In spring-powered airguns you calculate the effect in Joule.
I would like to change the spring in my airrifle but where I live the legal limit is 10 Joule and I bought a 18 Joule spring (there are no 10 Joule springs for sale).
This spring is 337 mm long and has 41 coils. I would like to calculate how many coils I have to remove to still be legal.
I tried to understand it myself from this page: http://hyperphysics.phy-astr.gsu.edu/hbase/pespr.html The potential energy in a spring is this: $$ PE=\frac{1}{2}kx^2 $$ where k is the spring constant and x is the displacement.
So I thought like this:
$$ 18=\frac{1}{2}kx^2 $$ $$ 10=\frac{1}{2}kx^2 $$
Assumption: $k$ is the same because it is the same spring (?). So in other words the $x$ must be a different value to make the equation system work if I assume that the displacement is numbered in coils and is the spring is zero length.
So I put the nr of coils (41) from the bought string as $x$ in the first and solve for $k$:
$$ 18=\frac{1}{2}k41^2 => 36 = k * 1681 => k = 36/1681 $$
$$ 10=\frac{1}{2}kx^2 => 20 = k * x^2 => \sqrt{\frac{20}{\frac{36}{1681}}} = x => x=30.559 => x = 30 $$
So I have to have 30 coils left (so I remove 11)
What do you think? are my assumptions and calculations correct?
Thank you in advance