
Spring rate can be confusing. While the spring rate (kg per mm of compression) increases in a shortened spring, it doesn't mean the spring is stronger. The shortened portion of the spring is still the same as it was and requires the same force to close its coils together (coil-bind compression). But it now operates over a shorter range. So it's a reduction in mm travel rather than an increase in compression force that makes the rate higher. Does that make sense? Probably not.
Maybe I can represent it graphically.
Let's imagine a hypothetical carby spring (I don't know the specs of a Laverda carb spring)
Free length 150mm
Coil bound length 25mm
Total spring range is 150 - 25 = 125mm
Compression force at coil bind 1kgf
Spring rate is 1kg/120mm or 0.008 kg/mm
Now let's cut 20% (30mm) off its length. The specs become:
Free length 120mm
Coil bound length 20mm
Total spring range is 120 - 20 = 100mm
Compression force at coil bind 1kgf
The spring rate is now 1kg/100mm or 0.01 kg/mm
Let's assume the carby slide travels 40mm between 50 and 90mm spring compression. This operating range applies to both springs because we haven't altered the carby.
Have a look at them on a graph, and the shortened spring has a steeper slope (higher spring rate) but always has less force on it over the entire range.
The left end of the graph is vertical because if the spring is loaded beyond coil bind, the force can go higher with no change in length.
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