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Eminence Legend B810 10" Bass Guitar Driver 150W 32 Ohm
Hello, I am seriously thinking of replacing my original 1970s 810 SVT speakers . All the speakers are in good shape, but they must be tired? I now have a SVT 4 Pro (1600 watts). The original cab is rated 800 watts. Will the 8B10s be an definite upgrade?
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Date published: 2014-09-11
I have a Gallien Krueger Goldline GLX410 and have 2 blown speakers. Can I replace them with the Eminence Legend B810 10"?
This is the GLX410 specs:Configuration: 4 x 10" Paragon Goldline loudspeakers.Power (RMS): 400WPower (Program): 800WImpedance: 8 OhmsSensitivity: 100dBMax SPL: 126dBFrequency Response: 30Hz to 7kHz
You have several options. You can recone/replace the voice coils on the two woofers - many shops perform this service. They also change woofer surrounds as they age and crack. Or you can change all 4 drivers, not just 2. You would need to measure the internal volume, divide by 4, and then find a woofer that matched that volume for performance. This is done with a box calculator using the T/S parameters with each woofer spec. If you have a port - it may need to be modified.
Date published: 2014-03-05
I have the original Ampeg bass cab that used NINE 10", 2 ohm drivers (in groups of three, in three separate chambers). How would I wire these?
I have the original drivers that need to be re-coned. Is that possible (2 ohm voice coils)? If the 32 ohm drivers could be wired up, that would be great. Any advice welcome.
This is a very odd impedance driver. It looks like it was designed to be used in groups of two, four, six, or eight in parallel. When designing a wiring for these, the first step is to determine the prime factorization of the number of drivers you want to use. Unfortunately, the factors of 9 are 1, 3 and 9. So you only have three possibilities. Using the factor of 1, you wire the drivers in series and obtain 288 ohms. Using the factor of 3, you wire three in parallel and 3 in series, and obtain your original 32 ohms, and using the factor of 9, you wire all drivers in parallel and obtain 3.555.... ohms. None of these solutions is very appealing.
Date published: 2011-09-12