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Posted: 4/19/2002 11:26:13 AM EDT
Hi, Would any of you know the answers to the following? Or perhaps you could help direct me to some technical resource that will answer the following? a. spring co-efficient of the recoil spring (for the M16 and/or M249 SAW) b. how much pressure it takes to cock an M16 and/or M249 SAW I know...its sounds ridiculous...I need it for a physics project. Thanks in advance.
Link Posted: 4/19/2002 12:22:49 PM EDT
Link Posted: 4/19/2002 12:36:24 PM EDT
[Last Edit: 4/19/2002 12:43:30 PM EDT by IMHO]
He means "spring constant", or "force constant", "k", I think. Also, I'm not sure why you've chosen "pressure" units for the other problem. Why not force?
Link Posted: 4/19/2002 12:38:16 PM EDT
[Last Edit: 4/19/2002 12:39:38 PM EDT by alphazulu]
Thanks for the reply...I think it should help a bit. In short, the spring co-efficient essentially means the stiffness of a spring (measured in pounds per inch or N/m). [edited for grammar]
Link Posted: 4/19/2002 12:46:50 PM EDT
"k" (lower case) is the "spring constant", not coefficient. "k" is the "strength" of the spring. The greater the spring constant the more force is needed to compress or stretch the spring a given distance. F=-kx
Link Posted: 4/19/2002 12:50:42 PM EDT
[Last Edit: 4/19/2002 12:52:36 PM EDT by IMHO]
Yep, Hooke's Law I: There's a direct relationship between the force applied, and the displacement, so linear: y=mx+b, in the case of the spring: F=kx+(Y intercept is zero, right?)
Link Posted: 4/19/2002 1:52:24 PM EDT
Please, everybody submit their resumes, NASA has openings at the Kenndey Space Center [:P]
Link Posted: 4/19/2002 2:14:11 PM EDT
Originally Posted By IMHO: Yep, Hooke's Law I: There's a direct relationship between the force applied, and the displacement, so linear: y=mx+b, in the case of the spring: F=kx+(Y intercept is zero, right?)
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Yes, No force is applied when x=0.
Link Posted: 4/19/2002 6:22:17 PM EDT
[Last Edit: 4/19/2002 6:22:47 PM EDT by IMHO]
Originally Posted By ECS: Please, everybody submit their resumes, NASA has openings at the Kenndey Space Center [:P]
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You might want to try out at The Second City if you haven't already. I recognize your potential. [:O)]
Link Posted: 4/19/2002 6:31:29 PM EDT
We just found the spring constant in physics by hanging any known weight from the spring and measuring it's displacement. That easily gives the spring constant. The force on the charging handle should simply be the necessary spring force at that displacement for full retraction of the charging handle. I can't imagine that the friction of the carrier to the reciever would be to high. I believe that it requires the most force at full retraction, more than is required to cock the hammer.
Link Posted: 4/20/2002 9:11:43 AM EDT
Originally Posted By Guncrazy223: We just found the spring constant in physics by hanging any known weight from the spring and measuring it's displacement. That easily gives the spring constant.
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Yep, you use the acceleration of gavity constant (a) with a known mass (m) to calulate the force (F) then you measure the delta or distance the spring "stretched" (x). Now you have all the variables to calculate "k". F=ma F=kx
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