# Quick maths problem.

Discussion in 'General Discussion' started by banja, May 9, 2017.

1. banja

# Posts: 2,242

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 (%)?

2. iamdjdz

Half?

3. Jonny L

# Location: Lincoln

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

4. banja

# Posts: 2,242

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.

5. banja

# Posts: 2,242

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.

Last edited: May 9, 2017
6. dirtybeatfreak

# Location: Eastbourne

I never try these because I am a thicko and will probably make myself look stupid, but here is my take on it looking at it as a numerical variable???

power is 1 - 4
volume 100 - 200

original perceived volume = 100 and power is = 1, to double the volume increase power to 4 = volume is now 200

reduce the power by 25% or 1 notch down to 3 which would reduce volume to 175. So it has been reduced by 25%

EDITED a few times! My poor Brain!

Last edited: May 9, 2017
7. SpeedFreak

# Location: Bristol

Perceived volume is relative to the distance from the source.

Specifically the square of the distance.

8. Avenged7Fold

# Location: Surrey

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

9. GeForce

# Location: Surrey, UK

You have reduced the perceived volume by 43.75%.

Edit: misread: below is correct, should be 1 - 0.866 = 13.4% reduction in volume.

Last edited: May 9, 2017
10. banja

# Posts: 2,242

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.

Last edited: May 9, 2017
11. wonko

# Location: Outside the asylum

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?

12. banja

# Posts: 2,242

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

13. Avenged7Fold

# Location: Surrey

I understood it as V=P^2

Where V is original volume and P is original power because:

To get 2*V OP says you can multiply P by 4, is how i read it.

So in my head it is V is proportional to P^2

So at 75% of P

(0.75P)^2=xV Where x is the fraction of the original volume

14. banja

# Posts: 2,242

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

15. wonko

# Location: Outside the asylum

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...

16. Avenged7Fold

# Location: Surrey

I am struggled to picture it from the OP (not the usual format of the maths questions i come across )but i will concede that my assumption of V=cP^2 shouldn't be the case as normally it takes significantly more power to increase the perceived volume and so my answer of increasing p by 1/3 nearly doubles volume is clearly wrong from a realistic point of view

Last edited: May 9, 2017
17. dl8860

# Location: Surrey

V^2 is proportional to P, not V proportional to P^2.

The first statement is "In order to double the perceived volume of an amplifier you need to multiply the power by 4 times."

V=P^2 means that if you increase P by 2, then V will increase by 4.

V^2=P means that if you increase P by 4, then V will increase by 2.

So I agree with a 13.4% reduction

18. banja

# Posts: 2,242

Well that's all greek to me

19. dl8860

# Location: Surrey

Actually I'm starting to think we don't have enough info. The fact that you need to quadruple the power to achieve double the volume isn't enough information to define the relationship between the two.

20. danlightbulb

# Posts: 2,127

Not sure where the squaring is coming from in the above examples, given the information in the question.

The op clearly describes a formula that looks like this:

4P = 2V (four times the power results in twice the volume). For the purposes of this question it could be being assumed to be linear.

Therefore is you reduce the left hand side by 25% you have to reduce the right hand side by 25%. Reducing by 25% is the same as having 75% remaining.

0.75 x 4P = 0.75 x 2V
3P = 1.5V

Is what the formula ends up as.

Although it's not clear what the exact question is, I would say the answer is that if you reduce the power by 25% you end up with one and a half times the original volume.

Last edited: May 9, 2017