Influence of gravity on ballistic trajectory?

Hi,

Can anybody tell me the influence of gravity on the trajectory of a bullet
when the aim is to shoot as far as possible?

To put it simple, if i shoot a bullet from a gun on earth or on the moon
(places with different gravity) does the angle, in order to shoot the as far
as possible, differ?

If you know any interesting web sites about this subject, i would very much
appreciate it!

thanks,
bjorn

"b. hotting" wrote in message
...
Hi,

Can anybody tell me the influence of gravity on the trajectory of a bullet
when the aim is to shoot as far as possible?

To put it simple, if i shoot a bullet from a gun on earth or on the moon
(places with different gravity) does the angle, in order to shoot the as
far
as possible, differ?


Yes, the angle would be different because the downward force due to gravity
is different.

Look in any physics textbook for a more in-depth explanation.

JD

"b. hotting" wrote in message
...
Hi,

Can anybody tell me the influence of gravity on the trajectory of a bullet
when the aim is to shoot as far as possible?

Gravity will pull the bullet down. Horizontal velocity will carry the bullet
further. Ideally there is a trade-off between horizontal to make the bullet
go *far* and vertical to make gravity take more time to pull the bullet back
down to ground level.

In the *absence* of air it turns out that you get maximum range when
horizontal and vertical velocities are perfectly balanced at the start - a
45-degree angle.

If you like calculus, this may help:
http://scienceworld.wolfram.com/physics/Trajectory.html

To put it simple, if i shoot a bullet from a gun on earth or on the moon
(places with different gravity) does the angle, in order to shoot the as
far
as possible, differ?

The 45-degree angle works very well on the moon, but the earth has air,
which resists the movement of the bullet. The net effect of air resistance
is a force that tries to slow the bullet down no matter which direction it
is going.

The bullet will be going slower after the peak of its arch, so for maximum
range we need to make better use of the higher horizontal speed it has
*before* the peak of its arch.

The exact angle will depend on various details such as the geometry of the
bullet, the weight, the density, the wind direction, altitude, humidity, and
so on, but a good rough guess will be that you will get your maximum range
on earth if you fire the bullet at an angle of 35-38 degrees above the
horizontal.

If you know any interesting web sites about this subject, i would very
much
appreciate it!

For general information, Google "ballistics".
For a nifty on-lint calculator for maximum ballistic range:
http://www.eskimo.com/~jbm/calculati...t/maxdist.html
For other related calculations (including a sunrise/sunset calculator):
http://www.eskimo.com/~jbm/calculati...culations.html


HTH

Tom Davidson
Richmond, VA

"Joe Delphi" wrote in message
news:mSaye.4073$Qo.1601@fed1read01...
"b. hotting" wrote in message
...
Hi,

Can anybody tell me the influence of gravity on the trajectory of a
bullet
when the aim is to shoot as far as possible?

To put it simple, if i shoot a bullet from a gun on earth or on the moon
(places with different gravity) does the angle, in order to shoot the as
far
as possible, differ?


Yes, the angle would be different because the downward force due to
gravity
is different.

Not quite:
http://scienceworld.wolfram.com/physics/Trajectory.html
The range of the projectile is given in equation 7, which depends on
gravity, but the *maximum* range occurs at a firing angle of theta = pi/4,
regardless of the (non-zero) magnitude of g.

It's 45 degrees on the earth OR the moon - barring air resistance (see my
other post in this thread).


Tom Davidson
Richmond, VA

In article ,
b. hotting wrote:
To put it simple, if i shoot a bullet from a gun on earth or on the moon
(places with different gravity) does the angle, in order to shoot the as far
as possible, differ?

If you (a) ignore the effect of air resistance on the bullet's path, and
(b) assume a flat world with gravity not changing with altitude (which is
reasonably accurate until ranges become quite long), then no, the angle
does not change: it's always 45 degrees. Mathematically:

For muzzle velocity V and angle x, the (upward) vertical component of
velocity is V*sin(x). The constant (downward) vertical acceleration of
gravity, g, reduces that to zero in time V*sin(x)/g; by symmetry, the
time taken to reach the ground again is the same, so the flight time
is 2*V*sin(x)/g.

The horizontal component of velocity is V*cos(x), and that does not
change. So the distance covered in that time is V*cos(x)*2*V*sin(x)/g.
A table of trig identities yields sin(x)*cos(x) = sin(2*x)/2, so we
simplify and get V^2*sin(2*x)/g as the distance.

Neither V nor g varies with the angle, so distance is maximum when
sin(2*x) is maximum. The largest value the sin() function can take is
1, which occurs at 90deg. So maximum range is at 90deg/2 = 45deg,
independent of the exact choice of V and g.

If memory serves, moderate air resistance calls for a slightly steeper
angle, but the math is ugly.

For long ranges, you cannot ignore the spherical shape of the planet
or the reduction in gravity with altitude. (Indeed, if muzzle velocity
exceeds the planet's escape velocity -- impractical on any of the planets
except possibly Pluto, but easy on the smaller moons -- then surface range
is not meaningful because the bullet never strikes the surface.) Then the
angle does become a function of both the planet's gravity and its radius.
More precisely, you can show that the range is a messy function of just
two quantities: the angle, and V^2/(g*R), where R is the planet's radius.
I believe a smaller planet, or weaker gravity, or both, means a lower
angle for maximum range, although I'd need some effort to confirm that.
--
"Think outside the box -- the box isn't our friend." | Henry Spencer
-- George Herbert |

There are a number of artillary games, which allow two players to shoot
cannons at each other. The speed and angle can be adjusted, so you can
see how these change the trajectory of the shell.

I wrote:
If memory serves, moderate air resistance calls for a slightly steeper
angle, but the math is ugly.

Oops, correction: a slightly shallower angle. Since the drag is steadily
eroding the horizontal component of velocity, you're better off with a
slightly higher initial horizontal component even if it means a slightly
shorter flight time -- you gain faster motion at the start and lose the
slowest part at the end.
--
"Think outside the box -- the box isn't our friend." | Henry Spencer
-- George Herbert |

Henry Spencer wrote:
I wrote:
If memory serves, moderate air resistance calls for a slightly steeper
angle, but the math is ugly.

Oops, correction: a slightly shallower angle. Since the drag is steadily
eroding the horizontal component of velocity, you're better off with a
slightly higher initial horizontal component even if it means a slightly
shorter flight time -- you gain faster motion at the start and lose the
slowest part at the end.

Correct. The drag force acting on a bullet is the biggest force acting
on it during the first part of its trajectory. Using a basic
mathmatical model, a 22 gram, 50 caliber bullet fired at 1500ft/s will
hit 'about' 1910m away when fired at a 45 degree angle. But at a 35
deg, it will hit about 2025m away. And at 30 degrees, it will hit
about 2035m away.

Dave

"b. hotting" wrote in message
...
Hi,

Can anybody tell me the influence of gravity on the trajectory of a bullet
when the aim is to shoot as far as possible?

To put it simple, if i shoot a bullet from a gun on earth or on the moon
(places with different gravity) does the angle, in order to shoot the as
far
as possible, differ?

If you know any interesting web sites about this subject, i would very
much
appreciate it!

This is in many science/physics textbooks. I'd suggest reading one. ;-)

Jeff
--
Remove icky phrase from email address to get a valid address.

In article ,
Jeff Findley wrote:
This is in many science/physics textbooks. I'd suggest reading one. ;-)

Actually, the textbooks tend to discuss only the flat-Earth no-air case,
which is the least interesting one. :-)
--
"Think outside the box -- the box isn't our friend." | Henry Spencer
-- George Herbert |