Understanding....rebound

JTT

Subscriber
Joined
Aug 25, 2000
Messages
1,407
Likes
0
#1
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


JTT

------------------
Logic Over Hype Coalition
 

JTT

Subscriber
Joined
Aug 25, 2000
Messages
1,407
Likes
0
#3
...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.
------------------
Logic Over Hype Coalition

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

John Curea

Lifetime Sponsor
Joined
Feb 29, 2000
Messages
177
Likes
0
#4
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



------------------
99 KX250
98 KTM50
88 LT250R
86 TRX70
 

JTT

Subscriber
Joined
Aug 25, 2000
Messages
1,407
Likes
0
#5
HI KXVET, I did read the threads you mentioned (most for second or third time
), but am still looking for specific answers to my questions (as in examples above)...I'm a little dense sometimes...but eager to learn


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

Thanks,

JTT

------------------
Logic Over Hype Coalition
 

will pattison

Sponsoring Member
Joined
Jul 24, 2000
Messages
439
Likes
0
#6
did you say "ANTI"-squat with regard to chain torque??



------------------
will pattison
engineer, racer
ignition
www.ignitioninc.com
 

Jeremy Wilkey

Owner, MX-Tech
Joined
Jan 28, 2000
Messages
1,453
Likes
0
#7
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
Joined
Jul 24, 2000
Messages
439
Likes
0
#8
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.



------------------
will pattison
engineer, racer
ignition
www.ignitioninc.com
 
Joined
Dec 10, 2000
Messages
1,490
Likes
1
#9
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.

------------------
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???
 
Joined
Dec 9, 1999
Messages
53
Likes
0
#10
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
 
Joined
Dec 9, 1999
Messages
53
Likes
0
#11
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.
 
Joined
Dec 10, 2000
Messages
1,490
Likes
1
#13
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
Joined
Dec 12, 2000
Messages
579
Likes
0
#14
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.


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

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

Hick

Member
Joined
Aug 15, 2000
Messages
224
Likes
0
#15
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.