I did do some calculating today from the specs of a 6.5" driver. That driver has Cms = 0.128 cm/ Newton. So here is how I proceeded.
(1) Calculate the pressure ( + or -) inside the box that results in a net force acting on the cone of the driver equal to 1 Newton. There is atmospheric pressure on the outside, so the answer is atmospheric pressure + 1 Newton divided by Sd. Call this pressure P2.
(2) Vb, or volume of air in the box with the driver at rest is made equal to Vas. From Boyle's Law, Vb' = (Atmospheric Pressure/ P2) times Vas. Solve for Vb' where Vb' is the compressed volume corresponding to 1 Newton of net force.
(3) Vb- Vb' is the difference between the interior volume of the box made equal to Vas and the volume enclosing the air with the driver compressing it (ever so slightly!). I calculated Vb - Vb' = 22.06 cubic cm.
(4) Divide Vb- Vb' by Sd which equals how far in the cone of the driver goes to give an acoustic force of 1 Newton. I got 0.174 cm which is somewhat greater than the mechanical compliance of 0.128 cm for 1 Newton of force. The distance corresponding to 1 Newton of force of mechanical and acoustic compliance should be the same.
I suppose that I could post the actual calculations, but I doubt that there is that much interest in seeing them. By the way,
Vb'/ Vb = 0.99923.
I can't believe that I made all of this effort.
Regards,
Pete
(1) Calculate the pressure ( + or -) inside the box that results in a net force acting on the cone of the driver equal to 1 Newton. There is atmospheric pressure on the outside, so the answer is atmospheric pressure + 1 Newton divided by Sd. Call this pressure P2.
(2) Vb, or volume of air in the box with the driver at rest is made equal to Vas. From Boyle's Law, Vb' = (Atmospheric Pressure/ P2) times Vas. Solve for Vb' where Vb' is the compressed volume corresponding to 1 Newton of net force.
(3) Vb- Vb' is the difference between the interior volume of the box made equal to Vas and the volume enclosing the air with the driver compressing it (ever so slightly!). I calculated Vb - Vb' = 22.06 cubic cm.
(4) Divide Vb- Vb' by Sd which equals how far in the cone of the driver goes to give an acoustic force of 1 Newton. I got 0.174 cm which is somewhat greater than the mechanical compliance of 0.128 cm for 1 Newton of force. The distance corresponding to 1 Newton of force of mechanical and acoustic compliance should be the same.
I suppose that I could post the actual calculations, but I doubt that there is that much interest in seeing them. By the way,
Vb'/ Vb = 0.99923.
I can't believe that I made all of this effort.
Regards,
Pete
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It is not an issue for small signal, but it can be for large signal as was shown in the Linkwitz writeup that I previously linked to.
Xmax of the 6.5" driver that I made my calculations from the specs of equals 2.55 mm. That would make my example large signal I believe, that is, I was looking at an excursion of 1.28 mm or about half of Xmax. I didn't intend for my calculation to reveal anything about non-linearity, but maybe it could? I'll have to take a look at the link to Linkwitz that you provided.
Thanks,
Pete
I can't say that I exactly follow what your graph shows. What are the units of the X and Y axes?
-Pete
Quiz
Let's say that besides being a speaker builder, you are also a mountain climber. This summer you plan to go to the top of Mt. Everest. To celebrate arriving at the peak, you want to play some of your favorite recordings through a pair of stereo speakers while sipping some champagne. Your preferred type of speakers are sealed or vented. What is it about Mt. Everest that would lighten your burden in wanting to listen to your recordings on the peak?
After giving a little bit of thought to it, this is a fairly easy quiz, so no hints. However I will say that there is a connection to some of what has been discussed in this thread.
Good Luck with the answer,
Pete
Let's say that besides being a speaker builder, you are also a mountain climber. This summer you plan to go to the top of Mt. Everest. To celebrate arriving at the peak, you want to play some of your favorite recordings through a pair of stereo speakers while sipping some champagne. Your preferred type of speakers are sealed or vented. What is it about Mt. Everest that would lighten your burden in wanting to listen to your recordings on the peak?
After giving a little bit of thought to it, this is a fairly easy quiz, so no hints. However I will say that there is a connection to some of what has been discussed in this thread.
Good Luck with the answer,
Pete
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You would have to adjust the values for air pressure and temperature in the tool you use to design the speakers.
Edit: ... and re-calculate the T-S parameters of the drivers.
That's close, but it doesn't answer the question how planning to listen to recorded music on the peak of Mt.Everest is made easier by something characteristic about being there.
More importantly, it doesnt answer the question of how you get the bottle of bubbly up the mountain intact.
Does wine freeze at -45 deg. C?
Can the glass bottle holding the (carbonated) bubbly withstand the increased pressure difference?
If it is still liquid, and the bottle is still intact, will you be able to transfer it to a glass?
Would you want to drink bubbly when it's at -45 deg. C?
I'll admit to not remembering much at this moment about exactly what is involved in a serious mountain climb such as to Mt. Everest. But one thing that I had in mind for the quiz is the idea that you have to severely limit the bulk and weight of what you carry with you in your backpack.
The density of air at Mt. Everest is about 1/3rd of what it is at sea level. What I'm trying to get across isn't enormous, but I think that it's kind of interesting.
-Pete
Answer to Quiz
Vas is directly proportional to air density. Air density on Mt. Everest is 1/3rd of air density at sea level. So the size (and therefore also weight) of the Mt. Everest version of your favorite closed-box or vented system would be 1/3rd smaller.
In making your climb up Mt. Everest, heavy and bulky loudspeaker systems on your back is the last thing in the world that you would want.
The total moving mass would be also lower due to 1/3rd air density, but off the top of my head, I don't know how much that would raise free air resonance Fs. I would hazard the guess that both sea level and Mt Everest versions would be very similar, other than the size.
With a vented system, maybe everything is not quite so straightforward. At the moment I can't say.
Regards,
Pete
Vas is directly proportional to air density. Air density on Mt. Everest is 1/3rd of air density at sea level. So the size (and therefore also weight) of the Mt. Everest version of your favorite closed-box or vented system would be 1/3rd smaller.
In making your climb up Mt. Everest, heavy and bulky loudspeaker systems on your back is the last thing in the world that you would want.
The total moving mass would be also lower due to 1/3rd air density, but off the top of my head, I don't know how much that would raise free air resonance Fs. I would hazard the guess that both sea level and Mt Everest versions would be very similar, other than the size.
With a vented system, maybe everything is not quite so straightforward. At the moment I can't say.
Regards,
Pete
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