# How fast Coil Springs are? Greatest mystery today?

• Darp
In summary: If the spring is anchored at one end, the other end is free to move and it will reach its maximum velocity pretty quickly.
Darp
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, let's say a standard 125cc motorcycle front fork coil spring, like this one http://www.ebay.com/itm/76-Honda-MT...uspension-Strut-Inner-Long-Coil-/321367298914

Maybe that one has twice as many coils as one be ideal (too close together). Let's say its got a 150 pounds per inch spring rate, is 20" extended and 10" compressed and 1.25" in diameter. Let's 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 Calculationsv 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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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

Darp said:
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.

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

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!

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.

Chestermiller said:
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).

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?
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/compression-spring-calculator-coil-calculator.php?id=compression

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

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.

$$v = u_0\sqrt{\frac{k}{m}}$$

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

1 person
Excellent, will do. Thanks

Chestermiller said:
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.

$$v = u_0\sqrt{\frac{k}{m}}$$

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 got to be at least 50 FPS in real world.

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

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

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 energy 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 $n$ smaller springs (of the equivalent spring rate $nk$), placed in series, which have their masses ($m/n$) 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:

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

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

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

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

Where $v_d$ is the speed at a distance $(d/n)\;l$ from the fixed end of a spring of length $l$. Of course, if $d=n$, 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.

Ranger Mike
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

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.

Chestermiller said:
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: $nk, m/n, u_0/n$

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

$$v_1 = \frac{u_0}{n}\sqrt{\frac{n}{n}\frac{nk}{m}}$$

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

$$v_2 = \frac{u_0}{n}\sqrt{\frac{n}{n-1}\frac{nk}{m}}+v_1$$
$$v_2 = \frac{u_0}{n}\sqrt{\frac{n}{n-1}\frac{nk}{m}}+\frac{u_0}{n}\sqrt{\frac{n}{n}\frac{nk}{m}}$$

$$v_2 = u_0\sqrt{\frac{k}{m}\left(\frac{1}{n-1}+\frac{1}{n}\right)}$$

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 $n$ gets larger, each spring-mass system gets stiffer ($\sqrt{nk/m}$), 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 $n$. I think it was an error on my part. Anyway, the harmonic series is still there and still goes towards infinity as $n$ gets larger.

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Ranger Mike
jack action said:
Each spring-mass system has the following value: $nk, m/n, u_0/n$

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

$$v_1 = \frac{u_0}{n}\sqrt{\frac{n}{n}\frac{nk}{m}}$$

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

$$v_2 = \frac{u_0}{n}\sqrt{\frac{n}{n-1}\frac{nk}{m}}+v_1$$
$$v_2 = \frac{u_0}{n}\sqrt{\frac{n}{n-1}\frac{nk}{m}}+\frac{u_0}{n}\sqrt{\frac{n}{n}\frac{nk}{m}}$$

$$v_2 = u_0\sqrt{\frac{k}{m}\left(\frac{1}{n-1}+\frac{1}{n}\right)}$$

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 $n$ gets larger, each spring-mass system gets stiffer ($\sqrt{nk/m}$), 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 $n$. I think it was an error on my part. Anyway, the harmonic series is still there and still goes towards infinity as $n$ 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:

$$(ρΔx)\frac{∂^2u}{∂t^2}=(kl)\left(\frac{∂u}{∂x}\right)_{x + Δx}-(kl)\left(\frac{∂u}{∂x}\right)_{x}$$
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:

$$m\frac{∂^2u}{∂t^2}=(kl^2)\left(\frac{∂^2u}{∂x^2}\right)$$
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:

$$\frac{∂^2ε}{∂t^2}=\frac{kl^2}{m}\left(\frac{∂^2ε}{∂x^2}\right)$$
This is a wave equation, with the backwards velocity of the wave equal to $v_s=l\sqrt{\frac{k}{m}}$
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:

$u=xε_0$ for x < l - vst

$u=ε_0(l-v_st)+(l-x)$ 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:

$v=0$ for x < l - vst

$v=-ε_0v_st$ 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:

$$v=u_0\sqrt{\frac{k}{m}}$$

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

Chet

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:

$$E_k = E_s$$
$$\frac{1}{2}mv^2 = \frac{1}{2}ku_0^2$$
$$v = u_0\sqrt{\frac{k}{m}}$$

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 $n=1$ 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?

...

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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jack action said:
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.

This assumption was incorrect. In your multiple spring-mass system, when the constraint is removed from the far end of the system, none of the masses have moved yet, and so they are all still in equilibrium, constrained on both sides by the adjacent springs. The only exception is the single mass at the very end end of the sequence, which is no longer in equilibrium, so it starts to move first. This then causes the spring between it and the next mass into start extending. The effect propagates backwards.

How can a spring in the middle stay still with no force holding it in place ?
There is a force on either side from the adjacent masses which haven't moved yet.
Actually, if you assumed $n=1$ 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?
The reason for this coincidence is that your solution for the multiple spring-mass problem is incorrect. Try setting it up as a finite difference problem, and you will see how it works.

Chet

Wes Tausend said:
...

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.
...

I gave the variable vs in my derivation the subscript s because it is the speed of sound for the "springy" matter.

Chet

THIS IS A CORRECTION TO A RELATIONSHIP I GAVE IN POST #16. IT DOESN'T CHANGE THE FINAL RESULTS OF THE ANALYSIS.

The displacement u as a function of time and position along the spring is given by:

$u=xε_0$ for x < l - vst

$u=ε_0(l-v_st)$ for x > l - vst

Chet

So I wanted to test out my model to see if it worked in other situations.

So what if I add a mass $M$ at the end of my succession of springs?

Then for every spring-mass system the mass becomes $i*m/n +M$. Then $v$ becomes:

$$v = \sqrt{\sum^{n}_{i=1}\left(\frac{1}{\frac{m}{M}i+n}\right)}\;u_{0}\sqrt{\frac{k}{M}}$$

If $n = 1$ (meaning that both $m$ & $M$ are localized at the end of a single massless spring):

$$v = u_{0}\sqrt{\frac{k}{m+M}}$$

Which is the right answer. But in your last post, you said that it was normal for my model to work with $n = 1$.

Now let's imagine that $M >> m$, then $\sum^{n}_{i=1}\left(\frac{1}{\frac{m}{M}i+n}\right)$ converges to 1 for any value of $n$ because $\frac{m}{M}i \approx 0$ and $\sum^{n}_{i=1}\left(\frac{1}{n}\right) = n\frac{1}{n}$, leaving us with:

$$v \approx< u_{0}\sqrt{\frac{k}{M}}$$

Which is again the right answer (it is sligthly "less than" as some energy has been used to accelerate $m$), no matter how many springs there are in my model.

I still think that the speed will be infinite just as the current in a shorted battery is infinite. We all know that in reality there is no such thing as an infinite current, that the rod between the two poles will heat up, some small resistance will exists and the current will be finite (although very large). We then need some more precise models that will assume something more complex than a simple constant stiffness for the spring.

My point was that with simple mechanics, the spring-mass model assumes a massless spring because it pushes a large mass $M$, so large that $m$ becomes irrelevant. By not having a large mass $M$, the spring force accelerates nothing, so no inertia, hence the acceleration becomes infinite and the velocity as well.

Another way to look at it is to see that when you take a coil spring and cut it in half, its stiffness increases and its mass is reduced, which means that by cutting it again and again it will become a spring with an infinite stiffness and mass = 0. And every spring is a succession of that spring. But if there is a very large mass $M$ at the end of the spring, then the fact that the spring mass tends toward zero becomes irrelevant and the velocity will not tend towards infinity.

Again, I understand that in such cases, nature finds a way to stabilize such unstable conditions; but I wanted to point out that unstable condition.

And I still try to understand what is that situation where someone needs a spring to push nothing at a certain velocity.

Hi Jack Action,

I'm going to formulate the model equations for the discrete spring mass system that you are looking at so we can try to get some consensus between us. I will set it up so there are n identical springs and n identical masses. The spring constant of each spring will be, as you indicated, kn, where k is the overall spring constant of the combination. The mass of each mass will be m/n, where m is the overall mass of the combination. Let the undeformed length of the system be l, and let l/n be the undeformed spacing of the masses. Let xi0 represent the location of the i'th mass in the undeformed system. Then xi0=il/n. Let ui(t) represent the displacement of the i'th mass at time t relative to its location in the undeformed configuration of the system, and let ui(0) be the displacement of the i'th mass of the initially compressed system at time 0 relative to its location in the undeformed configuration of the system. Initially, all the springs are compressed equally, so

$$u_i(0)=u_l(0)\frac{i}{n}$$

The initial amount that each of the springs is stretched in the initial compressed configuration of the system is then:

$$(u_i(0)-u_{i-1}(0))=\frac{u_l(0)}{n}$$

For compression of the system, ul(0) is negative.

The initial tension in each of the springs is given by

$$F = kn(u_i(0)-u_{i-1}(0))=ku_l(0)$$

This is, of course, the result we would expect.

Let's next do a Newton's 2nd law force balance on the i'th mass:

$$\frac{m}{n}\left(\frac{d^2u_i}{dt^2}\right)=kn(u_{i+1}-u_i)-kn(u_i-u_{i-1)}=kn(u_{i+1}-2u_i+u_{i-1})$$
where the term in parenthesis on the left hand side is the acceleration of the i'th mass.

I'm going to stop here to make sure we are in agreement regarding what I have written so far.

Chet

I'm following ...

Chet and Jack, Wow! you are so far beyond me in math. I will watch this to see if you come up with something in the 50fps range, I got 50 FPS on one spring and 42 FPS on another, empirically. The springs I have posted and in neighborhood of the 50 FPS.

jack action said:
I'm following ...
Excellent. So, to summarize, the set of force balance equations on the masses is:
$$\frac{m}{n}\left(\frac{d^2u_i}{dt^2}\right)=kn(u_{i+1}-2u_i+u_{i-1})$$
This is subject to the following initial conditions:

$u_i=u_n(0)\frac{i}{n}$ at t = 0

$\frac{du_i}{dt}=0$ at t = 0

Now, I would like to focus specifically on the very last mass i = n in the sequence. This is the mass that we release from constraint at time t = 0, which allows the combined system to expand. The force balance for this last mass both prior to release and after release is different from that of the other masses, because it is at the end of the sequence. Just prior to release of this mass, its force balance reads:

$$\frac{m}{n}\left(\frac{d^2u_n}{dt^2}\right)=F-kn(u_n(0)-u_{n-1}(0))=ku_n(0)-kn(u_n(0)-u_{n-1}(0))=ku_n(0)-kn\frac{u_n(0)}{n}=0$$

This means that, as expected, the net force acting on the mass at i = n is equal to zero prior to it being released, and its acceleration is zero.

Any time after it is released, the constraining force F is zero, and the force balance on the mass at i = n becomes:

$$\frac{m}{n}\left(\frac{d^2u_n}{dt^2}\right)=-kn(u_n-u_{n-1})$$

At the instant after the constraining force on mass i = n is released (i.e., at time zero), the force balance equation on this mass reduces to:

$$\frac{m}{n}\left(\frac{d^2u_n}{dt^2}\right)=-kn\frac{u_n(0)}{n}=-ku_n(0)$$

This means that, immediately after the constraining force is released, the mass at i = n instantly begins accelerating.

What about all the other masses in the sequence? Well, if we substitute the initial conditions on the displacements at time t = 0 into the force balance, we obtain:

$\frac{m}{n}\left(\frac{d^2u_i}{dt^2}\right)=0$ for i = 1,...,(n-1)

Thus, immediately after the constraint on the system is released at time t = 0, the accelerations of all the other masses in the sequence are zero. The only mass that is accelerating initially is the very last mass in the sequence (i = n).

What this shows is that, for the combined spring-mass sequence described by Action Jack, the disturbance at i = n does not affect the entire sequence of masses instantaneously. Because of the inertia of the masses, only the final mass at i = n moves first, and then, as time progresses, the masses further away from the free end of the spring feel the effect and begin to move.

There is a very close relationship between the force balance equation for the discrete sequence of springs and masses in Action Jack's model, and the continuous distribution of spring/mass in the continuum model I formulated and solved in Post #16. To see this relationship, let's substitute n = l/Δx into the force balance equation for the Action Jack model. By doing so, we obtain:
$$m\left(\frac{d^2u_i}{dt^2}\right)=kl^2\frac{(u_{i+1}-2u_i+u_{i-1})}{(Δx)^2}$$

In the limit as Δx approaches zero, the discrete sequence model and the continuous distribution model become identical. Therefore, their solutions must at least be qualitatively similar, and, of course, must match exactly in the limit of Δx->0.

The analytic solution I obtained to the continuous model satisfies the differential equation and initial conditions exactly, and, moreover, must be unique, since the equations are linear in the displacement. Therefore, its predictions that a disturbance propagates backwards along the spring, and that the strains, displacements, and velocities along the spring are spatially discontinuous must also be a feature of the discrete model (at least to a very close approximation). This has been verified at very short times, since, in this limit, only the final mass in the sequence i = n moves initially.

The continuous model predicts that the force on the first mass by the first spring in the sequence is not infinite, but, instead, is equal to F = ku_n(0). This force stays constant throughout the release until the disturbance has fully propagated backwards to the beginning of the sequence and the spring then loses contact with the constraint at x = 0. In the case of the discrete model, of course, the force at x = 0 must also be finite.

Chet

Trying to get an "exact" solution by setting up a discrete-mass system, solving it, and then letting the number of masses increase to infinity is doing it the hard way.

The easy way is similar to Post #16. You know that the motion of the continuous spring is governed by the wave equation, and you can easily find the wave speed as something analogous to ##\sqrt{E/\rho}## depending exactly how you choose to define the stiffness and mass.

So, use the d'Alembert solution which says the motion consists of two traveling waves moving in opposite directions. The initial conditions are that the spring is compressed linearly along its length, and its velocity is zero. If the displacements at time 0 are ##u(x) = Ux/L##, where U is the amount of compression of the free end, and L is the uncompressed spring length, the two traveling waves at time 0 are both equal to half the total displacement, i.e. ##u(x,t) = U(x\pm ct)/2L##. (Otherwise, the velocity would not be zero).

As the traveling waves move along the spring, they reflect off the ends, with a 180 degree phase shift at a fiixed end and no phase shift at a free end. So working out what happens when the spring is released can be done simply be drawing pictures of how the traveling waves behave.

I got lost reading post #16 so I'm not sure if that is exactly right or not (and I admit I'm not really interested enough in the problem to work out all the details for myself!).

Rough calculation:

Displacement = 5 "

Spring constant = 150 lb/in

Mass = 2 lbm

$$v=u_0\sqrt{\frac{k}{m}}= \frac{5}{12}\sqrt{\frac{(150)(12)(32.2\frac{lb_mft}{lb_fsec^2}) }{2}}≈70 ft/sec$$

Chet

Incidentally, a simpler way of solving this problem is to use conservation of energy: the kinetic energy of the spring after separating from the second constraint (at x = 0) must be equal to the initial stored elastic energy in the spring. Assuming that all parts of the spring are moving at the same velocity after total separation (as indicated by my detailed analysis in #16), we have:

$$\frac{1}{2}mv^2=\frac{1}{2}ku_0^2$$

This again gives:

$$v=u_0\sqrt{\frac{k}{m}}$$

Some additional comments: My analysis indicates that, at x = 0, the force on the spring doesn't change during the time that the spring is releasing. This force is equal to ku0. Since the release wave is traveling backwards along the spring at the speed $l\sqrt{\frac{k}{m}}$, the amount of time it takes the wave to reach the end at x = 0 is $\sqrt{\frac{m}{k}}$. Therefore, the impact of the force at x = 0 is equal to $ku_0\sqrt{\frac{m}{k}}=u_0\sqrt{km}$. This must be equal the change in momentum of the mass:
$$mv=u_0\sqrt{km}$$
So again:
$$v=u_0\sqrt{\frac{k}{m}}$$

Chet

Don't know if this helps any but as a machinist I have encountered a condition called "float" many times in race engines. The compressive strength of the spring steel has to be increased ( stouter springs installed) in high rpm racing applications because the valves can't close fast enough for the compression stroke. Near 8500 rpm with standard springs and 10000+ with high performance. So with the compressed installed length - maximum compressed installed length rpm factored and graphed and float being the indication fail @ maximum speed you could set up a mechanical experiment to check the "speed of a spring". Heat is an issue here too.

Don't know if this helps any but as a machinist I have encountered a condition called "float" many times in race engines. The compressive strength of the spring steel has to be increased ( stouter springs installed) in high rpm racing applications because the valves can't close fast enough for the compression stroke. Near 8500 rpm with standard springs and 10000+ with high performance. So with the compressed installed length - maximum compressed installed length rpm factored and graphed and float being the indication fail @ maximum speed you could set up a mechanical experiment to check the "speed of a spring". Heat is an issue here too.

I hope Darp (OP) does not mind that we explore this.

I kind of wondered if one couldn't empirically determine spring speed by resonance. Perhaps Darp has done this. A spring can't logically be any faster than its ability to return from a tension. The idea would be to drive the spring from one end with transducer and calibrated audio generator and observe when it reached maximum resonance, or "bounce" on the other end. Some lesser resonances would appear at harmonic frequencies, but the true primary resonance would be observed to be the most vigorous. Ideally one would also have a frequency analyser if great accuracy was needed.

In the practical cases of engine valve float, one could assemble the proposed valve train and poppet valve which should cause the effective resonant frequency to lower because of greater inertia. It could not be allowed to lower any more than the maximum design rpm of the engine which is within keeping of using stiffer springs.

This same principle should also apply to a vehicle suspension spring when the unsprung mass (suspension arms, wheel etc) is attached. The idea here is to return the wheel for continuous traction and could be best theoretically tuned to the known average bump frequency on a moto-cross track for instance.

Thanks,
Wes
...

Thank you Chet for taking the time to write the detailed equations. AlephZero might called it the "hard way", but I have to say that good old "F=ma" help me see the solution easier as I'm not familiar with the concept of wave equation.

Chestermiller said:
Incidentally, a simpler way of solving this problem is to use conservation of energy: the kinetic energy of the spring after separating from the second constraint (at x = 0) must be equal to the initial stored elastic energy in the spring. Assuming that all parts of the spring are moving at the same velocity after total separation (as indicated by my detailed analysis in #16), we have:

$$\frac{1}{2}mv^2=\frac{1}{2}ku_0^2$$

This again gives:

$$v=u_0\sqrt{\frac{k}{m}}$$

This relates to what I told you in post #17 as I recognized the equation. I tought you did a very simplistic evaluation.

This is what stumble me. This is the exact way to find the velocity of a mass attached to a single massless spring. In my post I was saying "Happy coïncidence ?" as I was under the impression that if spring mass was involved, it wasn't simply added to the end mass. IIRC, with problems concerning car suspensions, when time came to determine if a suspension component (i.e. spring, shock, a-arm) was supposed to be considered "sprung mass" or "unsprung mass", the rule of thumb was to divide the mass 1/3-2/3 between sprung and unsprung mass (or vice-versa). I never bother to do detailed calculations, but it seemed logic that it couldn't be simply put at one end of the spring (or split in half between the two masses in this case).

What is also puzzling me is that if we refer to the email in the OP, they are 2 equations: one that is basically yours and another similar one that reads:

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

v= F √ ( kg/P+1/3Ws)

I wonder what happens with those lighter springs that they don't "obey" the same rule ? Plus, the units don't fit, so it must be empirical, and there is that familiar 1/3 of spring mass that appears from I don't know where.

jack action said:
Thank you Chet for taking the time to write the detailed equations. AlephZero might called it the "hard way", but I have to say that good old "F=ma" help me see the solution easier as I'm not familiar with the concept of wave equation.

This relates to what I told you in post #17 as I recognized the equation. I tought you did a very simplistic evaluation.

This is what stumble me. This is the exact way to find the velocity of a mass attached to a single massless spring. In my post I was saying "Happy coïncidence ?" as I was under the impression that if spring mass was involved, it wasn't simply added to the end mass. IIRC, with problems concerning car suspensions, when time came to determine if a suspension component (i.e. spring, shock, a-arm) was supposed to be considered "sprung mass" or "unsprung mass", the rule of thumb was to divide the mass 1/3-2/3 between sprung and unsprung mass (or vice-versa). I never bother to do detailed calculations, but it seemed logic that it couldn't be simply put at one end of the spring (or split in half between the two masses in this case).

What is also puzzling me is that if we refer to the email in the OP, they are 2 equations: one that is basically yours and another similar one that reads:

I wonder what happens with those lighter springs that they don't "obey" the same rule ? Plus, the units don't fit, so it must be empirical, and there is that familiar 1/3 of spring mass that appears from I don't know where.
Hi Jack,

I can't really comment on the equations Darp posted from the "Institute", because they don't provide any details. All I can say is that I'm totally confident about the formulation and solution that I obtained. As far as my formulation is concerned, it is fully consistent with the multi-spring/mass model that you proposed, in the limit of a continuous distribution of springs and masses. And the resulting partial differential equation was straightforward to solve for the specified initial and boundary conditions. Further, the final result for the release velocity was consistent with the result obtained using conservation of energy.

As an aside, I might mention that this development is almost identical to what one can use to elucidate the details of an elastic collision between two cylindrical masses colliding head-on.

Chet

Perhaps looking at it from the perspective of a spring's natural frequency would help.

A spring, mass system oscilates at a natural frequency equal to √(k/m). As you decrease m, the frequency increases until you have only the mass of the spring itself operating. That frequency is the natural frequency of the spring as given by:

This is from efunda which provides the derivation.

This doesn't give velocity of any portion of the spring directly, but the motion of any point on the spring follows a sinusoidal path.

The highest velocity of any point on the spring then is at the free end of the spring when it passes through that point it would be at when the spring is stationary. That velocity is a function of how far the spring is initially compressed or the amplitude of the movement. The higher the initial compression, the higher the velocity as it passes through its unstressed shape.

I think velocity should be relatively straightforward to derive from this.

Last edited:
Great resource

I empirically solved the question, but am so impressed with the knowledge here. I am not where close to being able to do the math you folks have. And is case you missed it, the the empirical answer is about 50 mph, springs are slow, because they accelerate them own large mass.

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