The Q of a tuned parallel L-C circuit stands for quality factor and is a predictor to how much current flows in the tuned circuit at resonance. Here's the formula: -
Q = \$ \frac{1}{R_S}\sqrt{\frac{L}{C}}\$
\$R_S\$ is in series with the inductor, L
This basically means that if \$R_S\$ is zero then Q is infinite and the circulating currents in L and C are infinite for a non-zero voltage applied. Of course this doesn't happen because resistance is never 0 other than in a superconductor but that's another story!
Significantly for the question, if R stays the same and L is decreased by ten and C increased by ten (to retain the same resonant frequency), Q reduces by 10. In other words the high peaky tuned circuit you once had has become rather shallow and not so peaky in comparison.
Reducing inductance by ten does not usually mean \$R_S\$ reduces by ten. For a typical core wound with say 1000 turns, inductance might be (say) 1H. To reduce this to 0.1H means the turns reduce to 316 and the resistance therefore only reduces by about one-third.
Remember inductance is proportional to turns squared.
So after reducing inductance by 10 (resistance decreases by about 3) and increasing capacitance by 10, the net effect is a tuned circuit that has a Q that is 3x smaller than before and not as peaky or resonant.
Regarding the link - this is for an RLC parallel circuit where the resistance is represented as a parallel component and this is less-often used because the dominant loss in an RLC circuit is in the inductor (as series resistance). A parallel resistance could depict dielectric loss in the capacitor but this will be a tiny fraction (in most cases) of the inductor losses.
No force at all sounds like the bolt might be non-ferromagnetic. Simply test it with a magnet and make sure it is strongly attracted. 400-series stainless is only slightly ferromagnetic and 300-series stainless steel is almost totally non-magnetic under normal conditions.
If you sense the force is there (Luke?) but it is very weak, then keep in mind that the calculator you used assumes the total gap around the magnetic path is 1mm. If that sounds impossible to you- consider the following images from here:
(g = 1mm)
Or this one from the same site:
As you can see, the magnetic material of the core and armature forms a magnetic path that is only interrupted by a single small air gap (not counting the minimal clearance gaps required for the armature to pivot in the first image or the plunger to slide in the second image). The total reluctance around the path adds like electrical resistance, but the reluctance of the ferromagnetic materials is much lower than air, so the air gaps dominate.
Best Answer
Increasing inductance will affect the solenoid if you are feeding it PWM, but not necessarily in a bad way. PWM works precisely because the inductance averages the applied voltage. For the time that your PWM switch is on, current increases, but at a rate limited by the inductance. When the PWM switch is off, current decreases, but again at a limited rate.
The result of this is that the current (and thus the magnetic flux and force) approximates a steady value that would have resulted had you applied a fraction of the supply voltage according to the PWM duty cycle. For example, if the supply voltage is 12V, and you drive it with a 60% duty cycle, the current you get is essentially the same as you would get by applying a constant \$12V \cdot 0.6 = 7.2V\$ to the solenoid.
Increasing the inductance makes this average better, further reducing the ripple in the current between PWM cycles, or alternately, allows you to use a slower PWM frequency for the same current ripple.
Increasing the inductance also limits the rate of change of current, and thus force. However, it's quite difficult to make the inductance so high that this becomes significant in most applications. If you weren't worrying about it yet, I wouldn't start now.
The bigger problem with adding more turns is usually that the DC resistance of the solenoid also increases. Since the resistance converts electrical energy to heat but doesn't help you create mechanical force, less resistance means higher efficiency. However, more turns allows you to generate more flux with less current. Since resistive losses are given by \$P = I^2 R\$, reducing the current can reduce resistive losses more than reducing the resistance. Yet, more turns also means the induced EMF from the movement of the solenoid will be higher, which requires that you can supply a higher voltage to overcome that EMF to accelerate the mechanical load.
So, you are faced with many trade-offs. In the end, it comes down to optimizing the parameters that are important for your application. Adding more turns isn't good or bad, it's a trade-off. I'd suggest experimenting.