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How fast Coil Springs are? Greatest mystery today?

by Darp
Tags: coil spring, speed, spring
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Darp
#1
May8-14, 01:48 PM
P: 14
Hello,

Hope someone finds this a challenge.

Thought it would be a snap to find out, goggled it, nothing. Weeks later have not found one person in the world that knows of one example of how fast coil springs release. Just talked with Hyperco a major manufacturers head engineer, he eventually admitted he did not know, because they never have a need to know. Talked to the spring manufactures Institute, they gave me the formula (which I will copy below), but did not have a clue in real world of a single example. Have talked to one of the best experts in the world on airguns that are spring fired (piston generates air pressure that fires pellet, he had no idea how fast the spring is.

Have watched high-speed video of valve springs in engines at 10,000 rpm, but being valves go up and down 1/2" in 180 degrees, that is 20,000 inches per minute or 38 FPS, very slow, Heck people can throw things at 130 FPS.

Hope someone here can answer the question. If you need specifics, lets say a standard 125cc motorcycle front fork coil spring, like this one http://www.ebay.com/itm/76-Honda-MT1...-/321367298914

Maybe that one has twice as many coils as one be ideal (too close together). Lets say its got a 150 pounds per inch spring rate, is 20" extended and 10" compressed and 1.25" in diameter. Lets say the entire spring weighs two pounds. Plug in any numbers you want.

The problem I have is have no idea whether its 40 FPS or 400 FPS that coil springs can reach with no mass besides the spring itself to move. Have a feeling someone here can answer it. Am guessing they reach max velocity 2/3 of way to decompression. One end would be anchored, other end free released.

BTW I am not good enough at math to handle that equation.

Thanks!


INSTITUTE:
This as with most spring questions is best answered by your spring manufacturer who has the industry knowledge as well as experience in the unique action of springs and specific materials to clearly answer your questions. The basic information presented is available in the literature about dynamic actions of springs with regard in acceleration.


Basic Velocity Calculations


v velocity

vm ultimate attainable velocity

k rate of spring

g acceleration due to gravity (32.2 ft./sec.2



P force in lbs.

Ws spring weight

G shear modulus

p density

S stress

F deflection

√ squar root of



For springs with a mass ratio (P/Ws) >4



v= F √ ( kg/P+1/3Ws)



For springs with a mass ratio (P/Ws) between 1 & 4



v= F √ ( kg/P)



Lynne Carr

Executive Director

Spring Manufacturers Institute
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Darp
#2
May8-14, 04:33 PM
P: 14
Hi,

Since over 70 people have viewed this with no answers, thought maybe more info would help have found this in regard to some of the formula inputs




The below did not format well, here it is in a chart http://www.engineersedge.com/spring_general.htm

Spring Material and Types Data

Spring Design Menu | Spring Suppliers and Manufacturers

See Calculator Page for Spring Design Calculators!
Types of Springs

Compression Spring is an open-coil, helical spring that offer resistance to compressive loading.

Extension Spring is a close-coiled helical spring that offers resistance to a pulling force

Torsion Spring exert pressure along a path which is a circular arc, or, in other words, providing torque. Sometimes these springs are called torsion springs, motors springs, power springs.

Common Spring Materials and Properties
Material Tensile Strength min.
(psi x 103) Modulus of Elasticity
(psi x 106) Modulus in Torsion
(psi x 106) Max. Design Temp
(deg F)
Music Wire 229 - 300 30 11.5 250
Chrome Vanadium 190 - 300 30 11.5 425
Stainless Steel 302 125 - 320 28 10 550
Stainless Steel 17-7 (313) 235 - 335 29.5 11 600
AlephZero
#3
May8-14, 07:11 PM
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Quote Quote by Darp View Post
Weeks later have not found one person in the world that knows of one example of how fast coil springs release.
I guess you are haven't found the right people to ask. It would be very straightforward to get an answer for a particular spring by making a simple mathematical model of it and solving it numerically. Ignoring "small" effects like the exact geometry of the turns at the ends of the spring etc, the only relevant parameters would be the length, stiffness, mass, and how much it was initially compressed.

But it seems to be a rather theoretical question, because if your are using a spring to fire a projectile, you can't ignore the mass of the projectile. If the spring mass is small compared with the projectile mass, the answer is just a question about simple harmonic motion that any mechanical engineer should be able to solve, without making a computer model.

Have watched high-speed video of valve springs in engines at 10,000 rpm, but being valves go up and down 1/2" in 180 degrees, that is 20,000 inches per minute or 38 FPS, very slow, Heck people can throw things at 130 FPS.
The velocity is irrelevant there. Sure, people can throw something like a baseball at 130 FPS. But f you can find somebody who can throw something at 130 FPS and then stop it again within half an inch, that would certainly be impressive. The relevant quantity is acceleration, not speed.

BTW I am not good enough at math to handle that equation.
.. which suggests that even if somebody spent their own time solving the problem for you, you might not understand the answer,

FWIW I can't make any sense of the formulas from Lynne Carr, at least in the form you gave them here. A reference to the standard literature would have been more helpful than a garbled formula IMO. A screen-dump image of the original email might make more sense than your attempt to reproduce it.

Chestermiller
#4
May8-14, 08:49 PM
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How fast Coil Springs are? Greatest mystery today?

To expand on what AlephZero said, the spring constant is not going to be a constant because of the large deformation of the spring (it is initially half its unextended length). The local tensile force F in the spring is going to be a function of the local axial strain of the spring, du/dx, where u is the displacement, and x is location along the unextended spring. This behavior has to be measured in advance, particularly in the range of negative du/dx, before you one can make predictions. Then, the kind of model discussed by AlephZero can be developed and solved. The compression is going to release in the form of a wave traveling backwards from the released end to the fixed end.

Chet
Darp
#5
May8-14, 09:02 PM
P: 14
Hi Aleph,

Thank you for the response! The formula is as got it in email from her. If anyone just knows empirically, even a ball point pen spring. Anything would help.

I do not need to have a projectile, just how fast a spring on released side can get up to is all I need. So far it seems no one in the world has the knowledge of how fast coil springs can go, that I can find. Maybe later tonite here.


If you search Google for "coil spring release speed" this thread makes first page!

And none of the other linkson first page address the question.

It seems the question has never been asked on the internet. To my great surprise.

Acceleration is important but unless it reaches a certain velocity, coil springs are worthless for the app. That is about 180 FPS.

Thanks for the reply.
Darp
#6
May8-14, 09:15 PM
P: 14
Quote Quote by Chestermiller View Post
To expand on what AlephZero said, the spring constant is not going to be a constant because of the large deformation of the spring (it is initially half its unextended length). The local tensile force F in the spring is going to be a function of the local axial strain of the spring, du/dx, where u is the displacement, and x is location along the unextended spring. This behavior has to be measured in advance, particularly in the range of negative du/dx, before you one can make predictions. Then, the kind of model discussed by AlephZero can be developed and solved. The compression is going to release in the form of a wave traveling backwards from the released end to the fixed end.

Chet
Thanks CM, I am willing to deform less or more. Can not use something like car spring, but motorcycle, scooter spring or smaller is fine. Do not want to use titanium because of expense although it would be better.

So can adapt to any coil spring that weighs 3 pounds or less. The problem is have no evidence yet that coil springs are fast. It could be a 10" x 1" spring compressed to 7" just as long as it can get to 180 FPS (severely doubt that short of spring could do it).

This link might help https://www.efunda.com/designstandar..._frequency.cfm
Darp
#7
May8-14, 09:38 PM
P: 14
Okay, this is a thicker coil than thought would be necessary for a 100 pound/inch spring rate. But this has all specs for it, I think for a formula. And it weighs less that thought, that is good.

Does this help?



Loads & Rates
True Maximum Load, True Fmax : 440.120 lbF
Maximum Load Considering Solid Height, Solid Height Fmax : 440.120lbF
Spring constant (or Spring rate), k : 102.620lbF/in
Safe Travel
True Maximum Travel, True Travelmax : 4.289 inch
Maximum Travel Considering Solid Height, Solid Height Travelmax : 4.289 inch
Physical Dimensions
Diameter of spring wire, d: 0.250 inch
Outer diameter of spring, Douter : 1.500 inch
Inner diameter of spring, Dinner : 1.000 inch
Mean diameter of spring, Dmean : 1.250 inch
Free length of spring, Lfree : 20.000 inch
Number of active coils, na : 28
Number of total coils, nT : 30
Solid height, Lsolid : 7.500 inch
Type of ends: closed & ground
Spring index, C : 5.000
Distance between coils, Coilpitch: 0.696 inch
Rise angle of coils: 10.06
Material Type
Material type: Chrome Silicon A401
Weights & Measures
Weight of one spring, M : 1.677062 lb
Weight per one thousand springs, M : 1,677.061685 lb
Length of wire required to make one spring, Lwire : 117.810 inch
Stress Factors
Material shear modulus, G : 11,493,397.472psi
Maximum shear stress possible, tmax : 117,500.000
Wahl correction factor, W : 1.311
Suggested Part Number
Suggested Part Number : PC250-1500-30.000-CS-20.000-CG-N-IN

Got it here http://www.planetspring.com/pages/co...id=compression
Darp
#8
May8-14, 09:56 PM
P: 14
I found I can go shorter and get more travel length at same time, so that is better. less weight and more distance. We think of springs as fast but am fearing they are not.

True Maximum Load, True Fmax : 440.120 lbF
Maximum Load Considering Solid Height, Solid Height Fmax : 440.120lbF
Spring constant (or Spring rate), k : 95.778lbF/in
Safe Travel
True Maximum Travel, True Travelmax : 4.595 inch
Maximum Travel Considering Solid Height, Solid Height Travelmax : 4.595 inch
Physical Dimensions
Diameter of spring wire, d: 0.250 inch
Outer diameter of spring, Douter : 1.500 inch
Inner diameter of spring, Dinner : 1.000 inch
Mean diameter of spring, Dmean : 1.250 inch
Free length of spring, Lfree : 15.000 inch
Number of active coils, na : 30
Number of total coils, nT : 32
Solid height, Lsolid : 8.000 inch
Type of ends: closed & ground
Spring index, C : 5.000
Distance between coils, Coilpitch: 0.483 inch
Rise angle of coils: 7.02
Material Type
Material type: Chrome Silicon A401
Weights & Measures
Weight of one spring, M : 1.788866 lb
Weight per one thousand springs, M : 1,788.865798 lb
Length of wire required to make one spring, Lwire : 125.664 inch
Stress Factors
Material shear modulus, G : 11,493,397.472psi
Maximum shear stress possible, tmax : 117,500.000
Wahl correction factor, W : 1.311
Suggested Part Number
Suggested Part Number : PC250-1500-32.000-CS-15.000-CG-N-IN
Chestermiller
#9
May9-14, 06:47 AM
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OK. I did a quick analysis of the problem, and came up with a first order approximation you can use to ballpark the release velocity. This is based on the method alluded to by AlephZero, and it is similar in form to the equation given by the Institute.

[tex]v = u_0\sqrt{\frac{k}{m}}[/tex]

where k is the spring constant, m is the mass of the spring, and u0 is the initial compression displacement. When using this equation, you need to employ consistent units.

See what kind of number you come up with using this equation.

Chet
Darp
#10
May9-14, 09:35 AM
P: 14
Excellent, will do. Thanks
Darp
#11
May9-14, 12:24 PM
P: 14
Quote Quote by Chestermiller View Post
OK. I did a quick analysis of the problem, and came up with a first order approximation you can use to ballpark the release velocity. This is based on the method alluded to by AlephZero, and it is similar in form to the equation given by the Institute.

[tex]v = u_0\sqrt{\frac{k}{m}}[/tex]

where k is the spring constant, m is the mass of the spring, and u0 is the initial compression displacement. When using this equation, you need to employ consistent units.

See what kind of number you come up with using this equation.

Chet
It came out to 3FPS and using Lynn's it was 14 FPS, so something is off. A friend ran it who has engineering background. Its gotta be at least 50 FPS in real world.

Here is the spring we put into it, 227 pounds over 9 inches of compression.

Loads & Rates
True Maximum Load, True Fmax : 277.014 lbF
Maximum Load Considering Solid Height, Solid Height Fmax : 277.014lbF
Spring constant (or Spring rate), k : 30.534lbF/in
Safe Travel
True Maximum Travel, True Travelmax : 9.072 inch
Maximum Travel Considering Solid Height, Solid Height Travelmax : 9.072 inch
Physical Dimensions
Diameter of spring wire, d: 0.220 inch
Outer diameter of spring, Douter : 1.700 inch
Inner diameter of spring, Dinner : 1.260 inch
Mean diameter of spring, Dmean : 1.480 inch
Free length of spring, Lfree : 17.000 inch
Number of active coils, na : 34
Number of total coils, nT : 36
Solid height, Lsolid : 7.920 inch
Type of ends: closed & ground
Spring index, C : 6.727
Distance between coils, Coilpitch: 0.487 inch
Rise angle of coils: 5.98
Material Type
Material type: Chrome Silicon A401
Weights & Measures
Weight of one spring, M : 1.845217 lb
Weight per one thousand springs, M : 1,845.216501 lb
Length of wire required to make one spring, Lwire : 167.384 inch
Stress Factors
Material shear modulus, G : 11,493,397.472psi
Maximum shear stress possible, tmax : 119,850.000
Wahl correction factor, W : 1.222
jack action
#12
May12-14, 02:34 PM
P: 566
You will have to state more precisely the problem you are trying to solve, as what you are asking is a useless information.

A spring stores energy. You can release that enegy to do some work. If you just release the spring with no reaction force, what's the point ? The question you are asking - At what speed goes the tip of a released, unloaded, spring ? - is the same type of question as: If I short the two poles of a battery, what kind of current will I get?

The theoritical answer is an infinite current because we assume there is no resistance. For your case, it is the same answer as there is no "resistance", i.e. no mass coupled to your spring; So the theoritical speed that will reach the tip of your spring is infinite.

If we assume that one end is fixed, then the speed at that end is zero. This means that between both end, the speed can be anything between zero and infinity.

By dividing the spring into a succession of [itex]n[/itex] smaller springs (of the equivalent spring rate [itex]nk[/itex]), placed in series, which have their masses ([itex]m/n[/itex]) located at their free end, you can get an idea of the speed. The speed of the mass situated at the tip of the spring is:

[tex]v = n\sqrt{\sum^{n}_{i=1}\left(\frac{1}{i}\right)}\;u_{0}\sqrt{\frac{k}{m}}[/tex]

The larger is [itex]n[/itex], the closer you will be to infinity as the equation under the first squared root is an harmonic series if we set [itex]n = \infty[/itex].

To find the speed anywhere on the spring the equation is:

[tex]v_d = d\sqrt{ \sum^{n}_{i=(n+1-d)}\left(\frac{1}{i}\right)}\;u_{0}\sqrt{\frac{k}{m}}[/tex]

Where [itex]v_d[/itex] is the speed at a distance [itex](d/n)\;l[/itex] from the fixed end of a spring of length [itex]l[/itex]. Of course, if [itex]d=n[/itex], you get the previous equation (the speed of the free end).

Obviously, you will never reach an infinite speed in real life, just like you will not reach an infinite current by shorting two poles of a battery, because there are other resistances that you will never get rid of (like air friction for example).

All of this to say that if your spring doesn't do any work (like pushing a mass a lot larger than the spring itself), trying to determine its speed of deployment is a useless information and that is why nobody bother to determine it precisely.

Start by answering this question to help us help you: What are you trying to achieve?
Chestermiller
#13
May12-14, 04:49 PM
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Hi Jack Action,

Can you please show the derivation of your equation for the velocity? I solved the differential version of your model, but got a different result. Thanks.

Chet
Darp
#14
May12-14, 05:41 PM
P: 14
Hi Jack,

A very small weight such as 1 ounce can be used for the calculation. However a spring is accelerating itself, a 1.5 pound spring is accelerating about 1 pound of that quite a bit. So although beyond my capabilities the calculation can be solved with no weight/mass, just the spring because it is a real world practical thing. You can release a spring with no weight on it and judge the speed with a high speed camera.

I have empirically worked on this yesterday, and it looks like springs are slow, but they are powerful. It looks like 50 FPS is how fast spring is on average, or about 40 mph, slower than we can move our hands. They look fast, can have scary power, but they are slow. This was with motorcycle front fork spring released with 500 pounds compression.
jack action
#15
May12-14, 06:55 PM
P: 566
Quote Quote by Chestermiller View Post
Hi Jack Action,

Can you please show the derivation of your equation for the velocity? I solved the differential version of your model, but got a different result. Thanks.

Chet
Each spring-mass system has the following value: [itex]nk, m/n, u_0/n[/itex]

Starting at the fixed end, the first spring-mass system has a stiffness [itex]nk[/itex] and must push all masses [itex]n*m/n[/itex] (or just [itex]m[/itex]). So [itex]v_1[/itex] is:

[tex]v_1 = \frac{u_0}{n}\sqrt{\frac{n}{n}\frac{nk}{m}}[/tex]

For the next spring-mass system, the mass is modified to [itex](n-1)*m/n[/itex], but everything else stays the same. The "free" end of that is connected to the previous mass which already goes to [itex]v_1[/itex], so:

[tex]v_2 = \frac{u_0}{n}\sqrt{\frac{n}{n-1}\frac{nk}{m}}+v_1[/tex]
[tex]v_2 = \frac{u_0}{n}\sqrt{\frac{n}{n-1}\frac{nk}{m}}+\frac{u_0}{n}\sqrt{\frac{n}{n}\frac{nk}{m}}[/tex]

[tex]v_2 = u_0\sqrt{\frac{k}{m}\left(\frac{1}{n-1}+\frac{1}{n}\right)}[/tex]

And so forth for all other spring-mass systems until you reach the free end.

I know it's not an exact differential equation solution (the maximum velocities might not coincide), but as [itex]n[/itex] gets larger, each spring-mass system gets stiffer ([itex]\sqrt{nk/m}[/itex]), so inertia seems to become less relevant, meaning all spring-mass systems react in the same manner, at the same time (it also seems logical since all springs are equally preloaded).

That was just the time I had to put on this problem, so it is open to discussion and feel free to present a more complex solution to prove this one wrong.

edit: I just realized that in my previous post there is an extra [itex]n[/itex]. I think it was an error on my part. Anyway, the harmonic series is still there and still goes towards infinity as [itex]n[/itex] gets larger.
Chestermiller
#16
May12-14, 08:24 PM
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Quote Quote by jack action View Post
Each spring-mass system has the following value: [itex]nk, m/n, u_0/n[/itex]

Starting at the fixed end, the first spring-mass system has a stiffness [itex]nk[/itex] and must push all masses [itex]n*m/n[/itex] (or just [itex]m[/itex]). So [itex]v_1[/itex] is:

[tex]v_1 = \frac{u_0}{n}\sqrt{\frac{n}{n}\frac{nk}{m}}[/tex]

For the next spring-mass system, the mass is modified to [itex](n-1)*m/n[/itex], but everything else stays the same. The "free" end of that is connected to the previous mass which already goes to [itex]v_1[/itex], so:

[tex]v_2 = \frac{u_0}{n}\sqrt{\frac{n}{n-1}\frac{nk}{m}}+v_1[/tex]
[tex]v_2 = \frac{u_0}{n}\sqrt{\frac{n}{n-1}\frac{nk}{m}}+\frac{u_0}{n}\sqrt{\frac{n}{n}\frac{nk}{m}}[/tex]

[tex]v_2 = u_0\sqrt{\frac{k}{m}\left(\frac{1}{n-1}+\frac{1}{n}\right)}[/tex]

And so forth for all other spring-mass systems until you reach the free end.

I know it's not an exact differential equation solution (the maximum velocities might not coincide), but as [itex]n[/itex] gets larger, each spring-mass system gets stiffer ([itex]\sqrt{nk/m}[/itex]), so inertia seems to become less relevant, meaning all spring-mass systems react in the same manner, at the same time (it also seems logical since all springs are equally preloaded).

That was just the time I had to put on this problem, so it is open to discussion and feel free to present a more complex solution to prove this one wrong.

edit: I just realized that in my previous post there is an extra [itex]n[/itex]. I think it was an error on my part. Anyway, the harmonic series is still there and still goes towards infinity as [itex]n[/itex] gets larger.
This doesn't seem correct. When the leading end of the spring is released, the entire spring does not instantly accelerate. The only part of the spring that accelerates is the portion immediately adjacent to the released end. In your model, the only element of your system that accelerates it the mass immediately adjacent to the released end. The remainder of the spring has not responded yet.

Let ρΔx represent the amount of mass between spring cross sections x and x + Δx, where x is the distance parameter along the unstrained spring, and ρ is the linear mass density of the unstrained spring. Let u(x,t) represent the displacement of the cross section that is at location x in the unstrained spring at time t. Then a 2nd law force balance on the differential element of mass is given by:

[tex](ρΔx)\frac{∂^2u}{∂t^2}=(kl)\left(\frac{∂u}{∂x}\right)_{x + Δx}-(kl)\left(\frac{∂u}{∂x}\right)_{x}[/tex]
where k is the spring constant, and l is the unstrained length of the spring. If we divide this equation by Δx, and take the limit as Δx approaches zero, we obtain:

[tex]m\frac{∂^2u}{∂t^2}=(kl^2)\left(\frac{∂^2u}{∂x^2}\right)[/tex]
where m is the mass of the spring = ρl

The local axial strain in the spring is given by ε = du/dx. If we take the partial derivative of the above equation with respect to x, we obtain:

[tex]\frac{∂^2ε}{∂t^2}=\frac{kl^2}{m}\left(\frac{∂^2ε}{∂x^2}\right)[/tex]
This is a wave equation, with the backwards velocity of the wave equal to [itex]v_s=l\sqrt{\frac{k}{m}}[/itex]
Initially, the strain throughout the spring is uniform, and equal to ε=ε0 (where, for compression, ε0 is negative). Once the end at x = l is released, ε at x = l suddenly changes to ε=0. This change then propagates backwards along the spring as time progresses. So, at any time t, the local strain in the spring is given by:

ε = ε0 for x < l - vst

ε =0 for x > l - vst.

From this, it follows that the displacement u as a function of time and position along the spring is given by:

[itex]u=xε_0[/itex] for x < l - vst

[itex]u=ε_0(l-v_st)+(l-x)[/itex] for x > l - vst

From this, it follows that the spring velocity distribution as a function of time and position along the spring is given by:

[itex]v=0[/itex] for x < l - vst

[itex]v=-ε_0v_st[/itex] for x > l - vst

These equations indicate that the spring velocity is zero in the region where the compressive strain has not yet been released, and is equal to a constant value of -ε0vst in the region where the strain has been released. Since the initial strain is negative, the release velocity is positive. If we combine these equations, we get:

[tex]v=u_0\sqrt{\frac{k}{m}}[/tex]

where u0 is the initial compressive displacement at x = l.

Chet
jack action
#17
May12-14, 10:17 PM
P: 566
Wow! I never tought you got your equation with such an in-depth analysis!

Here is how I thought you got your equation (which is exactly the same as v= F √ ( kg/P), presented in the email):

I thought you assumed there was a massless spring with a single point mass fixed at the free end (which has a mass equivalent to the spring mass). Then:

[tex]E_k = E_s[/tex]
[tex]\frac{1}{2}mv^2 = \frac{1}{2}ku_0^2[/tex]
[tex]v = u_0\sqrt{\frac{k}{m}}[/tex]

I just applied this to a succession of springs which are all released at the same time since each one is restraint by the next.

The way I saw it (and maybe I'm wrong), if you remove the constraint at one end of the succession of springs, then all springs lose their constraints at the same time. How can a spring in the middle stay still with no force holding it in place ?

Actually, if you assumed [itex]n=1[/itex] in my equation, you will get your equation as well. That's why I found strange that a spring with its mass distributed along its length react the same way as one with its mass all concentrated at one end. Happy coincidence?
Wes Tausend
#18
May12-14, 10:24 PM
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...

It seems the end of the spring does not immediately accelerate when released. The motion transfer is done by the magnetic repulsion of the outer electron shell of the comprising atoms so therefore the speed of a spring minus the inertia is the speed of electromagnetism; the speed of light. The actual speed regarding mass acceleration is likely the speed of sound of the particular "springy" matter since sound propogates the same way. Since the force is purely magnetic, the speed of sound and that of light must be related and it seems they will be most likely related by a ratio of self-inertia (mass) vs the speed of light.

A good example of actual realtime stress is the operation of high speed valves in an internal combustion engine where the valves might "float" per over rev. Their own inherent inertia exceeds their otherwise reciprocal speed past certain limits.

I'm just guessing, an intuition if you will. I'm unable to set this to mathematics because of lack of training. But the suggested geometry should exist.

Wes
...


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