JTT

~SPONSOR~
Aug 25, 2000
1,407
0
I am trying to get a handle on forces effecting rebound on the shock and how these forces create HSR, MSR, and LSR...

Here's where I am so far...Rebound is effected by forces created by spring, chain torque (anti-squat effect) and wheel enertia.

Now how does this relate to rebound speed? It would seem to me that rebound is therefore effected by the position of the shock (for example, greater forces applied when spring is compressed fully, thereby creating a HSR situation? and along that same thinking, less spring compression would be more LSR?) meaning that, at least indirectly, rebound damping is position sensitive? ...small bumps, LSR, big hits, HSR? ...this just doesn't sound right to me...what factor am I missing?

Please help enlighten me...my brain hurts
biggrin.gif


JTT

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Logic Over Hype Coalition
 

JTT

~SPONSOR~
Aug 25, 2000
1,407
0
...so...small bumps, even closely spaced (such as acceleration bumps) that only move the suspension a small amount would be LSR?? (lessor forces exerted by spring therefore lower piston speeds)

JTT

Thanks Jer...I see a faint light...but it's very cloudy.....

I tried to find that thread but no luck.
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Logic Over Hype Coalition

[This message has been edited by JTT (edited 02-03-2001).]
 

John Curea

LIFETIME SPONSOR
Feb 29, 2000
177
0
JTT
Do a search on the thread "HSC or LSC"

Also do a search on " High Speed Comp", this will give you a bunch of excellent reading.

There is no place like this forum in the universe, trust me, I've done some serious surfing...........!!!!
Take Care, John



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99 KX250
98 KTM50
88 LT250R
86 TRX70
 

JTT

~SPONSOR~
Aug 25, 2000
1,407
0
HI KXVET, I did read the threads you mentioned (most for second or third time
smile.gif
), but am still looking for specific answers to my questions (as in examples above)...I'm a little dense sometimes...but eager to learn
biggrin.gif


I don't want anyone to give away "trade secrets", just want to understand better.

Thanks,

JTT

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Logic Over Hype Coalition
 

Jeremy Wilkey

Owner, MX-Tech
Jan 28, 2000
1,453
0
YES!!

Chain torque actually lifts the bike.. Want a quick test.. Land throtle off and throtle on.. The Squat of a bike is about weight transfeer not chain torque.. Cool stuff...

Jer
 

will pattison

Sponsoring Member
Jul 24, 2000
439
0
throttle off, i bounce into the weeds. throttle on, i track true to the trophy.

i don't buy it. rick johnson (the suspension guy) started this theory- as far as i know- and it's stuck, but a free body diagram and application forces and moments at the right places just don't support the theory. i would agree that weight transfer will cause a greater degree of squat, but i have a tough time with the idea that chain torque makes the bike stand up.

the bottom line is that when the bike is under acceleration, there is negligible tension on the bottom "side" of the chain. if you locked your back wheel up and tied a string to the top of the rear sprocket, then pulled on it from about the location of the countershaft, what would happen?

now, just to be safe, i AM willing to be proven wrong, but i ain't giving it much odds.



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will pattison
engineer, racer
ignition
www.ignitioninc.com
 

spanky250

Mod Ban
Dec 10, 2000
1,490
1
At full extension, the chain does pull the rear wheel down when you are on the throttle. If you draw a straight line thru the center of the axle and the countershaft, this line falls below the center of the swingarm pivot. The chain naturally wants to pull the rear wheel towards the countershaft, but it can't because the swingarm holds the wheel in place, so the force is directed in an arc around the pivot. Since the line of force is closer below the pivot than above it, the wheel is forced down. Granted, the wheel is only at full extension when there is no weight on it, like after a bump or jump, so this rarely comes into play. The opposite is true under most conditions. The rear wheel is up in the suspension travel, bringing our imaginary line above the center of the countershaft, causing the chain to pull up on the wheel. Suspension and chassis designers have to take these forces into consideration in the design. The influence of these forces on rear suspension are the reason your valving and spring rates result in the rear suspension working better under power, it is designed this way because you hit most obstacles under power. This is also why manufacturers try to locate the pivot as close to the countershaft as possible, to minimize this force. Eyvind Boyeson developed a rear suspension system that was supposed to eliminate all chain forces on the rear wheel and tried to interest the manufacturers in production, but got no takers for whatever reason, probably the cost, weight, and complexity of his design.

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1992 KDX 250-FMF porting,two-stage power reeds, Fatty pipe, Power Core silencer,titanium rod,Wiseco Ultra-lite, Pro-Action suspension...Why didn't I take the blue pill???
 

semi old guy

Member
Dec 9, 1999
53
0
well here we go again and i love it.

i made a scale model of the rear end with card board and twine and thumb tacks used for pivot points. the chain tension wants to compress the suspension as you would think by looking at it.

there is another force however. a force vector acts on the rear wheel at the point where it contacts the ground. it is directed forward (the direction of acceleration). that force is trying to extend the swing arm as long as the swing arm pivot point is above the axle. i think that this is the greater force and causes the rear end to stiffen under acceleration
 

semi old guy

Member
Dec 9, 1999
53
0
suppose you could ride your bike in reverse and while doing so you applied the rear brake. would the back end rise or fall? look at the bmw paralever.
 

spanky250

Mod Ban
Dec 10, 2000
1,490
1
SOG, I thing you are wrong about the tire producing greater force on the rear wheel than the drive chain. The tire is riding in dirt, a surface with decent traction
at best. The chain is pulling on the sprocket teeth, a definately non-slip surface. It is applying a force of thousands of pounds per square inch, and this force is transmitted to the bike through the axle, not around the circumference of the sprocket. The BMW Paralever is designed to counteract a different problem (although the problem nets the same effect), that being the gear on the end of the drive shaft trying to climb up the driven gear in the hub, which forces the rear weel down. If you have ever ridden a drive-shaft equipped bike, then I am sure you have experienced this "jacking up" of the rear end under power.

Any engineers in here that can shed some light on this?
 

Casper250

Motosapien
Dec 12, 2000
579
1
Let me take a crack at this. Some one before mentioned about the center of the rear axel being above the pivot point on the frame most of the time but when this is happening, the top of the rear sprocket is also above this pivot point. When this is true, the chain is at a diagonal sloping down from the top of the rear sprocket to the top of the motor sprocket. If you concider this as the angle of the force(ie the chain) and that this force is a vector, since it has a value and a direction, then it can be broken down into it's vertical and horizontal components. The horizontal component is a force in the direction from the rear axel to the motor. The vertical component is then from the top of the sprocket to the horizontal plane of the motor sprocket in the downward direction and is therefore pushing the wheel down.


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faster till the fear of death is outweighed by the thrill of speed...

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Hick

Member
Aug 15, 2000
224
0
Makes sense.

If you draw a perp. line through the axle and countershaft the motor will pull the axle up or down until the angle between our two lines and chain is equal.

So power on will usually mean pushing arm down via chain (unless you have an Amp link).

But I flunked Physics twice, so I should’ve probably left this one alone.
 

semi old guy

Member
Dec 9, 1999
53
0
i propose an experiment (i haven't tried this).

sit on your bike and apply the rear brake and slowly let out the clutch while giving it a little gas. i think the rear will squat.

now put your front wheel against a wall or something and do the same thing but don't apply the rear brake. you may have to take some weight off the bike with your legs in order to let the swing arm pivot rise above the rear axle. i'm guessing the rear will rise.

i'm going to try this tonight.

btw don't drink beer before attempting this
 

spanky250

Mod Ban
Dec 10, 2000
1,490
1
I do not have a degree in engineering, but it seems to me that the force of the chain is transmitted to the swingarm at the axle, not at the top of the sprocket. The chain does not pull only at the top of the sprocket, it applies pressure to all the teeth it is engaged in most of the way around the sprocket.

I see one flaw in your experiment, SOG. When you are sitting on the bike, the rear suspension will be compressed enough to raise our imaginary line from the axle center to the countershaft center above the center of the pivot, allowing the torque to pull the rear wheel up. Try this with no weight on the bike so the rear suspension is extended, bringing this line below the center of the swingarm pivot, and I will bet you the rear will jack up under power.

[This message has been edited by spanky250 (edited 02-06-2001).]
 

semi old guy

Member
Dec 9, 1999
53
0
i think thats what i suggested in the second part of the experiment. in the first part apply the rear brake and i bet the suspension compresses no matter if you're sitting on it or not
 

semi old guy

Member
Dec 9, 1999
53
0
ok just tried it

apply the brake and it squats

put the wheel against a wall, don't apply the brake and it rises

so what's it mean?

applying the brake gives zero force at the wheel/ground interface. in other words the wheel is not trying to accelerate the bike forward. the remaining force is the chain tension compressing suspension.

putting the front wheel against the wall and letting out the clutch while not applying the brake...the force at the ground/wheel interface exists and is stronger than the chain tension force thereby causing the suspension to extend.
 

spanky250

Mod Ban
Dec 10, 2000
1,490
1
OK, I guess I can't argue with that. Perfect example of the scientific method at work: come up with a problem, theorize an answer, test the answer. Good job SOG, I bow down before thee.
smile.gif
 

Casper250

Motosapien
Dec 12, 2000
579
1
"As the rear wheel spins it drives into the ground, the resultant force travels up the swingarm and into the frame, causing the front end to lift into a wheelie" from eric gorr's book. In your experiment, the force of the wheel didn't make you wheelie but it did push the frame up and therefore extend the rear suspension.



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faster till the fear of death is outweighed by the thrill of speed...

http://members.tripod.com/casper_12300/
 

mxneagle

Member
Jan 7, 2001
320
0
Eric Gorr did an article on this with all of the theory behind it in his early publications of MX Tech (me thinks thats that name) before MXA and the rest ran him out of town by threating TUFF. Anyhoo, the chain torque does tend compress the suspension. I saw a perfect example at Budds Creek in the mud when Emig cleared an uphill step with no lip by shutting off and then blipping the throttle on before take off. He was totally off the gas when he left the ramp but you could actually see the chain torque compress the shock and then the rebound (once he let off) throw him up to get the required lift.
 

will pattison

Sponsoring Member
Jul 24, 2000
439
0
well...alrighty then. this is what i call a stimulating discussion. it seems as if the question has been answered by a good experiment. one of my favorite engineering maxims is this:

"one test is better than 1000 expert opinions."

however, let me shed some more light, since, as spanky250 asked, i are an engunieer. masters degree in mechanical engineering, for whatever credibility that's worth. (i was actually a terrible student, i just hung around a few more years so i could keep drinking beer!)

spanky is partially right in saying that the force is applied to the swingarm at the axle. however, it has to get there via the sprocket and hub. at any given moment, it's totally valid to consider the combination of swingarm and wheel as a rigid body- an "l" shaped one that is attached to the frame via the chain.

it is also true that in analyzing the problem, one should break down the force vector extending from the top of the "l" to the frame (countershaft sprocket) into it's horizontal and vertical components. further, it's true that- at least when the swingarm is either centered with respect to the countershaft or above it- that the vertical component points at the ground. however, since we cannot consider the joint between the swingarm and the frame to be rigid, and the horizontal component pointing toward the frame is much larger than the downward vertical one, the swingarm has to rotate counterclockwise.

in other words, where you ultimately end up with all these force vectors is at a counterclockwise moment about the swingarm pivot.

to be fair, though, i don't know that chain torque is the most important factor in squat. weight shift certainly cannot be denied. i also think that, especially in the jump landing case, that what it does more than create squat is to counteract rebound, thereby creating a stiffening "effect".

jer?

wp.
 

will pattison

Sponsoring Member
Jul 24, 2000
439
0
i went back through and reread all these posts, and i picked up on something else: the question of the wheelie. i think it's appropriate to discuss it here, because the same phenomenon that causes that bit of fun also contributes to squat.

i've heard people say that weight shift is what causes wheelies. hogwash. while it certainly helps us wheelie on a motorcycle, it's not the actual physics that gets the wheelie started. i'm a farm boy, and i've spent many an hour on the back of a john deere 4010. i can tell you with no uncertainly that it's possible to pop a wheelie on one. it's all about torque.

the torque generated when we rack the throttle open generates angular acceleration...uh, that's acceleration about an axis. in this case, that is your rear axle. think about it this way. if you could skewer your bike about 10' off the ground with a rod welded to the sprocket, then some how get on it, put it in gear and go...well, you would- around in a clockwise circle.

when you ride like the rest of us (on the ground), and that angular acceleration takes place at a higher rate than your forward acceleration, what happens? boom! the bike tries to flip over backwards, your weight naturally shifts to the rear and then you are showing off for the girlies with a sweet wheelie...

fun thread for a geek like me.

wp.
 

semi old guy

Member
Dec 9, 1999
53
0
hey i'm a geek too

there's another force thing going on that causes a wheelie besides angular acceleration.

imagine you had a rope tied to rear wheel at the contact patch stretching out in front of the bike under the front wheel and you jerked it real hard. if you could pull it hard enough you could flip the bike over backward right.

it's because the center of mass is above contact patch
 

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