I agree that any given bullet fired on any given day won't likely see much of a magnus effect...and even if it does, it is very unlikely that it actually rises at any point, as opposed to simply resisting its decent. However, I can say that it is feasible, with the working knowledge I have, that it could occasionally happen in the right conditions that are possible under "natural" and real hunting conditions.

One thing you are mistaken on, the magnus effect has absolutely nothing to do with linear velocity, i.e. the speed of the bullet. It ONLY has to do with the fluid velocity (wind speed) and the rate of rotation or angular velocity...a 3000fps bullet of any diameter coming out of a 1turn/12" bbl is spinning at 180,000rpm, which is pretty quick. See below for details.

This is about to get REALLY long.

Want proof the magnus effect is real? Why do curveballs curve? Check out a golfball when you hit it some time, ever noticed how sometimes they'll climb very sharply if you hit them with a lot of spin?

Ok, if you want numbers, here goes: a 150grn .30-06 bullet at 3000fps from a 1/12" rifle in a 15mph crosswind in air at 40Deg. F:

L=lift exerted on a rotating cylinder PER length

L=pGV, where p=density of fluid (air), V=velocity of fluid (or , G=2*(pi)*radius*v, v=2*(pi)*radius*w, w is angular velocity=linear velocity*rate of twist


SO: L=pGV-->L=p*2*pi*radius*2*pi*radius*linear velocity*rate of twist*wind speed

Throw in a couple conversions, like the radius in inches-->ft, wind speed in mph-->ft/s, and pound(mass) must be converted to pound(force)....this conversion is 1lbf=32.174lbm*ft/second squared.....

Plug in all the numbers and you get 1.07lbf/ft.

I'm going to assign a value now for the average cylindricity of a 150grn .308" bullet. If you assign an average radius of .154" (1/2 of .308"), your spitzer type 3/4" long bullet is likely now only .5" long...so I'll use that length for my effective force length....that's 0.042ft in case you're wondering...

That comes up to 0.045lbf force exerted upwards due to magnus effect (or downward if the wind were switched).

The force of gravity can be calculated Force=mass*acceleration-->150grns*1lbm/7000grns*32.17ft/second squared/32.17lbm*ft/lbf*second squared

That gives a force of gravity of 0.021lbf. Roughly half of that exerted by the magnus forces.

However, this is for a constant 15mph wind, and assumes idealized boundary layer separation (i.e. it doesn't involve boundary layer/vortex shedding), and it doesn't include the boundary layer buffering effect that the linear velocity of the bullet causes (the bullet rushing through the air reduces the magnus moment: the boundary layer is kind of pushed off the back of the bullet). Nor does it account for likely a hundred other effects that algebraic models (which i used for simplicity) can not account for, differential and integral values for the vortex shedding and linear velocity terms.

Do a search on the web for "magnus effect" or magnus moment, and you'll find a wealth of information on the topic.

Here's a website I found interesting, but you'd have to have access to a LOT of constants to use this formula http://www.snipersparadise.com/wound...m#MagnusMoment

Like I said, it's not likely, but experimental and theoretical models have proven the validity of the case.