Thread: Springs?? (10 lbs. at 22 ft/s)
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Unread 09-01-2008, 16:08
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Re: Springs?? (10 lbs. at 22 ft/s)
I'll do the calculations for you, but you should be VERY careful about launching a 10lb ball at such a speed. FIRST may deem it a safety hazard.

You want to accelerate 10lbs to 22 ft/s. The potential energy stored in a spring is .5*k*x^2, where k is the spring constant (basically, the stiffness of the spring) and x is the deformation or extension of the spring.

10lbs traveling at 22 ft/s means a kinetic energy of .5*m*v^2, or 102 Joules (after conversion to SI units). Thus, you need at least 102 Joules of potential energy from the spring.

There are two ways to do this - either get a higher spring constant, or compress the spring more. So for a given spring constant or compression, you can alter the other factor to yield the potential energy you need.

Now let's look at McMaster's website.
http://www.mcmaster.com/param/dsc/dsc.aspx?dsc=Springs

All you have to do is find a combination of spring constant (also called rate on some pages) and compression distance that work for you. There are plenty of options there.