Barnsley Boy
Active Member
Pretty close! The full logarithmic formula goes:
power ratio = P1/P2 = 10^(decibels/10)
or, decibels = 10.log(P1/P2)
In this case
P1/P2 = 10 ^(-59.5/10) = 0.00000112
power is therefore 50W x 0.00000112 W = 56 microW
I expect you're finding a setting to use quite a few dB's louder, but if you engage this full attenuation, can you still hear it?
@JohnH ...... a slight tweak to the calculation
10 ^(-59.5/10) = 891250.9381.
Assuming P1 is the unattenuated power (in my case 50w), then the calculation should read:
P2 = 50/891250.9381 = 56 microwatts.
Just out of interest, I populated the table below showing the 17 increments of reduction (assuming the first one is with no attenuator connected) and the associated power reduction.
It's fascinating firstly how quickly the power drops, and secondly how bl**dy loud even 0.4 of a watt is!
With all stages engaged, what I would class as DEFCON 5, I would say that it is at a level where you could almost play the amp with everything cranked and it wouldn't wake a sleeping baby. Things get a lot hairier very quickly when the switches get flipped. It would be great to play at a volume more suited to the amp, and push the speakers to get that proper Marshall goodness, but without a soundproof enclosure and a decent pair of ear defenders, I'm going to have to wait!
My ears have never been the same since a Godfathers St. Valentines Day Masacre gig 5 years back. Tinnitus is a terrible thing, the ringing sound never stops!