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#1 Bubba Longshot

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Posted 31 July 2017 - 12:38 PM

Hi nerfers of the NIC. After researching all over the Internet, I can't seem to find the answer to this question: How do you figure out the kg force of a cut down spring?
My situation is this: I cut down the stock Magnus spring to put in my longshot. The Magnus spring is 6 1/2 inches long, and has a total of 4.5 kg in force (I know force is measured in Newtons, but the NIC does it in kg). If I were to cut down that spring to 4 1/2 inches, the total kg would change right? I don't know if this is correct, but a spring that's 6 1/2 in would have roughly 0.69 kg per inch (4.5/6.5). That means at 4 1/2 inches, it would be 3.1 kg (0.69x4.5). Along with the stock LS spring which is 4.5 kg, that would be a 7.6 kg spring load in my longshot correct? I want to know since I don't want to be putting too much stress on the internals.
By the way, I did in fact try to find out the spring constant myself [constant=force/(free length-compressed)] but I only had lifting weights not Dumbbells, so the results were really wild.

Thanks

Bubba Longshot

EDIT: If anyone is wondering how I knew the stock springs were 4.5 kg, it's because Captain Xavier of YouTube measured it in his "Can it take a [k26] episodes" and Magnus "shotgun" build.

Edited by Bubba Longshot, 31 July 2017 - 12:47 PM.

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#2 Meaker VI

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Posted 31 July 2017 - 11:25 PM

Spring Constant. The given force for a spring is it's expected free (uncompressed) length minus the compressed length (the amount the spring 'moves') times the spring constant.


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#3 IAmAPenguin

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Posted 01 August 2017 - 02:28 AM

Correct me if i'm wrong, but i was lead to believe that spring stiffness is not altered via the process of cutting the spring to the desired length.
An personal experience to back this up is that when i was building my homemade, i used 3 century c-836 springs (~12kg?) lined up for the desired spring length (30cm) and when i tested the draw weight, it definately didn't feel like 36kgs.
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#4 scruffynerfherder

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Posted 01 August 2017 - 02:42 AM

Cutting a spring can change the spring rate depending on spring. Some are progressive, getting stiffer towards an end, and some are linear staying the same stiffness throughout
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#5 Bubba Longshot

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Posted 01 August 2017 - 09:52 AM

Is my way was correct, or incorrect? I ran it through an online spring constant calculator (gave dimensions and coils, etc), and I did in fact get 3.1 kg. So that makes me wonder...
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#6 Meaker VI

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Posted 01 August 2017 - 11:53 AM

You need to know the compressed length to get the constant. Once you have that, your cut down spring will have a force equal to the new movement length times the constant.

 

For example, a theoretical 10" long spring with a 1# constant and .25" of coils/inch (coils/inch * wire diameter). Normally, you can compress the spring 7.5" for a spring weight of 7.5#. Cut 4" off, now you can compress the 6" spring 4.5" for a weight of 4.5#.

 

Your way sounds close. If you can measure your spring fully compressed, you've got the rest of the info you'd need to figure it out.


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#7 Bubba Longshot

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Posted 01 August 2017 - 04:16 PM

I know that the stock magnus spring had to move ~5.5 inches (6.5 to 1) inside the magnus plunger tube. When the spring had to compress 5.5 inches, it exerted 4.5 kg of power. Now in the Longshot plunger, the spring has been cut down to 4.5 inches, and has to move 4 inches to fully compress. That means I cut off 2 inches from the free length spring, and made it compress less by 1.5 inches.

If 5.5 inches of compression gave 4.5 kg, then that means that the spring gives out roughly 0.81 kg per inch of compression. Therefore at 4 inches of compression, the stock magnus spring will exert 3.24 kg of power; that number is roughly the same as my estimate. 

That means that the spring load in my Longshot is around 7.74kg. 

Is that correct?


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#8 Meaker VI

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Posted 01 August 2017 - 04:37 PM

I know that the stock magnus spring had to move ~5.5 inches (6.5 to 1) inside the magnus plunger tube. When the spring had to compress 5.5 inches, it exerted 4.5 kg of power. Now in the Longshot plunger, the spring has been cut down to 4.5 inches, and has to move 4 inches to fully compress. That means I cut off 2 inches from the free length spring, and made it compress less by 1.5 inches.

If 5.5 inches of compression gave 4.5 kg, then that means that the spring gives out roughly 0.81 kg per inch of compression. Therefore at 4 inches of compression, the stock magnus spring will exert 3.24 kg of power; that number is roughly the same as my estimate. 

That means that the spring load in my Longshot is around 7.74kg. 

Is that correct?

 

Sounds right!


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#9 Draconis

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Posted 01 August 2017 - 05:45 PM

I know that the stock magnus spring had to move ~5.5 inches (6.5 to 1) inside the magnus plunger tube. When the spring had to compress 5.5 inches, it exerted 4.5 kg of power. Now in the Longshot plunger, the spring has been cut down to 4.5 inches, and has to move 4 inches to fully compress. That means I cut off 2 inches from the free length spring, and made it compress less by 1.5 inches.

If 5.5 inches of compression gave 4.5 kg, then that means that the spring gives out roughly 0.81 kg per inch of compression. Therefore at 4 inches of compression, the stock magnus spring will exert 3.24 kg of power; that number is roughly the same as my estimate. 

That means that the spring load in my Longshot is around 7.74kg. 

Is that correct?

 

No. Also, the fact that you are mixing your units is a real pain in the ass, especially given that kg is not a measure of force.  I am going by your numbers, and have never bothered to measure the spring rate on any spring, but here goes.

Hooke's law is basically this equation: F= kx
If the force is 9.9lb (4.5kg * 2.2 lb/kg), and the displacement is  5.5", then our spring constant is:

k =  F/x = (9.9lb/5.5") = 1.8lb/inch of compression.

 

However, you are changing the constant k when you cut the spring, because the free length changes, but the spring modulus remains the same.  

k = spring modulus (λ)/ free length (l), so λ = k * l = 1.8 * 6.5 = 11.7

The new k = 11.7/4.5 = 2.6lb/inch

If you are compressing this shortened spring by 4 inches, then your new force F = 2.6 * 4 = 10.4lb


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#10 Bubba Longshot

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Posted 01 August 2017 - 06:22 PM

No. Also, the fact that you are mixing your units is a real pain in the ass, especially given that kg is not a measure of force.  I am going by your numbers, and have never bothered to measure the spring rate on any spring, but here goes.
Hooke's law is basically this equation: F= kx
If the force is 9.9lb (4.5kg * 2.2 lb/kg), and the displacement is  5.5", then our spring constant is:
k =  F/x = (9.9lb/5.5") = 1.8lb/inch of compression.
 
However, you are changing the constant k when you cut the spring, because the free length changes, but the spring modulus remains the same.  
k = spring modulus (λ)/ free length (l), so λ = k * l = 1.8 * 6.5 = 11.7
The new k = 11.7/4.5 = 2.6lb/inch
If you are compressing this shortened spring by 4 inches, then your new force F = 2.6 * 4 = 10.4lb

As stated in my first post, I know that kg is not a measurement of force (newtons is), but the vast majority of the NIC measures spring loads in kg so that's what I did.
From what I can understand from your post, the shortened spring requires 10.4 lbs to compress 4 inches; 10.4lbs is 4.7kg. Does that mean the shortened spring outputs more power then the stock one? Meaker VI confirmed that my calculations were correct, but you contradict that so now I'm confused... Does that mean my Longshot has a 9.2kg spring load? It sure doesn't feel like that, and the ranges/fps are only 7.5 kg LS level.
I've heard something along the lines of a shortened spring being stronger in some area, but most research I've done contradicts that. When people cut coils off springs, they loose power. Many mod guides recommend that you cut the most you can to get more power. Also logic says that the less you compress a spring, the less energy is required. It'd require less energy to compress 3inches of a [k26], then the whole 11inches. Another important fact is that I ran the cut down spring dimensions in an online spring constant calculator, and got 3.1kg for the total spring. Fully compressing a cut down spring should require less force then the full one. Say I needed x Newtons to fully compress 10 inches of spring. Then I cut the spring down to 5 inches, to fully compress that, I'd need less Newtons; it's basic physics. Since the constant is the same throughout, by cutting the spring I reduced the draw length, thus reducing power. Finally, the spring should have the same constant throughout. I determined that the Magnus spring is roughly 0.7kg/inch. This matches up with the full 6.5 inches being 4.5kg, and the way Meaker VI showed me. If my reasoning is in fact wrong, please correct me.

Edited by Bubba Longshot, 01 August 2017 - 10:43 PM.

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#11 Meaker VI

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Posted 01 August 2017 - 10:38 PM

Meaker VI confirmed that ...


First off, I'm a designer and not an engineer. I've got more engineering background than many in my profession, but I may be 100% off base on this so I'll defer to Drac, who IIRC is an engineer.
 

That said:

 

From what I can understand from your post, the shortened spring requires 10.4 lbs to compress 4 inches; 10.4lbs is 4.7kg. Does that mean the shortened spring outputs more power then the stock one?

This is where he looses me as well. If the spring constant is.. well, constant, shouldn't it output less force if you remove 1/3 of it?


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#12 Bubba Longshot

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Posted 02 August 2017 - 10:08 AM

 This is where he looses me as well. If the spring constant is.. well, constant, shouldn't it output less force if you remove 1/3 of it?

Yes, it doesn't make sense. I asked my dad this and he said that if you cut a spring, but the constant is the same throughout, there'll be less power at full compression (vs original full compression).
Keep in mind that I ran the cut down spring through a constant calculator, and got 0.81 kg/inch. That matches up with the original length being 4.5 kg, and the cut down one being 3.24 kg.
I also:
Figured it out through my estimate way
Figured it out with Meaker VI's method
Confirmed it with a spring constant calculator
Confirmed the numbers with my dad
Contacted an OMW employee about it

EDIT: After hunting for every spring related forum on Nerfhaven, the vast majority tends to use Meaker VI's method. Where constant=Force/(full length-compressed length)
So:
constant=4.5kg/(6.5-1)
constant=4.5kg/5.5
Constant=0.81/inch of compression
Therefore, 4 inches of compression would give 3.24kg.
For you politically correct people out there, I know that you don't measure force in kg, but the majority of the NIC does it like that. So, I measured in kg so I won't confuse or mislead anyone.

Edited by Bubba Longshot, 02 August 2017 - 10:40 AM.

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#13 Draconis

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Posted 02 August 2017 - 05:03 PM

As stated in my first post, I know that kg is not a measurement of force (newtons is), but the vast majority of the NIC measures spring loads in kg so that's what I did.
From what I can understand from your post, the shortened spring requires 10.4 lbs to compress 4 inches; 10.4lbs is 4.7kg. Does that mean the shortened spring outputs more power then the stock one? Meaker VI confirmed that my calculations were correct, but you contradict that so now I'm confused... Does that mean my Longshot has a 9.2kg spring load? It sure doesn't feel like that, and the ranges/fps are only 7.5 kg LS level.
I've heard something along the lines of a shortened spring being stronger in some area, but most research I've done contradicts that. When people cut coils off springs, they loose power. Many mod guides recommend that you cut the most you can to get more power. Also logic says that the less you compress a spring, the less energy is required. It'd require less energy to compress 3inches of a [[k26]], then the whole 11inches. Another important fact is that I ran the cut down spring dimensions in an online spring constant calculator, and got 3.1kg for the total spring. Fully compressing a cut down spring should require less force then the full one. Say I needed x Newtons to fully compress 10 inches of spring. Then I cut the spring down to 5 inches, to fully compress that, I'd need less Newtons; it's basic physics. Since the constant is the same throughout, by cutting the spring I reduced the draw length, thus reducing power. Finally, the spring should have the same constant throughout. I determined that the Magnus spring is roughly 0.7kg/inch. This matches up with the full 6.5 inches being 4.5kg, and the way Meaker VI showed me. If my reasoning is in fact wrong, please correct me.

 

Just because some people in this community, well, less here than elsewhere, refer to force in kilograms, does not make it correct.  The only reason they do is because the fucktards at OMW started using it because the wanted to be edgy and metric.  But they lack a theoretical understanding of how the springs they commission and sell actually operate.  And referring to a dynamic entity like a spring with a set value, makes it worse.  The only set value in this entire mess is the spring modulus, based on the material cross-section, orientation, and stiffness.  The force you are trying to label these springs with is only constant at a specific displacement distance relative to the full free length.  Say there is a hypothetical spring which has a free length of 10.0", and it requires a 10.0lbs force to compress it completely.  If you cut that spring in half, it will still require 10.0lbs of force to compress that half completely.  If the original spring required 4.0lbs of force to compress it 4.0", the half-length spring may also be able to compress 4.0", but that will take 8.0lbs of force.  The spring did not become less stiff because you made it shorter.  It can be confusing, I understand, but you have to be wary of nerf pages when you are trying to answer a physics question.  There is a great deal of misinformation which has been disseminated through this hobby, and it becomes worse when people use the incorrect nomenclature.

 

 

First off, I'm a designer and not an engineer. I've got more engineering background than many in my profession, but I may be 100% off base on this so I'll defer to Drac, who IIRC is an engineer.
 

That said:

 

This is where he looses me as well. If the spring constant is.. well, constant, shouldn't it output less force if you remove 1/3 of it?

 

Not yet, but getting there.  My progress through engineering school is slow, due to work and such.  I probably won't graduate for a few years.

 

The spring constant is only constant for the whole unit.  The spring modulus is constant for the material (ish).  If you cut a spring, the force required to compress the two to the same percentage of total length remains the same, but it acts over a shorter distance.  There is less stored elastic energy.  I would not be surprised if the "rate constant" supplied by McMaster-Carr is a usable approximation of the spring modulus, since they suggest finding the "rate" by dividing the constant by the length, and that is the same equation I outlined in the second portion of my first post.

 

 

Yes, it doesn't make sense. I asked my dad this and he said that if you cut a spring, but the constant is the same throughout, there'll be less power at full compression (vs original full compression).
Keep in mind that I ran the cut down spring through a constant calculator, and got 0.81 kg/inch. That matches up with the original length being 4.5 kg, and the cut down one being 3.24 kg.
I also:
Figured it out through my estimate way
Figured it out with Meaker VI's method
Confirmed it with a spring constant calculator
Confirmed the numbers with my dad
Contacted an OMW employee about it

EDIT: After hunting for every spring related forum on Nerfhaven, the vast majority tends to use Meaker VI's method. Where constant=Force/(full length-compressed length)
So:
constant=4.5kg/(6.5-1)
constant=4.5kg/5.5
Constant=0.81/inch of compression
Therefore, 4 inches of compression would give 3.24kg.
For you politically correct people out there, I know that you don't measure force in kg, but the majority of the NIC does it like that. So, I measured in kg so I won't confuse or mislead anyone.

 

 

Well, either your father is misremembering physics, or there was a miscommunication.  A shorter version of the same spring will store less energy, but that is because the distances involved changed, not because the spring did.  And I'm sorry, but OMW is a useless company.  From what I can tell, they do not actually employ engineers, only designers.  And they don't manufacture springs.  They order springs made to fit an inner and outer diameter, which will require a set amount of force to compress to a specific length.  These aftermarket parts supply companies would perform a much greater service to the community by adding more information for the people who care.  WIthout any excessive effort on their part.


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#14 Bubba Longshot

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Posted 02 August 2017 - 05:50 PM

  Say there is a hypothetical spring which has a free length of 10.0", and it requires a 10.0lbs force to compress it completely.  If you cut that spring in half, it will still require 10.0lbs of force to compress that half completely.  If the original spring required 4.0lbs of force to compress it 4.0", the half-length spring may also be able to compress 4.0", but that will take 8.0lbs of force.  The spring did not become less stiff because you made it shorter.  It can be confusing, I understand, but you have to be wary of nerf pages when you are trying to answer a physics question. There is a great deal of misinformation which has been disseminated through this hobby, and it becomes worse when people use the incorrect nomenclature. 

 

Ok so from what I understand from your very helpful post, spring output is dependent on the free length and how much you compress it. The theoretical 5 inch spring took 8lbs to compress it to 4 inches since that was 4/5 of the spring. If you wanted to compress 4/5 of the original spring (by 8 inches), that'll require 8lbs. The constant for the original spring was 1lb/inch of compression. When you cut it in half, that doubled...

 

So now I'm going to try to apply that in my situation. The stock 6.5 inch Magnus spring took 4.5kg to compress it by 5.5in, that means it has a constant of 0.81kg/inch. I changed the full length to 4.5 inches. 4.5 is roughly 2/3rds of the original spring length. That means the constant increased by 1/3 (0.27), so the  new constant is now 1.08kg/inch. Therefore, the new spring has an output of 4.32kg/inch?

I'm not sure this is right, so if I'm wrong, please correct me.

EDIT:THE ORIGINAL SPRING ACTUALLY COMPRESSED TO 0.5IN

HEHE, turns out I accidentally mismeasured the compressed length...whoops

 

So now I'm going to apply this to my situation.The stock 6.5 inch Magnus spring had to compress 6 inches, or 92% of the original spring. The new cut down spring is 4.75 inches long, and compresses by 4 inches; a.k.a 84%. It took 4.5kg to compress 92%, meaning 0.05kg per 1%. So to compress 84%, that'll require 4.2kg. 

Does that mean that my Longshot has a 8.7kg springload?!

 

EDIT #2: Now that I think about it, I really don't think the stock LS spring is 4.5kg. The NIC Spring database records it as 2lbs per inch of compression. It compresses by 4 inches + the 0.75inch of precompression; that's 9.5lbs, or 4.3kg. Which means, my LS springload is really 8.5kg total.

But if we really want to get technical, the nerfwiki lists the Longshot's spring at 3kg?! Finally, OMW lists their 8kg spring as 2.5x the original?! Maybe if I have time, I'll make an updated spring database through all of my knowledge gained...


Edited by Bubba Longshot, 02 August 2017 - 06:17 PM.

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#15 Draconis

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Posted 02 August 2017 - 07:07 PM

The only data you listed there that I would trust is the one in real units.  In general, Nerfwiki has always been terrible for false data and false history.  And OMW is no better for reliability.


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#16 Bubba Longshot

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Posted 02 August 2017 - 08:20 PM

The only data you listed there that I would trust is the one in real units.  In general, Nerfwiki has always been terrible for false data and false history.  And OMW is no better for reliability.

Thanks for clearing that up...
Also, may you please check if my new calculations are correct?

Thanks
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