Woo hoo!

I found a webpage which describes the mathematical foundation for the Magnus effect. It is located at the following URL, and describes the effect for rotating cylinders of infinite length:

http://www.mathpages.com/home/kmath258/kmath258.htm

I think it could be applied to our paintball, at least as a first cut approximation. After staring blearily at the equations, we come up with

L = - (rho) * v0 * (gamma), where

rho is the density of the fluid,

v0 is the velocity of the fluid (relative to the object in question), and

gamma is something called the circulation, defined as the line integral of the flow velocity around a closed loop.

What gamma basically means for us is "take the surface speed of the object and multiply it by the object's circumference". Thus,

(gamma) = vc * 2 * (pi) * r, where

vc is the circumferential (surface) velocity of the object,

pi is 3.14159 or thereabouts, and

r is the object's radius.

Let's plug in some of the data derived from the test 101 picture.

rho = 1.293 kg/m^3 (reasonable value for the density of air)

v0 = 85.34 m/s (280 fps)

vc = 1.25 m/s (4.1 fps surface speed)

r = 0.0086 m (radius of a paintball 0.68 inches in diameter)

So from this we get

L = -1.293 * 85.34 * 1.25 * 2 * 3.14159 * 0.0086

L = -7.45 (units?)

I'm not sure what the units should be. Doing a bit of unit analysis on the equation for L, we have the following:

L-units = (kg/m^3) * (m/s) * (m/s) * (unitless) * (unitless) * (m)

L-units = kg/s^2 ???

That sure isn't a unit of force I'm familiar with. We're missing a distance unit in the numerator somewhere, and darn if I can find it. Any help, anyone?

I also have happened across the following webpage, which gives a different equation (of similar form) for the Magnus effect:

http://carini.physics.indiana.edu/E1...ing-balls.html

Here they describe the effect as follows:

M = cM * (rho) * D^3 * f * v, where

cM is the Magnus force coefficient (1.23 works pretty well, according to the webpage),

rho is the density of the fluid,

D is the diameter of the object,

f is the object's rotational frequency (rotations per second), and

v is the velocity of the fluid (relative to the object).

If the Magnus force coefficient is unitless, then a unit analysis of this equation actually ends up with units of force coming out of it. Let's plug in some numbers derived from test 101:

cM = 1.23 (unitless?)

rho = 1.293 kg/m^3 (reasonable value for the density of air)

D = 0.0173 m (0.68 inch diameter paintball)

f = 23.1 rotations/s (15 degrees per strobe flash, one strobe flash every 1.8 milliseconds)

v = 85.34 m/s (280 fps)

M = 1.23 * 1.293 * 0.0173^3 * 23.1 * 85.34

M = 0.016 N

So for that amount of spin we end up with a Magnus effect force of 0.016 newtons. For a 3 gram paintball, this force results in an acceleration of 5.3 m/s^2, or right around 0.54 g's. Let's assume this equation has given us a correct answer, and see what the picture can tell us based on what we've calculated.

I am going to define the "first" strobe as the first image of the ball after it has exited the barrel, and incrementally name each successive ball image to the right of the first strobe (second, third, etc.).

An acceleration of 5.3 m/s should deflect the ball approximately 8.6 microns between the first and second strobes. This deflection is nearly two orders of magnitude smaller than the spatial sampling in the image (around 0.6 millimeters per pixel for the "bottom view" portion of test image 101). So even if the Magnus effect is at work, we simply can't see it from strobe to strobe. So let's look at the first and last strobes in the (strobes 1 and 5).

In this situation, the time interval is four times as large. Assuming the acceleration resulting from the Magnus effect is constant throughout the measured time, we should expect a deflection 16 times greater than that predicted between strobes 1 and 2, or around 140 microns (0.14 millimeters). This deflection is still smaller than the spatial sampling in the image.

If the ball is spinning at 15 degrees per strobe flash (and based on my other spin calculations and Tom's additional comments, I think we can safely assume the ball is not spinning at 195 or -165 degrees per strobe flash), and if the formula I used above is reasonable and accurate, then the high resolution pictures we have of the test are simply not sufficient to detect the resultant Magnus effect.

I looked at the "bottom view" portion of the test 101 picture to see if I could see any horizontal deviation in the ball's path, and I noticed something interesting... the laser aligned string is not straight. It curves slightly in the image. This leads me to believe we either have a camer/lens perspective effect going on here, or the string is vibrating during firing. The blast of air that escapes the barrel could be moving the string around.

I attempted to correct the slightly curved string by using Photoshop's transformation tools, but had little success. It seems that the transformation I'm looking for just isn't available in my version of Photoshop. Instead of measuring each ball's position from a single reference line, I generated local references corresponding to the string's location at each ball's position. I measured the following offsets:

Strobe 1: -4 pixels (-2.48 mm)

Strobe 2: -2 pixels (-1.24 mm)

Strobe 3: -2 pixels (-1.24 mm)

Strobe 4: -1 pixel (-0.62 mm)

Strobe 5: 0 pixels (0 mm)

These offsets are greater than the Magnus effect alone would suggest, unless the Magnus equation I used was incorrect. They may be the result of the escaping gas buffetting the ball around during the first few milliseconds of flight.

Well, that's about all I've got for now. Time to go do some work that actually fills up a paycheck.

BJJB