Sean Powell was kind enough to send me an explanation of how to convert the data from a test of bowling balls struck with rattan swords into joules of energy. As he points out in an earlier comment, unfortunately we don't have enought data to calculate from this how much kinetic energy was in the sword before it hit.

His explantion follows. While he provided tables for all the different post heights in the test, I was unable to align the data in a readable form in blogger, so I've reduced the data to the values for the highest and lowest post.

Converting Peter Van Dorns bowling-ball data to Joules of Energy

Peter has provided us with 3 pieces of data: Weight of a ball, distance the ball drops and how far the ball travels before it lands. This is a classic high school physics problem.

Step 1: Acceleration due to gravity, at the surface of the earth, will be a uniform 32ft/sec^2. (neglecting wind resistance, centripetal force from spin of the earth and the local gravitational effect of the balls mass) For the ball to fall a distance of H it will take a time T regardless of ball weight. H and T are related in the formula:

h = ½ * g * t^2

We can reorder this equation to get

t= (2 * h / g)^.5

(remember to convert h to feet first)

A spread sheet makes the following math easy:

h (in) 46- 58

t (s) 0.49-0.55

Step 2: We know how far a ball travels and how long it takes to get there. That means we can calculate ball velocity with the simple formula:

V = d / t

If we select one particular row to analyze: 12lbs, force Medium we get:

d (in) 38-42

t (s) 0.49-0.55

v (ft/sec) 6.47-6.37

Step 3: We can now calculate Joules of kinetic energy in the ball just after impact with the formula:

KE = ½ * m * v^2

We can also calculate ball momentum just after impact with the formula:

Momentum = m * v

But of course to keep doing this in Imperial units is a pain since we need to convert weight in pounds to mass in slugs and then re-convert slug-ft^2/sec^2 into Joules so first we will convert weight and distance to metric.

v (ft/sec) 6.47-6.37

v (m/sec) 1.97-1.94

Weight (lbs) 12

Mass (kg) 5.44

KE (kg-m^2/s^2)(J) 10.58-10.25

Mo (kg-m/s) 10.73-10.57

No both of these values seem rather consistent BUT for every different ball mass there is a different distance. If we look at the 8lb ball and the 16lb ball we will eventually get:

KE (J)=(kg*m^2/s^2) Medium

8lb 15.32-15.38

12lb 10.58-10.25

16lb 7.66-7.94

M (kg*m/s) Medium

8lb 10.55-10.57

12lb 10.73-10.57

16lb 10.55-10.73

And we can clearly see that kinetic energy of the lighter balls is higher then the heavier balls but the momentum of the various balls is NOT dependent upon weight.

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