weatherowl
Why is trajectory curvature important to the resulting path of an object?
Coriolis effect
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replied to: weatherowl
Switch2789
Replied to: Why is trajectory curvature important to the resulting path of an...
If your trying to get a projectile to go from point A to point B on a large scaled distance then you must factor in the rotation of the Earth and the strength of gravity on that object that is traveling in order to find out where said projectile would end up. Think of an object that would be traveling at greater distances than 1 mile.
I was having difficulty in believing the Coriolis effect could have any bearing on sniper aiming, I mean just a casual inspection would seem to suggest that the difference between two locations separated by a mile couldn't really have any bearing on a bullet which is fired at 1000 ms-1 or so, surely?
But I did the following calculations and came to an astonishing result, at least for me.
The velocity of a point on the earth's surface at a specific latitude due to its rotation is
Velocity = w * r
where
r = radius of circular motion of the point considered and is given by R * Cos[ Latitude ] where R is the radius of the Earth = 6356750 m
w = angular velocity of the earth in radians per second and is given by 2 * Pi/ ( 24 * 60 * 60 ) which reflects the earth completes 2*Pi radians of rotation in 24 hours
So the
Velocity = 2 * Pi/ ( 24 * 60 * 60 ) * R * Cos[ Latitude ]
= k * Cos[ Latitude ]
where
k = 2 * Pi/ ( 24 * 60 * 60 ) * R
The difference in velocity between two places of different latitude is simply
VelocityDifference = k * ( Cos[ Latitude1 ] - Cos[ Latitude2 ]
If we wish to consider the difference in velocities between 38 degrees North and a mile North of that point then the difference is
VelocityDifference = k * ( Cos[ 38 ] - Cos[ 38 + OneMileInDegreesOfArc ]
OneMileinDegreesOfArc = 1760/2000*1/60
as one nautical mile is 1/60 degree and 2000 yards and one statute mile is 1760 yrds
So expanding k as defined above
VelocityDifference = 2 * Pi/ ( 24 * 60 * 60 ) * 6356750 * ( Cos[ 38 ] - Cos[ 38 + 1760/2000*1/60 ] )
= - 0.073 ms-1 (7.3 cm s-1)
The flight time of a M16A2 firing a 5.56mm round to cover 1 mile at an angle of 5 degrees is about 5.5 seconds (determined from flight modelling ).
So the movement of the target over the period of the trajectory is
= 5.5 * 0.073
= 0.4 m ( 40 cm )
which is slightly different to mastersergeant's value of about 25cm ( 10 inches). My flight time models drag and the resultant effects on speed of the bullet. It takes a 975ms-1 muzzle velocity for the 5.56mm round. Presumably the difference is accounted for by the difference in the bullet and its launch conditions or mastersergeant's assumed flight time value does not include drag effects, or my calcs are wrong.
The results for me are quite astounding that the Earth's rotation has such a huge effect on trajectories over such small distances comparative to the size of the Earth.
I was having difficulty in believing the Coriolis effect could have any bearing on sniper aiming, I mean just a casual inspection would seem to suggest that the difference between two locations separated by a mile couldn't really have any bearing on a bullet which is fired at 1000 ms-1 or so, surely?
But I did the following calculations and came to an astonishing result, at least for me.
The velocity of a point on the earth's surface at a specific latitude due to its rotation is
Velocity = w * r
where
r = radius of circular motion of the point considered and is given by R * Cos[ Latitude ] where R is the radius of the Earth = 6356750 m
w = angular velocity of the earth in radians per second and is given by 2 * Pi/ ( 24 * 60 * 60 ) which reflects the earth completes 2*Pi radians of rotation in 24 hours
So the
Velocity = 2 * Pi/ ( 24 * 60 * 60 ) * R * Cos[ Latitude ]
= k * Cos[ Latitude ]
where
k = 2 * Pi/ ( 24 * 60 * 60 ) * R
The difference in velocity between two places of different latitude is simply
VelocityDifference = k * ( Cos[ Latitude1 ] - Cos[ Latitude2 ]
If we wish to consider the difference in velocities between 38 degrees North and a mile North of that point then the difference is
VelocityDifference = k * ( Cos[ 38 ] - Cos[ 38 + OneMileInDegreesOfArc ]
OneMileinDegreesOfArc = 1760/2000*1/60
as one nautical mile is 1/60 degree and 2000 yards and one statute mile is 1760 yrds
So expanding k as defined above
VelocityDifference = 2 * Pi/ ( 24 * 60 * 60 ) * 6356750 * ( Cos[ 38 ] - Cos[ 38 + 1760/2000*1/60 ] )
= - 0.073 ms-1 (7.3 cm s-1)
The flight time of a M16A2 firing a 5.56mm round to cover 1 mile at an angle of 5 degrees is about 5.5 seconds (determined from flight modelling ).
So the movement of the target over the period of the trajectory is
= 5.5 * 0.073
= 0.4 m ( 40 cm )
which is slightly different to mastersergeant's value of about 25cm ( 10 inches). My flight time models drag and the resultant effects on speed of the bullet. It takes a 975ms-1 muzzle velocity for the 5.56mm round. Presumably the difference is accounted for by the difference in the bullet and its launch conditions or mastersergeant's assumed flight time value does not include drag effects, or my calcs are wrong.
The results for me are quite astounding that the Earth's rotation has such a huge effect on trajectories over such small distances comparative to the size of the Earth.