MM cartridges and capacitance

If the inductance estimates are correct, it means one wants to have capacitance as low as possible in order to preserve the harmonics 'air' which extend beyond the audible band. It would be difficult to imagine why one would want to bring that mountain closer to the audible frequency.

No, not for MM's. As I indicated in my previous post, if the frequency of that mountain is brought down such that it occurs just a little above where the cartridge would otherwise be rolling off, it will compensate for that rolloff within the audible range, and have the effect of extending the overall treble response. Lower capacitance will move the peak to a point where it will no longer provide that compensation, and where the response will already have rolled off significantly, resulting in a less extended treble.

As an aside, I am looking at the formula just above the MM FrR calculator and I don't understand how a high SQRT(LxC) in the denominator would lower any ratio with a 1 in the numerator.

As with any fraction, if the denominator increases and the numerator stays the same, the resultant value goes down. The presence of the square root function simply slows down the rate of decrease as LC increases.

Best regards,
-- Al

Al,
Thanks.
With regard to the fraction... My math is OK. I misspoke.
I see frequency as being a number X greater than one. I see a higher frequency as being a number greater than X which is greater than 1. If I take one cycle, it's frequency (stated as a fraction) is a number Y less than one and greater frequency is an even lower number.

I expect it is the case that I do not understand where the decimal place is and how sqrt(faradsxhenrys) changes to frequency.

T-bone, if we use units of henries for L, and units of farads for C (which results in F being in units of hertz), then both L and C will be far less than 1. In fact C will be on the order of 0.0000000001 farads (or 100 pf). Therefore we will end up with a number in the denominator that is much smaller than 1, with the numerator being equal to 1. Therefore F, in units of Hertz, will be a number that is much larger than 1.

As for Farads x Henries resulting in frequency, after application of the square root and reciprocal functions and some constants, yes that is not intuitively obvious. But the derivation is as follows:

Resonance occurs at the frequency at which the magnitude of the impedance of L becomes equal to the magnitude of the impedance of C. Since the impedances of L and C have opposite polarities in the complex plane (don't ask!), the equal and opposite magnitudes will cancel each other out at that frequency, resulting in a net impedance of zero (apart from the resistance that is present).

The magnitude of the impedance of an inductor, aka its inductive reactance, is measured in ohms and is equal to 2 x pi x F x L.

The magnitude of the impedance of a capacitor, aka its capacitive reactance, is also measured in ohms and is equal to 1/(2 x pi x F x C).

So to find the resonant frequency we set those two formulas equal to each other, and solve for F.

2 x pi x F x L = 1/(2 x pi x F x C).

Re-arranging that equation:

1 = 2 x pi x F x L x 2 x pi x F x C

Therefore F x F = 1/(2 x pi x L x 2 x pi x C)

Therefore F = 1/(2 x pi x (sqrtLC))

Voila!

Best regards,
-- Al

Al,
You are a scholar and a gentleman.
Thank you very much for that explanation.

Al, Thanks so much. This is the first clear explanation for the apparently paradoxical effect of added capacitance on the treble response that I have come across. Makes perfect sense, now. Your post above should be archived somewhere.