Post by Knief on Mar 16, 2004 20:46:29 GMT -5
Right Luke, but the physics behind those materials don't change.
Regardless of what a spring is made of, it has a certain constant compression value, noted by "K". If "M100" is a measure of a springs "K" value, it doesn't matter if it's made out of steel, aluminum, lead, or those folded strips of paper with the holes in it that you tear off of the sides of the sheets of paper that are stuck to each other for old school printers. So if M100 is the spring constant as Arc suggests, then the displacement is the only other factor that can effect the amount of work the spring does. An M100 spring that is only compressed 1 millimeter isn't really going to get your bb moving, is it? Where as if you get a huge spring with an "M100" K and compress it a couple of feet, you're going to drive the bb through a wall, a barrel, and a car's engine block (Tremors 2, anyone?).
PDI springs are shorter, but they use a completely different measuring system. Their spring's K may be greater than a comperable velocity (i.e M100 vs. 130% or M120 vs. 170%) systema spring, but because it's shorter, it does the same amount of work.
Using easy numbers here, say you have a 20 cm spring with a spring constant of 100 Newton/Meters. The spring is compressed 10 cm from it's natural length down to 10 cm (converting to meters for the sake of matching units, it's .1 meter.
Ue=(1/2)(100)(.10)2
Ue=.5 Joules
Now, if we take a spring with the same spring constant, but it's only 15 cm long, and it's compressed down to the same 10 cm (the displacement used then in the equation is 5cm, or .05m), we'll find that the work done is less than that of the longer spring compressed to the same length.
Ue=(1/2)(100)(.05)2
Ue=.125 Joules
The difference is substantially less (1/4 lof the original, the ratio of the difference between the two displacements squared).
If you want to figure out what spring constant will result in the same work as the first spring with the length of the second spring, just set the equation as
.5=(1/2)(K)(.05)2
400N/M=K
So you need quadruple the spring compression constant to produce the same amount of work when it's only compressed half as far.
Did any of that make sense?
I gotta not know so much physics.
[edit]So Bigmack, yes, they would use a thicker or stronger steel to increase the spring constant and raise the energy transfered by the spring to the bb. The length of the spring wouldn't have to change in order to put more oomph into your shot.
Regardless of what a spring is made of, it has a certain constant compression value, noted by "K". If "M100" is a measure of a springs "K" value, it doesn't matter if it's made out of steel, aluminum, lead, or those folded strips of paper with the holes in it that you tear off of the sides of the sheets of paper that are stuck to each other for old school printers. So if M100 is the spring constant as Arc suggests, then the displacement is the only other factor that can effect the amount of work the spring does. An M100 spring that is only compressed 1 millimeter isn't really going to get your bb moving, is it? Where as if you get a huge spring with an "M100" K and compress it a couple of feet, you're going to drive the bb through a wall, a barrel, and a car's engine block (Tremors 2, anyone?).
PDI springs are shorter, but they use a completely different measuring system. Their spring's K may be greater than a comperable velocity (i.e M100 vs. 130% or M120 vs. 170%) systema spring, but because it's shorter, it does the same amount of work.
Using easy numbers here, say you have a 20 cm spring with a spring constant of 100 Newton/Meters. The spring is compressed 10 cm from it's natural length down to 10 cm (converting to meters for the sake of matching units, it's .1 meter.
Ue=(1/2)(100)(.10)2
Ue=.5 Joules
Now, if we take a spring with the same spring constant, but it's only 15 cm long, and it's compressed down to the same 10 cm (the displacement used then in the equation is 5cm, or .05m), we'll find that the work done is less than that of the longer spring compressed to the same length.
Ue=(1/2)(100)(.05)2
Ue=.125 Joules
The difference is substantially less (1/4 lof the original, the ratio of the difference between the two displacements squared).
If you want to figure out what spring constant will result in the same work as the first spring with the length of the second spring, just set the equation as
.5=(1/2)(K)(.05)2
400N/M=K
So you need quadruple the spring compression constant to produce the same amount of work when it's only compressed half as far.
Did any of that make sense?
I gotta not know so much physics.
[edit]So Bigmack, yes, they would use a thicker or stronger steel to increase the spring constant and raise the energy transfered by the spring to the bb. The length of the spring wouldn't have to change in order to put more oomph into your shot.
