- #1
Henk
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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:
[tex]F_{m}=2\pi \rho\ \omega v_{x}Lr^2[/tex]
Where r is the radius and L the length of the cylinder.
Then:
[tex]F_{x}=ma_{x}=0 \rightarrow v_{x} = v_{0x}[/tex]
[tex]F_{y}=-mg + 2\pi \rho\ \omega v_{0x}Lr^2[/tex]
[tex]\frac{dv_{y}}{dt}=g+\frac{2\pi \rho\ \omega v_{0x}Lr^2}{m}[/tex]
[tex]v_{y}=gt+\frac{2\pi \rho\ \omega v_{0x}Lr^2}{m}t+v_{0y}[/tex]
[tex]y=\frac{1}{2}gt^2+\frac{\pi \rho\ \omega v_{0x}Lr^2}{m}t^2+v_{0y}t[/tex]
[tex]y=t^2(\frac{1}{2}g+\frac{\pi \rho\ \omega v_{0x}Lr^2}{m})+v_{0y}t=0[/tex]
[tex]t=\frac{v_{0y}}{\frac{1}{2}g+\frac{\pi \rho\ \omega v_{0x}Lr^2}{m}}[/tex]
[tex]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}}[/tex]
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:
[tex]m\frac{v_{y}}{dt}=-mg+2\pi \rho\ \omega v_{x}Lr^2 -kv_{y}[/tex]
[tex]m\frac{v_{x}}{dt}=-kv_{x}[/tex]
from the second equation [tex]v_{x}[/tex] is easily solved:
[tex]v_{x}=v_{0x}e^\frac{-kt}{m}[/tex]
But then:
[tex]m\frac{v_{y}}{dt}=-mg+2\pi \rho\ \omega Lr^2 v_{0x}e^\frac{-kt}{m}-kv_{y}[/tex]
and I don't know how to solve this. Is there somebody who does?
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:
[tex]F_{m}=2\pi \rho\ \omega v_{x}Lr^2[/tex]
Where r is the radius and L the length of the cylinder.
Then:
[tex]F_{x}=ma_{x}=0 \rightarrow v_{x} = v_{0x}[/tex]
[tex]F_{y}=-mg + 2\pi \rho\ \omega v_{0x}Lr^2[/tex]
[tex]\frac{dv_{y}}{dt}=g+\frac{2\pi \rho\ \omega v_{0x}Lr^2}{m}[/tex]
[tex]v_{y}=gt+\frac{2\pi \rho\ \omega v_{0x}Lr^2}{m}t+v_{0y}[/tex]
[tex]y=\frac{1}{2}gt^2+\frac{\pi \rho\ \omega v_{0x}Lr^2}{m}t^2+v_{0y}t[/tex]
[tex]y=t^2(\frac{1}{2}g+\frac{\pi \rho\ \omega v_{0x}Lr^2}{m})+v_{0y}t=0[/tex]
[tex]t=\frac{v_{0y}}{\frac{1}{2}g+\frac{\pi \rho\ \omega v_{0x}Lr^2}{m}}[/tex]
[tex]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}}[/tex]
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:
[tex]m\frac{v_{y}}{dt}=-mg+2\pi \rho\ \omega v_{x}Lr^2 -kv_{y}[/tex]
[tex]m\frac{v_{x}}{dt}=-kv_{x}[/tex]
from the second equation [tex]v_{x}[/tex] is easily solved:
[tex]v_{x}=v_{0x}e^\frac{-kt}{m}[/tex]
But then:
[tex]m\frac{v_{y}}{dt}=-mg+2\pi \rho\ \omega Lr^2 v_{0x}e^\frac{-kt}{m}-kv_{y}[/tex]
and I don't know how to solve this. Is there somebody who does?
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