Some thoughts on Vincent Front Suspension. PART 1
Observations:
· A Thornton Front shock Absorber has a total travel of 3 inches without the bump stop in place. With the bump stop the travel is reduced to 2 ¾ inches
· Spring length, installed in spring boxes at maximum extension is 14 inches
· Spring maximum compression (travel) inside the spring box is 3 inches
· Thus spring length at full compression inside a spring box is 11 inches.
· At all times it is essential that there is some pressure applied by the springs inside the spring boxes. I.e. at full extension there MUST be some pre-load, just 2 or 3 Lb is sufficient, so that there is zero free play.
· Static sag (front) should be in the range of 28 to 32% of total available suspension travel for normal street use.
· Original front outer springs have a free length of 15 inches, C series and 16 ½ inches for the D series bikes. Both series springs have the same spring constant of 65 Lbs/inch. Plus the same springs were used in all models – Comets, Rapide, Shadow. The inner front springs (when used) are the same across both series with a free length of 15 inches and a spring constant of 9 ½ Lbs/inch. (Thanks to the VOC Spares co and David Dunfey for this information)
Let’s take a side trip so we can have some understanding of what we are dealing with.
When studying springs and elasticity, the 17ᵗʰ century physicist Robert Hooke noticed that the stress vs strain curve for many materials has a linear region. Within certain limits, the force required to stretch an elastic object such as a metal spring is directly proportional to the extension of the spring. This is known as Hooke's law and commonly written:
F=−kx
Where F is the force, x is the length of extension/compression and k is a constant of proportionality known as the spring constant which is usually given in Newtons per Metre (N/m) or in imperial measurement, Pounds per Inch (Lb/In) .
Though we have not explicitly established the direction of the force here, the negative sign is customarily added. This is to signify that the restoring force due to the spring is in the opposite direction to the force which caused the displacement. Pressing down on a spring will cause a compression of the spring , which will in turn result in an upward force due to the spring.
If you really want to know more about the physics of springs, this site is a good place to start: https://www.khanacademy.org/science/physics/work-and-energy/hookes-law/v/intro-to-springs-and-hooke-s-law
OK – lets get back on track.
It is my understanding that the role of a spring and that of a damper are totally different and should not be confused. The Spring is used to control the movement by the absorption and release of kinetic energy within the spring itself whereas the Shock Absorber controls the rate (or speed) of movement by the conversion of kinetic energy (movement) into heat.
In terms of our Girdraulic suspensions, if we were to remove the spring boxes, leaving just a front shock absorber in place, the shock absorber would slowly compress to its shortest length and that’s where it would stay. Just how quickly the shock absorber compresses is a function of its stiffness (the rate at which it can convert kinetic energy to heat); The stiffer the shock absorber, the slower it will be in settling. As the shock absorber settled absorbing the kinetic energy imparted to it by the weight of the bike, some heat would be generated within the fluids in the shock absorber. That’s it folks – you can then push down as much as you like – but there will be no further suspension movement as the shock absorber will have already reached the mechanical limit of its travel. Of course, you can also pull up on the bars, but without the springs in place you are attempting to lift the ‘dead’ mass of the bike. Given super human strength you could lift the bike but this time the suspension travel will be limited, again by the shock absorber reaching the other end of its available mechanical travel.
If we leave the spring boxes in place but this time removing the shock absorber, what happens is that the weight of the bike starts to compress the springs to the point where the upward force from the springs equals to downward force being applied. It is a point of equilibrium where the kinetic energy from the mass of the bike balances the kinetic energy stored in the springs. Now if we push sharply down (or pull sharply up) on the front of the bike the equilibrium is disturbed the suspension will now start to move up and down and will continue to do with the amplitude decreasing with each oscillation, eventually returning to the point (height) of equilibrium or as some think of it, the ‘sag point’. As an aside, this is where the Shock Absorber then comes into play, soaking up excessive kinetic energy and damping out the oscillations. Remember – ANY friction in the suspension system will also act as a shock absorber, with that friction converting the kinetic energy of movement into heat at the points of friction.
Key Point One. It is the springs alone that determine the ‘sag point’ of the front suspension.
Key Point Two. It is the static sag point with the bike stationary AND the rider (plus normal luggage) on board that counts.
Key Point Three. It is the travel range provided within the shock absorber that sets the total amount of available suspension travel.
Key Point Four. There should be zero (impossible!) friction within the suspension system. Realistically, there should be minimum friction.
Key point 5. It is ONLY the sprung weight (that’s the weight supported by the springs) that we need to consider when looking at front spring calculation around % of static sag. In this discussion I have assumed an unsprung weight of 50 Lb covering the front wheel assembly with tyre,
Remember what we are looking for is a static sag where the weight of the bike and rider supported by the spring happens at the 30% of suspension travel point. OK, now you realize you need to get the weight on the front wheel measured WHILE you are astride the bike less 50 Lb (being the assumed unsprung weight) to help you determine the equilibrium point for static sag.