Two questions, ones difficult

[Maximilian]

1. If i have a big speaker playing at 70 decibels in a sealed room, and i add two more speakers exactly the same as the first and they play the same sound at 70 decibels, so theres 3 speakers playing the same sound at 70db instead of just 1. Is it louder? Im thinking no, but i dunno....

2. Right, this is the difficult one. Air floats upwards in water ya? Say ive got a ball and i fill it with water, what percentage of that ball will have to be filled with air for the ball to float back to the surface of a body of water?

 

[damonpip]

1) Yes, it would be louder. I'm not sure exactly how much louder in db, but it would be louder.

2) It depends on the density and thickness of the ball. For something thin like a ping-pong ball, there would only have to be a very small percentage of air (maybe 5%).

 

[BigPoppa]

1)General rule in car audio: Everytime you double cone area: +3 dB. Everytime you double power: +3dB. Its a general rule, once you hit the extremes, power compresion comes into play and things change a lot.

So you tripled power and cone area. I'd say the system would be around 80dB, making it twice as loud as before. 10dB = twice as loud. So 80db = 2x 70db = 4x 60db, etc.

2) Once the density of the system (ball + air_inside +water_inside) < water, ball floats to the surface. Basically, you're compensating for the shell of the ball's density with the air's density. It would be a ratio depending on the density of the ball's shell.

 

[itachi]

this sounds like physics hw..

1. it's relative to whether the amplifier can handle the load or not. the amplifier supplies power to each of the transducers in the form of current. it works by changing the polarity of the current. once it reaches the transducer, its converted by the driver into energy. when you add a speaker to a set (assuming the amplifier has the power to supply multiple speakers with a consistent current supply), you're effectively doubling the power available to the system. that yields an increase by a few decibels, the exact amount can be found mathematically using logarithms.

2. a ping-pong ball would never sink. in order for it to sink, the density of the material has to be greater than water.. and a ping pong ball has a density of around 0.75 g/cm^3.
density = mass / volume. mass varies with depth.. f = m*g.. df = dm*g. this is where it could get difficult.. if the ball can't be compressed then df = dpVg, if it can then df = dpdVg, an equation with 2 variables. gonna ignore the latter case. when the sphere is in equilibrium at a depth of x meters.. f = pVg = xnAg, p is the effective density of the sphere, n is the density of water. f = (1/3)rApg = xnAg -> (1/3)rp = xn.. p = 3xn/r = 3n(x/r), and with the conditions- p < 3n*2 = 6n..
from before.. p = m / V = m_water / Vwater + m_air / Vair < 6n -> m_water*Vair/Vwater = 6n*Vair - m_air.. Vair/V = (6nVair - m_air)Vwater/(V*m_water).. and that's the percentage of air that you'd need. that's all assuming the thickness of the ball is negligible and the temperature is constant though.

 

[Maximilian]

Sweet.... im outta school now though, so these are just things i was wondering 😀

 

[Navid]

1- Adding another speaker that has 70dB of power doubles the amplitude (not power). So, the sound increases by 6dB. Adding 2 speakers increases it by 12dB. So, you would be at 82dB. The amplitudes add up since the sound from the speakers is correlated.

If it was white noise, the noise would be non-correlated. Then, it would add up in power instead of in amplitude. Then, it would go up to 76dB. But, since you said the speakers are playing, I assume that you mean it is music or audio. Then, the sound is correlated.

Another thing you have to consider is the phase of the sound coming from the three speakers at your ear. As long as we are talking about a small room, all the three sounds would be in phase and what I said would be correct.

However, if you are in a very large theatre, you need to consider where each of the three speakers are with respect to your ear and you may get different results. You may get distortion and beating.


2- The weight of the ball is reduced by the weight of the displaced water. So, if we assume that there is the right amount of air in the ball that keeps it floating, the volume of the ball that is under the water multiplied by the weight of the unit volume of water should be equal to the total weight of the ball (including the water in it).

If you pick a value for the displaced water like half the volume of the ball (the ball is half way under water), this turns into a single equation with a single unknown (volume of water inside the ball). A single equation with a single unknown can be solved.

Alternatively, you can solve this with the amount of displaced water as a parameter. Then, you will get a curve that tells you exactly how low the ball will be as a function of the amount of water in it.