Clipping Recoil Springs?

[benos]

db4,

Yea, between the two of us, I think we're saying the same thing.

[TDean]

I like to buy a spring, install it and go shooting.

[Duane Thomas]

My man! Two great minds with but a single thought and all that stuff.

[rooster]

I too got confused about the terminology, also I had a ISMI spring still in packaging with instructions on how to fit a spring.  It doesn't work out.  They say to remove spring and retract slide as far as it will go, and mark end of slide travel on frame, then install spring and do same, if marks do not line up then clip spring until they do.  I tried this and my marks never lined up, I had cut almost 2 inches off.  I beleive this was because the spring tunnel was stopping when it touched the guide rod head.   This distance is fixed and I beleive unless you have a really long spring you would always bottom out  with the tunnel before the spring coils touched.  After I tried Duannes method all my springs were correct.

[warpspeed]

A little thread drift.....

If you cut off coils, why not start with a commander length spring? Are they too short to begin with ?

[DogmaDog]

Ack Ack Ack!  

No no no nono!

Db4 was correct about the preload stuff, but coil springs don't become stiffer as they get shorter.

Torsion bars "get stiffer" when you shorten them becuase the moment arm becomes shorter...you exert the same number of pounds, times a smaller number of feet, and wind up with fewer ft-lbs of torque, so it's harder to bend the bar.  A coil spring has a moment arm equal to the diameter of the spring, regardless of how long it is, so cutting coils will have no effect.

OK...here goes:  a spring exerts a force equal to kX, where k = a spring constant in lbs per inch.  My recoil spring sitting on the table is about 8 inches long and compressed 0 inches, and exerts 0 force on anything.  When I put it in the assembled gun, it's length is now about 4 inches...X=4", so it exerts a force = 4 k (whatever "k" is)  If Mike Auger was correct, and the spring weight = the force exerted by the spring at max compression in the gun, then k for an 18# 1911 spring is about 3 lbs, so the force exerted by the spring when the gun is in battery is 12 lbs.  

When the slide is cycled, it moves back about 2 inches, compressing the spring that much further, for a total displacement (x) of 6 inches, resulting in a force of 6" times 3 lbs/inch = 18 lbs of force.  

If you cut coils off the spring, the initial displacement when you put the spring in the gun will be less, and so will the displacement when the gun cycles...the force will drop in a linear fashion as you remove length from the spring.

Now, the problem is that the spring coils have thickness, and therefore a maximum possible displacement (you can compress the spring only until neighboring coils are touching).  Once the coils touch, the spring acts more like a solid metal cylinder than a spring...the slide is just banging against the frame, and it's doing it prematurely, which means the spring isn't absorbing the slide's momentum anymore, and the gun is taking more of a beating than it absolutely has to.

So, if the total thickness of all the coils on your spring is greater than the total length of the spring tunnel in your slide when it is fully to the rear, YOU HAVE A PROBLEM...cut some coils off until they all can fit, with some space between them.  If the slide stop doesn't go up into its notch until the slide hits the frame, and your slide can lock, then your spring isn't too long.

OK, sorry.  I can be a cocky bastard when it comes to physics.  I don't mean this as a tirade.  Hopefully it makes some sense, and explains why you can't for e.g. use a commander spring in your full size 1911.  

Semper Fi,

DogmaDog

[db4]

DogmaDog...on further reflection, we're both wrong, about different things.

A coil spring does not get stiffer when it gets shorter- cut a 4 lb/in spring in half and you get 2 short 4lb/in springs. It does get stiffer when the coil spacing gets wider- take half as much wire and wind it to the same diameter and free length and you have to twist any given part twice as much for a given length change.

Torsion bar rate is expressed as deflection per unit of length/torque applied-degrees per foot per ft/lb. for example.

Take a 10' bar with a 3' lever and hang 10# on the lever, bar deflects n degrees. Cut the bar to 5' and it deflects n/2 degrees but the moment arm (3') doesn't change.

Ain't this fun?

[DogmaDog]

db4,

OK...guess I'm not really clear on what a "torsion bar" is...10' length with a 3' lever???  My mental picture was one where the length = the moment arm.  And, yeah...it is fun!  

I was gonna go look up stuff about atomic bonds in metal, and see how their strength varies with their length, and figure out how coil springs depart from the ideal f=kX based on that.  Instead, I'll just say something everyone can understand:

I checked out the total compression on the spring in my Kimber by pushing the guide rod forward in the slide with the spring installed.  The spring does indeed come very close to maximum compression with a shok buff installed.  

I fired it this weekend with the buff installed, and it suffered no visible damage, so I don't think the spring IS reaching max compression, but it looks like it could if some variables were changed within "normal" ranges (using a heavier spring, more than one shok buff, a somehow modified guide rod, etc.).  

Take home message:  You SHOULD check to make sure your spring doesn't reach maximum compression in your gun, and you should cut it just enough so that it doesn't (I would guess one or two coils would do it in most cases).

Semper Fi,

DogmaDog

[db4]

What I envisioned was a 10' torsion bar with a 3' lever arm attached at 90 degrees on one end, with the other end fixed. Then my 10# load @ 3' from center applies 30 ft/lb of torque to the torsion bar.

Are we on the same page yet?

In the immortal words of Anon.: "you may think you know what i said but what you heard is not what i meant!"

(Edited by db4 at 2:08 pm on April 30, 2002)

[DogmaDog]

Db4,

OK, I get it now.  Have to think about where the moment arm is in that setup (the moment arm of a lever doesn't have to be a physical piece of rigid material).  I'll just go away and do that for a while...  

Lata,

DogmaDog

[double_pedro]

Hi,

I realize this is an old thread but I thought I'd add a straightforward explanation about how cutting springs makes them stronger. I will discuss this in the context of a spring that stretches, but the same thing holds for springs that compress.

Typically, springs are characterized by a "spring constant", k, which expresses how much force it takes to strectch the spring a given distance. Let's say we have a spring with a spring constant of 10 lbs per inch....for each 10 lbs of force applied to the spring, it will stretch by one inch (so long as we remain within its range of linearity). The larger the value of k, the stronger the spring is. You can measure the k of a given spring by hanging a known weight on the spring and measuring how far it stretches.

Now, here's the example that shows how cutting spring makes it "stronger" - namely increases the spring constant, k. Suppose we have two identical springs that have a spring constant of 10 lbs per inch. Imagine taking one spring and hanging a 10 lb weight on it. It will stretch 1 inch. Now, imagine taking the two springs and hooking them end to end, effectively doubling the length of the spring. If you now hang the same 10 lb weight on this "new" spring, how far will it stretch? A little thought will convince yourself that this "new" spring will stretch 2 inches (each individual spring stretching 1 inch in response to the 10 lbs of tension applied). This new spring has an effective spring constant, k_eff, of 10 lbs divided by 2 inches = 5 lbs per inch. It has gotten WEAKER by a factor of two! If you simply reverse the above arguement (cutting a spring in half rather than doubling the length using two identical springs), you find that cutting the spring in half makes it twice as strong. Strange but true!

[Kory]

spring rate (lbs/in) is the average of the required force to compress between 20% compression and 80% compression*.

Cut a coil off and you just changed the compression lengths, and the K value will increase.

However it is mostly unrelated to recoil springs, as they are not rated in lbs/in, but instead rated in how much force they exert at full working compression.

Shorten the spring but keep the same installed length, the spring will exert less force at THAT length.

If you want to sort out your springs, measure the distance your spring has in your gun at full compression (and in battery), then use a homemade spring holder and a fish scale to determine how much force your springs are exerting at those two lengths.

* imagine a 5" spring rated at 10# It got that rating by averaging the force exerted when compressed to 4" and 1" OAL. The K value is approximates a linear constant through that 60% range.

[josh smith]

Hi,

I realize this is an old thread but I thought I'd add a straightforward explanation about how cutting springs makes them stronger. I will discuss this in the context of a spring that stretches, but the same thing holds for springs that compress.

Typically, springs are characterized by a "spring constant", k, which expresses how much force it takes to strectch the spring a given distance. Let's say we have a spring with a spring constant of 10 lbs per inch....for each 10 lbs of force applied to the spring, it will stretch by one inch (so long as we remain within its range of linearity). The larger the value of k, the stronger the spring is. You can measure the k of a given spring by hanging a known weight on the spring and measuring how far it stretches.

Now, here's the example that shows how cutting spring makes it "stronger" - namely increases the spring constant, k. Suppose we have two identical springs that have a spring constant of 10 lbs per inch. Imagine taking one spring and hanging a 10 lb weight on it. It will stretch 1 inch. Now, imagine taking the two springs and hooking them end to end, effectively doubling the length of the spring. If you now hang the same 10 lb weight on this "new" spring, how far will it stretch? A little thought will convince yourself that this "new" spring will stretch 2 inches (each individual spring stretching 1 inch in response to the 10 lbs of tension applied). This new spring has an effective spring constant, k_eff, of 10 lbs divided by 2 inches = 5 lbs per inch. It has gotten WEAKER by a factor of two! If you simply reverse the above arguement (cutting a spring in half rather than doubling the length using two identical springs), you find that cutting the spring in half makes it twice as strong. Strange but true!

I'm with dogmadog on this one.

k for 20" spring = k for 10" spring

spring constant will not change if you cut the spring in half.

Did you actually do the experiment for your calculation?

You're argument is like this, you're saying a 5 feet 10 year old boy will get to 10 feet tall by the time the boy turns 20.

Don't make since does it? Try doing an experiment on your calculation and you'll find out that you are wrong.

[josh smith]

spring rate (lbs/in) is the average of the required force to compress between 20% compression and 80% compression*.

Cut a coil off and you just changed the compression lengths, and the K value will increase.

However it is mostly unrelated to recoil springs, as they are not rated in lbs/in, but instead rated in how much force they exert at full working compression.

Shorten the spring but keep the same installed length, the spring will exert less force at THAT length.

If you want to sort out your springs, measure the distance your spring has in your gun at full compression (and in battery), then use a homemade spring holder and a fish scale to determine how much force your springs are exerting at those two lengths.

* imagine a 5" spring rated at 10# It got that rating by averaging the force exerted when compressed to 4" and 1" OAL. The K value is approximates a linear constant through that 60% range.

You're are both right and wrong.

Cutting a coil off changes the compression leght. but "k" stays the same. AND the "Force" lessen

[josh smith]

bountyhunter,

It's true, as the spring collapses and the coils touch and bind on each other, it does affect the spring rate - it gets stiffer.

The remaining spring gets "stiffer" simply because it gets shorter as you cut it. If "spring theory doesn't make sense, think of it like this - a coil spring is no different from a torsion bar type spring (on a car, for instance). (A straight piece of metal, fastened or fixed at one end, and attached to a moveable component at the other. like your car's control arm.) It is just a torsion bar, bent into coils. When you cut a torsion bar's length, it gets stiffer, or stronger. It has to. Imagine a fixed diameter rod or wire, one mile long, clamped in a vice at one end you are twisting on the other with a 1 foot long attached bar for leverage. Now imagine the same situation only the rod or wire is 1 foot long. Which do you thing you would be able to twist easier? Does any of this theory matter concerning what a particular spring does in a particular gun? I don't think so. So we just use approximations of spring rate and try to get the gun into the ballpark.

OK, I did pull some of that out of a previous post, as you may have  noticed.

be

Scientific theories always right, just the matter of having the tools to do the experiment.

You can't use the same formula for these two different things.

On your example, if using the SAME rod for 1 mile long and 1 foot long with 1 foot long on one end for leverage, the force required to twist the the rod until breakage is the same. If you're talking about longer leverage for 1 mile rod and shorter leverage for 1 foot rod then its a different story. you can twist same materials the same amount no matter how long.

if you change the length of the lever then its easy to twist with long lever.

Edited April 21, 2009 by josh smith

[Ray R.]

I read through this discussion twice, and I'm either missing something, or there is something wrong with the logic in all of this.

Thought experiment:

Take a constant rate 20# coil spring and compress it one inch. The force required to do that we'll call "P."

Now cut the spring exactly in half, and compress a short half one inch. The force (assuming a constant rate as the coils compress) should still be "P."

Why? Because each coil of the shorter spring has to travel twice as far to compress one inch as the coils in the longer spring. Thus each coil of the shorter spring exerts twice the resistance of each coil in the longer spring.

However, since no spring has in reality a constant rate of compression as the coils compress different distances, it is quite likely that the shorter spring will require more force to compress it one inch then the longer spring would. The real problem though, is that compressing either length of spring outside an enclosed area dosen't translate to compressing the springs the same distance in an enclosed length because the longer spring will start out compressed farther in an enclosed area then the shorter spring. And that is why springs get "lighter" (exert less resistance) as they are shortened.

It would seem then, that a recoil spring is "correct" if it allows function, and doesn't fully compress so that the coils reach the limit if their travel.

But I could be wrong....

[1SOW]

I think what we're missing here is the effect of spring pre-load: the amount of compression between free length and installed length. This determines how much pressure is exerted with spring installed and slide forward. Quick measure on a 5" Colt shows about 5.6" free and 3.6" installed, or 2" of preload (round numbers). So, an 18 lb/in spring gives us 36# of static preload. If we clip the spring 1", it does get stiffer (and no, I don't know how much) but we lose half the preload length, far an installed pressure of 19 or 20# (1 x whatever the new rate is).

  Now, at full stroke (about another 2in) our spring pressure is (3" x 19#) 57#, where the full spring was (4" x 18#) 72#!

 

Gotta take spring rate x total compression.

The BIG question is...did that make any sense?

DVC

Yes, it makes sense. Another thought has to do with inertia. Preload and inertia fights the movement initially. Suppose you install a 30# spring that is 1/2" away from contact--no preload and 1/2" of free movement. When the cycle starts, the slide would move fast gaining inertia. It might even compress the 30# spring and cycle.

When a spring is clipped a couple of coils, it may be stronger, but the slide will start moving faster and gain inertia faster.

P.S. When the hammer spring on a CZ is clipped two coils, the measured trigger pull "DECREASES.

Edited April 24, 2009 by 1SOW

[DougR]

Finding accurate, easy to understand info about compression springs can be difficult, but not impossible.

I got my information from the people that engineer and build springs.

My feeling is they know what they are talking about.

I'm being a bit pedantic here because of some of the posts from people that apparently have not studied springs or spring engineering.

Most spring engineering priciples are intuitive, such as increasing wire diameter increases spring rate.

Another intuitive compression characteristic is as the coil diameter is increased, the spring rate goes down.

However the compression and extension spring characteristic that most folks have a difficult time with is the concept of cutting coils and length to an existing compession spring of fixed wire diameter and fixed coil diameter.

The concept that cutting a coil from a spring increases the spring rate eludes people.

It doesn't matter who says it. I choose to believe the people that actually build springs.

The spring manufacturers must know their stuff.

Let's look at a standard 1911 recoil spring. The music wire diameter is .043", it has a free length of 6.55 inches, the coil outside diameter is .430", the insde diameter is specified as a minimum of .336". The spring is 30 coils of which 29 are active. The spring rate is 2.88 lbs/inch. The installed length is 3.72 inches yielding an installed initial force of 8 lbs.

The spring is rated at 13.55 lbs at a compression of 1.81 inches. The slide actually moves 2.1/2.2 inches from battery to full rearward.

The solid length for this diameter wire and number of coils and free length is 1.375.

This solid length is over .4 inches from the maximum rearward motion of the slide.

No chance of this spring at these specifications of reaching coil bind.

Now from the spring engineering site they say:

"as the number of active coils are cut from a given length spring the spring compression rate increases".

This has been mentioned in several posts here, but somehow has not soaked in or has been ignored.

Further the spring people say if you remove a coil the spring rate increase by the ratio of the original number of coils to the ratio of the new reduced number of coils.

In the case of a 1911 recoil spring with 29 active coils, cut out one coil and we now have 28 active coils.

Doing a little math, 29/28=1.0357. That is the spring rate went up 3.57% from 2.88 to a new rate of 2.98 lbs/inch.

The spring is also now 1/30th shorter. 6.2833 inches vs 6.5 inches.

Stay with me here we are getting to the meat of this subject.

So we install the newly cut spring and the new intial load is 6.2833 minus 3.72" = 2.5633 inches.

Multiply the 2.5633 times the new spring rate of 2.98 lbs/inch and we get 7.63 lbs. Less than the original 8.0 lbs.

Now we'll fire the pistol and make the slide move fully rearward. 2.1 inches from battery.

2.1 more inches times 2.98 lbs spring rate give us 6.258 additional lbs plus the original 7.63 gices us 13.888.

13.888 compares to the original specification of 13.55 lbs.

The force of the cut spring is greater than the longer uncut spring.

I'll bet you didn't see that one coming.

So is it just possible the folks that make these springs just might know what they are doing?

For those of you old enough to have cut coil springs to lower your car and then wondered why it rode so rough, you now know why.

Edited April 24, 2009 by DougR

[josh smith]

Finding accurate, easy to understand info about compression springs can be difficult, but not impossible.

I got my information from the people that engineer and build springs.

My feeling is they know what they are talking about.

I'm being a bit pedantic here because of some of the posts from people that apparently have not studied springs or spring engineering.

Most spring engineering priciples are intuitive, such as increasing wire diameter increases spring rate.

Another intuitive compression characteristic is as the coil diameter is increased, the spring rate goes down.

However the compression and extension spring characteristic that most folks have a difficult time with is the concept of cutting coils and length to an existing compession spring of fixed wire diameter and fixed coil diameter.

The concept that cutting a coil from a spring increases the spring rate eludes people.

It doesn't matter who says it. I choose to believe the people that actually build springs.

The spring manufacturers must know their stuff.

Let's look at a standard 1911 recoil spring. The music wire diameter is .043", it has a free length of 6.55 inches, the coil outside diameter is .430", the insde diameter is specified as a minimum of .336". The spring is 30 coils of which 29 are active. The spring rate is 2.88 lbs/inch. The installed length is 3.72 inches yielding an installed initial force of 8 lbs.

The spring is rated at 13.55 lbs at a compression of 1.81 inches. The slide actually moves 2.1/2.2 inches from battery to full rearward.

The solid length for this diameter wire and number of coils and free length is 1.375.

This solid length is over .4 inches from the maximum rearward motion of the slide.

No chance of this spring at these specifications of reaching coil bind.

Now from the spring engineering site they say:

"as the number of active coils are cut from a given length spring the spring compression rate increases".

This has been mentioned in several posts here, but somehow has not soaked in or has been ignored.

Further the spring people say if you remove a coil the spring rate increase by the ratio of the original number of coils to the ratio of the new reduced number of coils.

In the case of a 1911 recoil spring with 29 active coils, cut out one coil and we now have 28 active coils.

Doing a little math, 29/28=1.0357. That is the spring rate went up 3.57% from 2.88 to a new rate of 2.98 lbs/inch.

The spring is also now 1/30th shorter. 6.2833 inches vs 6.5 inches.

Stay with me here we are getting to the meat of this subject.

So we install the newly cut spring and the new intial load is 6.2833 minus 3.72" = 2.5633 inches.

Multiply the 2.5633 times the new spring rate of 2.98 lbs/inch and we get 7.63 lbs. Less than the original 8.0 lbs.

Now we'll fire the pistol and make the slide move fully rearward. 2.1 inches from battery.

2.1 more inches times 2.98 lbs spring rate give us 6.258 additional lbs plus the original 7.63 gices us 13.888.

13.888 compares to the original specification of 13.55 lbs.

The force of the cut spring is greater than the longer uncut spring.

I'll bet you didn't see that one coming.

So is it just possible the folks that make these springs just might know what they are doing?

For those of you old enough to have cut coil springs to lower your car and then wondered why it rode so rough, you now know why.

Hard to follow. Maybe because im not a spring person.

But I know simple physics. Coil springs obey Hooke's law.

And Hooke's Law is F=kx, where F is force, k is spring constant and x is the distance which the spring deformed

example 1) Two springs, S1=10", S2=20", k = 3lbs/inch.

Compress both springs to 5".

F=kx

F1=(3lbs/inch)(5inches)

F1=15lbs

F=kx

F2=(3lbs/inch)(15inches)

F2=45lbs

This example tells us that it takes only 15 lbs of force to compress 10" to 5" and it takes 45 lbs to compress 20" to 5"

example 2) (exaggeration)

Two springs, S1=10", S2=20", k=3lbs/inch

Compress both spring to 10 inches

For S2

F=kx

S2 =20"

x2=20"- 10" = 10"

k=3lbs/inch

so, F2=(3)(10) = 30lbs.

For S1

F=kx

S1=10"

x1= 0" since its only 10" long

k=3lbs/inch

so, F1=(3)(0) = 0 lbs.

It make since to me. longer spring is stronger than shorter spring

I can't tell you that you are wrong coz I can't follow your explaination, maybe bcoz im not a spring person.

but can you tell me why am I wrong?

[Ray R.]

josh has it correct. Earlier I said that it would take the same force to compress both springs the same distance "assuming a constant rate." But in reality the amount of force required to compress each coil changes according to the distance it must be compressed. Therefore, it takes more force to compress a 10" spring one inch then it does to compress a 20" spring one inch.

But again, when the spring is in the gun it is already under load, and the longer spring starts out under a greater load then the shorter spring. So it has to take more force to compress the longer spring a given distance. As josh notes: if you cut the spring short enough, no force is required to compress the spring because it isn't being compressed. And that's the limiting case.

Try it youself. Take an old recoil spring, put it in the gun sans other parts, and pull the slide back. Then shorten the spring about half the excess distance needed to close the slide, and pull the slide back again. Guess what? The shorter spring requires less force to fully retract the slide.

You can do the same experiment by cutting a coil or two off the spring in the mag release. Cut it too short, and it won't hold the mag in the gun under recoil.

End result is, if you cut coils from the recoil spring it takes less force to drive the slide fully rearward. And while this can indeed help ejection, it can also batter the hell out of the gun. Better to keep things in proper balance, and if you have ejection problems you may have to tune the ejector instead of the spring.

[DougR]

Hi Josh,

Thanks for your interest.

It looks like to me you are using the same spring rate for all three different springs.

If all three springs use the same diameter wire, wound to the same diameter coil, but different lengths then the spring rate will vary.

As has been said by others here, if you have three springs where everything is the exact same except the length, you will have three different spring rates. The relationship is linear. That is to say a spring that is twice as long will be half the spring rate. If a spring is exactly the same wire and diameter, just like the previous example, but half the length, the spring rate is double.

It is somewhat counter-intuitive.

Another way to look at it; Whatever the spring length, it doesn't matter for this example.

Spring rate is the amount of force to compress that spring one inch.

If you are compressing 10 coils, each coil is compressed 1/10th inch.

If you are compressing 20 coils each coil is only compressed 1/20th inch.

It only takes half the force to compress an individual coil half the distance.

Any individual coil doesn't know how many coils are in the spring.

It still takes the same force to compress that coil X dimension.

It takes twice the force to compress one coil 2X dimension.

It takes half the force to compress the coil 1/2X.

This is why the spring rate will vary as the length is varied for the the same exact wire diameter and coil diameter.

This is why cutting one or two coils increases the spring rate.

It's an inverse linear relationship.

Just for grins, buy a cheap screen door tension spring. Pull the spring as far apart as you can. Notice approximately the length you are able to pull.

Now cut the spring in half, then pull the same spring. Notice the spring is much harder to pull anything close to the same distance.

When you cut the spring in half, you doubled the spring rate.

Once the light comes on you'll wonder why it seemed strange.

When the light come on, the next challenge is to explain it to someone else.

It really is counter intuitive.

I see the making of a bar trick, with a spring gauge.

Edited April 24, 2009 by DougR

[josh smith]

Hi Josh,

Thanks for your interest.

It looks like to me you are using the same spring rate for all three different springs.

If all three springs use the same diameter wire, wound to the same diameter coil, but different lengths then the spring rate will vary.

As has been said by others here, if you have three springs where everything is the exact same except the length, you will have three different spring rates. The relationship is linear. That is to say a spring that is twice as long will be half the spring rate. If a spring is exactly the same wire and diameter, just like the previous example, but half the length, the spring rate is double.

It is somewhat counter-intuitive.

Another way to look at it; Whatever the spring length, it doesn't matter for this example.

Spring rate is the amount of force to compress that spring one inch.

If you are compressing 10 coils, each coil is compressed 1/10th inch.

If you are compressing 20 coils each coil is only compressed 1/20th inch.

It only takes half the force to compress an individual coil half the distance.

Any individual coil doesn't know how many coils are in the spring.

It still takes the same force to compress that coil X dimension.

It takes twice the force to compress one coil 2X dimension.

It takes half the force to compress the coil 1/2X.

This is why the spring rate will vary as the length is varied for the the same exact wire diameter and coil diameter.

This is why cutting one or two coils increases the spring rate.

It's an inverse linear relationship.

Just for grins, buy a cheap screen door tension spring. Pull the spring as far apart as you can. Notice approximately the length you are able to pull.

Now cut the spring in half, then pull the same spring. Notice the spring is much harder to pull anything close to the same distance.

When you cut the spring in half, you doubled the spring rate.

Once the light comes on you'll wonder why it seemed strange.

When the light come on, the next challenge is to explain it to someone else.

It really is counter intuitive.

I see the making of a bar trick, with a spring gauge.

I'm confuse with the term spring rate. Define spring rate?

I thought spring constant(k) depends on the material of the spring, diameter of the wire, diameter of the coil.

But we are not changing those when we cut the spring, we just change the lenght, so k stays the same.

Can you give me the formula that you use to get your calculation? I majored in physics and the formula I use is in the textbook and they never said that the spring constant changes when you cut the spring. And if the spring constant changes, Hooke Law(F=kx) will not work, will it? And I pay lots of money to go the university.

[josh smith]

I just googled "spring rate".

and I found this.

"As the number of active coil decrease, the spring rate increases"

TRUE if and only if the lenght of the spring stays the same.

and I also found this.

"Normal Springs has a fixed spring rate.

Step Linear Springs are springs which have 2 different spring rates.

Progressive Springs have a variable spring rate."

I do believe recoil spring is under Normal Spring, so spring rate is constant.

Spring rate is the same as Spring constant.

For people thinks that springs get stronger if you cut them. Prove it by giving us the formula that shows that its true.

If I'm wrong here I'm going to sue my university. lol

reference: http://www.tuninglinx.com/html/springs.html

[DougR]

Josh,

You are making this more complicated than necessary.

I defined spring rate already.

A physics whiz like youself should have no trouble understanding pounds per inch.

I'll discuss this further after you can tell me why it takes twice the energy to pull a screen door spring that is half the original length.

You don't need a laboratory to perform this experiment.

[josh smith]

Josh,

You are making this more complicated than necessary.

I defined spring rate already.

A physics whiz like youself should have no trouble understanding pounds per inch.

I'll discuss this further after you can tell me why it takes twice the energy to pull a screen door spring that is half the original length.

You don't need a laboratory to perform this experiment.

yes spring rate is same as spring constant.

and Normal springs has a fixed spring rate and recoil spring is a normal spring.

I gave you the formula thats proves I was right. (F=kx). Now give me the formula that

proves that you are right.

I don't believe you actually tried cutting the screen door spring in half and measure the force. I think you just wrote that because thats what you believe. You can convince that Im wrong if you give the formula of what you believe.

If you show me the formula of how you got twice the energy if you cut the door spring in half then I'll tell everything you want.

Edited April 24, 2009 by josh smith