And (2) Power = current Intensity times Voltage drop. (1) Ohms law says that voltage drop across a resistance is Voltage = current Intensity x Resistance. The fundamental results requires just two fundamental concepts. The first is the bass driver in free air or with one side in a box (0.08 ohms) and attached to the horn (8.0 ohms). We'll pick some numbers for acoustic resistance presented to the diaphragm in two conditions. We're going to look at the effect of improving the acoustic load on the front of the diaphragm. There is an electrical analogy to the acoustic loading of the driving system by the throat (say the opening of the tube) and the voice coil resistance. ![]() From this, you have the better pump and a bigger diaphragm.Īnd if you put the 2-foot or greater diameter virtual diaphragm in a corner, the trihedral corner forms a longer horn with a bigger diaphragm. ![]() Going further (my explanation), we can put the piston-like speaker diaphragm with a diameter of about 1 foot at one end of the pipe (throat of a horn), and then increase the area of the pipe by flaring it out along its lenght (the horn) of three or five feet until its diameter (mouth) is over two feet or more (call that a virtual diaphragm). (A bigger piston would do better though.) But if the put the piston in a tube, it is an effective pump. ![]() ![]() PWK analgized it to putting a piston in the middle of a lake and moving it back and forth. Yes, the horn really does increase the acoustic output.
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