Understanding spring geometry with respect to available droop and bump travel is not a simple thing. Let me try to explain. I think that some of the issues here are that Richard and Glenn aren't in agreement on vocabulary they are both using.
For this entire explanation I'm assuming that the car has zero caster and zero SAI (steering axis inclination). In other words, the strut/spring are perfectly vertical and act directly at the center of the tire on the ground. Making this assumption will just remove some confusing trigonometry from the discussion, but will not change the conclusions.
Bump travel, the easy one:
Spring deflection under static conditions is governed by the formula:
F is the force on the spring.
k is the stiffness of the spring
x is the change in spring length
Example #1) Consider a car with 1,000lbs of sprung weight on each front corner. Using an 8" spring with a rate of 450lbs/in (k). Assume the spring has a coil bind height of 3.95". Coil bind height is the height of the spring when all of its coils are stacked against each other. It can't ever get any shorter than this.
From this we know that the spring has a total available travel of 8"-3.95" = 4.05". When the car is placed on the ground the spring will compress 2.22" because the spring has a rate of 450lbs/in (k) and will push back with a force of 1,000lbs (F) when it is compressed (1,000lbs/450lbs/in) = 2.22" (x). Remember the car is at vertical equilibrium (not moving) so the vertical spring force must equal the vertical force from the car, by definition. We now know that there is (4.05"-2.22") = 1.83" of bump travel available before the spring coil binds.
Of course if the struts were badly designed, we could run out of bump travel before the spring coil binds. But this has nothing to do with the maximum available bump travel that this 8" spring can allow with this weight on it.
Notice that none of this has anything to do with ride height. If the car is too low, it is possible for the tire to hit the fender, the frame rail to hit the ground, etc. before the spring coil binds.
Example #2) If we use a 10" 450lbs/in (k) spring with a coil bind height of 5.00", then we have 5.00" of total travel available and 2.78" of bump travel available at ride height.
Going to a longer spring, with the same rate, gives you more available spring travel, which is ONLY ever available in bump. In example #1 we had 4.05", in example #2 we had 5.00". This is the advantage to a longer spring. It is important to note that it is possible to build a 10" 450lbs/in spring with LESS total travel available than an 8" 450lbs/in spring. This is a function of using the correct wire gauge, steel alloy and closing and grinding the spring ends properly. Within a given brand/line of springs, a longer spring of the same rate will almost always have more travel available.
Droop travel, the not so easy one:
Let me start with a definition of droop travel. Droop travel is the distance the tire can drop, from ride height, until the spring has come lost contact with the upper spring perch. Why not count the travel from that point down to where the strut has no more droop travel? Once the spring has come off of the upper perch, there is virtually zero force pushing the suspension downwards. Only the gas force in the strut or shock (about 100lbs per side) is going to pull the suspension down further. The strut damping forces are going to make that 100lbs take forever to pull the suspension to full strut droop.
In example #1 above, the front suspension had 2.22" of droop travel, because that is how much the spring was compressed. In example #2, the front suspension also had exactly 2.22" of droop travel available also. Huh? From this it should be clear, that the length of the spring has ABSOLUTELY nothing to do with whether it comes loose at full droop or not. The force on the spring and the spring rate determine how much the spring is compressed at ride height. If this amount of compression, 2.22” in the above cases, is less than the amount of droop travel in the strut at that ride height, then the spring will come loose at full droop.
It is possible for someone to build a strut with not enough total travel, and for this to artificially limit the droop travel in the suspension before the spring comes loose. This is essentially what all auto manufacturers do to keep the springs from coming loose. They make sure that the spring always has more available travel than the strut.
Bump resistance: Bonus points.
How much bump travel is needed on a given car to not have it bottom out? Higher wheel rates obviously make it harder to bottom out, but higher car mass makes it easier to bottom out. In addition, more bump travel makes it harder to bottom out. Which is most important? All three are. The formula below gives a normalized parameter which is directly proportional to how hard it will be to bottom out the suspension on a car.
g = Bump Travel/Droop Travel
If g is above 1, the car will be fine when used with a bumpstop. If g is below 1, it may be too easy to bottom out, especially if driven on curbs at a road race track.
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Last edited by Jack Hidley; 09-21-2013 at 09:32 PM.
Reason: Edited to make clearer