# Magnus effect

1. Dec 28, 2004

### Henk

For my article about long range artillery I was trying to calculate the magnus effect.
I first neglected air friction (which is ridiculous because without air friciton there is no magnus effect). For the magnus effect of a cylinder I used the formula:

$$F_{m}=2\pi \rho\ \omega v_{x}Lr^2$$

Where r is the radius and L the length of the cylinder.
Then:

$$F_{x}=ma_{x}=0 \rightarrow v_{x} = v_{0x}$$

$$F_{y}=-mg + 2\pi \rho\ \omega v_{0x}Lr^2$$

$$\frac{dv_{y}}{dt}=g+\frac{2\pi \rho\ \omega v_{0x}Lr^2}{m}$$

$$v_{y}=gt+\frac{2\pi \rho\ \omega v_{0x}Lr^2}{m}t+v_{0y}$$

$$y=\frac{1}{2}gt^2+\frac{\pi \rho\ \omega v_{0x}Lr^2}{m}t^2+v_{0y}t$$

$$y=t^2(\frac{1}{2}g+\frac{\pi \rho\ \omega v_{0x}Lr^2}{m})+v_{0y}t=0$$

$$t=\frac{v_{0y}}{\frac{1}{2}g+\frac{\pi \rho\ \omega v_{0x}Lr^2}{m}}$$

$$x=\frac{v_{0y}v_{0x}}{\frac{1}{2}g+\frac{\pi \rho\ \omega v_{0x}Lr^2}{m}}=\frac{2mv_{0y}v_{0x}}{gm+2\pi \rho \omega Lr^2v_{0x}}$$

and when m gets really large this becomes the equation for the distance without magnus effect and air friction. Is this correct?
Then I wanted to add air friction, I quickly got into trouble:

$$m\frac{v_{y}}{dt}=-mg+2\pi \rho\ \omega v_{x}Lr^2 -kv_{y}$$

$$m\frac{v_{x}}{dt}=-kv_{x}$$

from the second equation $$v_{x}$$ is easily solved:

$$v_{x}=v_{0x}e^\frac{-kt}{m}$$

But then:

$$m\frac{v_{y}}{dt}=-mg+2\pi \rho\ \omega Lr^2 v_{0x}e^\frac{-kt}{m}-kv_{y}$$

and I don't know how to solve this. Is there somebody who does?

Last edited: Dec 28, 2004
2. Dec 28, 2004

### dextercioby

Yep,it is.It is a first order nonhomogeneous linear ordinary differential equation of this type:
$$\frac{dy(x)}{dx}+Ay(x)=B+Cf(x)$$
,where f(x) is a function of "x",given.
Bring your equation to the form stated by me,and solve it following those steps:
1.Solve the homogenous equation:
$$\frac{dy(x)}{dx}+Ay(x)=0$$
through the method of variable separation.
2.Search for a particular solution for the nonhomogenous equation chosing y_{particular}(x) of this type:
$$y_{particular}(x)=D+Bx+C g(x)$$
,where the form of g(x) depends on the form of the function f(x),e.g.if f(x) is a "sin",then g(x) will be a combination of "sin" and "cos";for an exponential f(x),g(x) is an exponential as well.
In your case,assume the particular solution as a sum between a constant,a linear function of time and the exponential you have.Plug the particular solution in the DE and find the coefficients.

Daniel.

Last edited: Dec 28, 2004
3. Dec 28, 2004

### Henk

If I try:
$$v_{y}=D+Bt+Ce^\frac{-kt}{m}$$

$$\frac{dv_{y}}{dt}=B-\frac{Ck}{m}e^\frac{-kt}{m}$$

when I fill this in the DE:

$$m\frac{dv_{y}}{dt}=-mg+Ae^\frac{-kt}{m}-kv_{y}$$

I get:

$$mB-kCe^\frac{-kt}{m}=-mg+Ae^\frac{-kt}{m}-Dk-Bkt-kCe^\frac{-kt}{m}$$

But then the $$e^\frac{-kt}{m}$$ fall out

4. Dec 28, 2004

### dextercioby

I didn't solve the homogenous equation to find out that its solution was the same with the nonhomogeneity function,so that's why i supplied you with an erroneous solution/advice for the particular solution.
Okay,let's take it methodically:
Initial equation:
$$m\frac{dv_{y}(t)}{dt}+kv_{y}(t)=-mg+2\pi\rho\omega Lr^{2}v_{0,x}\exp(\frac{-kt}{m})$$
Divide by "m" and make the following substitutions:
$$\frac{k}{m}\rightarrow A;-g\rightarrow B;\frac{2\pi\rho\omega Lr^{2}v_{0,x}}{m}\rightarrow C$$
The initial eq.has been brought to an abstract form
$$\frac{dv_{y}(t)}{dt}+Av_{y}(t)=B+C\exp(-At)$$(*)
The general solution for the equation (*) is a sum from the general solution of the homogenous eq.and a particular solution for the nonhomogenous eq.,viz.
$$v_{y}(t)=v_{y,hom}(t)+v_{y,part}(t)$$ (^)
The homogenous part is simply found (by the method described in my prior post) to be equal to:
$$v_{y,hom}=D\exp(-At)$$ (^^)

Since the nonhomogeneity function and the general solution of the homogenous equation coincide,a particular solution is founf through the method of variation of constants,due to Joseph Louis Lagrange.
$$v_{y,part}(t)=E(t)\exp(-At)+F$$(**)
,where E(t) and F shall be determined by the request that (**) be a solution to unhomogenous eq.(*):
$$\frac{dv_{y,part}(t)}{dt}=E'(t)\exp(-At)-AE(t)\exp(-At)$$(***)
Multiplying (**) through A and adding the result with (***) and substituting everything in the LHS of (*) will give:
$$E'(t)\exp(-At)+AF=B+C\exp(-At)$$,from where,by identification u find:
$$E(t)=Ct+G;F=\frac{B}{A}$$ (****)
Substitute (****) in (**) and find the particular solution of the form:
$$v_{y,part}(t)=Ct\exp(-At)+G\exp(-At)+\frac{B}{A}$$(^^^)

The final solution is obtained from (^),(^^) and (^^^) and has the form:
$$v_{y}(t)=Ct\exp(-At)+(D+G)\exp(-At)+\frac{B}{A}$$
Relabel the sum of the 2 arbitrary constants D,G by H:
$$D+G\rightarrow H$$
,so that the final solution will depend only on one constant of integration,namely H (as it should,since it's first order in derivatives and u have't supplied initial conditions (Cauchy's problem)):

$$v_{y}(t)=Ct\exp(-At)+H\exp(-At)+\frac{B}{A}$$ (^^^^)

Make the substitutions back for the 3 coeficients A,B,C to finally get:
$$v_{y}(t)=(\frac{2\pi\rho\omega Lr^{2}v_{0,x}}{m})t[\exp(-\frac{kt}{m})]+H\exp(-\frac{kt}{m})-\frac{gm}{k}$$ (£,€,\$)

Daniel.

PS.Lagrange's method is very general and applies not only in the case encountered here.

Last edited: Dec 28, 2004
5. Dec 28, 2004

### Henk

By the way:

I understand why a projectile is deflected to the right by the Coriolis force when moving to the north or south on the northern hemisphere. I however do not understand why it is deflected when travelling east or west. Does it has anything to do with the fact that the speed of the projectile is less or more then that of the earth when travelling east or west?
This has nothing to do with the MAgnus effect nor with the mathematics above, I just wrote it down here because it's a waste to make a whole new thread about it.

Edit: We posted on the same time. Thanks a lot.

to determine H:
$$v_{y}(0)=v_{0y}=H-\frac{gm}{k}$$
thus:
$$H=v_{0y}+\frac{gm}{k}$$
right?

Then one more step: integrating for y.

$$\int(Ate^\frac{-kt}{m}+He^\frac{-kt}{m}-B)dt$$

first:

$$\int(Ate^\frac{-kt}{m})dt = A(\frac{-m}{k}te^\frac{-kt}{m}+\frac{m}{k} \int(e^\frac{-kt}{m} dt)) = \frac{-mAe^\frac{-kt}{m}}{k}(t+\frac{m}{k})$$

The other 2 terms are easy thus:

$$y=e^\frac{-kt}{m}(\frac{-mA}{k}(t+\frac{m}{k})-\frac{mH}{k})-Bt+c$$

and y(0)=0 thus:

$$C=\frac{m^2A}{k^2}+\frac{mH}{k}$$

Wow, that's a pretty complicated equation, I'm going to need to write a computerprogram so I can solve it for t when y =0 and then putting t in the formula for x. Then I can see the change in range caused by the Magnus effect.

Last edited: Dec 28, 2004
6. Dec 28, 2004

### dextercioby

IIRC,Gaspard Coriolis' force in an inertial one,hence appearing in noninertial reference frames.Since Earth provides a good noninertial frame for studying the movement of the projectile,i find natural the fact that Earh is rotating should have an impact on the projectile's movement.That $$\vec{v}$$from the expression of the Coriolis's force should be the relative velocity.( ).
However,i'm not sure,maybe someone with better knowledge of CM could provide with a "cleaner" picture...

Daniel.

PS.You're welcome...I enjoyed solving that ODE... :tongue2:

EDIT:It's okay with "H". Feel free to post any math troubles.

Last edited: Dec 28, 2004
7. Dec 28, 2004

### Staff: Mentor

If the projectile is fired east or west at the equator, there would be no Coriolis effect, but if fired at any other latitude, the projectile deflects toward the equator.

8. Dec 28, 2004

### Henk

But why?

I understand that it deflects to the right when shot to the North. When for example a particle is shot from the equator to Norway it deflects to the right because the eastward speed of the particle is greater then that of Norway.

But why does it deflect when it travels east?

9. Dec 28, 2004

### Gokul43201

Staff Emeritus
The coriolis force is proportional to $\Omega ~X~v$

Any projectile will feel a coriolis force since $\Omega$ is a fixed vector.

When the projectile is chucked east/west-ward from the equator, the cross product point radially, and hence causes no 'deflection'.

Simply use the right hand rule to determine what the Coriolis Force does under different circumstances.

10. Dec 28, 2004

### Henk

I understand the formula, I however don't understand why there is a deflection when a particle (not on the equator) travels east or west. I do understand why there is a deflection when a particle is traveling north or south.

11. Dec 28, 2004

### Staff: Mentor

Launching from any lattitude still provides a radial velocity component with respect to the earth's spherical coordinate system.

I have attached a figure, which I hope is clear.

#### Attached Files:

• ###### Artillery shell rad vector.gif
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12. Dec 28, 2004

### NateTG

It might help to think of it this way:
When the projectile is at high altitude, it has to go further to go east that if it's at the surface because it is traveling on a larger circle.

Let's look at conservation of angular momentum.
At the surface of the earth, the projectile's angular momentum is going to be
$$m\omega^2 r$$
where
$$\omega$$
is the angular velocity of the projectile on the earth.
$$r$$
is the surface velocity of the earth and
$$m$$
is the mass of the projectile.
Now, when the projectile is at some height $h$ above the surface, the angular momentum will be
$$m \omega'^2 (r+h)$$
since angular momentum is constant for a ballistic projectile we can set the two equal:
$$m\omega^2 r=m \omega'^2 (r+h)$$
so
$$\omega'=\omega \sqrt{\frac{r}{r+h}}$$
so the angular velocity of the projectile will be lower while it is in the air.

13. Dec 28, 2004

### dextercioby

Sorry,Nate,there's something 'stinky' round here.The angular momentum of the projectile is conserved,that's true,but i'm afraid it is given by
$$L=rp=rmv=mr\omega r=mr^{2}\omega$$
$$m\omega r^{2}=m\omega' (r+h)^{2}$$
$$\omega'=\omega\frac{r^{2}}{(r+h)^{2}}$$

Daniel.

PS.I think you still remember $m\omega^{2}r$ as the modulus for the centripetal/centrifugal inertial force...

14. Dec 29, 2004

### Henk

I understand this but how does this mean that the projectile is deflected north or south. It makes sense that $$\omega'<\omega$$ because of the greater circle. But when it comes down on the ground:
angular momentum is conserved so that the projectile regains his original $$\omega$$ but that means that it is on the same latitude where it started from.

One more question: does anyone know what $$\omega$$ is of a typical projectile. What order? is it around 50 (20 circulations/s) or much more or less?

Last edited: Dec 29, 2004
15. Dec 29, 2004

### NateTG

Thanks for the correction dex.

I didn't suggest that the projectile was being deflected north or south, but it is being deflected to the east or west. (Or depending on how you look at it, up and down or out and in.)

$$\omega=\frac{v_{tangential}}{r}$$
The angular speed of the surface of the earth, or a projectile that has been launched 'straight up' is
$$\frac{2\pi}{86400s}$$
To find the angular speed of a projectile that is launched at an angle, you need to add a term for the appropriate transverse component. Remeber that it's the rotational radius rather than the radius of a great circle that we're talking about here, so it's smaller in Alaska than in Ecuador.

Note that this is *not* the same rotational speed that you would be talking about for the magnus effect where you're talking about a spinning projectile.