Spring question

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.