Z3Jonathan

quote: Originally posted by justin_tx_16: I feel it is best to try them out, if you like them keep them, if not return them. Different people hear things differently. That is true what you say about the efficient nature of the speakers. Klipsch speakers are very efficient. What took 60 watts to do on a Mirage OM-5 speaker took only 40 watts on the Klipsch RF-7 Very true.... I'm a firm believer that the only oppinion that counts is my own and those of trusted friends . I usually read the personal reviews @ AudioReview.com to look for potential product problems (ie, does a product have a common failure?) -- info you usually can't find in profession/magazine reviews. When reviewing headphones, AR.com was useful to find out which ones will/won't work in a portable environment. AR.com isn't a substitution for my own personal testing, but rather a good way to get long-term user oppinions. Also, Headphone.com is making a tour of the US... its a great way to test out headphone/amp combos. They've got a lot of Sennheiser models in stock, including some of the newer models in the 400-Series. I'm probably going to pick up a pair of HD-457/477/490s tonite... I need a cheap pair of headphones for work.

quote: Originally posted by justin_tx_16: my first choice was the 570. I think I will go with my gut instinct on this. 64 ohm resistance means that Here is what the V=IR equation gets me... V=32I would mean to get 64 volts of power, I would need 2 amps, and to get 64 volts of power with 64 ohms, I would need 1 amp... haha, what is up with that? I must have something messed up here. How it should work is the higher the resistance, the more amps needed, right? So, lets turn it to I=V/R, I=V/32, vs I=V/64, so yes, half the power. If only I could find DanF's info on the control pod... But we want watts right? actually milli watts. So we turn to this equation, P=I^2R, which we got from P=IV and V=IR, substitute the V in P=IV for IR and you get P=I^2R, so we want 100mW, or .1 watts, so .1=I^2(64), divide by 64, you get 0.0015625, square root it and you get 0.03952 amps are required. So, just gotta remember the watts to amps equation. Well, lets look at a different equation. P=VI If I need 100mW's of power, .1 watts, that is .1=VI, .1=(j/c)(x c/s)........ i think that is the equation i need to work haha. Coulombs, the definition of fun! edit, i see some typos here, just don't grade my physics Justin: You also need to take into account the efficency of the headphones. In the end, its best to read online reviews (www.audioreview.com) and see what headphones are reccomended for portable/computer use. You could also alleviate the problem and buy a headphone amp, and achieve pure audio bliss ;-). Again, I've got HD-535's. Very nuetral and revealing headphones. They are very true to the source, so poorly recorded MP3s will sound horrible. On the plus side, its great to hear nuances in the music, especially live music. Jonathan