Springs - How important are they, really?

[Erik the Red]

Alright, I just gotta throw in my $0.02. I'm of the opinion that as long as the ends are the same (as mro suggested), it won't matter how the springs are installed.

If one has a spring compressed with ten pounds of force, it will exert a counterforce of five pounds in each opposing direction.

Pro- I disagree. Looking at it from a statics point of view, if ten pounds is applied to one end of a spring, the force will be transmitted through it to the other end. Lets say that at this other end is the scale that Ken referred to and it is pushing back with ten pounds. Now if you were to look at the spring/initial 10# force (or spring/scale) interface only, you would have 10# pushing on the spring end, therefore the spring end would have to be pushing back on that 10# force with 10# of its own. If it were any less (or more for that matter), that interface would continue to move. If we're applying statics, everything counteracts everything and nothing moves.

Now, I do agree with Pro that the coils which are wound looser will compress first. I also agree that the end of the spring to which the initial force is applied will begin to move first, since it does take time to transfer force through any material.

BUT, I say that the amount of time it takes to transfer the force through the spring is so infinitesimally small, that it won't matter. By the time the force reaches the other end of the spring, the spring will have compressed only an infinitesimally small amount. Thus, since that compression is so close to zero, is effectively zero, and what remains is where they compress first. That'll be where the spring is wound the loosest, whether that loosest place is at the top of the spring or the bottom.

This is coming from a statics point of view. I never took a full course in dynamics, which, in actuality, is what we're talking about, but when it comes down to it, the principles are very similar.

I have been wrong before, so I welcome any and all critiquing of my stance. Have at 'er.

[mro]

More than you wanted to know

While not for an XS valve spring, simular enough to apply.

http://www.dresser-rand.com/e-tech/P...iles/tp003.pdf

In practical terms, which end is up.... under normal operating conditions probably will make no "noticeable" difference , but could induce "floating" at lower RPM (spring compression cycles per second)



mro

[Ken Talbot]

Thanks, mro. I was googling for something like that the other night when the topic came up. I think I still need another read or two to 'get it', but not tonight. All going well, I'll be reassembling the head on the winter project bike soon. And yes, this will be the third winter on this winter project.

[prometheus578]

MRO... what the hell's the matter with you!"

You gotta be kidding me! All I wanted to know was why the tighter coils have to face downward, or something. Actually, I'm so confused now that I really can't recall just WHAT I was looking for!
Did you actually READ this drivvle?
(And you people think that "I" can talk for hours and not really be saying anything!)
...The governing equations were
derived using Hamilton Principle (Lin and Pisano, 1987). The equations were reduced to a finite difference form
using an implicit method and solved by using appropriate boundary and initial conditions for the two cases ...
WHAT??? HOW'S THAT???
This is garbage! Was this translated from Pig-Latin?
(My grandfather had patents for designing parts for a machine that cuts the threads on a screw. My father's patents are for designing oil seals for automatic transmissions. Oh, the conversations around the dinner table...)
"By an under-extension of the distal digits, proximal to the opposable phalangal counter-structure, can the quasi post-modern designed ceramic NaCl containment vessel be propelled in an ever increasing arc, it's radius not to excede "R+2", and it's velocity at a stable, yet ever ampifying rate of uniform acceleration, be brought to an alternate state of rest, the coordinates of X/Y..."
"Yes, dad, I'll pass the salt."
(Is it any wonder I ran away and joined the Infantry? I just had an urge to kill people, I don't know why)
"Measureable stress fluctuations along the helical structure during compression... INDEED!
About the only thing I gleaned from this, MRO, was
""During the initial phase of the spring motion (t = 0.0906 sec.), the spring remains close to its steady state
position and stress is close to the static case. During the subsequent phase (t = 0.0908 sec.), the coils near the
moving end is squeezed, while the stress near the fixed end has not changed yet.""
(If this supports one of my theories, then they are correct. If it doesn't, then obviously they are wrong)
What I think this says is that when the bucket pushes down on the valve spring, the bucket end of the spring starts to compress before the end sitting on the head does. This is why, with the progressively wound valve spring, the tighter coils are at the bottom, as they are wanted to compress last.
It's all so clear to me now...

[mro]

Cool, .............someone got it.............

Was not surprised that "statics point of view" was missing something. Now all we have to do is mount a cut away cam/valve asm. to a variable speed electric motor, paint some little white dots on spring and record what it does with a high speed cam. Winter Project??????


mro

[mro]

originally posted by prom
"By an under-extension of the distal digits, proximal to the opposable phalangal counter-structure

Ok....OK!......what's anatomy got to do with a "NaCl containment vessel" and wtf is a "NaCl containment vessel"


mro

[prometheus578]

(My hand grabbing the salt shaker.)

[mro]

Guess I'm getting a little slow

mro

[prometheus578]

"In reply to Erik, the Red"

(from one of my earlier statements)

If one has a spring compressed with ten pounds of force, it will exert a counterforce of five pounds in each opposing direction.

(ERIK, the RED wrote)

Pro- I disagree. Looking at it from a statics point of view, if ten pounds is applied to one end of a spring, the force will be transmitted through it to the other end. Lets say that at this other end is the scale that Ken referred to and it is pushing back with ten pounds. Now if you were to look at the spring/initial 10# force (or spring/scale) interface only, you would have 10# pushing on the spring end, therefore the spring end would have to be pushing back on that 10# force with 10# of its own. If it were any less (or more for that matter), that interface would continue to move. If we're applying statics, everything counteracts everything and nothing moves

(This might now be difficult to follow) Last night, Erik, I did these beautifully rendered artistist drawings. (suitable for framing). The idea was to demonstrate what you had written, but to also show that a spring, being a spring, also wants to spring back to shape, hence the "five pounds in each direction".
[IMG][/IMG]
[IMG][/IMG]
I never posted last night, as I fell asleep at the keyboard, again.
As I slumped here, dreaming of "World Conquest"(as I often do... it being my destiny, and all....), another vision overcame me, which brought out new questions.
(this was during the dream sequence concerning my "Wonder Weapon... a massive 'spring gun'. [patent to be apllied for])
' has to do with compressing the spring to register on the scale, or to move the ten pound block.
If I read you correctly, you assert that if ten pounds of force is applied to the spring, it will then apply ten pounds of force out the other side. I feel that this is incorrect, for reasons that I believe we have all overlooked.
Just what force is used, and taken up by the spring, to get it to compress in the first place? Is not a certain amount absorbed in this?
Referring back to my 'beautifully rendered artistic drawings', we see that the spring has to compress first before transferring the ten pound force to the left. What of the energy required to compress said spring? (Actually, the spring said nothing, just squeeked a little)
So... springing into action, I figgered the following: To have a ten pound force applied to a spring, so that it can transmit a ten pound force out the other end (to move a scale or a ten pound block) wouldn't TWENTY pounds of force need to be applied?(say... ten pounds needed for the spring's compression?
So... if you'd indulge me once again, good sir,(Or Ken, MRO, or the other contestants and viewers playing along at home) please answer, and explain(in terms I can understand, not like that techno-crap MRO tossed at my head) my following questions:
1: As per the 'beautifully rendered artistic drawings... when a spring is compressed with ten pounds of force, does it not exert an equal force out each end, that being five pounds each way?
2: As per the 'beautifully rendered blah, blah, blah'... energy(force) is lost, (as the act of the spring compressing must absorb some), so, is it logical to assume that in order to have ten pounds of force transferred through a spring, a force of more than ten pounds must first be applied? ((This makes my second drawing incorrect, where I wrote "when the spring has compressed to the point where it's stored ten pounds of force, the scale will then read ten.))

(Hey, I realize that this thread has gone a little farther than most people care to carry it. Everyone simply doesn't have the time to answer simple questions for the simple-minded..., but what good's a thread unless someone fully runs it into the ground?

[Douglas Money]

Here is a really easy test. Take your bathroom scale. Stand on it. Record your weight. Now insert spring on scale. Record weight. Stand on spring on scale. Record your weight. Should be your weight plus the weight of the spring.

[mro]

The answer is..............NO

Compress a spring with 10 pounds. Measure amount it has compressed. Now it does not mater what you use, one scale and a weight, your hand, or another scale to compress spring to the second scale, 10 pounds is 10 pounds and spring will compress same amount. In the case with two scales both scales will show ten pounds not 5 or 20.

Above does apply to a valve spring when engine is off.
When engine is running inertia and the mass of valve shim, bucket and retainers the and spring it's self come in to play.


mro
btw, sound good???? , could be right?

[prometheus578]

That is correct, Mr. Money. It will weigh the same.
I weigh 195, and am holding a spring that weighs 5 lbs. Combined weight is 200.
I put the spring on the scale, then stand on the spring. The combined weight will still be 200, though, through the event of the force needed to compress the spring(which we shall say is 10 pounds) my body is actually only weighing 185lbs.
200 lbs= 5 pound spring, potential energy stored in the spring of 10 pounds, one flabby individual of 185 lbs.

[Douglas Money]

I am getting confused now about the original question, but I think we are all saying the same thing only different.

But then again, 200 lbs is 200 lbs. Exerted force is dynamic where weight is static. If you are talking about force required to compress, 10 lbs is 10 lbs. If you are talking about 200=185 because of spring, If it requires 10 to compress, it has compressed and is no longer potential energy as it is when at rest. I understand what you are saying but I think we are comparing apples to oranges.

[mro]

I am getting confused now about the original question

Question
What question???

_______________
statics works for a constant force, but does not go far enough to explane the "dynamics" of a valve spring when engine running.
_______________

Hey prom,
pass the NaCl containment vessel..........


mro
btw, 78 million metric tons per cubic kilometer in sea water!!!

[prometheus578]

Apples to oranges... I suppose, yes.

I suppose I sort of misspoke(I know, MRO, not the first time)
Confusing force with weight.
I meant to say...
My body is no longer directly exerting 195 pound directly on the scale(as the spring has absorbed 10 pounds of my weight/force. so in effect, I only weigh 185. The spring still weighs only 5 pounds, but being under pressure(compressed), the force that is compressed is pushing down my extra 10 pounds for me.
If I was suspended from the ceiling by a rope with a fish scale on it, I would brush off the fish's scales and find a cleaner rope. I would then attach a scale used for measuring. Now I am hanging, and the scale reads 195 pounds. I now lower myself onto a spring, sitting on a bathroom scale. When the spring reaches it's maximum compression, the scale will read 200 pounds. Remember... the compressed spring is trying to propel me upward, while at the same time pushing my 'extra' weight downward onto the bathroom scale. Still 200 pounds all around, just distributed somewhat differently.