# Storing energy in a compression spring

1. Dec 6, 2014

### bobie

I'd like to learn more about the way energy is stored in a springhttp://en.wikipedia.org/wiki/Spring_(device) . I found this site http://www.leespring.com/int_learn_compression.asp

I learned that a 50mm free-length spring can be compressed for 20mm (at most by 40%) to solid length (30mm) and that 10% clash allowance must be left. So if the spring is compressed by 18 mm and the stiffness k is 12 N/mm the load is 18*12 216 N.

Can you give me a good link where I can find the answers to the following questions?
1) how do I find the PE that can be stored in that spring when the load is 216N? what is the conversion formula?
I found: .5*k*d^2 is the PE = K = .5*12*18^2 ?
2) supposing the positive work done to compress the spring is K J, how much of it can be actually stored, can we get values as high as 97-98%?
3) is the lost energy all transformed in heat, is the percentage linearly dependent on the mass?or on what?
4) if the stored energy is 97% K, will it be totally retrieved or some extra energy will be dispersed, and the actual energy restituted will be 94% K

Lastly, if insead of a force applied there is an impulse, an impact with a body with KE,v and p, does anything change? can we find the released energy just applying the rate , say, 94%?

2. Dec 6, 2014

### Muti

I do not know you Qualification (mechanical engineer, studying physics or college), but if you check any control system book's chapter on modelling you will find how spring is used as model. There you may find answer to impulse input to a spring. The link below
http://ctms.engin.umich.edu/CTMS/index.php?aux=Home
has excellent tutorial on control system ad modelling (better then some books)

3. Dec 6, 2014

### bobie

Thanks for the link, I am just a layman, before I dip into the tutorial can you tell me if the values I gave are realistic? can a 50mm spring have a load o 216 N? What energy corresponds to that 1994 J? and what percentage of it is actually restituted?
Thanks

4. Dec 6, 2014

### Muti

The values can be realistic because you have some "margins" for improvements or enhancement, that is, you have not specified spring material, its thickness, spring diameter etc. You can visualize 216N as roughly 21kg weight (F=W=m.g) being put on the spring so pretty realistic. 1994J can be compared to any thing like for example, I found over an internet quick search "Every mole of methane (16 g) releases 810 KJ of energy on burning". But you need to differentiate between energy and power. Formula for PE of spring is also correct. I donot know how one arrive at the efficiencies involved in energy conversion for spring. Perjaps someone else can answer. Thank you

5. Dec 6, 2014

### Staff: Mentor

that formula is what you get if you apply the rule that work equals force times distance, so it's a good start. But be sure to pay attention to the units...
google for "Spring damping coefficients" and you will find values for springs made of various materials. Some materials will allow a spring to give back better than 99% of the energy stored in them.
in just about any mechanical system, "lost" energy ends up as heat. Flexing springs are no exception, but the exact amounts involved will depend on the details of the system.
Again, that depends on the details of the system, but in general both the compression and expansion will dissipate some energy.

No. We can't even start to think about the problem this way until we know more about the problem that you're trying to solve.[/quote]

6. Dec 6, 2014

### Danger

Thanks for that link, Muti. I've bookmarked it into my "Technical Research" folder. Every time someone provides something like that, it just makes my life better (even if I never use it).

7. Dec 6, 2014

### Muti

Thanks Danger.....

8. Dec 7, 2014

### bobie

[/QUOTE]
Thanks for your exhaustive replies, Nugatory. ( I did not understand your PM as my 'follow up' questions are the same as in my OP)

I am not trying to solve any specific problem. I am trying to understand how energy is actually stored in a spring.
I'd appreciate if you told me how to upload an image from my PC so that I can show what I have elaborated, because , when I click the 'image' icon I am allowed to upload only images from the web, and not from my own PC.

Anyway: supposing we have a 50 mm (free length) spring with k = 1.5, solid length = 30mm, and therefore a clash of 2mm and a working range of 18 mm. The max load of the spring is 1.5*18 = 27 N and the max stored PE is 27*18/2= 243 J.
Now, suppose we apply the max force: 27 N, do positive work and spenf 243 J, we cannot compress the spring by 18 mm (deflection = delta), as some of this energy will be turned into heat (because me must deform the steel lattices to higher energy levels) , if this is correct,
1) delta will not be 18mm , and the load will not be 27 N, but the formula L = k * delta says it is 27 N because we spent 243J
2) Is the same damping coefficient (Dc , say, = 99.4, can we call it CoR?) involved now, too? shall we apply the Dc twice to get the energy actually released #243*99.4^2/10^4 = 240 J# ?
3) Should we also consider that some energy is lost in pushing thw table the floor on which the spring is lying?

Now, I was then thinking of what happens if, instead of applying Pressure, a Force, we apply an Impulse: suppose a ball with mass = 20Kg and v = 4 m/s, p= 120, KE = 240 J hits the spring fixed on a block of concrete with mass = 30 t.
If the ball were perfectly elastic (and therefore the spring were inside the ball) it would bounce back with KE = 249.4, p = -120, and the wall would get 0.96 J and would budge at 0.008 m/s
In this case the spring is not in the body but on the wall. The spring will release the 240 J stored, but I am not sure it can restore velocity and momentum to the ball. I read that it depends on hysteresis, but,why doesn't it depend simply on the speed, or acceleration the spring reaches when it relaxes?

9. Dec 7, 2014

### Staff: Mentor

bobie, please try to focus on one or two questions about one scenario at a time. I see here in this last post something like 5 questions about 3 different scenarios. Please pick one scenario and one scenario only, and ask at most two questions about that scenario.

One general comment. You need to keep better track of your units. Don't just write numbers without writing their units. Here you have two important equations:
$F_{elastic}= -k x$
$E_{elastic}= \frac{1}{2} k x^2$

So for $k=1.5 N/mm$ and $x=18 mm$ then we get:
$F_{elastic}= -(1.5 N/mm) (18 mm) = 27 N$
$E_{elastic}= \frac{1}{2} (1.5 N/mm) (18 mm)^2 = 243 N*mm$

But $1 J = 1 N*m = 1000 N*mm$ so this is only $0.243 J$, not $243 J$.

10. Dec 7, 2014

### Staff: Mentor

Not correct. The total work done in compressing the spring will be the sum of the potential energy that ends up in the spring and the energy that ends up as heat.

The potential energy is indeed given by the $kd^2/2$ formula you dug up; if you calculate it correctly (as DaleSpam shows above) you will get .243J as he did. Thus, the total amount of work done compressing the spring will be .243J plus whatever is lost as heat.

The total work could be as much as 4.86 Joules if we were applying the full 27N for the entire length. In fact it is less than 4.87 J, because the resistance of the spring starts at zero when it's not compressed so the force is not 27N over the entire distance.

Now, for your next question: As DaleSpam says, pick one scenario, and ask not more than two questions about it. And before you ask these two questions please carefully read the last two posts.

11. Dec 7, 2014

### bobie

Thanks for your explanation, I do not understand that, how you got from .273 to 4.86 J

I have not even one scenario, I gave some examples as you asked me what problem I want to solve. As I said in OP, I was looking for general information about how springs work.
I have understood that if I compress a 5 cm spring with k = 1,5 N/mm by 18 mm, which is roughly the maximum recommended value, I can load 27 N and store 0.243 J of which 99.4 % will be given back. Is this correct? I wanted to know how much work done is lost to heat (as an average) from what you say it seems a lot.
Then I wanted to know at what speed a ball will bounce back when it impacts the spring and if this speed is fixed or can vary.

That is what I'd like to know, numbers and scenarios are not necessary. I just was trying to make a comparison with an elastic ball that can bounce back from a wall at any speed, according to the rule of conservation of momentum.

12. Dec 7, 2014

### Staff: Mentor

Sorry, that's .486J, and I calculated it from W=Fd, a force of 27N acting over a distance of .018 meters. That's the work done by a force of 27N acting over a distance of .018 meters. This is of course more work than is actually done to compress the spring, as the force starts at zero and is less than 27N for most of the distance.

The point here is that whatever is applying the force (presumably you, compressing the spring?) is capable of exerting a force of up to 27 N and therefore doing as much as .486 J of work on the system. The amount of work it actually does is the .243 J that ends up as potential energy in the spring plus whatever is ends up as heat when as the spring is compressed.

13. Dec 7, 2014

### Staff: Mentor

Unfortunately, no, it is nowhere near that simple. A damped spring follows a much more complicated differential equation than an undamped one. In many cases it does not even oscillate once.

Here is as gentle an introduction as possible. http://hyperphysics.phy-astr.gsu.edu/hbase/oscda.html

14. Dec 8, 2014

### bobie

I imagine you are thinking of springs as shock absorbers. Can you answer these specific questions, that is almost all I'd like to know

1) to the best of your knowledge what is the highest percentage of work (done in compressing) a spring can give back when it relaxes? Or , from another perspective what is the minimum percentage of heat a spring can absorb?
2) does the speed of relaxation vary in relation to the deflection? or, if I compress a spring by 5 or 10 mm, will the speed of relaxation be doubled?
As I said I'd like to confront the property of a steel ball (which can rebound at same speed on a hard wall), with the properties of a steel spring fixed to a wall when it is compressed by a ball with speed v. Will the ball always rebound with speed -v , if we find the right parameters of a spring?

Thanks

15. Dec 8, 2014

### Staff: Mentor

100%

The period is $T=2\pi\sqrt{m/k}$. It does not vary as a function of the deflection.
http://hyperphysics.phy-astr.gsu.edu/hbase/shm2.html

I think what you are calling relaxation would be 1/4 of that.

16. Dec 8, 2014

### bobie

Thanks a lot DaleSpam, That is all I was looking for,

1) can you tell me or give me a link to learn what materials and what specifications are needed to achieve 100%? How can you completely eliminate heat loss?
2) so if a ball hits a spring attached to a wall, what happens? In theory if might get back 100% of its former KE, but in practice, if the spring cannot give it the adequate v it cannot. What happens then to remaing PE?

17. Dec 8, 2014

### jbriggs444

I do not believe that Dale was trying to say that 100% is practically achievable, only that there is no fundamental physical principle that prevents you from getting as close as you please.

Heat in the spring. Vibrations in the spring. Noise transmitted to the air. Heat transmitted to the air, etc. It seems to me that there are three main transactions involved in the energy loss budget.

1. Impact of the ball with the spring. To the extent that the material in the spring is soft, this may result in some local deformation and prompt loss of energy to heat. To the extent that the material in the spring is hard the spring may bounce off and subsequent collisions will occur. In addition, energy will be lost to vibrations in the spring. A compression wave will travel up the spring to its anchor and back down the spring to the mass. Unless the entire transaction is timed perfectly, this compression wave will eventually die out and the associated energy will be lost to heat.

2. Asymmetry between force conveyed to spring during compression and force recovered during expansion as a result of hysteresis-like effects in the spring material. This energy loss manifests as heat in the spring material.

3. Release of the ball from the spring. The spring will be moving when the ball flies away. It will still be moving afterward. This results in a rarefaction wave travelling up and down the spring which is eventually damped into heat.

You can minimize 1 and 3 by using spring material with a very high strength to weight ratio or (possibly) by tapering the spring so that it is light and weak toward the ball end and heavy and strong toward the anchoring end.

You minimize 2 by using a a very elastic material. This may conflict with the best material to optimize for 1 and 3.

18. Dec 8, 2014

### bobie

Thanks for you excellent reply, jbriggs. What I'd like to know is not the abstract possibility, I'd like to know some details about the material that gives the best performance and what the actual value of this c is.
As to the impact, what is the relation between the deflection and the speed of relaxation? I suppose that it increases with it and, likewise, when the spring is released, it gets immediately the greatest speed and this decreases as it approaches the position of equilibrium, that's why you say that the ball will lose contact before it is fully extended.
Any idea why this does not happen in an elastic tennis ball.
The hysteresis is important, too, but I suppose it can be regulated, as it happens with sad and happy balls, is that right?

19. Dec 8, 2014

### jbriggs444

The ideal situation would be with a massless spring. That can be modelled as simple harmonic motion. The motion during the catch (first quarter cycle) is the mirror image of the motion during the release (second quarter cycle). The spring starts the release at zero speed and its speed increases as the equilibrium point is approached. At the equilibrium point, the spring is no longer pushing on the ball with positive force so the ball flies away -- right when the spring and ball is at its greatest speed.

This same thing does happen with an elastic tennis ball. But that is a different scenario with a different question and it is best that we stay focused.

20. Dec 8, 2014

### bobie

Does that contrast with
.

If the spring has and gives indeed constant acceleration, then (if there was no heat loss) the ball should reach the original speed and lose contact at that speed, and the spring will not extend anymore and cannot oscillate

21. Dec 8, 2014

### jbriggs444

When the ball loses contact with the spring, both ball and spring are moving at (near to) the ball's original speed. If the spring were massless then the amplitude and energy of its remaining oscillation would be zero. Real springs are not massless.

22. Dec 8, 2014

### bobie

So, if we can wind this up, a ball (m=1,v0 =1, E = 0.5 J) has the force of g (10N), if it impacts a spring with k = 1N/mm (c = 0.994) it will deflect it by 10 mm. The spring will store 0.497 J and will restitute 0.494 J, and the speed of the bouncing ball will be roughly 0.994 m/s. (v0 *c is this a general rule :v1 = v0 * c)

Is this realistic? Can you give me a link where I can find the max c achieved by technology to date?
That would be great!, Thanks

Last edited: Dec 8, 2014
23. Dec 8, 2014

### Staff: Mentor

100% is a limit that cannot be achieved, but can be approached arbitrarily closely.

24. Dec 8, 2014

### bobie

You do not know the state of the art, now? Can I buy a spring with c 0 0.99?

25. Dec 8, 2014

### Staff: Mentor

I don't know the state of the art, but again c is not generally the same as the percentage of energy that you will get back. The actual relationship is far more complicated.