Understanding....rebound

(address future rebound questions to Dennis Rodman)

i thought of a better example.

imagine you have no wheels at all but instead a rocket motor at the rear axle. you're sittin on the thing (on some launch rail ala wile e coyote) and you lite it off.

you will look like a bottle rocket without the stick.

you wheelie for the same reason. the propulsion force/direction is not lined up with the center of mass

Hi:

I was reading this tread and thought that could jump in to share some info.
About shaft speed a conventional hydraulic shock absorber is not position sensitive but speed sensitive (a non linear suspension leverage system is position senstive).
No matter how deep the shaft is in its travel it will give the same amount of absorption (sorry it is not the term), but in conjunction with the spring and talkin about rebound the system becomes more or less position sensitive as spring rate becomes greater when is compressed (it will give faster rebounding speed when highly compressed because it has more force stored in its coils).
As the problem of squat arised i would like to add something.
Squat is a product of weight transfer, we have to see where is the force of thrust applied. In the case of a motorcycle, the force is transmitted to the chassis throug the swingarm axle. Now geometry, the force will became a couple at the center of gravity, so first draw a line representing the swingarm and suppose that the cg is located (with rider on) about where the carburettor is (i have never checked this but seems possible) draw a line perpendicular to the ground through the cg and you will find that the intersection point between the two lines is well below cg thus creating a couple that pushes the back of the bike down. "anti" squat would put this intersection over cg but this would make the rear end to lock extended while on the throttle. This is vehicle dynamics theory and practice. I will recomend Paul Van Walkenburg book or my favourite from Carrol Smith to clarify this (and the best of this book is that Mr Smith wrote it for the "normal" guy willing to learn).
If you apply the throttle with the bike on a stand you will see the end of the bike coming down, this due to rotational inertias and chain "torque?".
The answer to the "stiffening" suspension seems more related to the forces that applied to the ground by the rear wheel contact patch tries to make the bike go on a wheelie, that is as stated pure torque. I also was a farm boy and we used to wheelie our JD's (small vineyard models).
I have made some sketches of this to share with you but don't know where or how to put them on the post. Maybe someone can help me on this.
I seem to be always late on the posts, but hope it helps. If i didn't make myself clear let me know.
Please excuse my english.

Pedro

This seem to be getting to technical..

I think of it like this

Compression is how quickly the tire moves to the fender.
High speed being the speed of the shock and not the bike.Example breaking bumpe and acc bumps.

low speed is just what it is.example landing a jump.


Rebound is how quickly the tire returns to the ride hight.

I believe that is why springs can give you the most adjustment off the bat for the least amount of money.

You must also make small adjustments and not to many al one time.Make a change ride a little and come back and also keep a log on what you change so you can see what you have done what helped and what did not....

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it dont hurt till the bone shows.

2001 YZ 250

at the risk of sounding like i know what i'm talking about:

the change of the rpm (acceleration) of the spinning wheel will cause a little rotation about the axle forward or backward depending if the wheel is accelerating or decelerating.

the extension force doesn't exist while the bike is in the air. it requires the ground to resist the accelerating wheel. the instant the rear wheel hits the ground after the jump the rear wheel tries to "take off" with the axle and swing arm while leaving the rest of the bike behind...thereby extending the suspension.... because the swing arm pivot is located above the axle.

semi old guy,

Here is another experiment for you on the rear torque and squat question.

Rather than blocking your front wheel and slipping the clutch to look for the rise in the rear, tie or block the bike at the handlebars or triple clamps to simulate a higher C.G. and see if the rear doesn't go down instead of up.

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Something else to keep in mind when landing throttle on is the rotational inertia of a fast spinning wheel hitting the dirt will push the bike forward and up without ANY chain at all!!. Something like the rocket analogy, this is part of the reason the bike doesn't sink or squat as much. The upward slope on the swingarm will try to lift. It is similar to the physics of a drive shaft (but less pronounced) on the initial impact.

James

[This message has been edited by James Dean (edited 02-09-2001).]

Pedro MX
You make reference to a book by Carroll Smith. Which book are you talking about because he has several out in printing.
Thx, John

hey guys...sorry i had to leave the thread for a few days. business trip, races this weekend, blah, blah, blah.

anyhow, to answer the question about why the rear end "stiffens" with more throttle on a jump landing: i think part of this is percieved rather than actual. since landing with the throttle on keeps us from rebounding like pogo sticks, we feel that the package is stiffer because it's more controlled. i also agree that rear wheel inertia plays a role, but the truth is, that with all the thought-experiments i've been doing since we started this thread, i can't come up with a coherent theory that would explain an actual increased resistance to compression with the throttle on. there just don't seem to be any physics that would explain it.

what we do have, however, is physics that indicate that chain torque does cause at least some squat.

maybe we can figure out together what the real answer is regarding landing witht the juice on...

wp.

i was thinking about this last week, and i think you may be on to something. it certainly seems logical that the fixed length of chain available between the countershaft and rear sprocket at the moment of impact would impede or even completely override the rear wheel's tendency to move upward. so, what it appears we should say is that while chain torque always TRIES to compress the rear suspension, it can only actually do that after the swingarm has moved past the midpoint.

i'm comfortable with that.

wp.

Hi:

KXVET, if you can get them all, do it! I may sound like Mr. Smith's publisher but it is worth it.
Specificly the books that I mentioned are Tune To Win and Engineer To Win, they are about racing cars (real ones) but a good bit of information can be translated to bikes (chapters about suspension geometry, shocks, spring, etc.)
The other books about preparation are pure gold. You may think that they are slightly dated, but they are not for all practical purposes.

I hope you find them as enlightening as I did when I fist read them.
Good luck.

Pedro.

If you really want to understand chain force and its contribution to anti-squat read "THE RACING MOTORCYCLE: A CONSTRUCTOR'S GUIDE" by JOHN BRADLEY. Although aimed at roadrace technology, this book has the best description of the physics involved in motorcycle design that I've ever seen. Bradley explains how many GP "works" roadracers have adjustable swingarm pivot heights--to allow the tuning of anti-squat characteristics for different track conditions. Bradley even explains how the "average Joe" can alter/tune his anti-squat forces by simply changing the rear sprocket size (while also changing the c/s sprocket to retain the same final drive ratio.) Bottom-line: Will Patterson and Semi-old guy are on the right path. Its the combination of 1) Chain pull on the rear sprocket and 2) Rear wheel drive (traction) forces transmitted from contact patch to the swingarm pivot that produce the resultant anti-squat vector for the overall rear suspension system. As has been previously mentioned, the resultant anti-squat vector will continuously change as the suspension moves through its stroke. To learn more get Bradley's book. It costs $70 but it you enjoy learning about motorcycle engineering this book is without peer and well worth the money.

Forget to mention how to buy the BRADLEY book. Go to www.eurospares.com, see "Books."
Enjoy!

VZXN, GOOD TIP!!

THANKS!!

I was watching a semi take off from a stop sign today. Whenever he let out the clutch and gave it power the suspension became full extended. It would actually lift up the whole cab and trailer about 3 inches.

I am not sure what this means, but I thought I might add this.

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motor86
85' KDX200(mine-for sale) 96' CR250R(mine)  81' YZ125(Dad's)  81' CM400 (my street bike)
www.geocities.com/motor86_cr250/home.html

i just got mcn in the mail yesterday and there's all the info we have been hypothisising about in an article by tony foale. i 'm not saying it's perfectly accurate but it looks right to me. jtt...to answer your sprocket question.. the larger sprokets would inrease the squat forces thereby decreasing the antisquat. the suspension should by softer under acceleration with the larger sprokets

motorcycle consumer news....I think

SOG is right about this, but the book "The Racing motorcycle" has an overly complicated explanation. To find out what will happen to the rear suspension under acceleration, all you have to do is add up the forces that are trying to rotate the swingarm about its pivot point. In this case we have the chain force and the tire contact force. You find the perpendicular distance of the force from the pivot point and multiply it by the force to get the torque that will try to rotate the swingarm. For the chain, you just measure the distance from the center of the chain to the center of the pivot point. For the tire contact force ,on level ground, it is the distance from the ground to the center of the pivot point. We can't easily get the forces, but we can get the ratio of them, and that is the ratio of the sprocket diameter to the tire diameter.
As an example, on my bike, the tire is 26" and the sprocket is 10" so the chain has 2.6X as much force as the tire contact patch.
The chain is 2" from the pivot point, and the pivot point is 16" from the ground, so the tire contact force has 8 times as much leverage as the chain.
You divide 8 by 2.6 to find that the tire contact force has 3 times as much effect as the chain force, so under acceleration the rear wheel will try to go down, which will make the rearend go up. This is one of the reasons you should get on the gas when landing from a jump, it will effectively stiffen the rear suspension.
If the rear tire were trying to go over a basketball size rock, the tire contact force would be adding to the chain force in trying to raise the rear tire up and move the rear end down.

ET,

Your explanation of calculating anti-squat forces is interesting and easy to visualize. However, its premise that the wheel drive force is applied to the swingarm at a point 16" below the swingarm pivot is valid only if the wheel is not free to rotate around the axle. As long as the wheel is free to rotate around the axle, the swingarm can never be rotated by simply by applying leverage factors to either the wheel or sprocket--no matter how much force is used. However, the sum of the rotational forces applied to the wheel/sprocket will result in the wheel trying to move the axle in a particular direction (vector). In other words, all wheel forces are transmitted to the swingarm at only one point--the axle centerline--not from the perimeter of the wheel or sprocket. The trick is to understand the force vector's direction in relation to the swingarm pivot. If the vector passes through the swingarm pivot, there will be no anti-squat; if it passes below the swingarm pivot this (the distance between the vector and the swingarm pivot) will provide the wheel axle a lever arm with which it can rotate the swingarm downward (anti-squat) in respect to the frame. If the vector is above the pivot you have pro-squat.

This explains why SOG saw such dramatic differences in his bike's reaction to feeding power with the rear brake locked. In this situation the wheel is no longer free to rotate and the swingarm/wheel become a rigid member. This causes the chain to apply rotational force to the entire swingarm through the lever provided by the top edge of the sprocket--similar to your description. In this case, there is no drive force at the rear wheel and chain pull developed by the engine is applied to the rigid swingarm/wheel through the lever provided by the rear sprocket. This will always pull the swingarm upward--compressing the suspension--as long as the top chain run passes above the swingarm pivot.

However, as SOG observed, when the wheel is free to rotate separately from the swingarm everything changes. In this case, you basically have a lever between the chain and the ground, with the rear axle as the fulcrum. The chain is where the work is applied and the ground is the load. As forces flow through this lever, driving the motorcycle forward, the lever pushes against the fulcrum (axle) attempting move it in a particular direction. However, these forces can never apply torque to rotate the fulcrum/axle--they can only impart a force vector against it. In SOG's case, this vector was passing below the swingarm pivot and was thus imparting a rotational torque around the swingarm pivot that extended the suspension.

We introduce errors if we fail to view wheel forces as transmitted through the axle and instead think of forces coming from points around the wheel. Take a current paralever-equipped BMW street bike for example. This bike has no chain pull (its shaft-drive) and has a parallelogram linkage connecting the rear drive hub to the frame, which effectively cancels all driveshaft induced torque reactions that are fed into the frame. Using your example this bike's rear contact patch would have a huge lever (the distance from the contact patch to the swingarm pivot) with no counter lever provided by chain pull force. If this were true the Beemer would have a huge amount of anti-squat, when in fact it does not. This is because--like all vehicles--the beemer feeds its wheel drive forces into the rear suspension linkage from the rear axle centerline, not the wheel perimeter.

ET, thanks for your ideas, this is a fun discussion!