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.
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.