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  1. #1
    Freak of Nature
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    Problems with Full Suspension...

    I've heard stuff mentioned about problems with FS bikes, something to the effect of "chain slack" and/or braking lock? I'm not sure what people were talking about when they mentioned these things? Can someone fill me in on the downsides of certain FS designs?
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  2. #2
    Telecommunication Tweek's Avatar
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    Ehh, I've never heard of chain slack or brake locking problems with full spension bikes.

  3. #3
    Freak of Nature
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    Quote Originally Posted by Tweek
    Ehh, I've never heard of chain slack or brake locking problems with full spension bikes.
    I think it may have been single pivot designs (not quite sure exactly what that is). I'm not exactly sure what it is exactly, but I've seen things like floating brake controls or whatnot to combat problems. That help anybody?
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  4. #4
    Giggity giggity! Dirtbike's Avatar
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    Single pivot bikes suffer from brake jack, and depenending on pivot location, pedaling forces can jack(raise), or squat(lower) the suspension under load.

    There are ways to get around these problems (floating brake arm, chain torque eliminator), but it still cant make up for the fact that its still single pivot.

    Suspension designs such as FSR have little or no brake jack, due to the way the "seat stay" link moves throughout the travel. It acts like a floating brake. It is also very neutral from pedaling forces too.
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  5. #5
    Freak of Nature
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    Quote Originally Posted by Dirtbike
    Single pivot bikes suffer from brake jack, and depenending on pivot location, pedaling forces can jack(raise), or squat(lower) the suspension under load.

    There are ways to get around these problems (floating brake arm, chain torque eliminator), but it still cant make up for the fact that its still single pivot.

    Suspension designs such as FSR have little or no brake jack, due to the way the "seat stay" link moves throughout the travel. It acts like a floating brake. It is also very neutral from pedaling forces too.
    why do braking forces jack the suspension? i'm having a hard time understanding the physics of it.
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  6. #6
    The geek on two wheels.
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    Quote Originally Posted by SquatchCO
    why do braking forces jack the suspension? i'm having a hard time understanding the physics of it.
    When you hit the brakes, the momentum keeps your bike going forward. Since your center of gravity is higher than the contact point with the ground, you'll be forced forwards. This puts more weight towards the front and takes it off the back, causing the rear shock to decompress.

  7. #7
    Giggity giggity! Dirtbike's Avatar
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    Its something that is hard to understand at first, thats hard to explain in typing, and is hard to explain without getting too in depth. I really wish there was a Flash movie explaining it or something.

    Here are bikes with a Floating brake arm installed. Note that the caliper is not mounted on the swingarm or suspension, but rather the brake arm.
    http://braketherapy.com/floatmodel.htm

    Brake jack or chatter is when rear suspension acts differently when you apply the rear brake. The suspension will skip on stutter bumps when under braking instead of tracking over them. On a bike with a floating brake, the brake caliper's position relative to the ground is not influenced by the position in the travel the rear suspension is at. The caliper's position stays the same from the beginning to the end of the travel. The braking forces remain seperate from the rear suspension... it goes to the frame. When the caliper is attached to the swingarm on a single pivot, (no floating brake) the forces go to the swingarm, making the swingarm want to turn with the wheel.

    Hopefully that gives you a vague idea of brake jack and brake forces. I need to get some sleep.
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  8. #8
    Freak of Nature
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    Quote Originally Posted by edf825
    When you hit the brakes, the momentum keeps your bike going forward. Since your center of gravity is higher than the contact point with the ground, you'll be forced forwards. This puts more weight towards the front and takes it off the back, causing the rear shock to decompress.
    Seems like this is a problem that could be rectified by shifting your body position backward a bit (something that seems like a good idea regardless of brake jack). Does it really affect braking power a lot?
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  9. #9
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    Quote Originally Posted by Dirtbike
    Its something that is hard to understand at first, thats hard to explain in typing, and is hard to explain without getting too in depth. I really wish there was a Flash movie explaining it or something.

    Here are bikes with a Floating brake arm installed. Note that the caliper is not mounted on the swingarm or suspension, but rather the brake arm.
    http://braketherapy.com/floatmodel.htm

    Brake jack or chatter is when rear suspension acts differently when you apply the rear brake. The suspension will skip on stutter bumps when under braking instead of tracking over them. On a bike with a floating brake, the brake caliper's position relative to the ground is not influenced by the position in the travel the rear suspension is at. The caliper's position stays the same from the beginning to the end of the travel. The braking forces remain seperate from the rear suspension... it goes to the frame. When the caliper is attached to the swingarm on a single pivot, (no floating brake) the forces go to the swingarm, making the swingarm want to turn with the wheel.

    Hopefully that gives you a vague idea of brake jack and brake forces. I need to get some sleep.
    I checked out the link and read your explanation. They seem to say different (and exclusive) things. (Actually the link didn't really say much at all.) It seems like the "brake caliper's position relative to the ground" would change a lot, even with a floating brake setup.

    Anyways, I haven't seen an explanation that passes the smell test.
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  10. #10
    Throw the stick!!!! LowCel's Avatar
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    The truth is you will never completely understand or believe it until you ride a few different full suspension designs. I rode a Santa Cruz Superlight for years and never even knew that it had brake jack until I started riding my Ellsworth Truth. Now that I am riding a bike that doesn't suffer brake jack I can definately tell the difference. It is amazing.
    I may be fat but I'm slow enough to make up for it.

  11. #11
    Ride bike or bike ride? Hopper's Avatar
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    Ok let us begin, chain suck is a problem where the chain grows through it's travel, this causes tension and it can be pulled, this forces pedal feed back through the cranks, this can be imprved by having a pivot design that pivots around the BB.

    Now brake jack is a much more complicated thing. I am going to copy a post from another website, it is very VERY long so bare with me. It was written by uni student and is a 'basic' model and also discusses the pro's and cons of designing a bike so that it squats or jacks under braking.

    Warning: very, very long posts.

    Since there’s been a lot of incorrect explanations and theories on brake induced suspension interference (BISI), here’s a full explanation with diagrams. If you’re just interested in what happens but not so much “why” then skip to the end, most of this is just a reiteration of commonly-understood physics and its application to bicycles in relation to braking.

    Disclaimer: this article is purely intended for demonstration of the basic concepts of BISI. It is not a full quantitative method of calculation, nor should it be used as such. Calculations shown deliberately omit various less-significant/more complex factors such as moments of inertia, weight shift, centres of mass etc. The article remains a fair approximation of the basics of BISI, but if you wish to calculate the precise reaction to braking that your bicycle has then you must be aware of the other factors, and know how to incorporate them. Keep in mind too, that motion characteristic generation (ie designing for a characteristic) is much more difficult than simply analysing an existing system.

    The physics:
    Imagine you have a random rigid structure/body attached to another, fixed rigid structure/body by means of a simple pivot, so that when the fixed (stationary) body is taken as a point of reference, the other one is free to rotate around the pivot that joins them. (Fig 1)



    Now imagine if you had the same thing, but you applied a force at any random point on the rotating body. The force can be broken up into its vector components relative to the pivot, which are called the normal (directly towards/away from the pivot) and tangential (perpendicular to the normal) components, as shown. (Fig 2)



    The concept of a moment is fairly basic – it is the product of a force and its perpendicular distance from another reaction force (which prevents the body from translating, ie moving in a straight line) at a fixed point (in this case the pivot), and measures their coupled tendency to cause a body to rotate. In other words, the tangential component of force F in figure 2 (since it is perpendicular to the normal force at point A), multiplied by the distance between point A and point B (the pivot), gives the moment acting on the rotating body, about point B. The tangential component of force is given by F x cos(q). So the moment about point B is Fcos(q) x dist(AB). Not too complicated. This is pretty easy to visualise – imagine lifting your front wheel off the ground (brakes not being applied) and pushing against the tyre with your hand. If you push directly in at the axle (perpendicular to the tyre/rim at that point; this is a purely “normal” force) then the wheel will not spin, and you will cause an equal and opposite reaction force on the wheel from the axle. However, if you push parallel to the tyre’s surface at that point (tangent to its path) then you will cause the wheel to spin. Anywhere in between will give a combination of the two – pretty obvious.

    Newton’s second law of motion: SF = m x Sa
    (the “S” means “sum of”, or “net”). That is to say, if there is a net force (that is not zero), there will be a net acceleration (in the same direction as the force is acting, obviously), directly proportional to the force applied and inversely proportional to the mass (due to the mass’s inertia resisting movement). The same applies to moments – the sum of all moments is equal to the body’s mass moment of inertia (the rotational equivalent of normal inertia) multiplied by the angular (rotational) acceleration. For the moment (no pun intended) you only need to understand the basic concept of this, not the technicalities (which are fairly complex).

    Newton’s third law: For every action (force) there is an equal and opposite reaction (force). This gives us the concept of static equilibrium – if a body is supported in such a way that it cannot move (relative to your reference frame) then any force acting on it will generate reaction forces at its supports. Depending on the supports holding a body, it will be considered either statically determinate or statically indeterminate. If the body is statically determinate you can work out reaction forces based on basic equilibrium equations (forces acting in the X direction, forces in the Y direction, and moments about any given point), in other words it’s fairly simple to deal with. If the body is statically indeterminate, you will burn it at the stake and write several books specifically concentrating on denying that it ever existed (it’s much easier than doing the calculations). Fortunately, bicycle suspension components are (in the plane which they are intended to move, ie discounting lateral flex etc) nearly always statically determinate (and those which technically are not, can be approximated as being determinate anyway). If this sounds complicated, that’s because you’re not reading my mind well enough.

    First we will look at a basic bicycle (single pivot) swingarm, (initially) having three points of attachment to the outside world. These points are point A (the axle), point B (the pivot), and point C (the shock mount). Note specifically that all three points are pivots and are free to rotate relative to the bodies to which they are attached (the front triangle, the shock, and the wheel). If you’re wondering why the axle is a pivot, it’s because the rear wheel is attached via bearings, and cannot exert a moment on the axle itself (because if you spin the wheel, it will simply rotate freely around the axle – leaving braking forces and chain interference out of the picture, and assuming that there is negligible friction in the wheel bearings). So the wheel and the swingarm can rotate independently of each other at this time. Assume the bicycle is on flat ground.

    At sag, when the suspension is not cycling (moving), the bike’s swingarm is in a state of static equilibrium (all forces and moments sum to zero). See fig 3 – blue lines are forces acting at the various points, pink lines and letters are to show the notation used for the distances from C to B (vertically), and A to B (horizontally). Note that the force at A (the axle) can ONLY be vertical because any non-vertical (ie tangential) force at the tyre will simply rotate the wheel around the axle, transferring no force to the axle other than the vertical (normal) component. Also note that the force at C is only horizontal (axial to the shock) as shocks can only transfer load ALONG their axis (quite obviously – any force on the shock that is not along its axis will simply rotate it about its other mounting point). For ease of calculation, we will neglect the shock’s rotation relative to the front triangle, and assume it stays horizontal at all times.


    Ok, so now into some basic calculations. Assume that A(y) is a given force on the axle. Since the swingarm is in equilibrium, moments about ANY point on the swingarm must sum to zero. For ease of calculation we will start with point B. Positive moments are taken to be clockwise:
    SM(B) = 0
    = [A(y) * -AB(x)] + [C(x) * CB(y)] Note that AB(x) has a negative sign in front of it
    [A(y) * -AB(x)] + [C(x) * CB(y)] = 0
    [A(y) * AB(x)] = [C(x) * CB(y)]

    If we let AB(x) = 0.45 metres (realistic measurement), A(y) = 500N (again not unrealistic, this is equivalent to about 50kg), and CB(y) = 0.15 metres (we will assume that this distance does not change significantly over the stroke of the suspension, for ease of calculation), then we get:
    [500 * 0.45] = [C(x) * 0.15] Divide both sides by 0.15 to calculate C(x)
    [500 * 0.45]/0.15 = C(x)
    C(x) = 1500N (this is roughly equivalent to 330lb – if you had a 330lb/inch spring you would get 1 inch shock stroke sag, which given the 3:1 shock leverage ratio would be 3” of sag at the axle – again, a realistic example).

    So now we have calculated C(x) using basic equilibrium formulae. The only unknowns left are B(x) and B(y). Note that in Fig. 3 these forces are drawn in the opposite directions to C(x) and A(y) respectively. To calculate these we have two choices – balancing moments about either A or C, or balancing forces in the X and Y directions. Balancing the forces is easier since in this example all the forces are either horizontal or vertical, and we have no measurements for some of the distances necessary to sum moments about A or C.

    SF(x) = 0
    = C(x) – B(x) Minus B(x) because it is acting in the negative direction
    \ C(x) = B(x)
    B(x) = C(x) = 1500N

    SF(y) = 0
    = A(y) – B(y)
    \ A(y) = B(y) = 500N

    Hence we now have the reaction forces (Fig. 4)



    Now consider a free body diagram of a wheel with a disc brake rotor attached, under braking forces. Assume the caliper is directly above the axle for ease of calculation. (Fig. 5)



    Assume that the bike, despite braking, is maintaining a constant speed (as though it’s rolling down a hill with the brakes applied to a level where constant speed is held). Due to the wheel’s axisymmetric structure (it is identical all the way around, ie everything is concentric about the axle – brake rotor, rim/tyre etc), if we are to neglect the wheel’s own rotational momentum (because relative to the other forces/momentums at play, it can be considered insignificant) then the wheel can be modeled as though it is in static equilibrium (since we’re not worried about loads internal to the wheel, only the external forces and reactions) – all (external) forces and moments sum to zero. Assume the radius of the whole wheel (to the outside of the tyre) is 0.33m (13”) and the radius of the disc rotor is 100mm (4”). Let us choose an arbitrary force for T(x), which we will say is 200N. Take positive rotation to be clockwise.
    Thus:
    SM(A) = 0 summing moments about the axle
    = [T(x) * -0.33] + [D(x) * 0.1]
    = [200 * -0.33] + [D(x) * 0.1]
    200 * 0.33 = D(x) * 0.1
    \ D(x) = 660N

    SF(x) = 0
    = D(x) + T(x) – A(x)
    = 660 + 200 – A(x)
    \ A(x) = 860N (pointing to the left in the diagram)

    SF(y) = 0
    = T(y) – A(y)
    \ A(y) = T(y)
    If we take axle force to be 500N once again, then
    A(y) = T(y) = 500N


    Now we will look at the reaction forces acting on the swingarm. Note that the forces are identical in magnitude at all points of attachment/contact with the wheel (that is, the axle and the caliper) but act on the swingarm in the opposite direction to the directions they act on the wheel (Newton’s action/reaction pairs). (Fig. 6)



    Note that the force acting horizontally on the axle (with regards to the frame, not the wheel) is the sum of the frictional force on the tyre, and the frictional force on the brake rotor, and that it is always higher in magnitude than the braking frictional force between the caliper/rotor – if it wasn’t, braking would actually accelerate the bike forwards due to a net forwards force acting on it (which obviously is impossible).

    From here we have two options to analyse the effect of this braking force on the swingarm. Firstly and most obviously, we can simply use each point force (X and Y components acting on the axle, and the X component on the brake caliper) and their respective normal distances from the main pivot (at B) to calculate the moment about B and thus the reaction force C(x) (as well as the reactions B(x) and B(y) if we so desire). If the reaction force C(x) increases, then that means that the shock’s resistance/spring force has to increase (and the opposite is obviously also true). Given the nature of springs, we know that to increase the shock’s reaction force, it has to sit further into its stroke in order to compress the spring to suit.

    The second method, and one which will prove more useful later on when considering linkage bikes/floating brakes, is to consider the forces acting on the swingarm from the caliper and the axle as a single couple moment (generated by the caliper force pushing forwards, and an equal force pushing backwards at the axle so as to have no net force in any direction) plus a horizontal force at the axle instead of two separate forces. In the diagrams shown, this couple moment would be anti-clockwise. This method has the advantage of being able to separate these two components so that we can recognise the following things:

    - Regardless of the orientation of the caliper or its distance from the axle, the couple moment will remain the same for any given frictional force on the tyre; this is because that frictional force multiplied by the radius of the wheel is what gives the magnitude of the moment and it is independent of rotor size (note though that the caliper/axle forces change inversely proportional to rotor radius however).
    - There are two elements which can semi-independently cause BISI, and thus are separable: these are axle path/horizontal axle force and the couple moment. Couple moments can easily be removed from the equation of by parallel linkages (including floating brakes) which will be covered later, and the horizontal force acting on the axle needs a moment arm (perpendicular to the line of force, so in other words a vertical distance) between the axle and the swingarm pivot, in order to generate a moment about the pivot. As such, both factors need to be taken into account when considering brake interference. It is possible (but not necessarily desirable) to create a setup where these two factors cancel each other out (this will be elaborated on later).
    - What determines the amount of squat (or jack, but that will be dealt with later) is determined by two things: the height of the pivot (or instant centre, explanation of instant centre to follow shortly), and the length of the swingarm. The height of the pivot/IC is what determines the vertical distance (which is the moment arm as mentioned before) between the pivot and the axle, and so a higher pivot = longer moment arm = more squat generated by horizontal axle forces. The length of the swingarm changes the effects of the couple moment (which is effectively a constant for whatever calculation you’re doing) by altering the moments generated by the vertical force on the axle and the force of the shock on the swingarm, about the main pivot. For example, if, as above, the axle was 0.45m from the main pivot and the vertical force on the axle was 500N, the moment generated by that (which has to be equaled by the shock’s force multiplied by its normal distance [0.15m] from the pivot) would be 225N.m. A couple moment of say 22.5N.m would be an increase of 10% in this case, which if the shock rate was perfectly linear, would make the bike sit 10% further into its travel in order to regain equilibrium. Now consider that the swingarm was twice the size in every dimension; ie axle to pivot is 0.9m, and pivot to shock mount is 0.3m. The shock ratio is still the same so the shock doesn’t have to supply any additional force at equilibrium. However, the moments (note that these are not “net” moments and as such we are not concerned with acceleration) are now doubled also due to the doubled moment arms – that is to say, the axle force/swingarm length would yield a moment of (0.9 * 500 = 450N.m) that the shock would obviously balance out in the opposite direction. However, an addition of the 22.5N.m couple moment via braking is only a 5% increase now, half of what it was before. From this we can see that swingarm length is inversely proportional to squat generated by the couple moment alone.
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  12. #12
    Ride bike or bike ride? Hopper's Avatar
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    Instant centre theory:
    In a true 4-bar linkage (this includes FSR, VPP, DW-link, Lawwill etc, as well as floating brake linkages), ie not linkage-driven singlepivots, the instant centre (IC) of zero velocity is a point about which everything in the “isolated” link (the rear triangle/seatstays) is moving tangent to. A good explanation can be found here http://www.mtbcomprador.com/pa/engli...#InstantCenter, I highly recommend reading it. Please note the mention of a “constant centre” also.

    With respect to braking performance, forces/reactions on a 4-bar linkage bike can be calculated at any instant in the same way as a singlepivot (as demonstrated above) by using the distance from axle to IC as the length of the swingarm when trying to determine the effect of the couple moment. If the linkage is a parallelogram (as many floating brakes are) or the links are parallel at that point, then the IC will be at an infinite distance away, and as such the couple moment will have no (direct) effect on the suspension’s state (it will not attempt to compress or extend the suspension). If the linkage is a parallelogram, then you will note that the axle’s tangent path is parallel to the tangent path of each of the two “arms” of the linkage. From this we can see that it must have a centre of curvature that is fixed, which can be placed relative to the forward pivots, identically to how the axle is placed relative to the two rear pivots. (Fig. 7)



    In order to calculate the moment acting on the suspension due to the horizontal force on the axle, the vertical displacement between axle and main pivot (on a singlepivot) can be substituted for the vertical distance between the axle and the centre of curvature here. Calculating the reaction forces at the shock and pivots however, is more complex due to the linkage setup. If you are proficient with basic vector calculations and trigonometry (or instant centre theory in its entirety), you should be able to calculate these reactions. However, the calculations are too variable and numerous to demonstrate here.

    A few notes:
    - The vast majority of bikes have some amount of brake squat. Very few bikes actually have brake “jack”, but apparently that doesn’t stop every man and his dog misusing the term “brake jack”.
    - Generally the instant centre of any bike is in front of the axle. Some bikes have ICs that are behind the axle (such as Lawwills) which due to the brake’s couple moment effect may tend to develop a next extension force under braking – but, and this is a big but, having an IC behind the axle does not automatically necessitate that a bike will “jack” (extend), due to the horizontal axle force giving an equal or greater moment trying to create squat.
    - All demonstrations of calculations above are for conceptual explanation only – they make assumptions for the sake of simplicity which make them mathematically inaccurate (by which I mean not exact, not that they give a “wrong” answer/idea). They are used only to give the reader some basic understanding of how the main calculations are performed. The inertia of the wheel, brake, linkage/swingarm etc have been ignored thus far, and all calculations have been assumed to be at equilibrium – this is not precisely the case in the real world, but it will suffice purely for the purpose of conceptual demonstration.
    - It is possible, although not necessarily desirable, to have a suspension setup which does not have a squatting or extending tendency under brakes (at a given instant or instances). This, I believe, should be kept separate in terminology, from suspension extension due to weight shift. It is fairly easy to understand that if the brake torque/axle reaction force doesn’t exhibit a compressive or extensive force/moment on the suspension AT ALL, then under any braking the suspension will extend due to weight shifting forwards. For this reason, it may be useful to have some amount of pro-squat (tendency to compress). I know of no situations where it is helpful to have anti-squat (a net extension force, in other words actual “brake jack”) under brakes, as this only exaggerates the forwards weight transfer.
    - The “couple moment” and “horizontal force on axle” effects can be treated as independently as above in all situations – note that in some cases, one can be negative whilst the other is positive, reducing either the pro- or anti-squat reaction under braking (depending which one has a greater effect).
    - All calculations above work for all situations that I know of. That is, to my understanding, this explanation is universally applicable.
    - There are other factors which I have not bothered to mention due to their relative consistency between bicycles, such as centre of mass, shock rate etc. These do have an effect on braking, believe it or not, but their contribution is not significant enough (as well as being much more complex) to be worth mentioning in this article.
    - Parallelogram linkages and floating brakes do remove the couple moment component of BISI, but they do not (necessarily) make any change to the effect of the horizontal force on the axle. On many (singlepivot) bikes a floating brake makes a considerable difference to the braking characteristics because of said removal of couple moment component, however this does not completely eliminate all brake-induced effects on the suspension (for better or worse).
    - There are other ways to explain braking characteristics, as far as I know these are all in agreement with what is written above. A common way is, with acceptance of the internal forces within the wheel/swingarm, to simply consider the rear wheel contact patch (and the frictional force acting on the wheel at that point) with relation to the IC, and nothing else. This is a perfectly viable method of calculation/comprehension, but I have chosen to explain it differently because separation of the two main components of brake interaction shows more clearly how the forces internal to the suspension can separately affect the suspension and how they can be manipulated to give the desired effect.
    - The claim that bikes will “brake jack” and/or “lock up/out under brakes” should always be treated with suspicion. This is nearly always untrue or inaccurate – braking does not “lock out” any bikes, and in no circumstances that I am aware of is braking capable of bottoming out a properly set-up bike. True brake “jack” (extension) can give the rider the impression that the bike has stiffened suddenly, but unlike a rigid/locked out bike, the braking extension tendency is not a reaction to a vertical (bump) force, and as such will not be immovable or rigid, as those forces can be overcome.



    The end result: what does it mean?
    Most bikes squat to some degree. This is not necessarily a bad thing, as mentioned above. This includes most FSR bikes, despite Specialized’s “fully active in all circumstances” claim (with the understanding that "fully active" means "shock absorption completely unaffected by braking"). Some bikes do actually “jack” but these are few and far between, and it’s not always bad enough to even be noticeable, let alone a problem. There is a definite placebo effect surrounding brake systems, and it is not unlikely that this is due mainly to lack of education/understanding on the subject. Some people will swear black and blue that singlepivots are nearly unrideable due to perceived “brake jack”, others will simply state that they’ve never even noticed it. From this we can make a logical conclusion: BISI does exist, and that it is not necessarily a problem – in fact in some forms and to some degrees it can even be useful. However it is hard to believe that any common amount of brake squat can make a bike unrideable, or anything to that end. Notably, the bike on which Fabien Barel won the 2004 world DH championships on had a brake linkage designed specifically to increase the level of (pro-)squat far beyond what normal bikes generate. Riding the production version of this bike, you can feel a huge tendency for the rear end to dive when the rear brake is applied. Given that no owners of those bikes seem to have any problem with the extreme brake setup, one might logically assume that it’s not actually that bad, and that other bikes with considerably less brake induced squat can hardly be any worse off, and thus are perfectly fine to ride – although not necessarily as comfortable as they could be.

    The moral of the story is that almost any pro-squatting braking setup is usable. That is not to say that there is no reason to dislike certain degrees of brake interaction – that varies with riding style, terrain and personal preferences. Another important point to note is that true “neutrality” under any acceleration (positive [pedalling] or negative [braking]) is not necessarily an optimum setup – certain reaction forces under braking/pedalling can help stabilise the bike as well as offer greater comfort and traction. It is also useful to know that it’s not hard to make stuff perform worse, so be wary of playing with your bike’s braking characteristics unless you know what you’re doing – that incorporates more than is written in this article.


    All content remains copyright to Kenneth Sasaki, to whose work I linked regarding instant centres. Feel free to reproduce it as long as full credit is given. This permission from him
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  13. #13
    Freak of Nature
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    thank you for an incredibly informative response. Long, but based entirely in facts and math/physics, which I appreciate.

    A quick summary, if I may: torque forces on the brakes correspond (in certain designs) to compression forces on the shock, with nothing to do with momentum/center of gravity of the rider. Do I have it?

    And I guess going to "uni" as you put it is more of an accomplishment there than it is in the states...at the college I go to it'd be some stoner saying "look, man, it's not that the bike brakes...you just slow down the earth, man."
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    Ride bike or bike ride? Hopper's Avatar
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    These calculations are very simple, it is pretending as though the rider's weight is is in the centre of mass of the bike. Yes there will be huge differences in how the brakinf will effect the rear end. This is a very simple model!

    Basically yes, the torque created on the rear end can send feed back through the swing arm into the shock, it will either "squat" (compress into the trave) or "Jack" (Extend the travel) Both of these effects will partially lock the shock up in one way or another. However this isn't always a bad thing, some riders prefer this. Fabien Barrel rides a bike nptorious for having brake problems, however he has had a floating caliper put on his bike to increase the amount of Squat he gets.

    Companies like Specialized are always going to be saying how much of aterrible thing it is, and how their bikes are totally neutral. This is a gross exaggeration, it can be good for the rider and the horst link hasn't totally eliminated the "problem"
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  15. #15
    I drink your MILKSHAKE Raiyn's Avatar
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    Quote Originally Posted by Hopper
    Basically yes, the torque created on the rear end can send feed back through the swing arm into the shock, it will either "squat" (compress into the trave) or "Jack" (Extend the travel) Both of these effects will partially lock the shock up in one way or another. However this isn't always a bad thing, some riders prefer this. Fabien Barrel rides a bike nptorious for having brake problems, however he has had a floating caliper put on his bike to increase the amount of Squat he gets.

    Companies like Specialized are always going to be saying how much of aterrible thing it is, and how their bikes are totally neutral. This is a gross exaggeration, it can be good for the rider and the horst link hasn't totally eliminated the "problem"
    Granted I'm a strong advocate of Specialized, but I'm setting that aside for now.
    How exactly is the suspension locking up under braking or pedaling forces a "good" thing?

    One would think (at least I would) that you would want pedal and brake forces to be completely seperate from suspension action. (Specialized may or may not be completely perfect but it's a damn sight closer than anyone else)
    Also if it's not a "bad" thing why are the other companies trying to eliminate it through other designs and shock technologies?

  16. #16
    Ride bike or bike ride? Hopper's Avatar
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    There are companies trying to eliminate it because there are people with lots of money who listen to the hype that brake induced suspension actions are bad. Yes I agree that pedal feed back is bad, but this can be fixed with a simple chainwheel pulley positioned correctly, ala BB7. I personally hate brake jack and don't know anyone who thinks it is good however I do prefer a bike that has a bit of squat in braking This gives the bike the feeling that it is sinking into the corner and then shoots you out again as you come out. Also when braking, most big forks are going to dive, if your rear end squats at th same time it is going to keep geometry and balance of the bike much more similar to what it would be like without this squat effect.

    One thing to note, that the less weight on the rear will create bigger issues, so the worse time there will be brake problems is when the rear end is un weighterd and why would you brake then? The motorcycle industry used to use floating disc calipers but realised that it did bugger all difference now I can't think of any MX bikes with floaters.

    As for specialized, yes they have a system that is "neutral" under braking, I found this nice but I still prefer being able to feel the rear end, not feel like I'm floating on a pillow of air, while braking. Yes their Horst link makes lots of difference but it is in no way neutral, it is a slightly Squating system. There are many other systems that are designed to change how the rear acts under braking however no shock comapny has ever tried to design out brake forces on the rear of a bike. Mojo have even publically said that brake jack and squat is an evil term that has been coined and will not have any effect on a bike being ridden by a competent rider that knows how to brake!

    Finally most riders will not feel the difference between a bike that supposedly has brake jack or not, axel paths, pedaling forces..... are all things that a rider will notice the difference about. Then when there is a rider that actually has enough skill to tell what is going on, they will have their own personal feelings, Barrel obviously wantys something that squats and has it custom made too, Peat could run a floater but CHOOSES no too, but then again some riders prefer the feeling of an active bike, (god who is that commencal guy???) He runs an after market floater trying to eliminate the problems. And we cannot go buy pros riding on bikes designed to get rid of it, they have very little choice o nthe type of bike they ride.

    Lastly, if you want a bike that feels ultimate under braking, pedaling and over bumps........ *cough* DW link *cough*

    What I am trying to say is there are different strokes for different folks, ride what feels best to you and don't let comapnies that have spent lots of time and money on a thing such as brake squat think not to buy a bike, it is a marketing thing. Oranges are light and hand built, Specialized have active suspension, Cannondale's are built in America...... it is all marketing, ride what you think feels best, not what you are told feels best!

    *edit* sorry for all these long posts if you don't want to read this whole post, just read the last paragraph
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    http://www.specialized.com/sbc4Bar.jsp?minisite=10081

    Examine the animations from the link down the page

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    mmm... chicken! Funkychicken's Avatar
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    Hopper's point should be broadcast around a bit more. When i first got into the sport, i was obsessed with how efficient a suspension design can be and was confused by the vast amount of seemingly conflicting information coming from different manufacturers.

    Now i feel that an absolutely efficient suspension design is less important than the overall bike's feel. Things such as brake jack and pedal feedback can be compensated for when riding, in fact that's something that a rider should do to adapt and grow with the bike.
    That's a lie.

  19. #19
    Senior, Senior Member ExMachina's Avatar
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    Holy crap, Hopper, that was quite a little treatise! Nicely done.

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    Quote Originally Posted by Hopper
    Ok let us begin, chain suck is a problem where the chain grows through it's travel, this causes tension and it can be pulled, this forces pedal feed back through the cranks, this can be imprved by having a pivot design that pivots around the BB.

    Erm, that's pedal kick.

    THIS is chainsuck.

  21. #21
    Ride bike or bike ride? Hopper's Avatar
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    Quote Originally Posted by Funkychicken
    Hopper's point should be broadcast around a bit more.
    DAMN STRAIGHT

    Quote Originally Posted by Funkychicken
    Now i feel that an absolutely efficient suspension design is less important than the overall bike's feel. Things such as brake jack and pedal feedback can be compensated for when riding, in fact that's something that a rider should do to adapt and grow with the bike.
    If what you are trying to say is forget about all the bs advertising and ride what YOU think feels best then my point has come accross

    Quote Originally Posted by Big Tommy C
    Erm, that's pedal kick.

    THIS is chainsuck.

    Whoops my bad Yes that is pedal suck, but I always call pedal kick pedal suck, just a habit I suppose
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  22. #22
    Wood Licker Maelstrom's Avatar
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    Quote Originally Posted by edf825
    When you hit the brakes, the momentum keeps your bike going forward. Since your center of gravity is higher than the contact point with the ground, you'll be forced forwards. This puts more weight towards the front and takes it off the back, causing the rear shock to decompress.
    To continue this is dependant on pivot location. Something your back jacks and sometimes they squat. The affect is the same, the suspension stops moving.

    Sorry if that is mentioned above but I didn't want to read through every post. Its to early in themorning

  23. #23
    Wood Licker Maelstrom's Avatar
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    Quote Originally Posted by Funkychicken
    Now i feel that an absolutely efficient suspension design is less important than the overall bike's feel. Things such as brake jack and pedal feedback can be compensated for when riding, in fact that's something that a rider should do to adapt and grow with the bike.
    While I won't disagree with this assumtion, and hopper is DEAD on on all points that I could see. Training yourself to outride your bike is not easy. The ability to NOT brake in rough conditions, turning etc is not that common an ability. Your brains instinct goes against that. FSR for example, lets you not worry about braking perfectly and lets you maintain speed. There is good reason why it seems the best in the world use single pivots and the people below them use vpp, fsr etc...

    Again not disagreeing, I totally 100% agree with the point above, but I find meeting a lot of riders, most people will not like a single pivot vs an fsr once they ride an fsr because they can't let themselves not brake. I am a good example of someone who sometimes messes up my braking because my brain panics. I am learning not to. But I would hate to see what would happen if I had been riding a single pivot all season. Ouch...

  24. #24
    Bandolero Bandrada's Avatar
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    Great thread! This is the thread that brought me here. It would make a great sticky.

    I currently have a few of the various 4bar design offerings. I like 'em for different things. I'm not a big fan of high maintenance mini-link designs as I like to fiddle and tune.

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