.

Isn't that the answer rather than the question ?

ass faced fuck wit at work who insists he knows everything says get no more than a 30watt amp or I'll fry the speakers - my highly uneducated guess is that an amp of about 200 watt is going to make sure the peak times dont distort and still sound OK, but to be aware that this is the limit so maybe a 120-150 watt amp. 30watt amp for 100watt rms Cerwin Vegas, i dont think so.

Hmmm, I think you need to define the question and then start with some basic physics.

 

However, an amp that puts out say 50w RMS per channel is likely to peak at double that, or more. So in theory, so as not to blow your speakers, they should have a rating of something like 100w. An over simplification - there are books written about this sort of thing.

AndyP, was looking at your web page/s and noticed that you have a switch just under the sterring wheel and to the right - next to a small recess - What are they for ? I cant see that the switch on mine does anything apart from light up ???

 

Hmmm, you are observant ! It's for the fog lights. Dumb place to put it but it was done by the importer when they did the conversion. One day I'll change it to standard UK switchgear.

Hi just to help on this one with some info for you guyz.

 

The rms is the real power, and peak power is the theoretical maximum.

 

The speakers were given in rms as 100watts, and so an amp which gives 100watt rms is OK.

The peak power is double the rms, so if your given only one you can work out the other.

 

Leigh.

 

 

The three most basic units in electricity are voltage (V), current (I) and resistance ®. Voltage is measured in volts, current is measured in amps and resistance is measured in ohms.

A neat analogy to help understand these terms is a system of plumbing pipes. The voltage is equivalent to the water pressure, the current is equivalent to the flow rate, and the resistance is like the pipe size.

 

There is a basic equation in electrical engineering that states how the three terms relate. It says that the current is equal to the voltage divided by the resistance.

 

I = V/r

 

Let's see how this relation applies to the plumbing system. Let's say you have a tank of pressurized water connected to a hose that you are using to water the garden.

 

What happens if you increase the pressure in the tank? You probably can guess that this makes more water come out of the hose. The same is true of an electrical system: Increasing the voltage will make more current flow.

 

Let's say you increase the diameter of the hose and all of the fittings to the tank. You probably guessed that this also makes more water come out of the hose. This is like decreasing the resistance in an electrical system, which increases the current flow.

 

Electrical power is measured in watts. In an electrical system power (P) is equal to the voltage multiplied by the current.

 

 

P = VI

The water analogy still applies. Take a hose and point it at a waterwheel like the ones that were used to turn grinding stones in watermills. You can increase the power generated by the waterwheel in two ways. If you increase the pressure of the water coming out of the hose, it hits the waterwheel with a lot more force and the wheel turns faster, generating more power. If you increase the flow rate, the waterwheel turns faster because of the weight of the extra water hitting it.

 

In an electrical system, increasing either the current or the voltage will result in higher power. Let's say you have a system with a 6-volt light bulb hooked up to a 6-volt battery. The power output of the light bulb is 100 watts. Using the equation above, we can calculate how much current in amps would be required to get 100 watts out of this 6-volt bulb.

 

You know that P = 100 W, and V = 6 V. So you can rearrange the equation to solve for I and substitute in the numbers.

 

 

I = P/V = 100 W / 6 V = 16.66 amps

What would happen if you use a 12-volt battery and a 12-volt light bulb to get 100 watts of power?

 

 

100 W / 12 V = 8.33 amps

So this system produces the same power, but with half the current. There is an advantage that comes from using less current to make the same amount of power. The resistance in electrical wires consumes power, and the power consumed increases as the current going through the wires increases. You can see how this happens by doing a little rearranging of the two equations. What you need is an equation for power in terms of resistance and current. Let's rearrange the first equation:

 

 

I = V / R can be restated as V = I R

Now you can substitute the equation for V into the other equation:

 

 

P = V I substituting for V we get P = IR I, or P = I2R

What this equation tells you is that the power consumed by the wires increases if the resistance of the wires increases (for instance, if the wires get smaller or are made of a less conductive material). But it increases dramatically if the current going through the wires increases. So using a higher voltage to reduce the current can make electrical systems more efficient. The efficiency of electric motors also improves at higher voltages.

 

This improvement in efficiency is what is driving the automobile industry to adopt a higher voltage standard. Carmakers are moving toward a 42-volt electrical system from the current 12-volt electrical systems. The electrical demand on cars has been steadily increasing since the first cars were made. The first cars didn't even have electrical headlights; they used oil lanterns. Today cars have thousands of electrical circuits, and future cars will demand even more power. The change to 42 volts will help cars meet the greater electrical demand placed on them without having to increase the size of wires and generators to handle the greater current.

 

Not that this has anything to do with anything lol

Well that's the basic physics lesson done and I'm delighted to see that your spelling and grammar have improved immeasurably !

Just one point guyZ,

 

The one thing that blows speakers is the amps flat-topping (distortion). The amp should be rated higher than the speakers really if you are gonna push the system. It's difficult to explain why the amp distorting causes the speakers to blow cos I'd have to go in to waveforms etc and run the risk of sounding like warren

 

A 120 Watt RMS per channel amp would be about right and use the gain control to bring it down to 100W if that makes any sense what-so-ever

 

CheerZ,

 

Andy

which is what I was thinking anyways............... nice one,

shopping and wiring time !

 

and time to work out how the hell to get 'em in the front doors - plastic adapters ? mdf collars ? sod it, will just leave them loose on the floor and finish decorating first, god it's quite at work tonight - any interesting site suggestions welcome - not thehun.com damn firewalls !

 

[This message has been edited by zach (edited 05-10-2001).]

I have a tip for late nights - MSN Firewall may block it though

 

andrewduff60@hotmail.com

 

CheerZ,

 

Andy