
who has a single pivot bike near by put bike in big ring and biggest rear put the front wheel up against wall balence on bike and load pedals see how much rear end rises now try the same on smallest rear see if there is any differenc dont use the brakes
just been on t'internet on the forum for linkage
a1 curve is suspension's anti-squat curve. Anti-squat is suspension's resistance to compression due to acceleration. Over 100% of anti-squat (AS) means suspension will extend under acceleration. With 100% AS suspension won't neither extend nor compress. Under 100% AS means tendency to compress under acceleration. It is determined based on actual linkage IC, chain force line and center of mass height. I would say that pedal-kickback effect is a byproduct of anti-squat characteristics of the suspension.
a2 curve depicts suspension reaction under rear-braking without front-braking. Again, value > 100% means suspension will extend under rear-braking, value < 100% means it'll compress under rear-braking. This curve depends on actual IC position and center of mass height.
this is realy good,
yes it is, thanks 🙂
This needs A level physics as a minimum I would say, not to discourage anyone or put anyone down in the slightest
Oh dear, which what I teach for a living. I think it hasn't been helpful to over simplify things in my case. I was imagining spinning the rear wheel with it off the ground to isoloate the effect of chain tension
there is a torque otherwise there would be no tension in the chain, and the chain line is offset from the line linking the axle and pivot as the chain leaves the rear sprocket at a distance from the axle even if the pivot is on the point where the chain meets the front ring
There is a torque applied to the sprocket/wheel. The 2 forces in the couple at the reaction at the axle and the tension the chain. These two forces roatate the wheel. But the the only 2 forces acting on the swing arm are a force at the axle and a force at the pivot. If the line of action of the reaction is through the pivot then it has no moment. So I believe, but may be wrong, that if the chain stay pivots at the BB and the chain is parrallel to the chain stay then spinning up the rear wheel with it off the ground will not change the force in the shock
you don't get any chain tension effects with a BB-concentric-single-pivot.
That is my current conclusion. I'd be greatful if you pointed me toa supoorting argument
At some point I'll try and do all the forces with the wheel on the ground
you don't get any chain tension effects with a BB-concentric-single-pivot.i don't care how many long words, diagrams, or *****ng screen grabs from linkage we can link to.
you just don't.
so there.
X
nice well reasoned point sir
ampthill - MemberThat is my current conclusion. I'd be grateful if you pointed me to a suporting argument
easy:
it's *****ng obvious!
you can quote me on that.
thankyou.
X
(you will get mass-of-rider's-body/legs-accelerating-up-and-down-effects)
I,m not very good at maths etc so i try to look at things diferently
imagine a piece of wood laid on the ground with a piece of string tied to one end,, the wood is hinged to the ground at the other end now pull the string in line with the wood towards the hinge ,, waht is the result the wood is in compression but it stays on the floor
now lift the end of the string up to about the radius of the front ring and pull the string and what will happen now ,,