Quick maths problem.

[banja]

This should be simple with some basic algebra, but I can't work it out (maths never my strong point)

In order to double the perceived volume of an amplifier you need to multiply the power by 4 times.

If I reduce the power by 25%, how much have I reduced the perceived volume by (%)?

 

[banja]

Half?

No because that would mean if increased the power back by 25% it would double the perceived volume, which isn't correct.

It's just basic a maths equation, but I can't work out how to construct it.

 

[banja]

If you are talking percentages then both have been reduced by 25%. I think?

No I think it's an exponential relationship, not a direct correlation.

Edit - although it could be 25%, if that is what the mathematical result is.

 

[banja]

Are you saying that the volume is directly proportional to the square of the power?

If so, then whatever you reduce the power to, volume will be reduced to a square of that:

V proportional to P^2

If P is 25% less then

(0.75P)^2

So reducing power by 25% will make volume be 56.25% of the volume at full power

It's a loose approximation in actuality (not interested in the physics of the issue etc) but as a numerical value I just want the figure that corresponds to the equation.

So you think it's just over half? Is that right?? Because then if I were to increase that power back to full power (+44%) I wouldn't be quadrupling the perceived volume back to the original 100%.

I think. I'm confusing myself

My back of an envelope exponential X/Y graph puts it at around 85%, but I've no idea if that's right.

 

[banja]

Isn't it the other way around... V ^2 is proportional to P

Reducing power by 25% gives 0.75 of original power.

So reduced perceived volume would be sqrt(0.75) of original volume = 0.866 or a 13.4% reduction

Or am I having a bad day?

That's what my crappy graph suggested, but not sure it's right.

 

[banja]

To put it another way, reducing the power by 75% (to 25% of original) will halve the perceived volume.

 

[banja]

Mmm, let's try to put it another way.
If we start with with V = c P^2, and we quadruple the power then the revised volume would be = c (4P)^2, or 16x the original volume (not the 2x expected).

But if P = c V^2 and we double the volume then we get a revised power = c (2V)^2 which gives us the 4x we expect...

Well that's all greek to me

 

[banja]

Its not the exact science I'm looking at and I'm not looking for the proper physics of the amp situation - I just wanted to get the approximate % via the maths. So, assuming four times the power equals twice the perceived volume, what % of 100% volume would 75% power equal. Simples!

 

[banja]

The only relationship that fits your description of 'four times the power equals twice the volume', for all values of V and P is P=V^2, as below


So, let's take a certain value of P. You want to take 75% of that P, and work out the corresponding Volume decrease.

Thanks to everyone for contributing. Sorry if my explanations were crap. And props actually to Geforce for being ti first to suggesting 86%/13%

If we start at P = 4096, the corresponding V is 64. If you take 75% of 4096, that gives you 3072. P = V^2, so V is root(P) = 55.4. That's 0.86% of the original value, a decrease of 13.4%.

If we start at P = 16, then V here is 4. If you take 75% of 16, that's 12. Here, V is 3.46. That's 0.86% of the original value again, confirming the decrease is 13.4%.

This is basically the same as Wonko said in post #11, but just a bit more illustrated

Yes, thanks! As I said, my crude X Y plot of power vs volume came out at about the 85 mark, so I think this is definitely correct

Not sure who got to the answer first, but thanks all for contributing. Sorry if my explanations were crap.