The ideal spring type, or minimizing the effect of Cms on distortion

454Casull · 7th October 2008, 12:21 AM

Is the ideal restoring force applied by the spring on the moving assembly:

1) constant => F(restoring) = c for all x, c is a constant
2) linear => F(restoring) = kx for all x
3) parabolic => F(restoring) = kx^2 for all x (might be another coefficient in there)

I think Klippel covered this in one of his papers but either I can't find it or it disappeared off the Internet.

LaMa · 7th October 2008, 10:12 AM

ideally, the mechanical compliance of the moving system or (1/stiffness) is constant for each position of the cone. So answer (1). This is also the assumption in the most simple electromechanical models.

Unfortunately, this is not the case for real world drivers as can be seen in Klippel measurments of Cms kindly borrowd from npdang on diyma. The measured Cms values can be reasonably approximated by an quadratic fit, so (3).

sreten · 7th October 2008, 12:57 PM

Hi,

I'm sorry but 1) is a non-starter and not the definition of a spring.

2) Is of course a classic ideal spring, theoretically zero distortion.

3) you can reword to as non-linear, not necessarily parabobic.

Practically all suspensions are 3).
The air in box is very nearly 2 except at very high compression.

/sreten.

phase_accurate · 7th October 2008, 01:02 PM

Which 4) leads to the conclusion that the combination of 3) and the linear behaviour of the air spring is also quite linear IF the effect of the air spring is dominating.

Regards

Charles

454Casull · 7th October 2008, 05:09 PM

Quote:

Originally posted by sreten
Hi,

I'm sorry but 1) is a non-starter and not the definition of a spring.

2) Is of course a classic ideal spring, theoretically zero distortion.

3) you can reword to as non-linear, not necessarily parabobic.

Practically all suspensions are 3).
The air in box is very nearly 2 except at very high compression.

/sreten.

A spring need not be linear (follow Hooke's Law), so I don't see how a constant force means that the device is not a spring.

2) - is F=kx really the ideal for zero distortion? This conflicts with what LaMa said.

sreten · 7th October 2008, 05:44 PM

Hi,

A constant force is exactly that - a constant force, no springyness.

LaMa did not read your list slowly enough, constant compliance = 2).

/sreten.

454Casull · 7th October 2008, 11:28 PM

So a negator spring is not a spring? I think you are reading too literally into the relationship between linear springs and Hooke's Law.

Rybaudio · 8th October 2008, 07:22 AM

A linear spring is one that obeys Hooke's law!

spring = force is a function of displacement
linear = force is a linear function of displacement (f = kx)

454Casull · 8th October 2008, 11:02 AM

Quote:

Originally posted by Rybaudio
A linear spring is one that obeys Hooke's law!

So?

phase_accurate · 8th October 2008, 01:34 PM

Just one question: If a linear restoring force were the way to go, how would you want it to behave within a loudspeaker driver (which should be able to move in two directions) ? How about the situation at zero displacement ? Equal restoring force in two directions at once ?

A spring following Hooke's law as closely as possible is still the way to go. There are possibilities to implement progressive restoring force at high excursions in order to achieve mechanical protection at high excursions though.

Regards

Charles

7th October 2008, 12:21 AM	#1
454Casull diyAudio Member Join Date: Jan 2004 Location: Toronto, ON, Canada	The ideal spring type, or minimizing the effect of Cms on distortion Is the ideal restoring force applied by the spring on the moving assembly: 1) constant => F(restoring) = c for all x, c is a constant 2) linear => F(restoring) = kx for all x 3) parabolic => F(restoring) = kx^2 for all x (might be another coefficient in there) I think Klippel covered this in one of his papers but either I can't find it or it disappeared off the Internet.

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7th October 2008, 10:12 AM	#2
LaMa diyAudio Member Join Date: Sep 2002 Location: NL	ideally, the mechanical compliance of the moving system or (1/stiffness) is constant for each position of the cone. So answer (1). This is also the assumption in the most simple electromechanical models. Unfortunately, this is not the case for real world drivers as can be seen in Klippel measurments of Cms kindly borrowd from npdang on diyma. The measured Cms values can be reasonably approximated by an quadratic fit, so (3).

7th October 2008, 12:57 PM	#3
sreten diyAudio Member Join Date: Nov 2003 Location: Brighton UK	Hi, I'm sorry but 1) is a non-starter and not the definition of a spring. 2) Is of course a classic ideal spring, theoretically zero distortion. 3) you can reword to as non-linear, not necessarily parabobic. Practically all suspensions are 3). The air in box is very nearly 2 except at very high compression. /sreten.

7th October 2008, 01:02 PM	#4
phase_accurate diyAudio Member Join Date: May 2002 Location: Switzerland	Which 4) leads to the conclusion that the combination of 3) and the linear behaviour of the air spring is also quite linear IF the effect of the air spring is dominating. Regards Charles

7th October 2008, 05:44 PM	#6
sreten diyAudio Member Join Date: Nov 2003 Location: Brighton UK	Hi, A constant force is exactly that - a constant force, no springyness. LaMa did not read your list slowly enough, constant compliance = 2). /sreten.

7th October 2008, 11:28 PM	#7
454Casull diyAudio Member Join Date: Jan 2004 Location: Toronto, ON, Canada	So a negator spring is not a spring? I think you are reading too literally into the relationship between linear springs and Hooke's Law.

8th October 2008, 07:22 AM	#8
Rybaudio diyAudio Member Join Date: Mar 2003 Location: State College, PA	A linear spring is one that obeys Hooke's law! spring = force is a function of displacement linear = force is a linear function of displacement (f = kx)

8th October 2008, 01:34 PM	#10
phase_accurate diyAudio Member Join Date: May 2002 Location: Switzerland	Just one question: If a linear restoring force were the way to go, how would you want it to behave within a loudspeaker driver (which should be able to move in two directions) ? How about the situation at zero displacement ? Equal restoring force in two directions at once ? A spring following Hooke's law as closely as possible is still the way to go. There are possibilities to implement progressive restoring force at high excursions in order to achieve mechanical protection at high excursions though. Regards Charles

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