|
Audio Asylum Thread Printer Get a view of an entire thread on one page |
For Sale Ads |
46.140.171.83
In Reply to: RE: in a word: acceleration. posted by Bill Fitzmaurice on June 10, 2023 at 08:03:45
No, it is not a red herring. Acceleration is the rate of change in velocity and this has a direct impact on the amplitude of the wave generated.
If I have a driver that moves 1mm in 1 millisecond vs. the same driver that moves 1mm in 10 microseconds, which one do you think will have the higher amplitude? The force of compression of air is much greater with traveling the same distance in a much shorter time frame.
A larger surface area will also give a higher sensitivity for a given force as clearly more air is being moved. A large e-stat panel will therefore be able to make a reasonable SPL level without having a particularly strong motor and therefore somewhat slower acceleration.
You can make ribbon speakers with high sensitivity as well...but it takes very powerful magnets to do so to generate the acceleration needed for high sensitivity. Again, back to acceleration.
Follow Ups:
A diaphragm accelerates as fast as it needs to for the required excursion and frequency. In your example of 1mm in 1 millisecond vs. the same driver that moves 1mm in 10 microseconds they would not reproduce the same frequency. A wave period of 1 millisecond is 1kHz. A wave period of 10 microseconds is 100kHz.Acceleration only enters the picture when comparing two different excursions. A given tone at 1mm excursion would require ten times the acceleration as the same tone at 0.1mm excursion. If the transducer lacked the necessary acceleration to realize the given excursion within the wave period the result would be a lower frequency, not lower sensitivity.
Edits: 06/12/23 06/12/23
"It is not a very long post, but if you are really busy, here is a TL;DR summary:
Displacement in speaker diaphragm does not generate the pressure wave in air that is the sound we hear. Acceleration does.
What causes the diaphragm to accelerate (and therefore the air in front of it to pressurize) is force. A lower acceleration only results in lesser sound pressure level (SPL), i.e. less loud, not how quickly it appears, i.e. the lack of speed.
The speed at which we can modulate "force" is unrelated to the mass of the diaphragm.
I'll expand a little further on Purifi's blog post, since someone will inevitability ignore the last point above and will insist that acceleration is force divided by mass, and therefore lower mass gives higher acceleration. So how do we find how much acceleration we need?
The late Siegfried Linkwitz (RIP) gave us a very handy formula to predict the free field SPL generated by a speaker driver, given its size, diaphragm travel, and frequency. [Link, see the box "Theory Behind the Nomographs"] It is:
SPL = 94.3 + 20 log10(x) + 40 log10(f) + 40 log10(d) - 20 log10(r)
where: x is the peak-to-peak diaphragm travel in meters,
f is frequency in Hz,
d is the effective diameter of the diaphragm in meters (d = sqrt(4 * Sd / pi), with Sd = effective area in m^2)
r is the listening distance in meters
Now, say we want to generate the same SPL at two different frequencies, f1 and f2, what will the diaphragm travels (x1 and x2) be?
SPL1 = 94.3 + 20 log10(x1) + 40 log10(f1) + 40 log10(d) - 20 log10(r)
and
SPL2 = 94.3 + 20 log10(x2) + 40 log10(f2) + 40 log10(d) - 20 log10(r)
Since we want SPL1 = SPL2 ,and "d" and "r" remain the same, we have:
20 log10(x1) + 40 log10(f1) = 20 log10(x2) + 40 log10(f2)
or
log10(x1) + 2 log10(f1) = log10(x2) + 2 log10(f2)
or
x1 * f1^2 = x2 * f2^2
or
x2 = x1 * f1^2/f2^2
So, the amount of travel the diaphragm needs to produce the same SPL in inversely proportional to the ratio of the frequencies squared (i.e. with the same diaphragm travel, SPL goes up/down by 12 dB/octave).
How about acceleration? Well, given the displacement amplitude x, acceleration = x * (2*pi*frequency)^2. Which means acceleration goes up by frequency squared. Since, for the same SPL:
x2 = x1 * f1^2 / f2^2
and
a2 = x2 * (2*pi * f2)^2
= x1 * (f1^2 / f2^2) * (2*pi * f2)^2
= x1 * (2*pi * f1)^2
= a1
The acceleration amplitudes are the same! And therefore forces. Amazingly we need the same force amplitude to produce the same SPL regardless of frequency. Of course, the rate of fluctuation of force is higher with higher frequencies, but the force magnitude is independent of frequency. We need to wiggle the diaphragm more frequently, but that is completely countered by the fact that we need to wiggle it less far.
There are plenty of other reasons why a woofer is not suitable to produce treble. Mass of the diaphragm ain't one."
All this math very few people will bother to try and understand is a waste of time and effort. Modern hifi is much simpler than that with the tools that do all the math for you.
Go listen to some music on whatever you have and be happy.
more simply:Acceleration = Force/Mass
Same force, high mass → low acceleration = low SPL
Same force, low mass → high acceleration = high SPLThe relationship (for first order effects) is independent of frequency.
Or High force, same mass - high acceleration = high SPL
Low force, same mass - low acceleration = low SPL
Edits: 06/12/23
FAQ |
Post a Message! |
Forgot Password? |
|
||||||||||||||
|
This post is made possible by the generous support of people like you and our sponsors: