# Calculate Distance for 5ft-lbs Impact w/ 1.18lb Sphere

• mrb112103
In summary, the conversation discusses the process of calculating the distance needed to create a 5ft-lbs impact with a 1.18lb sphere being dropped vertically onto a surface for an impact test. The most direct method is to equate potential and kinetic energies, using the mass and impact force to determine the necessary distance. The conversation also mentions converting from English units to metric units and clarifies the formula for calculating kinetic energy.

#### mrb112103

Hello,

I am probably going to suffer some embarassment, once I see the answer to my question. Anwyay, I need to calculate the distance needed to create a 5ft-lbs. impact with a 1.18lb. sphere being dropped vertically onto a surface.

How would I go about calucating the vertical distance needed?

This is for an impact test from UL, where the mass and impact force are defined, but the distance is not.

(well I abhor those english units, but I reckon the principles are the same). what i like even less are words like "impact." since its in ft-lb, I assume they are looking for energy.

there are a few ways to go about this, but usually the most direct is to equate potntial and kinetic energies

wt*h=1/2Mv^2=impact energy

Still unclear...

So if the Mass of the sphere (M) = 1.18 lbs = .535kg & the impact force = 5 ft. lbs. = 6.8 Joules, then is the velocity = to acceleration by gravity ~ 9.8m/s^2?

So it's .5(.535kg * (9.8 m/s^2)) = 6.8 Joules ??

If your you want your final kinetic energy to be E, then use E=mgh. Or like denverdoc said, E=F*d (force times distance). Gravitational force is 1.18 lbs, you want 5 ft-lbs of impact energy, so d=E/F=(5/1.18) ft.