10 replies to this topic

[adama]

I know that they are the golden spring for getting extra range. However, I am wondering what the spring constant was on them so I could order something of similar power through small parts or mcmaster. At the moment I am only finding springs that will require almost 40 pounds of force to draw back and obviously this is a one shot deal and no good. I don't want to order blindly.

thanks
- adama

Edited by adama, 12 September 2008 - 11:42 PM.

[Draconis]

I know that they are the golden spring for getting extra range. However, I am wondering what the spring constant was on them so I could order something of similar power through small parts or mcmaster. At the moment I am only finding springs that will require almost 40 pounds of force to draw back and obviously this is a one shot deal and no good. I don't want to order blindly.

thanks
- adama

I don't remember the McMaster number off hand, but you can find it by SEARCHING through the +Bow build write up in Homemades. The spring that Captain slug specifies is nearly identical in design. And less expensive than most AR-15 springs too.... You just have to buy five at a time.

Edited by Draconis, 13 September 2008 - 12:40 AM.

[CaptainSlug]

Which spring is cheaper depends on whether or not you have a decent gun shop near you with AR-15 springs in stock.

The spring constant of the AR-15 spring is around 120.

[adama]

Which spring is cheaper depends on whether or not you have a decent gun shop near you with AR-15 springs in stock.

The spring constant of the AR-15 spring is around 120.

What are the units of K?

thanks

i ordered the plus bow springs

thanks everybody

[Split]

The spring constant k is generally in newtons per meter, or pounds per inch I believe.

[mocky]

Could you clarify the spring constant please. If it is
120lbf/inch = 533.786 N/inch
533.786 x 39.370 N/m = 21kN/m

Then I did the conversion for 120N/m, which equates to:
26.977 lbf/m = (26.977/39.370)lbf/inch
= 0.685lbf/inch.

So using the standard imperial and SI units i get either 0.685lbf/inch, which is only half as strong as the weakest strong nerf spring, or its 21kN/m which is strong enough to be used as a motorbike suspension spring.

So can someone clarify the spring constant or point out where my calculations are wrong.

Edited by mocky, 20 September 2008 - 09:23 PM.

[CaptainSlug]

To determine how many inches long your spring should be, take the spring constant and divide it by the number of coils per inch. Then take this value and divide it by your desired spring rate in lbs./inch.

Also see Hooke's Law: http://en.wikipedia....iki/Hooke's_Law
The formula being F = -kx

Edited by CaptainSlug, 20 September 2008 - 09:45 PM.

[Split]

A quick search on Mcmaster showed that they use lbs/inch, which is what you were looking for I believe. If not, feel free to ask again.

[mocky]

I'm familiar with the applications of Hooke's law, I'm aware that the spring constant has the units of force per length, commonly N/m or lbf/inch.

The spring constant of the AR-15 spring is around 120.

120 what exactly? Is it 120N/m, 120lbf/inch, 120N/inch, 120lbf/m, 120lbf/mm, 120N/cm and etc....


Splitlip said its usually lbf/inch or Newtons/meter, however I've calculated for both these units and as stated in my previous post, 120 (N/m) or 120 (lbf/inch) is not the correct spring constant for any Nerf applicable spring. So the units of the spring constant that CaptainSlug has given is the non-conventional units, so can you clarify what the units that have been used.

Also it seems my previous post was overlooked, but I've calculated 120N/m and 120lbf/inch and both are clearly not the correct spring constants for an AR-15. Which means CaptainSlugs' "120" spring constant is not using a common unit of measure for spring constants.

Edited by mocky, 20 September 2008 - 10:40 PM.

[CaptainSlug]

I was only reporting the "Constant" listed on McMaster for a spring I know is almost identical to the AR-15 spring. I don't know the unit of measurement the number indicates, only how to use it to tell me if that particular spring is powerful for it's length and what lengths I can cut it to to get the performance parameters I want.

I HATE algebra and math in general. If I want to know if a spring is powerful I'll look at the wire gauge and the coils per/inch then compare those figures to other springs I have tried. From my experience you don't want a wire gauge larger than .080" and you want a minimum of 2 coils per inch or it won't provide the stroke length you want.

When you shop for continuous length springs they don't tell you the total compression load of the spring because they assume you are going to cut it to match the desired load. So the constant and the formula they provide is all you have to tell you want kind of force can be applied at various cut lengths.

I can't give you terribly exact figures for the AR-15 spring because I never did any real tests on it. The spring I buy from McMaster in packs of 5 to use in the +Bow (part# 9637K26) I have measured with a postal scale as delivering a total compression load of 32lb when fully compressed. I wouldn't recommend a spring any stronger than this one be used in anything Nerf related because it's ridiculously powerful. Anything stronger would also require a great deal more leverage in the priming action in order to be compressed manually.

Edited by CaptainSlug, 21 September 2008 - 01:30 AM.

[mocky]

AAAA, I see the way they've done it at McMaster. The constant is used in their formula for calculating for your application. The units of the constant is in units of coils*lbs/inch, this is not the spring constant however.Also I found this quote from their page:

Then take this value and divide it by your desired spring rate in lbs./inch.

Not sure how you can choose the desired spring rate considering the spring constant is, well... constant. Anyhow as long as the spring works then that's all.

From what Slug has given me and the spring specs from Mcmaster:
Max deflection (Full compression stroke) = Overall length - Overall length*Coils per inch*wire size

For #9637K26
Max deflection = 11" - 2.7192"
Max deflection = 8.2808"

If the maximum compression load 32lb from the postal scale, then spring constant = load/ Max deflection
Spring constant = 32/8.2808 lbf/inch
k (Spring constant) = 3.864lbf/inch
SI units
k = 676.981 N/m or 0.67698 N/mm

This value seems a rather average for mod nerf spring, I'm guessing the spring wasn't compressed to its solid length. Hope this rant might be useful for modders.

Edited by mocky, 21 September 2008 - 05:57 AM.